SMSG (School Mathematics Study Group) Geometry: Student’s Text, Part II

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GEOMETRY PART II

S C W L MATHEMATICS STUDY GROUP

YALE UNIVERSITY PRESS

School Mathematics Study Group

Geometry

Unit 14

Geometry St1id(wtJs Tr,\-t, P'7rt I1

thc t'ancl

o t Sanlplc ~

7'cxtbrwks

of thc School Mathcr~~atics SruJv I;rnup: Frank B. Allcn

LL-LCIIS T n \ \ - l ~ s h HiS]> i~ Sc110t,l

Ediv~tlC. Douglns Dotlald E. Rich~nond Charles E. Rickart Heiir): Sxvain itbbcrr J. Walkcr

Tofi School

Williams Collcgc Yalc L l n i v r r a r ~ NEWTr1c.r T o ~ t - ~ l s h iHigh p Sclic)ol ~ o r i i c I I~nivcrsit!-

Ncw Havcn and Lundun. Y ~ l cUniversity t'rcss

Copvright o 1960, 1961 by Yale University. Printed in thc Urlircd States of Arncrica.

All rights reserved. This book may not bc. ccpraduccd, in whole or in part, in a n y iortn, without writtcn permission from the Fublishcrs. Financial support for thc School Mathctnaticj Study Groiip has brun provided by thc N a t i n t ~ a l Scicncc Foundation.

Chapter 11

AREAS OF POLYGONAL REGIONS

11-1. Polygonal R e ~ l o n s . A triangular region is a figure that c o n s i s t s of a t r i a n g l e plus i t s i n t e r i o r , l i k e t h i s :

A polygonal regLon is a figure in a p l a n e , like one of these:

that can be " c u t up" i n t o triangular regions.

To be e x a c t :

D e f i n i t i o n s : A t r i a n g u l a r region is the union of a t r i a n g l e and its interior. A polygonal r e g i o n is the union of a f i n i t e number of coplanar triangular regions, such t h a t if any t w o of these i n t e r s e c t the intersection is e i t h e r a segment o r a point.

The d o t t e d lines in the figures above show how each of the two figures can be c u t up in t h i s way. Here are more examples:

In the last two examples the figures have "holes" i n them.

This

possibility is n o t excluded by the definition, and these figures

are p e r f e c t l y good polygonal regions. On the o t h e r hand, the region

APDFQC

cannot be "cut up" into

r e g i o n s ABC and DEF even though it is the union of these two t r i a n g u l a r regions. The intersection of t h e two t r i a n g u l a r regions is the quadrilateral region EPBQ, which is c e r t a i n l y not a segment or a p o i n t . This does not mean that APDFQC is n o t a polygonal region, b u t merely t h a t its description as a u n i o n of t r i a n g u l a r regions ABC and DEF I s n o t enough to

show this. below.

APDFQC

is in f a c t a polygonal region, as is shown

The polygonal regions form a rather large c l a s s of f l g u r e s . Of course, t h e r e are simple and important figures that a r e n o t polygonal r e g i o n s . For example, t h e figure formed by a c i r c l e t o g e t d e r w i t h its i n t e r i o r is n o t o f t h i s type.

If a f i g u r e can be c u t up into triangular regions, then this can be done in a g r e a t many ways. F o r example, a parallelogram p l u s i t s i n t e r i o r can be c u t up i n many ways. Here a r e t h r e e of thes e ways .

In thls c h a p t e r we will s t u d y t h e areas of polygonal regions, and learn how to compute them. The sixteen p o s t u l a t e s t h a t we have introduced so f a r would enable us to do this, but t h e treatment would be extremely d i f f i c u l t and quite unsuitable f o r a beginning geometry c o u r s e like t h i s o p e . Instead we shall i n t r o d u c e measure of area in much the same way we d i d f o r measure of d i s t a n c e and angle, by means of a p p r o p r i a t e postulates.

P o s t u l a t e 17. To every polygonal reglon t h e r e corresponds a unLque p o s i t i v e number.

The area of a polygonal region is the number assigned to it by P o s t u l a t e 17.

Definition:

-

We designate the area of a reglon R simply by area R. En the f o l l o w i n g p o s t u l a t e s , when we speak of a region, f o r s h o r t , it would be understood that we mean a polygonal region. Our i n t u i t i o n tells us t h a t two reglons of the same shape and size should have the aame area, regardless of theZr positions in space. This fundamental f a c t is t h e motivation of the next postulate

.

Postulate 18. ~f t w o t r i a n g l e s are congruent, then the t r i a n g u l a r regions have t h e same area.

If a region 1s c u t into two pieces it is clear that the area of t h e r e g i o n should be the sum of the areas of the pieces. This is what o u r next p o s t u l a t e says, Let us s t a t e t h e postulate and then consider i t s meaning. Postulate 19. Suppose t h a t the region R is the union of two regions Rl and R2. Suppose that RI and Re i n t e r s e c t at most i n a f i n i t e number of sements and p o i n t s . Then the area of R is the swn of the areas o f R1 and R2.

The three figures below show examples of the, application of

this P o s t u l a t e .

In each figure the Intersection is heavily marked, and consists of a segnent in the f i r s t f i g u r e , three segments in the second, and t w o segments and a p o i n t in the t h i r d . On the o t h e r hand, the next figure is the union of two tri-

angular regions, R1 and R2, but t h e i r i n t e r s e c t i o n is not made up of a f i n i t e number of segments and points. Instead I t 1s the quadrilateral reglon in the middle. Thus Postulate 19 cannot be applied to t h i s case. If we tried to c a l c u l a t e t h e area of the whole region by adding the areas of R1 and R2 the area of the q u a d r i l a t e r a l region would be counted twice. It was in anticipation of this situation that we insisted, In the definftion of polygonal region, that t h e triangles determining the r e g i o n m u s t n o t overlap. As was the case w i t h distance and angle, t h e " u n i t of areat' can be s p e c i f i e d a t will. However, It is convenient and customary t o choose t h i s u n i t t o be c l o s e l y associated with t h e unit of distance. If we are to measure d i s t a n c e in inches, we measure area in square Inches; and in general, whatever u n i t of d i s t a n c e we use, we use the corresponding square unit to measure area. One way t o ensure t h i s would be to state aa a postulate that the area of a square Is t o be t h e square of t h e length of an edge.

( B t~h e "area of a squaret1 w e mean, of course, the area of the polygonal region which is the union of the square and i t s i n t e r i o r . We will speak in the same way of the area of any quadrilateral, meaning the area o f the corresponding polygonal r e g i o n . ) The statement A = e2 is, however, a little too s p e c i a l t o be convenient. The d i f f i c u l t y is that if we establish o u r u n i t of area by t h e p o s t u l a t e A = e 2 , then we would have the problem o f proving t h a t the corresponding formula holds a l s o f o r rectangles. That i s , we would have to prove that t h e area of a r e c t a n g l e is the product of t h e l e n g t h o f its base and t h e length of i t s altitude. Of course, if we know that t h i s holds f o r r e c t a n g l e s , t h e n it f o l l o w s immediately t h a t for s q u a r e s we have A = e 2, because every s q u a r e is a rectangle. The converse can a l s o be proved, b u t t h e p r o o f is harder t h a n one m i g h t think. The most convenient t h i n g t o d o , f o r the p r e s e n t , is to take as a p o s t u l a t e the more general formula, t h a t is, the one f o r rectangles:

Postulate

20. The area of a r e c t a n g l e is the

p r o d u c t of the length of i t s base and t h e l e n g t h of

its altitude.

Notice that i n the previous paragraph and I n Postulate 20 we were very c a r e f u l t o say, " l e n g t h of its base" and " l e n g t h of its altitudet'. In using Postulate 20 from now on, we w i l l j u s t say, "The area of a rectangle is t h e product of its base and i t s altltudet'. This means t h a t we use "base" and " a l t i t u d e " sometimes t o indicate l i n e segments and sometimes t o indicate t h e i r lengths. From now on we will do t h i s fairly g e n e r a l l y , trusting i n your a b i l L t y t o tell from the c o n t e x t which meaning of a word we i n t e n d . If we "bisect a side of a t r i a n g l e ' ' the word " s i d e t t w i l l have i t s o r i g i n a l meaning, as a set of points. If we "square the a i d e of a t r i a n g l e " we are using the word " s i d e " as an a b b r e v i a t i o n f o r "length of t h e s i d e " , Such a b b r e v i a t i o n s w i l l be very convenient I n this and l a t e r chapters. On the basis of the f o u r area postulates we can c a l c u l a t e t h e areaa of triangles, parallelograms, and a variety of other figures.

Problem Set 11-1 1.

Show t h a t each of the regions below i s polygonal by indicating how each can be c u t into t r i a n g u l a r regions such that i f two o f them intersect t h e i r intersection is a point or segment of each of them. T r y t o find the s m a l l e s t number o f t r f a n g u l a r regions i n each case.

C sac. 11-1 ]

2.

Find the area of a rectangle 50 ft. long and 1% 1 ft. wide.

3.

a.

If you double the altitude of a rectangle and l e a v e the base the same, how is the area changed?

b.

If b o t h the altitude and t h e base of a rectangle are doubled, how is the area changed?

4.

How many tiles, each 6 inches square, does it take t o cover a rectangular floor 37 ft. 6 in. by 12 ft.?

5.

The figure shown is a face o f a c e r t a i n machine p a r t . In order t o compute the c o s t of p a i n t i n g a great number of these parts It is necessary to know the area o f a face. The shaded regions a r e not t o be p a i n t e d . Find the area t o be painted.

+6.

A r e the following statements t r u e or f a l s e ?

Give a reason for

each answer. a.

A triangle is a polygonal region.

b.

P o s t u l a t e 17 says that f o r every p o s i t i v e number there corresponds some polygonal region R.

c.

Every polygonal region has an unique area.

d.

If two triangles are congruent, then the triangular regions have t h e same area.

e.

The union of two polygonal regions has an area equal to the swn of the areas of each region.

f.

P o s t u l a t e 20 assures us t h a t t h e area o f a square having s i d e e is A = e 2

A

.

7.

g.

The i n t e r i o r of a trapezoid Is a polygonal region.

h.

A triangular r e g i o n IS a polygonal r e g i o n .

A rectangular region h a v i n g base

6 and altitude 4 can be

d i v i d e d up i n t o squares having a base 2, as in Figure 1. Notice that a square with base 2 is the l a r g e s t square possible which will divide the rectangular region I n t o an exact number of congruent squares

.

F i g u r e 1.

[ sec

. 11-1 ]

Similarly, a square w i t h base 1 is the l a r g e s t square possible which wfll exactly divide a rectangular region w i t h 1 as in Figure 2. base 4 and altitude 17,

Figure 2. Determine t h e s i d e of t h e largest square which will exactly divide rectangular regions having the following measures:

What d i f f i c u l t y do you find in parts ( e ) and (f)? Do you see t h a t this relates to the d i ~ c ~ S 5 i oof n the text preceding Pos tula t e 20?

*8.

Inthefollowlngfigure, A , B, C, D, E, F, - - - c a l l e d v e r t i c e s , the segments AB, BC, CD, DE, - - EF, FD, FB are c a l l e d edges, and the polygonal ABE, FED, BCDF are called faces. The exterior figure uLll a l s o be considered as a face.

-

-

G are - EG, GA, regions of the

L e t the number of faces be and t h e number of edges be

the number of v e r t i c e s be V, e . In a theorem originated by a famous mathematician, Euler, the following formula occurs: f - e + v , which r e f e r s t o figures of which the above f i g u r e is one possibility. Using t h e figure, let's compute f - e + v . You should see that f = 4, v = 7 , e = 9 9 and this gives us f - e + v =- 2. f,

Using the two figures below, compute f - e the edges a r e n o t necessarily segments. b.

+

v.

Notlce t h a t

Suppose this f i g u r e t o b e a s e c t i o n of a map showing counties :

c.

What pattern do you observe in the r e s u l t s of the three computations?

d.

In p a r t (a) take a point in the interior of t h e quadril a t e r a l and draw segments from each of t h e four v e r t i c e s to t h e p o i n t . How does t h i s a f f e c t the computation o f f - e + v? Can y o u explain why?

e.

Take a p o i n t i n the e x t e r i o r o f t h e figure of p a r t (a) and connect it to the t w o nearest vertices. How does t h i s a f f e c t the computation?

f.

I f you a r e interested i n this problem and would like to pursue it f u r t h e r , you will f i n d it discussed in he Enjoyment of ~ a t h e r n a t i c s "by Rademacher and T o e p l i t z and

in "~undamentalConcepts of ~ e o m e t r y "by Meserve. [ s e c . 11-11

--

11-2. Areas of Triangles and Quadrilaterals. L e t us now compute some areas, on t h e basis of our p o s t u l a t e s .

Theorem 11-1. The area of a right triangle is half the product o f its legs.

A = - I ab.

2A = ab.

2

Proof: Given A PQR, w i t h a right angle a t R . L e t A be the area o f A FW. L e t R t be t h e i n t e r s e c t i o n of the parallel t o f3 e PR through Q and the p a r a l l e l to QR through P. Then QRIPR is a rectangle, and A PQR A QFR1 By Postulate 18, this means that t h e area of A QfR1 is A . By Postulate 19, t h e a r e a of the rectangle is A + A , because the two triangles i n t e r s e c t only in the segment PQ. By P o s t u l a t e 20, the area of the rectangle is ah Theref o r e 2A = ab, and 1 A = 7 ab, which was to be proved. From this we can g e t the formula f o r the area of any t r f a n g l e . Once we g e t t h i s formula, it will include Theorem 11-1 as a

.

special case.

Theorem 11-2. The area of a t r i a n g l e is half the product of any base and t h e a1titude to t h a t base.

Proof: L e t A be t h e area of t h e given t r i a n g l e . The t h r e e figures show the t h r e e cases that need to be considered. (1) If t h e foot of the altitude Fs between t h e t w o end-points then the altitude d i v i d e s the given triangle i n t o two

right t r i a n g l e s , with bases b l and b2, as indicated. By t h e preceding theorem, these two triangles have areas $blh and ; b 2 h . By Postulate 19, we have

Since

bl

+

b2 = b, we have

which was t o be proved. (2)

If the f o o t of the a l t i t u d e is an end-point of t h e base, there is nothing l e f t to prove: we already know by the 1 preceding theorem that A = Fbh.

(3)

In the t h i r d figure, we see the given triangle, w i t h area and two right t r i a n g l e s ( a b i g one and a l i t t l e o n e . ) A, We have =b h 7 1

+

1

A = T(bl

+

b)h.

The s t u d e n t should supply t h e reason F o r this s t e p . 1 S o l v i n g a l g e b r a i c a l l y f o r A, we g e t A = Tbh, which was to be proved,

N o t i c e t h a t Theorem 11-2 can be applied to any triangle in three ways, because any side can be chosen as t h e base; we then multiply by the correeponding altitude and divide by 2, t o g e t t h e area. The figure below shows t h e t h r e e choices f o r a single triangle. 1 The three formulas ?blhl, 1 must give -b h and lb h 2 2 2 t h e same answer, because a l l t h r e e of them give the right answer f o r t h e area of the triangle. /

Notice also that once we know how t o f i n d t h e area of a t r i a n g l e , there is not much l e f t of the area problem f o r polygonal regions: all we need to do is chop up the polygonal regions into triangular regions (which we know we can do) and then add up t h e areas of the triangular regfons. For parallelograms and trapezoids t h i s is fairly t r i v i a l .

Theorem 11-3. The area of a parallelogram is t h e product o f any base and the c o r r e s p o n d i n g a l t i t u d e .

Proof: Draw diagonal SO. By Theorem 9-14 divides the parallelogram i n t o two congruent t r i a n g l e s . Postulate 18 t e l l s us t h a t congruent t r i a n g l e s have equal a r e a . Now the area o f I A P S Q = Tbh. Hence t h e area of parallelogram P&RS = bh, which was t o be proved.

I sec . 11-21

Notice that Theorem 11-3 can be applled to any parallelogram in two ways, because any side can be taken as the base, and can then be multiplied by t h e corresponding altitude.

In the f i r s t case we g e t A = bh, and in t h e second case we g e t A = b'hm . These t w o expressions bh and b t h ' must give t h e same answer, because both of them glve the r i g h t answer f o r the area of the parallelogram. The area of a trapezofd can a l s o be obtained by separating it into two triangles.

Theorem 11-4. The area of a trapezoid is h a l f the product o f its a l t i t u d e and the sum of I t s bases, b9

Proof: Let A be t h e area of t h e t r a p e z o i d . Either diagonal divides the trapezoid i n t o two triangles, w i t h areas $blh and h he dotted lines on the r i g h t l n d l c a t e why the second triangle has t h e same altitude h as the f i r s t . ) By P o s t u l a t e 19 1 1 A=-bh+ 2 1 2b 2h . Algebraically, t h i s I s equivalent to the formula

h2h.

A

=

+(bl

[ sec

+

be).

. 11-21

The formula f o r the area of a t r i a n g l e has two usef'ul con-

sequences, b o t h of whlch are e a s y t o see: Theorem 11-5. If two triangles have e q u a l altitudes, then the ratio of their areas is equal t o the r a t i o of their b a s e s .

Given:

Prove :

A ABC

and A DEF

w i t h equal altitudes.

Area of, A ABC - bl Area of A DEF' -

6;'

This is easy to establish once we have the f o r m u l a

A =

$bh

1

because it simply means t h a t

-= -

Theorem 11-6. If two triangles have equal altitudes and e q u a l b a s e s , then they have e q u a l areas.

The proof o f t h i s is clear because the formula t h e same answer I n each case.

A

=

T'bh

gives

Problem -S e t 11-2 1.

In r i g h t t r i a n g l e RC

2.

=

24,

AB

=

C,

AC =

7,

25.

.

Find the area o f A ABC

b,

Find the altitude to the hypotenuse.

The hypotenuse o f a right triangle is 30, one l e g is 18, and t h e area of the triangle is 216. Find t h e l e n g t h of the altitude to the hypotenuse and t h e length of the a l t i t u d e to H

a.

If AE

b.

If

BC

-

C D ~ A B and

In A A B C ,

El%K

4.

with right angle a t

a.

the given leg. 3,

ABC,

A B = ~ , =

cD=g,

6, find BC. AE=5, find CD.

AB=11, =

15,

c.

If C D = 1 4 , A E = 1 0 , BC = 21, f i n d AB.

d.

If AB = C , CD = h, BC = a, find AE.

In this figure C Q = Q D , Prove that the Area A ABC = Area A ABD .

5.

If ABCD is a s q u a r e , f l n d the area of the star pictured here in terms of s and b. The segments formlng the boundary of the s t a r a r e congruent.

A

6

In paralleloaram e D ,e AF BG DA.

ABCD,

lG, and

1

a,

If AE = 7 , DC = 12, BC = 14, then AF =

GB C .

=

15, then

If AF AE = 8 ,

7.

AD

=

AF

=

=

6,

DC = 14, then

16, then

Prove that the diagonals of a parallelogram dlvide it I n t o f o u r triangles which have equal areas. A

0

Find the area of trapezoid ABCD ,

9,

10.

Find t h e area of a trapezoid if i t s altitude has length 7 and i t s median has length 14. (~int: See Problem 10 of Problem Set 9 - 6 , ) A triangle and a parallelogram have equal areas and equal

bases. 11.

How are t h e i r altitudes related?

Compare the areas of

a'.

Parallelogram ABCD and triangle BCE.

b.

ABCF

c.

A A B F and A F C D , If F is the mid-point of AD.

d'.

a n d A B C E and para1 l e lagram ABCD, if F is the mid-point ACFD

of

AD.

and ABCE.

B

C

12.

In surveying the field ahown here, a surveyor laid off f* north-and-south line NS through B and then located C, t h e east-and-west lines CE, C, e DF, AG. He found that CE = 5 rods, DF = 12 rods, AG = 10 rods, BG = 6 rods, BF = 9 rods, FE = 4 rods. Find the area of the f i e l d .

13.

Prove the theorem: If quadrilateral ABCD has perpendicular diagonals, i t s area equals one-half t h e product of the lengths of the diagonals.

D A

8 14.

Write a corollary to the theorem of Problem 13 r e l a t i n g to the area of a rhombus.

15.

The area of a quadrilateral is 126 and the length of one diagonal is 21. If the diagonals are perpendicular, find the length of the o t h e r diagonal.

16. The diagonals of a rhombus have lengths of 15 If an altitude of the rhombus I s length of one side.

its area. +17.

Would the theorem of Problem 13 s t i l l be true if the poly- 0 gonal region ABCD was n o t convex, as in this f i g u r e ?

Find 12, find the and 20.

C

.

18. Prove that a median of a triangle d i v i d e s the triangle i n t o two t r i a n g l e s each having an area equal t o one-half the area of the original t r i a n g l e .

19, a,

b

.

A

AD

and BE a r e two medians of A ABC i n t e r s e c t i n g at G, prove that Area A AEG Area A BDG.

If

=

Determine what part Area A BDG is of Area A ABC ( ~ i n: t Use o t h e r median G. )

.

20.

If is a f i x e d segment in plane E, what other positions of P In plane E will l e t the area of A ABP remain constant? Describe the location of all possible positions of P In plane E which satisfy the candltlon. Describe the location of a l l possible p o s i t i o n s of P in space which s a t i s f y the condition.

0

A

B

"21.

The f i g u r e at t h e right I s formed from f o u r right triangles and f o u r r e c t angles. Notice t h a t there is a square hole one u n i t on a s i d e . a.

b.

Total t h e areas of the eight parts. (omit the hole.) Show that the same r e s u l t is o b t a i n e d by taking one-half

the product of the length o f the base and t h e length o f t h e altitude t o it. c.

*22.

A

Explain why the r e s u l t s in ( a ) and (b) come o u t t h e same i n s p i t e of the hole.

N

-

L

M

F

E

A line c u t s a r e c t a n g u l a r

r e g i o n into two regions of equal area. Show that

it passes through the intersection of the diagonals of the r e c t -

II a

D

s/ b

C

d

0

h

angle. A

c

/R

The Pythagorean

Theorem. Now that we know how to work w i t h areas, the Pythagorean Theorem is actually r a t h e r easy to prove. 11-3.

.

Theorem 11-7. h he Pythagorean he or em) In a right triangle, the square of the hypotenuse is equal t o t h e sum of t h e squares o f the l e g s .

Proof: We take a square f o r which the length of each s i d e is a + b. In t h i s square we draw f o u r r i g h t triangles w i t h legs a and b, like t h i s :

then

Each of the four right triangles is congruent to the given triangle by t h e S . A . S . Postulate. Therefore t h e i r hypotenuses have length c, as indicated in t h e figure above. [ sec

. 11-3 1

(2) The q u a d r i l a t e r a l formed by the f o u r hypotenuses is a square. We can show this in the following way:

LZ + rnl

is a r i g h t angle because mLy + rnL z + mLx = 180, and mL y x = 9 0 . h he a c u t e angles of a right triangle are complementary). Since all f o u r sides are each equal to c, the quadrilateral is a square.

(3) The area of the large square is equal to the area o f t h e small square, plus the areas of the four congruent right triangles. Theref o r e

The ref ore a2

and finally, a2

+

+

b2 = c

2ab 2

+

b2 = c 2

+

, which was t o

2ab,

be proved.

The converse of the Pythagorean Theorem is also true.

Theorem 11-8. If the square o f one side of a t r i a n g l e is equal t o t h e s u m of the squares of the o t h e r two s i d e s , then t h e t r i a n g l e is a right triangle, with a right angle o p p o s i t e the first s i d e .

Proof: Glven A A B C , as I n the f l g u r e w i t h c 2 = a2 L e t A A ' B I C 1 be a right triangle with legs a and b.

d

+

b2.

L e t d be the hypotenuse of t h e second triangle. Pythagorean Theorem,

By the

.

Therefore d2 = c 2 Since c and d are b o t h positive, this means that d = c . By the S.S.S. Theorem, we h a v e A A 1 B t C f S A A E C . Therefore L C C1 Therefore C is a r i g h t angle, which was t3 be proved.

.

1

--

Problem S e t 11-3a 1.

A man walks due n o r t h 10 miles and then due e a s t 3 miles, How far is he from h i s s t a r t i n g p o i n t ? ("AS t h e crow flies".)

2.

A man walks

3.

A

4.

In the rectangular solid indicated in the diagram, find the length of AC; of AD.

5.

Which of the following s e t s of numbers c o u l d be t h e lengths of the sides of a r i m t t r i a n g l e ?

7 miles due n o r t h , 6 miles due e a s t and then 4 miles n o r t h . How f a r is he from his starting p o i n t ?

man travels 5 miles n o r t h , 2 miles e a s t , 1 mile n o r t h , then 4 miles east. How f a r is he from h i s starting point?

6.

a.

Show by the converse o f the Pythagorean Theorem t h a t i n t e g e r s which represent lengths of s i d e s of right triangles can be found in the following manner.

Choose any positive i n t e g e r s m and n, where rn > n . Then rn2 - n2 and 2n-m will be t h e lengths of t h e legs o f a r i g h t triangle and rn2 + n2 will be t h e length of I t s hypotenuse.

7.

b.

Use t h e method o f p a r t ( a ) t o l i s t integral l e n g t h s of s i d e s of r i g h t triangles w i t h hypotenuse less t h a n o r e q u a l to 25. There are six such t r i a n g l e s .

a.

With right angles and lengths as marked in t h e figure, find AY, AZ and AB.

b.

If y o u continue the p a t t e r n established in this f i g u r e making BC = 1 and

mL CBA

= 90,

-

what would be the l e n g t h of AC? mat would b e the l e n g t h of the n e x t segment from A? You s h o u l d find an i n t e r e s t lng p a t t e r n developing.

8.

In the rectangular s o l i d at the r i g h t AD = 2.

A W = 1, Flnd

AY.

AB=2,

cC0/ \ \

\

\

'

\

Y \ \

A

X

"9.

In A

AB = 1 4 ,

ABC,

BC = 15,

AC = 13.

*lo.

a,

Find t h e l e n g t h of the altitude, h c , to AB.

b.

Find t h e length o f t h e altitude, ha, to E.

1

A A B C has obtuse angle B, and AB = 6, BC = 14, AC = 18.

Find the length of t h e altitude, # hc, to AB.

11.

One angle of a rhombus has a measure of 60 and one side has length 8. Find t h e length of each diagonal. C

12. In rhombus ABCD, AC = 6 and BD = 4. Find the length of t h e perpendicular from any vertex to e i t h e r opposite s i d e

.

13.

In the f i g u r e BC BC = 5 , CA = 12, Find CD.

1-CA, CD

IAB.

l?7c

A

0

14.

The lengths of t h e legs o f

right triangle ABC are 1 5 and 8. Find t h e l e n g t h of

A

the hypotenuse. Find the length o f t h e ~ l t i t u d eto t h e hypotenuse. 15.

If t h e lengths of the l e g s of a r i g h t triangle ABC are a and b, find t h e length of t h e altitude t o t h e hypotenuse

A

.

16.

D

c

O

0

B

C

A ABC is i s o s c e l e s w i t h CA = CB. Medians and BQ are p e r p e n d i c u l a r t o each o t h e r a t S If SF = n , f i n d the length of each segment and the areas of polygonal regions AS&, ASB, ABC and @PC in terms o f n. (DO not change r a d i c a l s to decimals )

.

.

B

A

17.

A proof of t h e Pythagorean Theorem

making u s e of the following figure

was discovered by General James A , G a r f i e l d s e v e r a l years b e f o r e he became P r e s i d e n t of the United States. It appeared about 1875 i n the " ~ e wEngland Journal of ducati ion." P P O V ~t h a t a 2 + b2 = c 2 by stating algebraically that t h e area o f the trapezoid equals t h e sum o f t h e areas of t h e t h r e e triangles. You must include proof t h a t EBA is a r i g h t

1

angle. [sec

. 11-3 I

D

a

:yo=---

E

; I !

\

*18.

ABCD is a three-dimensional "pyramid-like" s o l i d .

Note t h a t p o i n t s A , B, C, and D are not coplanar. We a r e told t h a t BD = BC = BA = AC = CD = DA = 2.

*19.

a.

and S a r e m i d p o f n t s o f BA and E , respectively. Prove RS isperpendicular to both BA and CD.

b,

Find the length of

c

6 -------------A

0

R

In AABD,

RS.

L ABD

is a right angle, AB = BC = 1, AC = CD. Find AD. Find rnL ADC and m i DAB.

A

0

[ sec

. 11-3 1

C

D

The Pythagorean Theorem also gives us i n f o m a t i o n about the shapes of c e r t a i n simple triangles. Two very u s e f u l relationships are stated in the following two theorems. W e give figures which suggest their p r o o f s . Theorem 11-9, he 30-60 Triangle Theorem.) The hypotenuse of a r i g h t t r i a n g l e is twice as long as a l e g if and only if the measures of the acute angles are 30 and 60.

h he

Isosceles Right Triangle Theorem. ) r i g h t t r i a n g l e is isosceles if and only if t h e hypotenuse is times as long as a l e g . Theorem 11-10,

A

fl

Problem S e t 11-3b 1.

The lengths of two sides of a triangle are 10 and 14 and the measure of the angle included between these sides is 30. What Is the length of the altitude to the side 14? What l a the area of the triangle?

2.

The measure of the congruent angles of' an i s o s c e l e s triangle are each 30 and the congruent sides each have length 6 . How l o n g Is the base o f the t r i a n g l e ?

[sec

. 11-3 1

3.

The measure of one acute angle of a ri&t triangle i a double the measure of the other a c u t e a n g l e . If t h e l e n g t h of the l o n g e r l e g Is 5 J 3 , what i s the l e n g t h of the hypotenuse?

4.

Show that in any 30' - 60' right t r i a n g l e with hypotenuse t h e length of the side o p p o s i t e the 60' angle is given by h

5.

s

=$A.

In parallelogram ABCD, AB = 2 and AD = 3, mL B = 60. Find the length of the altitude from A to

G.

80'

B

C

6.

If an a l t f t u d e of an e q u i l a t e r a l triangle I s 15 inches long, how long is one s i d e of t h e triangle?

7.

In a right t r i a n g l e with acute angles of 30' and 60°, what is the ratio of the shortest slde to the hypotenuse? Of the hypotenuse t o the s h o r t e s t s i d e ? O f t h e shortest side to the side opposite the 60' angle? Of the a i d e opposite the 60' a n g l e t o the shortest side? Of the s i d e o p p o a i t e the 60' angle t o t h e hypotenuse? Of the hypotenuse t o the s l d e opposite the 60' angle? Are these r a t i o s the same f o r every 3 0 ' - 60' r i g h t t r i a n g l e ? If you have done this problem c a r e f u l l y , you should find the results very h e l p f u l i n many of the following problems.

8.

What is t h e area of the isosceles triangle whose congruent sides have lengths of 20 inches each and whose base angles have measures of:

9.

What is the area o f the isosceles triangle whose base has a length of 24 inches and whose base angles each have measures of:

10.

Use the information given in the figures to determine the numerical values called f o r below:

1

11. In this figure plane A BFY lies in plane E. HF 1E. A!3 = BH = 6 .

mL

E.

A

FHB = 3 0 .

Give the measures of as many o t h e r segments and angles of the f i g u r e as you can determine

2

In AABC, AB = 3

rnL

.

A = 30,

a.F l r d

BC

.

AC = 4,

Is

LC

a r i g h t angle? A

343

B

"13.

In A

as shown in t h e figure,

ABC

.

BC ( ~ i n t : D r a w the altitude from C ) find

14.

.

The base of an i s o s c e l e s trfangle is 20 inches and a l e g is 26 inches. Find the area. F

15.

In this figure FD = FC, DB = C A , DF 1 E, and -

1

CF FA. prove A FAB is isosceles,

16.

-

and CB are both perpendicular to in this figure. AE = FB and DF = CE. Prove DA

Lx=Ly. 17.

D

B

A

C

[3

bd

A

E

0

F

Prove the theorem: The area of an equilateral t r i a n g l e with s i d e s is given by 2 Area

=

&.

A

D

B

18. Find the area of an equilateral triangle h a v i n g t h e length of a side equal to:

19. The area of an equilateral t r i a n g l e is

9

fi*

Find its side

and its altitude.

16 2/3. Determine its

20.

The area of an equilateral t r i a n g l e is side and its altitude.

21.

area is 81 has i t s perimeter of length equal to the length of the perimeter o f an e q u i l a t e r a l triangle. Find the area of the equilateral triangle. A square whose

This figure represents a cube. The plane determined by p o i n t s A, C and F is shown. If AB is 9 i n c h e s , how long is A C ? What is the measure of FAC? What is t h e area of A PAC?

1 23.

24.

I n t r a p e z o i d ABCD, base angles of 60' include a base of length 12. The non-parallel slde AD has l e n g t h 8. Find the area o f the trapezoid.

Find the area of the trapezoid.

A

B

In the figure, plane E and H plane F i n t e r s e c t in AB,, forming d i h e d r a l angle F-AB-E. CG p l a n e E, DG and CD E-. D is the mid-point of AB. - -

1

LAB,

BCZAC.

1

~f

A B = ~ & ,

AG = 6 ,

mL CBG = 45, and mL CAG = 45, f i n d CG and rnL F-AB-E.

2

Figure ABCD is a r e g u l a r tetrahedron (its faces a r e e q u i l a t e r a l ) . L e t any edge be e . NM AB and -

1

NM

1DC.

a.

Show that t h e length of a bl-median, that is, t h e segment, NM, joining the mid-polnts o f opposite edges, is

b.

Show that the length of the altitude,

-

AH,

of the

t a t r a n e d r o n is e. (~int: Draw and E. Does H lie on Recall that the medians o f a 2 triangle a r e concurrent at a p o i n t 7 of the distance from each v e r t e x . ) J

m?

27.

ABXY is a square. mL X-AB-E = 6 0 .

A3

=

6.

Rectangle ABCD is the projection o f square ABXY on plane E. What is t h e area of rectangle ABCD?

*28.

Given any two rectangles anywhere In a p l a n e , how can a s i n g l e line be drawn which will s e p a r a t e each rectangular region into two r e g i o n s o f e q u a l area?

Review Problems

1.

If t h e side of one square is double t h e side of a n o t h e r square, then the area o f t h e f i r s t square is the area of the second square.

In AABC, Find

BC.

3

1

, AB = 8,

CD

=

times

9

and

AE =

6.

A man walks

5 miles north, then 2 miles

then 1 mile How f a r will he be f r o m his

north, then 6 miles e a s t . starting point?

east,

If the diagonal of a square is 15 f e e t long, how long is each s i d e ? Find the area of an i s o s c e l e s t r i a n g l e in which the base is

12 and each congruent s i d e is 10.

In the figure, parallelogram, H C* and SV QR.

1

P&RS is H QT 1 SR,

a

a.

If S V = 7 and P S = 5 , find the area o f P a .

b.

If

SV-8,

SR = 10,

QT=4 f i n d &R.

Q

P

t ,

and

T

/ / 1 '

% ,' '

/

In an e q u i l a t e r a l t r i a n g l e the l e n g t h of t h e altitude is 6 inches. What is the length of each side? The s i d e of a rhombus is 13 and one o f its diagonals is 24. Find its a r e a .

In A ABC base AB = 12, median The area of A ABC is

Derive a formula f o r the area of the f i g u r e a t t h e r i g h t in terms of t h e indicated lengths .

CD

=

8,

and

mL ADC

=

30.

11.

Find the area of the shaded region of the Figure at the right.

12.

13.

-

Diagonal AD of the pentagon ABCDE ahown Is 44 and the perpendiculars f r o m B, C, and E are 24, 16, and 15 respectively. AB = 25 and CD = 20. What is the area of the pentagon?

Parallelogram ABCD w i t h X and E midp o i n t s of' AB and AD

Given:

respectively

To prove: 1 AECX = 2

ABCD

0

E

.

C

Area of region

area parallelogram

.

A

14,

E Prove t h a t the area of an isosceles r i g h t triangle is equal

to one f o u r t h the area of a square having the hypotenuse of the triangle as a side. *15, An e q u i l a t e r a l triangle has one s i d e in a given plane. The plane of the t r i a n g l e is inclined to the given plane at an angle of 60'. What is the ratio of the area of the triangle to t h e area of i t s projection on t h e plane?

+16. Explain how to d i v i d e a trapezoid i n t o two parts that have ,equal areas by a line through a vertex. *I?.

Find t h e length of the dlagonal of a cube whose edge is 6 units long.

*18, In t h i s rectangular s o l i d AE = 5, AB = 10 and AD = 10.

a.

Find

AC.

b.

Find

AG.

Given: Square ABCD with p o i n t s E and F as shown, so that A r e a ABCD = 256 sq. ft Area of A C E F = 200 sq. ft.

1z. .

Find

BE.

X, Y and 2 are mid-points of sides of square ABCD, as shown in the f i g u r e , compare the area of t h i s square w i t h t h a t of aquare WPQ.

If W,

21.

The f i g u r e shows two isosceles r i g h t t r i a n g l e s . The f i r s t of these has a horizontal s i d e of length 10 units and the second has a horizontal hypotenuse of length 14 u n i t s .

a.

Draw t w o such triangles on graph paper. Cut out t h e second one and place it on the first to show that their areas are apparently e q u a l .

b.

In t h e f i r s t figure count the number of small squares and the number of small half squares (right isosceles triangles). Use these n u m b e ~ sto compute t h e area.

c.

Do the same f o r the second figure.

d.

Explain t h e discrepancy.

Chapter 12 S IMILARITY

The Idea ---

of Similarity. Proportionality Roughly s p e a k i n g , two geometric figures are similar if t h e y have exactly t h e same shape, but n o t necessarily the same size. F o r example, any two c i r c l e s a r e similar; any two squares a r e similar; any two equilateral triangles a r e similar; and any two segments are similar. 12-1.

.

Below a r e two t r i a n g l e s , w i t h the lengths of t h e s i d e s as indicated:

These figures stand in a vem s p e c i a l kind o f r e l a t i o n t o each o t h e r . One way to descrFbe t h i s r e l a t f o n , speaking v e r y r o u g h l y , is to say t h a t t h e triangle on the left can be t t s t r e t c h e d " , o r t h e

one on the r i g h t can be 'rshrunk",s o as to match up w i t h t h e o t h e r t r i a n g l e , by the correspondence ABCMA'BtC'

.

.

Of course, this correspondence is not a congruence, because each side of t h e r i g h t - h a n d t r i a n g l e is twice as long as the corresponding side o f t h e other. Correspondences of t h i s t y p e are called simflarltles. The exact definition of a similarity will be given l a t e r in this chapter. Notice that the lengths of the s i d e s o f o u r two triangles form two sequences o f p o s i t i v e n u m b e r s , a, b, c and a', b l , c', s t a n d i n g in a very s p e c i a l relation: each number in t h e second sequence is e x a c t l y twice t h e correspondlng number in t h e f i r s t sequence; o r , s a i d a n o t h e r way, each number in t h e first sequence is exactly half the corresponding number in t h e second sequence Thus

.

Another way of putting t h i s is to w r i t e

Sequences o f p o s i t i v e numbers which are r e l a t e d in this way are called proportional.

Definition:

and

P,

q,

r,

Two sequences of numbers,

...,

-a = - b= - =c

P

q

r

a,

b,

c,

...

none of which is zero, a r e p r o p o r t i o n a l if

...

. 0

f

=

+

=

-rC = " ' .

The s i m p l e s t proportionalitlea are those i n v o l v i n g only four

numbers, and these have s p e c i a l properties t h a t a r e w o r t h noting. We list some of them for l a t e r reference.

Csec

. 12-1 1

of 2

A l ~ e b r a i cProperties

a

If with

Simple Proportion.

a,

(1)

then

ad = bc,

ca = bz ,

(2)

(3

a-tb

c + d

a - b

c - d

b = T '

(4)

Proof:

5,

all d i f f e r e n t from zero,

d

c,

b,

=

b

=

d

j

a

Taking the original equation

(1) Multiply b o t h s i d e s by (2) Multlply b o t h sides by (3)

Add

(4)

Subtract

1

-

ad = bc;

a

to g e t

to b o t h sides to get

1

c 3,

to get

bd b

=

=

b -. a'

a + b = -c; + d

f r o m both s i d e s to g e t

a - b =7 c - d . 7

Other relations can be derived, but these are the most u s e f u l . a = b Definition: If a, b, c are p o s i t i v e numbers and 6

then b

c,

-

is the geometric mean between

a

and

c.

From Property ( 1 ) above, it f o l l o w s that the geometric mean between

a

and

c

is

@.

--

Problem Set 12-1 1.

Complete each statement:

If -ba --

then

7a =

b.

7 x = F1 If 3

then

4x =

c.

If

then

6y =

a.

- = -

5

Y

2.

In each of the following proportionalities, f i n d

3.

Complete each statement:

4.

"5.

6.

a.

If

b.

If

c.

If

d.

If

xa -- -

,

and

4m, then g4 = -

,

and

Ea = -

,

and

X 5=- ,

and

3a=2x,

5-3

=

7b = 4 a ,

5.9

=

then

then

6x, then

x.

-. m = -. b = -. a 5 = -. X a =

In each of the following proportionallties, express the number a i n t e r n of the numbers b, c and d .

Complete each statement: 3

a.

If

=

b.

If

TX = $ ,

c.

If -a= c

d.

If

+

then

a - b

-.

,

and

-aC = -,

and

a - c = -. C

and

b - a - -. a

=

then - = -

l1 ' then T

:=5,

a-kb

-=-

then -b=+a- a

Here are three sequences of numbers, sequences p r o p o r t i o n a l ?

[ sec

. 12-1 1

,

Are any two pairs of

One can tell at a glance that the sequences a and b a r e proportional since each number in b is 3 times the corresponding number In a. The comparison of a and c is not such a simple matter. An efficient way t o make such a comparison m i g h t be to change each to a new proportional sequence beginning with 1, that is,

7,

In the following list of sequences of numbers, which p a i r s of sequences are proportional? Make a complete l i s t of these p a i r s of sequences.

/

8.

If

g.

ff

10.

*=-==, W

v

20

-3= -4= - 1= X

Y

1 Z

what are the values of

4 T,

w

what a r e the values of

and x,

v? y

and

Which of the following a r e c o r r e c t f o r a l l values of the l e t t e r s involved assuming that no number in any sequence shown is zero?

[ sec

. 12-13

Z?

-& & g,

11.

If 16

12.

The geometric mean of two p o s i t i v e numbers

=

=

=

what a r e t h e v a l u e s o f a

p,

q

and

and

c

is

t?

a + c b =&. The arithmetic mean of a and c is d = 2 . F i n d t h e geometric mean and the arithmetic mean of the following pairs :

a.

4

and

9.

d.

2

and

24.

b.

6

and

12.

e.

2

and

3.

c.

8

and

10.

12-2. Similarities between Triangles. W e can now s t a t e t h e definition of a similarity between two triangles, Suppose we have given a correspondence ABC -A'

B TC

'

between two triangles

B

As i n d i c a t e d in the f i g u r e ,

a

is the l e n g t h of the s i d e opposite A, b is the length of the s i d e opposite B, and so on. If corresponding angles are congruent, and

then the correspondence

AEC@A'BICt

is a similarity, and we

write A ABC- A A T B ' C ' .

D e f i n i t i o n : Given a correspondence between the v e r t i c e s of two triangles. If corresponding angles a r e congruent and the corresponding sides are proportional, then t h e correspondence Is a similarity, and the triangles are said t o be similar.

Notice t h a t this definition requires two t h i n g s : (1) corresponding angles must b e congruent, and ( 2 ) corresponding s i d e s must be proportional. In putting both of these requirements i n t o t h e definition, we are making sure that t h e d e f i n i t i o n may be. applied t o polygonal figures of more than t h r e e s i d e s . T o s e e what t h e p o s s i b l e t r o u b l e s might be, if we used only one of o u r two requirements, l e t us look at t h e situation for quadrilaterals.

F i r s t consider the correspondence ABCD*AqBhCTD1, between t h e two rectangles in the figure. Corresponding angles a r e congruent, because a l l of t h e angles a r e right angles, b u t the t w o rectangles don't have the same shape, by any means. Now conslder a square and a rhombus, w i t h edges of l e n g t h 1 and 2, like t h l s :

Under t h e correspondence ABCD- A1 Bt C D3, corresponding s ides a r e proportional, w u t the shapes are quite d i f f e r e n t .

We shall

the case of correspondences between t r i a n ~ l e s ,if either one of o u r c o n d i t i o n s holds, then so does the o t h e r . That is, if corresponding angles a r e congruent, then corresponding sides are proportional; and conversely, if corresponding sides are proportional, then corresponding angles are congruent. These f a c t s a r e given in the A . A . A . Similarity Theorem and the S.S.S. Similarity Theorem, which will be proved l a t e r in this chapter. s e e l a t e r t h a t for

--

Problem Set 12-2 1.

Given a similarity A ABC

-

A DEF,

A

0

write down the proportionality between corresponding sides, using the notation AB, AC, and so on, Then: a.

Express

AB

in t e r n of

AC,

DE and DF.

b.

Fapress

BC

i n t e r m a of

AB,

DE

&nd EF.

c.

Express

AC

i n t e r m s of

BC,

EF

and

d.

Express

AB

in terms of

BC,

DE and EF.

e.

Express

BC

in t e r n of

AC,

EF

and

DF.

f.

Express

AC

in terms of

AB,

DE

and

DF.

DF.

2.

Below are l i s t e d five s e t s of 3 numbers. Point out which pairs of s e t s of numbers ( n o t necessarily in t h e order given) m i g h t be lengths of sides of similar triangles. Wrfte o u t the equal ratios in each case. For example, a , b;

3.

Two p r i n t s of a negative are made, one a contact p r i n t and one enlarged. In the c o n t a c t p r i n t an object has a length of 2 inches and a height of 1.6 inches. In the enlarged print the same object has a length of 7 . 5 inches. Find i t s height in the enlargement.

4.

AABC = " A A I B t C l , Why or why n o t ?

5.

Prove:

If

does it follow that

A ABC

A'BqCt?

The t r i a n g l e whose vertices are the mid-points of the sides of a given triangle is similar to the given triangle.

12-3.

-The Basic

Similarity

Theorems.

Consider a t r i a n g l e A ABC. L e t D and E be d i f f e r e n t * C, p o i n t s on the s i d e s AB and E, and suppose that DE and BC a r e parallel.

f t looks as if the correspondence ABC-

ADE

[ sec , 12-31

ought to be a similarity, and it is, as we shall presently see, We prepare t h e way w i t h a series of theorems. Theorem 12-1. th he B a s k Proportionality Theorem.) If a line p a r a l l e l t o one s i d e of a triangle intersects the other two sfdes in d i s t i n c t p o i n t s , then it cuts off segments which are proportional to these sides. Restatement: such that %

let Then

In A A B C

11

D and

E

be p o l n t s o f

AD

and

a

Proof: (1) In A ADE and A BDE thlnk of and w the bases and the altitude f r o m E to A 3 as their common altitude. Then by Theorem 11-5,

as

area A BDE BD area A ADZ= XIS'

AE

In h AED and A CED think of and CE as C-* the bases and t h e altitude from D t o AC as t h e i r common altitude. Then by Theorem 11-5, (2)

area A CDE area A A I E

_- -.AE CE

( 3 ) A BDE and A CDE have t h e same b a s e , DE, and congruent altitudes, since t h e lines and %8 are parallel. Hence by Theorem 11-6,

area A BDE = area A CDE.

( 4 It f o l l o w s from (l), ( 2 ) and ( 3 ) t h a t

Applying Algebraic Property (31, from S e c t i o n 12-1,

The converse of Theorem 12-1 is a l s o t r u e (and is e a s i e r t o

prove).

That ts, we have:

If a llne i n t e r s e c t s t w o s i d e s of a t r i a n g l e , and c u t s c f f segments proportional t o these two s i d e s , then it is Theorem 12-2.

parallel to the third s i d e . Restatement: L e t A ABC between A arLd 8 , and l e t

be a triangle, L e t D b e a p o i n t E be a p o i n t between A and C . If

AC

AB

m 4+8

then BC

and

@

DE

=

TITy

a r e parallel. A

i I

B Proof: L e t and i n t e r s e c t i n g

C'

H

BCT be the l i n e through 3, p a r a l l e l t o * AE in C 1 . By Theorem 12-1, A3

AC'

A C ' = AE

a

f3

DE,

m=TJ AB AD '

But the equation given in the h y p o t h e s i s o f t h e theorem means t h a t

Therefore A C T = A C . Therefore 4--* DE, which was t o be p r c v e d .

C l = C,

and

--

P r o b l e m S e t 12-3a

1.

I n t h i s f i g u r e the lengths 3 f segments are a, b, x and y as shown, a + b = a

-,

a

6=-.

x

-a -+- b - x + b

2.

-a= -

x

I n this f i g u r e if HT FA = FH

3.

I I AB,

TB

FT=

FA = HA

-FT =

m = H P.

BTAH -

FH

In the f i g u r e ,

A R

4, HF = 7 , then AB =

6,

HF = 10, then BF =

5,

RF = 20, t h e n BF =

f,

BC

is parallel to

4,

5.

6.

11 E .

Inthe figure,

P

a.

If

AC=12,

b.

If

AD=6,

c.

If B C = 2 2 ,

B = 6 ,

e.

If AC

CE

= 15,

CE=8,

find

BC.

BE = l o , CD = 4,

find

CE.

find

AC.

CD=4,

=

6,

CD=8,

BC

In the figure let the aegments have measures as indicated. Can MN I [ EL? ~ u s t i f yyour answer.

Which of the following s e t s of data make FG 1 I E? a.

AB-14, AG = 3 .

AF=6,

AC=7,

b.

AB=12,

FB=3,

AC=8,

AG = 6 .

-

18, find

AD.

L

7.

I f , in the figure,

5 1 E,

prove

Hint: Use Theorem 12-1 and subtract 1 from each fraction.

8.

Given the figure, one person handled the problem of finding w in t h i s way: 7 - 19

7-

-

w-

w

Propose a more convenient equation. Do you g e t the same r e s u l t ?

9.

C

Place conditions upon x such that 11 E , given that CD = x - 3 , DA = 3~ - 19, CE = 4 , and EB = x - 4.

B

A

10.

In t h i s figure i f EF ) I E , FG (I=, HE I I Must the f i g u r e be planar?

z.

I sec . 12-31

and

a

11 E,

prove

11.

12.

Prove: If t h r e e or more parallels are cut by two transversals, the intercepted segments on t h e two tranaversals are proportional. Restatement: If the l i n e s L and L2 are transversals M of the parallel linea AD, w M BE, and CF, then

Three l o t s extend from Packard Street to State Street as shown in this drawing. The side lines make right angles w i t h S t a t e S t r e e t , and t h e t o t a l frontage on Packard S t r e e t is 360'. Find the frontage of each l o t on Packad S t r e e t .

such t h a t meet in 0 and

G I ~ ~ : ~ A B C XYZ, , @

XA,

-

-

YB,

ZC

E I I ?~ ~ ,I I V . Prove:

%? II%?.

1

a

In

60'

120' State Street

90'

14.

-

A printer wishes to make a c a r d

6 inches long and of such width that when folded on the dotted l i n e as shown it will have the same shape as when unfolded.

1

I I

x

I

I

I

What should be the width?

Theorem 12-3. (The A . A . A . Similarity Theorem. ) Given a correspondence between two triangles. If correspondfng angles a r e congruent, then the correspondence i s a a i r n i l a r i t y .

Restatement:

-

Given a correspondence ABC

between two triangles. then

If

LA

DEF

ZLD, L B

ZL E

and

C Z

L

F,

A ABC * A DEF. Notice that to prove t h a t the correspondence is a similarity, we merely need to show t h a t corresponding sldes are p r o p o r t i o n a l . (We don't need t o worry about the angles, because corresponding angles are congruent by hypothesis). The proportionality of the sides means that

-

It will be sufficient do prove that t h e f i r s t of these equations always holds. (~xactlythe same proof could then be repeated to show that t h e second equation a l s o holds).

Thus we need to prove t h a t

AB = AC

+

Proof: that

+

Let E1 and F1 be p o i n t s of AB and AC, such AE1 = DE and A F 1 = DF. By the S . A . S . Postulate, we have AAEtF'

Z L B.

ADEF.

*

u

Therefore E t F l and BC are parallel, or coincide. If they coincide then A A E t F ' = AABC, and so A ABC Y ADEF; In t h i s case,

Therefore

L A EIFr

AE3 = DE

and

AC = DF,

w t* If EIFb and BC a r e parallel, then by Theorem 12-1,

But

AE1 = DE

and

AFt =

we have

DF. Therefore

which was to be proved. The theorem j u a t proved allows us to prove a corollary which, it turns out, we quote oftener than t h e theorem I n showlng that two triangles are slmilar. Recall from Corollary 9-13-1 t h a t If two pairs of corresponding angles of two triangles are congruent, the t h i r d pair must be also. Thus from Theorem 12-3 we Immediately get the followfng c o r o l l a r y :

-

Corollary 12-3-1. ÿÿ he A . A . Corollary. ) Given a correspondence between two triangles. If two pairs of corresponding angles a r e congruent, then t h e correspondence is a similarity. For example, if

L A Y LD

and

B

G L E,

then

A ABC *ADEF.

If

1A Z L D

and

LC

L F,

then the same conclusion f o l l o w s .

for the t h i r d case. W e can now j u s t i f y our statement at t h e beginning of this section by proving the following corollary: And similarly

Corollary 12-3-2. If a l i n e parallel to one side of a triangle intersects the o t h e r two sides in d i s t i n c t p o l n t s , t h e n I t cuts off a t r i a n g l e aimilar to the given triangle.

C,

M

For if DE 11 BC then by correaponding angles L ADE Y L B 1 AED G L C. A l s o L A Y L A . Hence A A D E .u A ABC, by Theorem 32-3 or Corollary 12-3-1.

-

and

Theorem 12-4. he S. A . S . Similarity Theorem.) Given a correspondence between kwo triangles. If two pairs of correapondlng sides are proportional, and the included angles are congruent, then the correspondence is a aimilarity.

Restatement:

Given ABC-DEF.

and A ABC

then

Proof:

that

AEt =

Let

DE

-

A DDF

.

El and F' be p o i n t s o f and AF1 = DF. Then

-

---*

AB

and

w

+ AC,

such

By Theorem 12-2, t h i s means that E f F 1 and BC are parallel. When two parallel lines are c u t by a transversal, correaponding angles are congruent Theref ore

.

and

But we know, by t h e S. A . S. P o s t u l a t e , t h a t A A E i F 1 Z A DEF.

Therefore and

Therefo r e

and

We already knew by hypothesis that

LAELD.

-

T h e r e f o r e , by t h e A . A . A , Similarity Theorem, We have

A ABC ADEF, which was to be p r o v e d .

We have one more basic similarity theorem f o r triangles, Theorem 2 2 - 5 . h he S . S . S . S i m i l a r i t y Theorem.) Given a correspondence between two t r i a n g l e s . If c o r r e s p o n d i n g sides a r e p r o p o r t i o n a l , then the correspondence is a similarity.

Restatement : Given

A ABC

then

.

ABC-DEF

A

-

A DEF.

E B

-+

AC,

C

Proof: As before, let Ef and F * such that AE' = DE and AF' = DF.

be p o i n t s o f

AB AC DE = m'

2-

AB - AC z v -AFT' -

3.

4. 5.

7.

E1F 1

and

AB

and

R e as ons

Statements 1.

+

I

BC

are parallel

L ~ ~ and L LB f ' L c . AABC A AEIF' .

-

' EIFl = B %AE =B%.

DE

1.

Hypothesis.

2.

Substitution.

3.

Statement 2 and Theorem 12-2.

4.

Theoremg-9.

5. A .

I

A . Corollary.

6. D e f i n i t i o n of similar triangles.

'7,

Statement 6 and substitution.

Isec. 12-31

E I F 1 = EF.

AAEIFf

9

ADEF.

L ~ ~ and L LE f A B L E and L C AABC

-

L F. L F.

ADEF.

9.

Statements 7 and 8.

10.

The S . S . S . Theorem.

11.

Corresponding p a r t s .

12. Statements 4 and 11. 13.

The A . A , Corollary.

P r o b l e m Set 12-3b

Given a correspondence ABC-DEF between two triangles. Which o f the following cases are sufficient t o show t h a t the correspondence is a similarity?

c.

Corresponding sides are p r o p o r t i o n a l .

d,

Both triangles are e q u i l a t e r a l .

e.

Both triangles a r e i s 3 a c e l e s , and

f.

m L C = m L F = 90, and

4A

=

mLD.

AB = D E .

Which of these similarity theorems do n o t have related congruence theorems: S . A . S . , S . S . S . , A . A . A . , A . A . ?

Is there any p o s s i b i l i t y of a.

two angles of

angles of

AI

A I befng similar

A I have measures of

to

A 1

if:

60 and 70 while two

have measures of 50 and 80?

b.

two angles of A I have measures of 4 0 and 60 while two angles of A XI have measures of 60 and 80?

c.

is a right A , while angle of measure 40?

d.

I

A I1

I s isosceles w i t h one

A I has sides whose lengths a r e 5, 6 , 7 , while A 11 has a perimeter of 36,000.

.

E sec 12-3I

5.

Given the figure shown w i t h

ZlE a. b

.

c.

and

C

ElE.

Name an angle which is congruent to L ACB. N a m e an angle with the same measure as

LB.

A

X

Name a triangle which is similar to A ACB.

- -

If the l e n g t h s of DX, XE, and PX a r e p , q and r respectively, what length of XG will assure similarity o f the triangles? If p = 3q, must rnL D = 3mL E?

D

B

7.

Below is a series of statements givlng the lengths of s i d e s of a number of triangles. Decide f o r each pair whether the triangles are similar and then make a statement as follows:

A

is similar to

A

is not similar t o

A

9

Or

A

For each p a i r t h a t a r e slmilar w r i t e a statement showing the proportionality of t h e s i d e a

.

8.

Given:

Prove:

Fig. a .

LBELD. CD = 4AB. BD

=

5BL.

Fig.c

Fig. b.

Fig. d

In each figure a segment has been drawn parallel to the base of a triangle, and the lengths of c e r t a i n segments have been indicated.

Csec. 12-31

c.

Prove that x = r ( ~ i n t : Write a p r o p o r t l o n . ) Prove t h a t x = mp. Prove that x = k 2

d.

Prove:

e.

Part c is a s p e c i a l case of which o t h e r part?

f.

Part d l a a special case of which other part?

g.

Do the results depend on the s i z e of t h e vertex angle?

a.

b.

.

1 x = -f;,

10. Explain how two triangles can have f i v e parts (sidea, angles) of one triangle congruent t o f i v e p a r t s o f the other triangle, but n o t be congruent triangles. 11.

Given: M

OD

II

OID1

Prove:

*12.

a.

In the diagram

*

OB OD q = q .

Ifm, = a n d m a r e perpendicular to BE, name the p a i r s of similar triangles.

b.

c.

R

Which is correct:

-Y = eq ,, EY = PP +? q Which is correct:

2=9 X

P

Z x

=

S

P + q

? -

B

C

D

e.

long does it take two men t o complete a task which one alone can complete i n 6 hours and the o t h e r alone in 3 hours?" can b e answered by s o l v i n g

The problem,

"HOW

E + - J = ~ .Solve t h i s equation geometrically. s e e p a r t (d) and

13.

the figure.)

Glven parallelogram ABRQ w i t h diagonal and segment AF i n t e r s e c t i n g in B as shown.

Prove:

14.

(~int:

QH

. HF

= HI3

AH,

A

1

In this figure if and DQ = BQ = 2AQ = 1 QC. Prove:

a . AAQD

.v

A DQC.

b. ABQC - A A Q D .

..

ElZ.

Prove the following theorem: the b i s e c t o r of sn angle of a triangle divides the opposite side Into segments p r o p o r t i o n a l t o t h e adjacent aides.

Given: A ABC, s e c t o r of L A Prove:

CD 158

=

AD

the bimeeting BC

CA m.

A

C

6

Given A angles at A Prove that

L e t the b i s e c t o r s of the internal and external meet in points D and D t respectively.

& .

"17.

=

%8

(~int: Make

11 FA.

If we have an e l e c t r i c a l circuit consisting of two wires in parallel, with resistances RL and R2, then t h e resistance R of the c i r c u l t l a given by the equation

The following scheme has been used to f i n d and

R,

given

R2.

Fig. 2

R1

Numerical scales are marked off on three rays as in Figure 1. A straight-edge is placed 80 as t o pass through R1 and R2 on the two outer scales, and R I s read off on the t h i r d scale. Using the scales of the figure, select values f o r R1, R2, f l n d R from the figure and check y o u ~result to see that the equation above I s s a t i s f i e d . a.

Prove that the method really works,

b.

Could the same diagram be used to flnd 1 1 equation - = - - I ? R1

=*

In t h i s figure WS and L& are medians RW = RT - WS and AL AM -

n*

Given in R A l Z , E -1- AF.

t h i s figure t h a t -

and

-

Prove t h a t A HRA A BAF and HR'BF = BA-HA.

-

See Figure 2.

R

in the

386 20.

A method

of enlargement.

The figure AIBICIDl an arbitrary point

P,

the rays

locating

Cg

and D2

PB2

A2,

= 2PBl,

A$I2 etc.

B2, etc.;

-

has been enlarged by introducing from _

PAl,

so

j

that

*

PBI,

-

PC1

and

PD1;

PA2 = 2PA1,

and finally drawing aegnents

A&,

a.

D r a w a sfmple o b j e c t , a block or a table, for example, and enlarge It by the method shown. Is it necessary t h a t PAL, PB1, etc. be doubled?

b.

How could the method be modified t o draw a figure with s i d e s h a l f the length of those of AIBICID1?

c.

Prove:

A P A ~ BAPA2B2 ~ ~

Prove :

A ALBIDl

d

.

e.

-

A

and

*92 PA2 K;B;=Y~~;.

A&J12.

Could the enlargement be carrled out if P selected on or inside the given figure?

.

C sec 12-3 1

were

Given: Quadrilateral as in the figure with and A QXR A TXS

-

Prove:

QR =

.

TS

.

BFRQ is a square with Q on AW and R on

as shorn in the figure.

Prove: and

23.

ABmWR = QW-BQ, ABmFM = RFaBQ.

Prove the following theorem: In slmilar triangles corresponding medians have t h e same r a t i o as corresponding s i d e s . F

24.

Prove the following theorem; In similar triangles corresponding altitudes have the same r a t i o as corresponding sides. F

25.

Prove t h a t i f the s i d e s of two triangles are respectively parallel, the t r i a n g l e s are similar.

.

11

Given:

AFJlrn.

BF 1 1

A

Prove:

m..

ABF

Given: Show

LA

2

a HRW .

Case IT

Case I

26,

. I .

1B

and

AC =

BD.

1)

[ aec

.

12-31

+27, It is known { s e e Chapter 5) t h a t if two triangles correspond so that two sides and the angle o p p o s i t e one of them in one triangle are congruent r e s p e c t i v e l y to two s i d e s and the angle opposite the corresponding side of the o t h e r (S S A . ) , the triangles need n o t be congruent. ( s e e diagram.

. .

P

1s the following statement true o r false?

Explain.

If two triangles correspond such t h a t two sides of one triangle are p r o p o r t i o n a l t o two s i d e s of the o t h e r , and t h e angles opposite a pair of corresponding sides a r e congruent, then the t r i a n g l e s a r e similar. *28.

AF is i s o s c e l e s w i t h DE = DF. A ABC is such t h a t E and F l i e between A and

a.

What t r u e statements con-

cerning similarity and proportions can be made concerning

b.

1.

A ABC

and

A ADE?

2

A ABC

and

A ADF?

C

E

F

Is the following statement true or f a l s e ? F z p l a l n . Given

,

AABC

such t h a t

parallel.

-

D on segment AB, X on segrnent AB BC ! and %? must be = m, then %

with

m

tennis ball is served from a h e i g h t o f 7 feet t o clear a net 3 feet h i g h . If it is served from a line 39 feet behind the net and travels in a s t r a i g h t path, how far from the n e t * does it hit the ground?

"29.

A

"30.

In the parallelogram ABCD shown in t h e f i g u r e t h e line f-) H BF i n t e r s e c t s AC at E, 8 at G , and a t F. Prove that FZ is the geometric mean of EG and EF .

M

Given

A ABC

w

XYZ

and

M

such

C*

AX, BY and CZ are parallel and a l s o %? 11 H BA and %? meet in D and %! and %? meet in E. that

Prove:

v.

Xi? 11

5Z

11

Z.

h

The angles in the figure marked w i t h small squares are r l g h t angles AD a. Show t h a t BF = z.

.

b.

Then show t h a t

-

.

12-4. S irnilarities in R i & t Trianales Theorem 12-6. In any right triangle, the altitude to t h e hypotenuse separates t h e triangle i n t o t w o triangles which a r e similar to each other and to the o r i g i n a l triangle.

Restatement: angle at

AB.

C

.

Let

Let A ABC be a right t r i a n g l e with its r i g h t CD be the altitude from C to the hypotenuse

Then

A ACD

-

A ABC - A CBD.

Notice that the restatement is more explicit than the first statement of the theorem; it t e l l a us exactly how the v e r t t c e s should be matched up t o g i v e the similarities. Nctice a l s o what the scheme is in matching up the angles: (1) The right angles match up with each o t h e r , as t h e y have to in any elmllarity of

r i g h t triangles.

(2)

Each l i t t l e t r i a n g l e has an angle i n common

with t h e b i g triangle, and so the angle matches i t s e l f . remaining angles a r e then matched.

( 3 ) The

Proof: In t h e proof, the notation f o r the angles will be as shown in the f i g u r e . S i n c e L C is a r i g h t angle, we know that a r e complementary. That is,

mL a

1d

Also, s i n c e

+

rnL b

+ rnL

b

b t = 90,

L b q b ' .

Trivially,

La =La;

and

L c q d , d

L

= 90.

The re f o r e

L

and

is a right angle, mLa

because we have

La

is a right angle.

By the A . A . A .

AACD - A

Similarity Theorem,

ABC.

The proof of the o t h e r half of the theorem is precisely analogous, w i t h t h e p o i n t B behaving like the p o i n t A .

C o r o l l a r y 12-6-1. Given a right t r i a n g l e and the altitude from t h e right angle to the hypotenuse: (1) The altitude is the geometric mean of the segments into which it separates the hypotenuse. ( 2

E i t h e r leg I s the geometric mean of the hypotenuse and t h e segment of t h e hypotenuse adjacent t o the l e g .

Restatement: Let A ABC be a r i g h t t r i a n g l e w i t h its r i g h t angle a t C, and let D be the foot o f the altitude f r o m C to .-

AB.

Then

Proof:

(1) By Theorem

Hence, ( 2)

By Theorem

Hence,

and so

BD

12-6, AD

2 6 ,

AD

-A . /\ -A

CDB

ADC

CD

= ~ j 6

ADC

AC

m=mBC

m=mm

--

Problem Set 12-4

1.

Given r i g h t h ABC with a l t i t u d e dram t o t h e hypotenuse and lengths as shown, f i n d the unknown lengths.

.

Y

ACB

.

3.

In this right t r i a n g l e w i t h the altitude drawn to the hypotenuse it Is possible to flnd a numerical v a l u e f o r each segment a, x , y , Flnd them.

4.

In a right t r i a n g l e if the altitude to t h e hypotenuse I s 12 and the hypotenuse is 25, f i n d the l e n g t h of each l e g and of the segments o f the hypotenuse.

5.

In right A ABC, with right angle a t C and altitude m, a.

i f A D = 2 and D B = 8 , find AC, CD and CB.

b.

if

c.

If CB = 12 and AD = 10, what are the lengths of the o t h e r segments?

d.

A C = 8 and D B = 1 2 , what are t h e l e n g t h s of the o t h e r segments?

C D = g and A D = 3 , find AC, CB and AB.

iP

B

--

12-5. Areas of S lmllar Triangles. Given a square of s i d e a, and a square of side 2a, it is easy to see that the area of the second square is 4 times the area of the f l r s t . (This is because (2a12 = &a2 . ) In g e n e r a l , if two aquares have s i d e s a and ka, then the r a t i o of the areas is 2 k , because

An analogous result holds f o r similar triangles :

Theorem 12-7. The ratio of the areas of two similar triangles is the square of the r a t i o of any two corresponding s i d e s .

Proof :

kt k

Given

A ABC

.v

b, A

B' C , Then

be the oornmn value of these ratios, so t h a t

-

a t = ka,

bl = kb , cl = k c . L e t be the altitude from B to AC l e t BIDI be the altitude from B t to Since A ABD A A ' B ' D t are r i g h t triangles, and L A L A ! , we have

Let

A1

and

A2

be the areas of the two t r i a n g l e a . 1

Al = *h,

and

A,

=

Then

.

and

and

Therefore

*2 = *l

2

= (

a

=

bt

ci 2 = (b) ,

which was to be proved.

-

Problem S e t 12-5 1.

What ia the r a t i o of t h e areas of two similar triangles whose bases are 3 inches and 4 inches? x Inches and y inches?

2.

A side of one of

3.

In the figure if H is the mid-point of and K i a the mtd-point of , the srea of A ABF La how many times as great as the area of A A I M ? If the area of AABF is 15, find the area A of A m .

4.

two similar triangles i a 5 timea the c o r r e s ponding side of the other. If the area of the f i r s t is 6 , what is the area of t h e second? F

K

The area of the larger of two similar triangles is 9 times the area of the smaller. A s l d e of the larger is how many

times t h e corresponding s l d e of the smaller?

5

The areas of two similar triangles are 225 s q . in. and 36 a q . in. Find the base of the smaller if the base of the larger is 20 inches.

6.

The areas of two similar triangles are 144 and 81. If a side of the former i a 6 , what l a the corresponding s l d e of the

latter?

7.

In A A B C , the point D i t 3 o n s i d e AC, and AD is twice CD. Draw parallel to i n t e r s e c t i n g BC at E, and compare t h e areas of triangles ABC and DEC.

8.

9.

10.

The edges of one cube are double those of another.

a.

What I s the ratio of the sums of their edges?

b.

What I s the ratio of their t o t a l surface areas?

How long must a side of an e q u i l a t e r a l triangle be in order that its area s h a l l be twfce t h a t of an equilateral triangle whose a i d e is lo?

If stmllar triangles are drawn on the s i d e and on the altitude of an equilateral triangle, so that the side and altitude are corresponding s i d e s of the triangles, prove t h a t their areas are t o each o t h e r as 4 is t o 3 .

11.

Two pieces of wire of equal length are bent t o form a square and an equilateral triangle respectively. What is the ratio C of the areas of the two figures? A triangular lot has a i d e s

with lengths 130 ft., 140 ft., and 150 ft. The length of the perpendicular from one corner to the side of 140 ft. is 120 ft. A fence is to be erected perpendicular to the side of 140 ft. so t h a t the area of the lot 18 equally divlded How f a r from A along A a hould thia perpendicular be drawn?

.

13.

Prove the theorem: The midpoint of the hypotenuse of a r i g h t triangle is equidistant from the vertices.

[ sec

. 12-5 1

I 3 14 0'

8

14.

Prove Theorem 11-9 by using the following diagram and problem 13.

15. In this t r i a n g l e Prove that A ABC triangle

1 6

AR = RC = RB.

is a right

.

Prove: The geometric mean of two p o s i t i v e numbers I s leas than their arithmetic mean, except when the t w o numbers are equal, i n which case the geometric mean equals t h e a r i t h metic mean. (~int: Let the two given numbers be the d l s tances AH and HB, let HC be perpendicular to AB,

HC =, Band let M be t h e mid-point of E . Prove L ACB is a r i g h t angle and u s e the preceding two problems. ) with

17. Given: P-ABC is a triangular pyramid w i t h a aection RST p a r a l l e l to the base ABC. PY is perpendicular to the plane of the baae, and X is the intersection of PY w i t h the plane of A RST. 2 Prove: area area A RST ABC =

(g).

A

I

?4

H

B

P

C

*18.

In the figure,

A ABC

is a

r i g h t triangle, with hypot-

-

AB, and 5 is the altitude from C.

enuse

Let the areas of A ABC, A A C H , A CBH be K1, K2, K3 r e s p e c t i v e l y .

A

B

The following sequence of statements constitutes a d i f f e r e n t proof of the Pythagorean Theorem. Give a reason for each of the following statementa:

Preamble. In the following problem, the lengths of two sides and the included angle of a t r i a n g l e are given, and it is required to find the length of the thLrd s i d e . By the S .A .S congruence theorem, the third a i d e is uniquely determined, so there should be a method of finding it numerically. Another way of g i v i n g t h e , included angle is to give a represehtative right triangle in which the angle (or i t s supplement) is one of the a c u t e angles. Actual$, only the nwnber k = RS la needed. For nmerlcal work, t h i s number, which depends on L R , has been tabulated; and if this t a b l e is readily available t h e computation of the length of the third side ia q u i t e straightforward. The number k is called the cosine of L R, abbreviated k = cos L R, and the table is c a l l e d a table of cosines. For this reaaon the formula f o r a2 t h a t we f l n d is called the cosines. You will meet it again in trigonometry.

.

400

*19.

In t h e two triangles shown In the diagram, L A L R , AC = b, AB = c, RS = k and L S is a right angle. Find a In terms of b , c, and k.

B

(~int: L e t D be the foot of t h e altitude t o E, and let x, y, h be as indicated in the figure. Express a' in t e r n of h and y; express h and y i n terms of x, b, and c ; then, from the similari t y AADC A FIST, express x in terms of b and k.)

-

*20.

In t h e two triangles shown in the diagram, L BAC is the supplement of L PI, and AC = b, Db x \A AB = c, RS = k and L S is a right angle. Find a in terms of b , c and k.

a =?

B Y

Y

T

(~int: L e t D be t h e foot of the perpendicular to AB from C. Then A ADC - A RST.) f,

'21:

a.

L e t ma AABC,

b.

Let

A

ma,

ABC,

be the length of t h e m e d i a n to the side of and l e t BC = a, CA = b, AB = c. Prove t h a t

mb,

m,

be t h e lengths of t h e medians of

w i t h sides of length

.

1sec 12-51

a,

b,

c.

Prove that

Review Problems -

1.

2.

In the figure

HQ

I(E.

d.

If H A = 6 , FB=12, m = 3 , QB= ?

a.

Are the two triangles

pictured here, similar if AB = 4, AF = 9 , &R = 3 , and AT = ?

A3=5, AT=3, 4 A& = 4--, what must AF 5 be to make A TAQ .v A BAF?

Give the geometric mean and the arithmetic mean f o r each of the following: 8

a. 4.

5.

0

If

b.

3.

T

and

10

6&

b.

and

3&.

Sketch two flgurea which are not similar, but which have the sides of one proportional to the corresponding s i d e s of t h e other. B I n r i g h t A A B C , if FC is the altitude to the hypotenuse, AF 12 and BF = 3 , find

-

AC,

PC

and

BC.

A

C

6,

If CD = x 4- 3 , DA = 3x + 3, C E = 5 and E B = x + 5, what must be the value of x to assure that % (1 E?

7,

Given in this figure,

LB*LD, Prove BD

8,

A side of one e q u i l a t e r a l triangle is congruent

to an altitude of another e q u i l a t e r a l triangle. What 1s the ratio of t h e i r areas?

9.

A

CD=4AB. = 5BE.

-

In A A B C , A C l E , AB = 20 and FC = 8, a, b , x, and y.

A DEF

ZlG,

and 4 A C B -

If

11.

Given rectangle ABFQ as shown in the figure w i t h WX AF.

1

Prove:

F

Find

10.

AABC-

x

A DEF,

ahow t h a t

AB=AC.

The t a l l e s t trees in t h e world are t h e redwoods along the c o a s t of' n o r t h e r n C a l f f o r n i a . To measure one of these g i a n t s you move some distance from the tree and drive a s t a k e i n the ground. Then you hold a small m i r r o r at ground level and sight it in, moving away from the stake u n t i l t h e top of the stake and t h e t o p of the tree are in a d i r e o t l i n e . If your stake 1s 5 feet t a l l and is 520 feet from the base of the tree, and if your m i r r o r is 8 feet from the stake when the t o p of the stake and the t o p of the tree are in a s t r a i g h t line, how t a l l is the tree? 13.

*14.

I n right A ABC with CF t h e altitude to the hypotenuse, and l e n g t h s as indicated in the figure, find x, y, and w.

A

9

Join the v e r t i c e s of A ABC to a point R outside the triangle. Through any point X of - -draw XY 11 AX3 meeting BR at Y. Draw YZ 1)BC meeting RC at 2. Prove AXYZ .v A ABC.

AR

15.

When we photograph a t r i a n g l e , is t h e p i c t u r e always similar to the o r i g i n a l triangle? When can we be s u r e t h a t it is?

Chapters 7 t o 12

REVIEW EXERCISES

Write (1) if the s t a t e m e n t is true and ( 0 ) If it is false. able to explain why you mark a statement f a l s e .

Be

a triangle is larger than any interior angle of the t r i a n g l e .

1.

An exterior angle of

2

In space there is only one perpendicular to a given line through a given external p o i n t .

3.

The angle o p p o s i t e the longest side of a t r i a n g l e is always the largest angle.

4.

In

5.

If

6.

A triangle

AABC,

if r n L ~ < r n L B , then

A?; 1 E,

then

P.B


r = F .

(4)

I and

From ( 2 ) and (3)

Therefore A A I B I C 1 *- A ABC by the S.S.S. S i m i l a r i t y Theorem, 2 a r e a A AtBIC1 = by Theorem 12-7. area ABC

(i)

Theorem 16-5. In any pyramid, the ratio f the area of a S c r o s s - s e c t i o n and the area of the base is ( ) , where h is the altitude of the pyramid and k is the distance from the vertex t o the plane of the cross-section.

------------

I

I I

I I

I I

I I' h

1 I I I

t I I I

Let us c u t up the base into triangular r e g i o n s w i t h areas A , , , (1n the figure, n = 4. ) Let Al'j A 2 I , ... , %t be the areas of the correspondfng triangular regions in the cross-section. Let A be the area o f the bane, and l e t A ' be the area of the cross-section. Then Proof:

... . A

= A 1 + % + . . . + % ,

and A ' = AI1

+

A2'

i-

.,. + An'.

By the r e s u l t which we have just proved f o r t r f a n g u l a r pyramids,

we know t h a t

Alt

=

k 2

A1,

%'

=

k

$,

and so on.

merefore

Theref ore

A'

2

=

(El ,

which was to be proved.

Theorem 16-5 has the following consequence.

-

Theorem 16-6. he Pyramid Cross -Sec t i o n Theorem. ) Given t w o pyramids with the same altitude. If the bases have the same area, then cross-sections equidistant from the bases also have

In the f i g u r e , for the sake of simplicity, we show triangular pyramids, but the proof does n o t depend on the shape of the base. L e t A be the area of each of the bases, and l e t A1 and A2 be the areas of the cross-sections. Let h be the a l t i t u d e o f each of the pyramids, and l e t d be the d i s t a n c e between each c r o s s - s e c t i o n and the corresponding base. Then the vertices of the two pyramids a r e at the same distance k = h - d from the planes of the cross-sections Therefore

.

by the previous theorem. Slnce the denominators on the l e f t and i r f g h t are equal, so also are the numerators. Therefore, A1 = A2, which was to be proved.

i

Problem -S e t 16-2 1.

If the base of a pyramid is a square, each cross-section wlll be a . If the base of a pyramid i s an e q u i l a t e r a l triangle whose s i d e is 9 , each cross-section will be and the length of a side of the croas-sectlon one-third of the distance from the vertex to t h e base w i l l

2.

Given two pyramids, one triangular, one hexagonal, with equal base areas. In each t h e altitude is 6 inches. The area of a cross-section of t h e triangular pyramid, 2 inches from the base, is 25 square i n c h e s . What is the area of a cross-section 2 inches from the base of the hexagonal pyramid?

3.

A r e p l a r pyramid is a pyramid whose base is a regular

polygonal r e g i o n having for its c e n t e r the foot of the perpendicular from t h e vertex t o the base. Prove that the l a t e r a l faces of a regular pyramid are bounded by congruent i s o s c e l e s triangles.

*4.

5.

Given a t r i a n g u l a r pyramid with vertex V and base B C , find a plane whose i n t e r s e c t i o n w i t h the pyramid is a parallelogram.

Show t h a t the l a t e r a l area of a regular pyramid is g i v e n by 1 A = 2 ap In which p is the perimeter of t h e base and a is the altitude of a l a t e r a l face.

6,

is parallel to base ABCDE in the pyramid shown here, with altitude VS = 7 inches and altitude VR = 4 inches. If the area of ABCDE is 336 square inches, what Is t h e area of FGH JK? FGHJK

A regular pyramid has a square base,

10 inches on a s i d e , and is one foot t a l l . Find t h e l a t e r a l area of the pyramid and the area of the cross-section 3 Inches above the b a s e ,

Prove: In any pyramid, the r a t i o of the area of a cross-section to the area of the base 1s

(E)2 ,

where a 1s the length of a l a t e r a l edge of the smaller pyramid and b is t h e corresponding l a t e r a l edge of the larger pyramid. (~int: D r a w altitude F. )

.

Volumes of Prism and Pyramids, CavalieriTs P ~ i n c i p l e A vigorous treatment o f volumes requires a careful definition o f something analogous t o polygonal regions in a plane (polyhedral regfons is t h e name) and t h e i n t r o d u c t i o n of p o s t u l a t e s similar t o t h e four area p o s t u l a t e s . We will n o t give such a treatment, but instead will r e l y on your I n t u i t i o n to a considerable extent, p a r t i c u l a r l y when it comes to c u t t i n g up solids o r f i t t i n g them t o g e t h e r . However, we wlll s t a t e explicitly the two numerical postulates we need. One of them is t h e analog of Postulate 20, which gave the area of a rectangle. 16-3.

-

Postulate 21. The volume of a rectangular parallelepiped is t h e product of t h e altitude and the area of the base.

To understand what is going on in our next p o s t u l a t e , l e t us first think of a physical model. We can make an approximate model of a square pyramid by forming a stack of thin c a r d s , c u t to the proper size, l i k e t h i s :

The figure on the l e f t represents the exact pyramid, and the figure on the right is the approximate model made from cards.

Now suppose we drill a narrow hole in the model, from the top to some point of the base, and i n s e r t a t h i n rod so that it goes through every card in the model. We can then tilt the rod in any way we want, keeping its bottom end f l x e d on the base. The shape of the model then changes, but its volume does not change. The reason is that I t s volume is simply the total volume of the cards ; and t h i s t o t a l volume does n o t change as the cards s l i d e along each o t h e r . The same principle applies more generally. Suppose we have t w o solids with bases in a plane which we shall think of aa h o r i z o n t a l . If a l l h o r i z o n t a l crosa-sections of the two s o l l d s a t t h e same l e v e l have the same area then the two s o l i d s have t h e same volume.

The reason is that if we make a caM-model of each of the s o l i d s , then each card in the first model has exactly the same volume as the corresponding c a r d in the second model. Therefore the volumes of the two models are exactly the same. The approximation given by the models is as close as we please, if only the cards a r e thin enough. Therefore the volumes of the two s o l i d s t h a t we s t a r t e d with are the same.

The principle Involved here is called Cavallerils P r i n c i p l e . We haven't proved it; we have merely been explaining why it is reasonable. L e t us t h e r e f o r e s t a t e it in t h e form of a postulate:

-

Postulate 22. (cavalierils p r i n c i p l e . ) Given two s o l i d s and a plane. If for every plane which i n t e r s e c t s t h e a o l l d a and is parallel to t h e given plane the t w o Intersections have equal areas, then the t w o s o l i d s have the same volume.

Cavalierits P r i n c i p l e is the key to the calculation of volumes, as we shall soon see. Theorem 16-7. The volume of any prism is t h e product of the altitude and the area of the b a s e .

Proof: L e t h and A be the altitude and the base area of the given prism. Consider a rectangular parallelepiped with the same altitude h and the base area A , and with i t s base in the same plane as the base of the given p r i s m . We know by the P r i s m Cross -Sec tion Theorem t h a t all c r o s s -sec t i o n s , f o r b o t h prisms, have the same area A . By Cavalierits P r i n c i p l e , this means t h a t they have t h e same volume. Since the volume of t h e rectangular parallelepiped is Ah by P o s t u l a t e 21, the theorem follows.

Theorem 16-8. If two pyramids have the same altitude and the same base area, then they have the same volume.

Proof: By the Pyramid Cross-Section Theorem, corresponding cross-sections of the two pyramids have the same area. By C a v a l i e r i t a P r i n c i p l e , t h i s means t h a t t h e volumes are the same.

Theorem 16-9. The volume of a triangular pyramid I s onethird t h e product of i t s altltude and its base area.

Proof: Given a triangular pyramid with base PQR and vertex S , we take a t r i a n g u l a ~ p ~ i s r n PQRTSU w i t h the same base and altitude, like t h i s :

W e next c u t the prlsm F n t o three triangular pyramids, one of them being the o r i g i n a l one, l i k e t h i s :

P

Think of pyramids I and TI as having bases P W and PRU, and comon vertex S . The two triangles b FTU and A PRU lie in t h e same plane and are congruent, s i n c e they are the two triangles i n t o which the parallelogram PTUR is separated by the diagonal UP. Hence pyramids I and f I have the same base area and the same altitude (the distance from S to p l a n e PTUR),

and so by Theorem 16-8 they have the same volume, In t h e same way, thinking of pyramids I1 and 111 as having bases S U R and S&R and common vertex P, we see that I1 and I11 have t h e same volume. Therefore the volume of all three pyramids is the same number, V, and the volume of the prfsm is 3V. If area A PQ.R = A and the a l t i t u d e of SPQR = h , then 3V = Ah, whence V = 1 Ah whlch was to be proved. I

The same result holds for pyramids in general: Theorem 16-10. The volume of a pyramid is one-third t h e product of its altitude and i t s base a r e a .

Given a pyramfd of altitude h and base a r e a A . Take a triangular pyramid of the same altitude and base area, w i t h it3 base in the same plane. By the Pyramid Cross-Section Theorem, cross-sections a t the same l e v e l havd the same area, Therefore, by Cavalierl's P r i n c i p l e , the two pyramids have the same volume. 1 Therefore the volume of each of them is 5 Ah, which was to be proved.

Proof:

Problem -S e t 16-3

5' x 4 '

is f i l l e d with water t o a d e p t h of 9 How many cubic feet of water a r e in t h e tank? How many g a l l o n s ? (1 gallon = 231 cubic inches . )

1.

A rectangular tank

2,

A lump of

3.

If one fl s h requires a g a l l o n of water for good h e a l t h , how many f i s h can be kept in an aquarium 2 feet long, ;1 1 feet wide, and 1 feet deep? C

4.

If one edge of the base of a regular hexagonal pyramid Is 12 inches and the altitude of t h e pyramid is 9 inches, what is the l a t e r a l area? What Is the volume?

metal submerged in a rectangular tank of water 20 Inches long and 8 inches wide raises the level of the water 4.6 Inches. What is the volume of t h e metal?

A

5.

The volume of a pyramidal t e n t w l t h a square base is 1836 cubic feet. If the side of the base is 18 feet, find the height of the t e n t . A plane b i s e c t s the altitude

o f a pyramid and is parallel t o i t s base.

What is t h e ratio of the volumes o f t h e solids above and below t h e plane?

*7.

A monument has the shape of an

obelltsk -- a square pyramid c u t off a t a c e r t a i n height and capped w i t h a second square pyramid. The vertex of the small pyramid is 2 f e e t above i t s base and 32 feet above the ground. If t h e base pyramid had been continued to i t s vertex It would have been 60 feet t a l l . Find the volume o f the o b e l i s k if each side of the base, at the ground, is 4 feet long.

*8.

S t a t e and illustrate a p r i n c i p l e , corresponding to Cavalierits

Principle, having the conclusion that two plane regions have equal areas.

Cylinders -and Cones, Note that in the d e f i n i t i o n of a prism, and of a s s o c i a t e d terms l n Section 16-1, it is not necessary to r e s t r i c t K to be a polygonal region. K could in f a c t be any p o i n t s e t i n El, Such tremendous generality is n o t needed, but we certainly can consider the.case i n which K is, a circular region, the union of a c i r c l e and its i n t e r i o r . In this case we c a l l the r e s u l t i n g s o l i d a c i r c u l a r cylinder. You should wrfte out a definition of a c i r c u l a r cylinder for y o u r s e l f . You can use the following f i g u r e t o help you. 16-4.

We c a n have c y l b d e r s w i t h o t h e r kinds of bases, such as e l l i p t i c c y l i n d e m , but the c i r c u l a r cylinder is by f a r the most common and the only one considered in elementary geometry. J u s t as t h e d e f i n i t i o n of a c i r c u l a r cylinder is analogous to t h a t of a prism, the d e f l n i t i o n of a c i r c u l a r cone la analogous to the deflnition of a pyramid. Check your understanding of t h i s by writing out a definition of' a c i r c u l a r cone. You c a n use the notation of the following figure to h e l p you.

D e f i n i t i o n : If the center of the base c i r c l e is t h e f o o t o f the perpendicular from V to E, the cone is c a l l e d a right circular cone. The f o l l o w i n g analogs of the theorems on prisms and pyramids are provable by the same general methods. We omit t h e details.

Theorem 16-11. A cross-section of a c i r c u l a r cylinder is a circular region congruent to t h e base.

I

Idea of proof: L e t C be t h e center and the base. Then, by parallelograms, PICl = PC

the r a d i u s of

r =

r.

Theorem 16-12. The area of a c r o s s - s e c t i o n of a c i r c u l a r cylinder is equal to the area of the base. Theorem 16-13. A cross-section of a cone of altitude h, made by a plane a t a distance k from the vertex, is a c i r c u l a r region whose area has a r a t i o to the area of the base of

(k)2.

Idea o r p r o o f :

Let

VU = h .

(I) A VQT * A VPU.

(2)

A VW*A WW.

Since PW has a c o n s t a n t v a l u e , regardless o f the p o s i t i c n o f W, then QR has a constant v a l u e . Thus, a l l p o i n t s R 1 I e o n a ctrcle. Thc corresponding c l r c u l a r region is the cross-section.

('1

area of c i r c l e with c e n t e r Q - k area oi c i r c l e with c e n t e r P - (8)

We can now use Cavalleri's Principle to find the volumes of cylinders and cones . The volume of a c i r c u l a r c y l i n d e r is t h e the altitude and the area of t h e base.

Theorem 16-14,

product of

Proof like t h a t of Theorem 16-7.

-

Theorem 16-15. The volume of a c i r c u l a r cone is one-third the product of the altitude and the area of the base.

Proof l i k e t h a t of Theorem 16-10.

--

Problem Set 15-4

1.

Find t h e volume of t h i s r i g h t c i r c u l a r cone.

(J-0

--I-

---6---

2.

'3.

4.

-

Find the number of gallons of water which a c o n i c a l tank will hold if it is 30 i n c h e s deep and the radius of the c i r c u l a r t o p is 14 inchea. h here are 231 cubtc inches in a 22 22 gallon. Use as an approximation of n. Why is Ta more convenient approximation than 3.14 in problems c o n t a i n i n g the number 231?) A drainage t i l e is a cylindrical s h e l l

16 inchea long, The inside and o u t s i d e diameters are 5 i n c h e s , and 5.6 Fnches. Find the volume of clay n e c e s s a r y t o make t h e t i l e . A c e r t a i n cone has a volume of

rl

cubic inches. I t s h e i g h t is 5 inches. A second cone is cut from the f i r s t by a plane parallel to the base and two inches below t h e vertex. Find the volume of the second cone. [sec.

16-41

5.

On a shelf in t h e supermarket a tand two cans of imported o l i v e s . The first is twice as tall as the second, but the second has a diameter twice that of the f l r s t . If the second c o s t s twice as much as the firat, which i s t h e better buy?

6.

In t h i s figure we are looking down upon a pyramid, whose base Is a square, inscribed in a right c i r c u l a r cone. If the altitude of the cone or pyramid is 36 and a baae edge of t h e pyramid is

20,

find the volume of each.

7.

Flgure 1 represents a cone

in a cylinder and Figure 2, two congruent cones in a cylinder. If the cylinders are t h e same s i z e , compare the volume of the cone in Figure I with the volume of t h e two cones in Figure 2. Would your conc l u ion be changed if the cones in Figure 2 were n o t congruent? F i g . I.

8.

A right

Fig. 2.

c i r w l a r cone stands f n s l d e a r i g h t c i r c u l a r cylinder

of same base and height. Write a formula f o r the volume of the space between the cylinder and the cone.

*9.

If a plane parallel to t h e base o f a cone (or pyramid) c u t s off another cone (or pyramid) then t h e solid between the parallel plane and the base Is c a l l e d a frustum, A frustum of a cone has a lower radius of 6 inchea, an upper radius of 4 inches and a height of 8 inches. Find its volume.

16-5.

Spheres; Volume -and Area. By the volume of a sphere we mean t h e volume of the s o l i d which is t h e union of t h e sphere and i t s i n t e r l o r .

--

-

Theorem 16-16. The volume of a sphere of' radius

r

is

Proof: Given a sphere of radius r, let E be a tangent plane. In E take a c i r c l e of radius r and c o n s i d e r a right cylinder with t h i s c i r c l e as base, altitude 2 and lying on the same s i d e of E as the sphere.

F i n a l l y , c o n s i d e r two cones, with the two bases of the cylinder as t h e i r bases, and t h e i r common vertex V at the mid-point of the a x i s of the cylinder. Take a cross-section of each s o l i d by a plane parallel t o E and at a distance s f r o m V. The cross-sections will look l i k e this :

The area of t h e s e c t i o n of the sphere is

by the Pythagorean Theorem. We wlsh t o compare t h i s with t h e s e c t i o n of the s o l i d l y i n g between the cones and t h e c y l i n d e r , that is, o u t s i d e the cones, but Inaide the c y l i n d e r . This s e c t i o n is a c i r c u l a r r l n g , whose o u t e r radius is r and whose inner radius is a . (Why?) Hence, i t s area is

Thus,

A1 = A2,

and by Cavalierirs Principle the volume of the

sphere is equal t o the volume between the cones and the cylinder. Therefore the volume of t h e sphere is the difference of the volume of the cylinder and twice the volume of one cone, that is, 4 3

Using t h e formula f o r the volume of a s p h e r e , we can g e t a formula for the area of the surface o f a sphere. Given a s p h e r e of radius r, form a s l i g h t l y larger sphere, of radius r + h . The s o l i d lying between the two s p h e r i c a l surfaces is c a l l e d a spherical s h e l l , and looks l f k e this:

L e t the surface area of the Inner sphere be S . The volume V o f v t h e s h e l l I s then approximately IS. Thus, approximately, S = h' As the shell g e t s thinner, the approximation g e t s better and better. Thus, as h gets smaller and smaller, we have

+s. But we can calculate exactly, and see what it approaches when h becomes smaller and smaller. This w i l l t e l l us what S is. The volume V is the d i f f e r e n c e of the volumes o f t h e two spheres Theref ore : 4 3 4 3 V =~ ( + hr) -

.

(You should check, by multiplication, t h a t 3 3 2 3 = r +.3r h + 3 r h + h .) Therefore

r2

+

3rh

+

(r

+

h) 3

h2]

I Here

the e n t i r e second term approaches zero, because h--+0. 2, and so S, = 4 n2 Therefore 4 Thus re have the theorem:

.

Theorem 16-17. The surface area of a sphere of radius

r is

S = 4 m2 . Thus we end t h i s chapter with the interesting f a c t that the surface area of a sphere of radius r is 4m2 Have you n o t i c e d that t h e surface area is exactly 4 times as great as the area of a great circle of t h e sphere?

.

Problem S e t 16-5

1.

Compute the aurface area and the volume of a sphere having diameter 8.

2.

The radius o f one sphere is twice as great as the mdiua of a second sphere. State a r a t l o expressing a comparlson of t h e i r surface areas; their volumes. If t h e radiua of one sphere is three times as great as the radius of another sphere, compare their surface areas; their volumes.

3.

A spherical storage tank has

gallons will it hold?

4.

a radius of (use n =

A large storage shed is in

7.1

7

feet.

How many

the

shape of a hemisphere. The shed is to be painted. If the f l o o r of the shed required 17 gallons of p a i n t , how much p a i n t will be needed to cover the exterior of the shed?

5.

It was shown by Archimedes (287-212 B . C . ) that the volume of 2 a sphere is 3 that of t h e smallest right c i r c u l a r cylinder which can c o n t a l n it, V e r i f ' y thfs.

6.

An i c e cream cone

5 inches deep and 2 inches in t o p diameter has placed on top of it two hemispherical scoops of ice cream alao of 2 inch diameter. If the i c e cream melts l n t o the cone, will it overflqw? [aec.

16-53

a,

Show that if the length of a a i d e of one cube is f o u r times that of another cube the ratlo of t h e i r volumes is 64 to 1.

b.

The moon has a diameter about

that of the earth.

How do their volumes compare?

In the figure, the sphere, with radlus r, is inscribed in the cone. The measure of the angles between the altitude and the r a d i i to p o i n t s of tangency are as shown. Find the volume of the cone In terms of r. The c i t y engineer who was a F x feet t a l l walked up to inspect the new spherical w a t e r tank. When he had walked to a p l a c e 18 feet from the point where the tank rested on the ground he bumped h i s head on the tank. Knowing that the c i t y used 10,000 gallona of water per hour, he immediately figured how many hours one tank full would l a s t . How did he do it and what was h l a result?

Half the air is l e t out of a rubber balloon. If it continues to be spherical I n shape how does the resulting radlus compare with the original radlus? Use the method by which Theorem 16-17 was derived to ahow that the l a t e r a l area of a right c i r c u l a r cylinder 1s 2ma where a Is its altitude and r the radius of its base.

Review Problems

If the base o f a pyramid I s a region whose boundary is a rhombus w i t h s l d e 16 and an angle whose measure Is 120, then

a. b

.

c

.

any c r o s s -see t l o n I s a region whose boundary is a and whose angles measure and the l e n g t h of a s i d e of a c r o s s - s e c t i o n midway between the v e r t e x and the base is

the area of a cross-aectlon midway between the vertex and the base is

spherical b a l l of diameter 5 has a bollow c e n t e r of diameter 2. Find the approximate volume of the shell. A

Find the altitude of a cone whose radius is volume is 500.

5 and whose

an altitude of 12 inches and volume of 432 cubic inches. What is the area of a cross-section 4 inches above the base? A pyramid has

Given two cones such that the a l t l t u d e of the f i r s t is twice t h e altitude of t h e second and the radius of the base of the f i r s t is half the radius of the base of the second. How do t h e volumes compare?

can with radius 12 and h e i g h t 20 is f u l l of water. If a sphere of radius 10 Is lowered into the can and then removed, what volume of water will remain In the can? A cylindrical

A s p h e r e is i n s c r i b e d in a right circular cylinder, so that

it is tangent to b o t h b a s e s . What is t h e ratio of the volume of t h e sphere to t h e volume of t h e c y l i n d e r ?

*8.

9.

#lo.

The altitude of a r i g h t c i r c u l a r cone is 15 and t h e radius of i t s base is 8, A cylfndrical h o l e of diameter 6 is drilled through the cone w i t h the center of the d r i l l following the axls of the cone, leav3mg a s o l i d as shown in the figure. What is the volume o f t h i s s o l i d ? Prove: If the base of a pyramid is a parallelogram region, the plane determined by the vertex of the pyramid and a dfagonal of the base d i v i d e s t h e pyramid i n t o two pyramids of equal volume. Prove that a sphere can be circumscribed about a rectangular parallelepiped.

Chapter 17 FLANE COORDINATE GEOMETRY

17-1. Introduction. Mathematics is the only science in which practically n o t h i n g ever has to be thrown away. Of course, mathematicians are people, and being people, they make mistakes. But these mistakea usually get caught p r e t t y quickly. Therefore, when one generation has learned something about mathematlca, the next generation can go on to learn some more, without having to s t o p to c o r r e c t seriow errors in the work t h a t was supposed to have been done already. One symptom of thia a l t u a t l o n is the fact that nearly everything that you have been learning about geometrg, so f a r in t h i s course, was lmown t o the ancient Greeks, over two thousand yearn ago. The f i r s t r e a l l y big step forward in geometry, a f t e r t h e Greeks, was in the seventeenth century. This was the discovery of a new method, c a l l e d coordinate aeometq, by Rene Descartea (1596-1650). In t h i s chapter we will give a s h o r t i n t r o d u c t i o n to coordinate geometry j u s t about enough to give you an Idea of what it is like and how it works.

--

17-2.

Coordinate Systems cIn a- Plane.

In Chapter 2 we learned how to s e t up coordinate systems on a line.

Once we h a v e s e t up a coordinate system, every number describes a p o l n t , and every point P is determined when its coordinate x is named.

I n coordinate geometry, we do t h e same sort of t h i n g in a plane, except t h a t in a plane a point is described n o t by a s i n g l e number, but by a pair of numbers. The scheme works llke this:

F i r s t we take a line X in the plane, and s e t up a coordinate system on X . T h i s line will be c a l l e d the x-axis. In a figure we u s u a l l y use an arrow-head to emphasize the p o s i t i v e d i r e c t i o n on the x-axis. Next we let Y be the perpendicular to the x-axis through the point 0 whose coordinate is zero, and we s e t up a coordinate syatembn Y. By the Ruler Placement P o s t u l a t e this can be done so that p o i n t 0 also has coordinate zero on Y. Y will be c a l l e d the y-axis. As before, we indicate the p o s i t i v e direction by an arrow-head. The h t e r s e c t i o n O of the two axes is c a l l e d

the o r i ~ i n . We can now describe any point In the plane by a p a i r o f numbers. The scheme I s t h i s . Given a point P, we drop a perpendicular t o the x-axis, ending at a point M, with coordinate x. We drop a perpendicular to the y-axis, ending at a point N, with coordinate y. (1n accord with S e c t i o n 10-3 we can call M and N t h e projections of P into X and Y.)

The numbers x and y are called the coordinates of t h e point P; x is the x-coordinate and is the s-coordinate. Definitions :

y

1 In the figure x = l1 and y = 9. The point P therefore 7 has coordinates lT1 and We wrlte these coordinates in the

4.

form

1 1

t h a t point

1 ,

P

giving the

x-coordinate f i r s t .

has these coordinates we write

To l n d l c a t e 1 1 ~(1,,%) or

L e t us look at some more examples. Y b

4

P3r---" I

--

3

I

.--I

2-.---

I

I I

I

5:------I I

-5

1

I

----+--I

I

I

I

I

I

I

4

:P

1

-4

-3

1-2 I

I

I

I I I I

-1

I

I I

;2

I1

-I--2-.-------

1 I

I ps'---------

P

TI '

-3

I 1

re3

t 4

3

I

I I

14

5

X

II I 1 I

!I I

I I i

-4

-.-------------- A P T

We read off the coordinates of the p o i n t s by following the dotted l i n e s . Thus the coordinates, in each case, are as follows:

Notice t h a t t h e o r d e r i n which the coordinates a r e w r i t t e n makes a difference. The p o i n t with c o o r d i n a t e s ( 2 , l ) l a n o t the same as point 1 2 Thus, the coordinates of a p o i n t are really an ordered pair of r e a l numbers, and you canlt tell where the p o i n t is unless you b o w the order in which the coordinates a r e given. The convention of having the f i r s t number of t h e ordered pair be t h e x-coordinate, and the second the y-coordinate, is h i g h l y important. J u s t as a s i n g l e l i n e separates the plane i n t o t w o p a r t s ( c a l l e d half-planes) so the t w o axes separate the plane i n t o f o u r p a r t s , called quadrants. The quadrants a r e identified by number, like this:

.

Y

We have shown that any p o i n t of o u r plane determines an ordered pair of numbers. Can w e reverse the p r o c e s s ? That is, given a pair of numbers (a, b) can we find a p o i n t whose coordinates are ( a ,b ) ? The answer is e a s i l y seen t o be "yes". In fact, there I s exactly one such point, obtained as the i n t e r s e c t i o n of t h e line perpendicular to t h e x-axis a t the p o i n t whose c o o r d i n a t e is a and the line perpendicular to the y-axis at the p o i n t whose c o o r d i n a t e is b. Thus, we have a one-to-one correspondence between p o i n t s in t h e plane and ordered p a i r s of numbers. Such a correspondence is called a coordinate system in the plane. A coordinate system is s p e c i f i e d by choosing a measure of distance, an x - a x f a , a y-axis perpendicular t o it and a p o s i t i v e d i r e c t i o n on each. As l o n g as

we s t i c k to a s p e c i f i c coordinate aystem, which will be the case in all our problems in t h i s book, each point P i s a s s o c i a t e d with exactly one number pair (a,b), and each number pair w i t h exactly one point. Hence, it will cause no confusion if we s a y the number pair is t h e p o i n t , thus enabling us t o u s e such convenient phrases as "the p o i n t (2,3)I' or "P = (a,b) "

-

.

Points on Graph Paper. As a matter of convenience, we ordinarily use printed graph paper for drawing figures i n c o o r d f n a t e geometry. The horizontal and v e r t i c a l l l n e s are printed; we have t o draw e v e r y t h i n g e l s e for ourselves.

17-3.

How to Plot ---

In the f i g u r e above, the dotted lines represent the lines t h a t a r e a l r e a d y printed on t h e paper. The x-axis and t h e y-axis s h o u l d be drawn w i t h a pen or a pencil. Notice that t h e x-axis is labeled x rather than X; t h i s is customary. Here the symbol x is n o t t h e name of anything, but merely a reminder t h a t t h e c o o r d i n a t e s on t h i s axis are going t o be denoted by the l e t t e r x. S i m i l a r l y , [ s e c . 17-31

.

f o r the y-ax1 s Next, the p o i n t s with coordinates (1,O) and (0,l) must be l a b e l e d in o r d e r to indicate the unit to be used. This is the usual way o f preparing graph paper f o r p l o t t i n g p o i n t s . We could have Indicated a l i t t l e less or a lot more. For your own convenience, it 1s a good idea to show more than t h i s , But if you show l e s s , then your work may be actually u n i n t e l l i g i b l e . Note t h a t we could draw the axes in any of t h e following pos itions :

--

and so on. There 1s n o t h i n g l o ~ i c a l l ywrong with a- of these ways of drawing t h e axes. People f i n d it e a s i e r to read each othepla graphs, however, i f they agree at the o u t s e t t h a t the x-axfs is to be horizontal, with coordinates increasing from left t o right, and the y-axis l a to be vertical, with coordinates increasing from bottom to top.

Problem 1.

Set

Suggest why the kind o f coordinate system used in t h i s c h a p t e r

is sometimes c a l l e d "cartes ian"

.

2.

What are the coordinates of t h e o r i g i n ?

3.

What is the y-coordinate o f the pofnt

4.

Name the point which ia the proJection of

(7,-3)? (0,-4)

into t h e

x-axis.

5.

Which p a i r of points are c l o s e r together, or ( 2 , l ) an6 (2,0)?

6.

In which quadrant is each of t h e f o l l o w i n g p o i n t s ?

7.

What a r e the coordinates of a point which does n o t l i e in any quadrant?

8.

The following p o i n t s are projected Into the x-axis.

10.

(1,2)

Write them In such an o r d e r t h a t t h e i r p r o j e c t i o n s will be in o r d e r from left t o right. A: ( 6 , - 3 ) .

9.

(2,l) and

B: (-2,5)

.

C:

(0,-4).

D: I-'j,~).

If the p o i n t s in the previous problem are progected i n t o the y-axis arrange them ao t h e i r proJections will be in order from bottom t o t o p . If s is a negative nwnber and r a p o s i t i v e number, in what quadrant will each of the following p o i n t s lie?

Csec. 17-31

11.

a coordinate system on graph paper. Using segments draw some simple picture on the paper. On a separate paper list in pairs the coordinates of the end points of the segments in your picture. Exchange your list of coordinates w i t h another student, and reproduce the p i c t u r e suggested by his list of coordinates.

S e t up

*12. A t h r e e dimensional coordinate system can be formed by considering three mutually perpendicular axes as shown. The y-axis, while drawn on t h i s paper, represents a l l n e perpendicular t o the plane of the paper. The negative portions of the A2 x, y and z axes extend to the left, t o t h e rear, and down respectively. Taken in / / p a i r s the three axes determine (O,OJl*~ / three planes c a l l e d the yzbX plane, the xz-plane, and t h e xy-plane. A point (x,y,z) is located by i t s three coY ordinates: the x-coordinate is the coordinate of i t s projection into the x-axis; the y and z coordinates are defined in a corresponding manner.

-----

a.

On which axis will each of these points lie? (0,530);

b.

L LO,^);

(0,0381

On which plane will each of these p o i n t s lie? (2,0331;

c.

,'

(0,5,-7);

~1,190)

What is the distance of the point ( 3 , - 2 , 4 ) from the xy-plane? from the xz-plane? f r o m the yz-plane?

The Slope of-a-Non-Vertical Line. 17-4. The x-axis, and a l l l i n e s parallel to it, are called h o r l z o n t a l . The y-axfs, and a l l llnes parallel to it, are c a l l e d vertical. Notice that these terms a r e defined in t e r n of the

coordinate system t h a t we have s e t up.

the h o r i z o n t a l l i n e L1, all p o i n t s have the same y-coordinate b, because the point ( 0 , b ) on the y-axis is the foot of all the perpendiculars from p o i n t s of L1. Far the same s o r t of reason, a l l p o i n t s of the v e r t i c a l line L2 have t h e aame x-coordinate a . Of course, a segment 1s.horizontal ( o r v e r t i c a l ) I f the l i n e containing it is h o r i z o n t a l (or v e r t i c a l ) . Conslder now a segment PIP2, where = (xl,yl) and

On

P2

=

(x2,yp), and suppose that

PIPB is not- v e r t l c a l .

Deflnltion:

P1P2

The slope of

is the number

m =

Y2 X2

-

Yl XI'

This r e a l l y is a number: sfnce the segment is not v e r t i c a l , PI and P2 have d i f f e r e n t x-coordinates, and so the denominator is not zero. Some things about the s l o p e a r e easy t o s e e . (I) It is important t h a t the o r d e r of naming the coordinates is t h e same in t h e numerator -as in the denominator. Thus, if we wish t o f i n d t h e slope of PQ, where P = (1,3) and Q = [4,2) we can either choose P1 = P, xl = 1, y1 = 3, P2 = Q, x2 = 4, 1; y2 = 2, g i v i n g slope of PQ = 2 -- 3 = - T

-

giving slope of

1* PQ = 3 - 2 = - T

What we cannot say is

slope of

-

PQ =

-

or

2 = .

3

Notice t h a t if the p o i n t s are named in reverse order, t h e slope is the same as before. Algebraically,

Thus the value of m depends only on t h e segment, n o t on the order in which the end-points are named. (2) If m = 0 , then the segment is horizontal. (~lgebralcally, a f r a c t i o n is z e r o only if i t s numerator I s zero, and this means that yp = yl.) (3) If t h e segment s l o p e s upward from left to right, as in the left hand figure on page 578, then m > 0, because the numerator and denominator a r e both positive ( o r both negative, i f we reverse the o r d e r of the e n d - p o i n t s . )

If the segment s l o p e s upward f r o m r i g h t to left as in the r i g h t hand figure below, then m < 0. This is because m can be wrltten as a f r a c t i o n with a p o s i t l v e numerator y2 yl and a negative denominator x2 - xl ( o r equivalently, a negative numerator yl y2 and a p o s i t i v e denominator xl - x 2 ) (4)

-

.

-

(5) We do n o t trg to write the s l o p e of a v e r t l c a l segment, because t h e d e n o m t o r would be zero, and so the f r a c t i o n would be meaningleas. In either of the two figures above, we can complete a r i g h t triangle A P1P2R, by drawkg horizontal and v e r t i c a l lines through

PI

and

P2,

like t h l s :

Since oppoaite sides of a rectangle are congruent, it 1s easy to see t h a t

(1) if rn

>

0, then

m

=

md

Once we know t h i s much about s l o p e s , I t is easy t o get o u r f i r s t b a s i c theorem.

Theorem 17-1.

On a non-vertical line, all segments have the

same slope,

Proof:

There are three cases to be considered.

Case ( 3 )

Case (2)

I n e i t h e r of the other cases i l l u s t r a t e d above, a E 1 a t, and s i n c e the triangles are right trfangles, t h l s means that

A P1P2R Therefore, in e t t h e r case,

.v

A PI P2' R t

.

I n Case (2), these f r a c t i o n s are the slopes of

PIP2 and P l t P g f and therefore the sements have the same s l o p e . In Case (31, the slopes are the negatlves of t h e same f r a c t i o n s , and a r e t h e r e f o r e equal. Theorem 17-1 means t h a t we can talk not only about the slopes of segments but a l s o about t h e slopes of lines: the s l o p e of a non-vertical l i n e is t h e number m which is the s l o p e of every segment of the line.

Problem -S e t 17-4 1.

2.

3.

I

Replace the "?" in such a way that the line through the t w o points will be horizontal. a.

(5,7)

b.

( 0 - 1 and

and

(-3,?).

(4,?).

Replace the "?" in such a way that the line through the two p o i n t s will be v e r t i c a l ,

and

a.

(?,2)

b.

( - 3 1

and

(6,-4). (?,0).

By visualfzing the points on a coordinate system i n parts (a), (b), and ( c ) , g i v e the distance between: (7901

a.

(5,0)

b. c.

(591) and (791)(-3,-4) and (-6,-4)-

d.

What is a l i k e about parts (a), (b) and

and

4

(c)?

e.

S t a t e a rule giving an easy method f o r finding the

d i s t a n c e between such pairs o f p o i n t s .

f.

Does your rule apply to the d i s t a n c e between

(6,5)

(3,-5)?

4.

By visualizing the p o i n t s named i n parts ( a ) , (b), ( c ) and (d) on a coordinate system, give the distance between the p o i n t s in each part.

and (-3,-1).

b.

(-3,l)

d*

( x ~ , Y ~ and )

e.

What is alike about p a r t s (a), (b),

f

.

(x~,Y~)* ( c ) and ( d ) ?

S t a t e a rule giving an easy method for f i n d i n g the

dfstance between such pairs of p o i n t s .

5.

With perpendiculars drawn as shown below, what are the coordinates of A , B and C?

6.

Determine the distances from and C in Problem 5 .

7.

Compute t h e s l o p e of

8.

A road goes up

distance.

P

and

t o points

A,

B

PQ f o r each figure in Problem 5 .

feet f o r every What is i t s slope? 2

Q

30

feet of horizontal

and

9.

Determine the s l o p e of the segment joining each of the followi n g p o i n t pairs

.

(6,~).

and

a.

(0,O)

b.

(0,0) and

(2,-6).

c.

(3,5) and

(7,12)

d.

(0,0)

e.

(-5,7>

(-by-3).

and

(3,-8).

and and

10.

"11.

g.

(-2.8,3.1)

h.

( r1n , o )

and

(2.2,-1.9)

1

(o,~).

and

Replace the " ? " by a n u m b e ~so that the l i n e through the t w o p o i n t s will have the s l o p e given. (~int: Substitute in the s lope formula.)

- a.

( 5 , ~ ) and

b.

( - 3

and

(?,GI. (

m = 4. ?

m

=

1 7.

w

w

w

PA and PB are non-vertical l i n e s . Prove that PA = P 9 if and only if they have the same s l o p e ; and consequently t-* if PA and PB have d i f f e r e n t s l o p e s , then P y A and B,

cannot be collinear. 12.

a.

Is t h e p o i n t ~ ( 4 , 1 3 ) on t h e line j o i n i n g ~ ( 1 , l )to ~ ( 5 , 1 7 ) ? (~int: is t h e s l o p e of AB the same as that o f BC?)

b.

Is the p o i n t to (6,-a)?

( 2 - 1 ) on the segment joining

(-5,4)

13.

Determine the s l o p e of a segment joining:

c.

(a + b , a )

and

( a - b,b).

14.

Given A:(101,102), B : 6 c:(-95,-941, determine w u whether or n o t lines AB and BC coincide.

15.

~:(101,102), ~:(5,6), ~:(202,203), ~:(203,204). Are W and %? parallel? Could they possibly coincide?

6

D r a w t h e part of the first quadrant of a coordinate system having coordinates leas than or equal to 5 . Draw a segment through the o r i g i n which, if extended, would paks through

Given

P (~OOOOOOO, 6ooooooo)

.

-

and Perpendicular Lines. 17-5. Parallel It is easy to see t h e algebraic condition f o r

two

non-vertical

lines to be parallel.

-

If the lines are parallel, then A PQR A P I Q q R 1 , and it follows, as In the proof of the preceding theorem, that they have the same slope.

Conversely, I f two d i f f e r e n t lines have t h e same s l o p e , then they are parallel. We prove t h l s by the method of contradiction.

Ll and L2 are not parallel. If as ,shown in the figure P1 is t h e i r p o i n t of intersection, and P2 and Pg have the same x-coordinate x2, t h e s l o p e of

Assume as in t h e figure above that

Lf.

ml

is

=

Y2 2

Since

ml

#

yg

m2.

#

yg,

-

-

Y1 , and the s l o p e of X~

L2 is m2

=

Y3 2

-

Y1 X ~ '

the f r a c t i o n s cannot be equal, and hence

Thus our i n i t i a l assumption that the t w o l i n e s

L1 were n o t parallel has l e d us to a c o n t r a d i c t i o n of the and L hypothesis that ml = rn2. Hence the two lines L1 and L2 must be parallel. Thus we have the theorem:

Theorem 17-2. Two non-vertical lines are parallel If and only if they have t h e same s l o p e .

Now t u r n i n g to the condftion f o r two l i n e s to be perpendicular,

l e t us suppose t h a t we have given two perpendicular lines, n e i t h e r o f which is v e r t i c a l . f

P be their p o i n t of intersection. As lm the figure, let Q be a point of one of t h e lines, l y i n g above and t o the r i g h t of P. And let Q' be a point of the o t h e r line, lying above and to t h e left of P, such that PQ1 = PQ. We complete the right triangles h PQR and A Qt PRt as i n d i c a t e d in the f i g u r e . Then Let

Theref ore

Q f R t = PR

and

R I P = RQ.

and hence

Let

m

be the slope of

C*

PQ,

and let

rnl

be the s l o p e of

Then and

The ref ore

Q'R'

rnr

PR

=-m=-mm mt

= - -1

rn ' That is, the slopes of perpendicular l i n e s are t h e negative reciprocals of each other.

3.

-

1 Suppose, conversely, that we h o w that m 1 =- rn We then construct A PQR as before, and we construct the right triangle A Q1PR1 makfng R 1P = RQ. We can then prove that QIRl = PR; t h i s glves the same congruence, A PQB E A QlPRl, as before, and it follows that L QtPQ is a right angle and hence PQ PQ These t w o facts are s t a t e d together in the following theorem:

-

-.

Theorem 17-3. Two non-vertical l i n e s are perpendicular if and only if t h e i r s l o p e s are the negative reciprocals of each

other .

Notice that whfle Theorems 17-2 and 17-3 tell us n o t h l n g about vertical lines, they d o n ' t really need t o , because t h e whole problem of parallelism and perpendicularity is trivial when one of t h e lines is vertical. If L is v e r t i c a l , then L1 is parallel to L if and only if L' is also vertlcal (and different from L,) And if L is v e r t i c a l , then L' is perpendicular to L if and only if L1 is horizontal.

Problem Set 17-5 1.

Four p o i n t s taken in pairs determine six segments. Which palrs of segments determfned by the following four p o l n t s are p a r a l l e l ? ~ ( 3 , 6 ) ; ~ ( 5 , 9 ) ;~(8,2);~ ( 6 , - (caution: Two segments are not necessarily parallel if they have the same s l o p e ! )

2.

Show by considering slopes that a parallelogram is formed by drawing segments joining in order ( 15 , 3(5,1), ~(6,-2) and D(o,~).

3.

Llnes and

L1, 1

L2,

Lg and Lq

respectively.

have slopes

2 5,

-4,

-1i

Which pairs of lines are perpendicular?

It is asserted that b o t h of the quadrilaterals whose v e r t i c e s are given below a r e parallelograms. Without p l o t t i n g t h e points, determine whether or not t h i s is true. (1) A : I-5,-2),

B: ( - 4 , 2 ) ,

C: ( h Y 6 ) ,

D: (3,l).

(2) P

9:(4,2),

R: (9,1),

5 : (3,-3).

-

2

,

The vertices of a t r f a n g l e a r e

~ ( 1 6 , )~

( 9 ~ 2and )

c(0,o) a.

What are the s l o p e s of Its s i d e s ?

b.

What are the s l o p e s of its altitudes?

Show that the quadrilateral joinlng A ( - 2 , 2 ) , ~ ( 2 , - 2 1 , ~ ( 4 , 2 ) , and ~ ( 2 , 4 ) is a t r a p e z o i d w i t h perpendicular diagonals.

Show that a line through {3n,0) and a line through (6n,0) and (0,211). Show that a line through to a l i n e through ( 0 , 0 )

(0,0)

and

(0,n) is parallel t o

I s perpendicular

and ( a , b ) (-b,a)

.

Show t h a t if a triangle has vertices x(r,s), ~ ( n a - t r , n b + s) and z(-mb+ r,ma+ s ) it will have a right angle a t X.

Given the p o i n t s P ( ~ , P ) , Q ( 5 , - 6 ) and R(b,b); determine t h e value of b so t h a t L PQR is a r i g h t angle.

-P

= (a,l),

PQ

#

FE,

Q = (3,2), and t h a t I f

R = (b,l),

P& 11

S = (4,2).

then

a

=

b

-

Prove that 1.

17-6. The Distance Formula. If we b o w the coordinates of two p o i n t s PI and P2 then we know where the points are, and so the distance PIP2 is determined. L e t us now f i n d out how the distance can be calculated. What we want is a formula that gives P1P2 in terms of the coordinates xl, x2, yl and y2.

. I

1 I

I I MI!

p 2

.X

2 0 Let the projections MI, M2, N1 and N2 be as in the figure By the Pythagorean Theorem, ( P ~ P ~=] ( P ~ R ) + ( R P ~ ) XI

*.

*

PIR

also

=

MIM2

RP2 = N1N2,

and

because opposlte sLdes of a rectangle are congruent. Theref ore

( p , ~ , ) ~= ( M ~ M 4-~ ( I N~ ~ N ~ ) ~ ~

But we b o w t h a t

and Theref ore

(p1p2)

2

M1M2 ' 1 5

- X1l

N1N2

-

=

= 1y2

Ix2 -

xll

2

Y1l-

+ 1y2

-

yll

2

.

Of coume, the square of the absolute value of a number Is the same as the square of the number itself.

.

2

The ref ore

and since

P1P2

2

0, t h i s means that

T h i s is the f o m l a that we are looking f o r .

Thus we have t h e

theorem:

Theorem 17-4. the p o i n t s (xl,yl)

h he

Distance or mu la .) The distance between and ( x ~j , 2) is equal to

For example, take ~y formula,

P,P,

PI

=

= J(2

(-1,-3)

+

1l2

Pa = ( 2 , k ) .

and

+

(4

+

312

Of course, if we plot the p o i n t s , as above, we can g e t the same answer d i r e c t l y from the Pythagorean Theorem; the legs of t h e right trlangle A P1RP2 have lengtha 3 and 7 , so that

Jm

= as before. If we find the distance this way, P1P2 we are of course simply repeating the derivation of the distance formula in a s p e c i f i c case.

Problem -S e t 17-6 1.

2.

3.

4.

a.

Without using the distance formula s t a t e the distance between each pair of the pofnta: ~ ( 0 , 3 ) , ~(1,3), c(-3,3) and ~ ( 4 . 5 , 3 ) .

b.

Without using the d i s t a n c e formula s t a t e the distance between each pair of the points: ~(2,0), ~(2,1), C(2,-3) and ~(2,4.5).

a.

Write a simple formula f o r the distance between and (xp,k). ( ~ i n t : The p o i n t s would lie on a horizontal line.)

(xl,k)

b.

Write a simple formula f o r the distance between and ( k , ~ ~ )

(k,~~

Use the d i s t a n c e formula to find the distance between:

(-5,-7).

a.

( 0 ~ 0 )and

(3,k).

e.

(3,8) and

b.

( 0 , ~ )and

(3,-4).

f.

(-2,3) and

(-1,Q).

c.

(1,2) and

(6,14).

g.

(10,l) and

(49,811.

d.

(8,111

h.

(-6,3) and

(4,-2).

a.

Write a formula f o r the square of the distance between

and

the p o i n t s b.

(15,35). (xl,yl)

and

(x~,Y~).

Using coordinates write and simplify the statement: The square of t h e distance between (0,0) and (x,y) is 25.

5.

Show t h a t the t r i a n g l e with vertices R ( O , 0), s ( 3 , 4 ) and T l ) is isosceles by computing the lengths of I t s s i d e s .

6.

Using the converse of the Pythagorean Theorem show that t h e triangle joining D ( 1 ~ ( 3 , 0 ) and ~ ( b , 7 ) is a right t r i a n g l e w i t h a right angle a t D.

7.

Given the points A - 6 , ~ ( 1 , 4 ) and ~(7,-2). Prove, w i t h o u t p l o t t i n g the p o i n t s , t h a t B is between A and C.

8.

Suppose the streets i n a city formcongruent square blocks with avenues running east-weat and s t r e e t s north-south.

a.

If you f o l l o w walk from the the corner of length of 1

b.

What would be the distance "as the crow flies" between

the sidewalks, how f a r would you have to corner of 4th avenue and 8th s t r e e t t o 7th avenue and 12th street? (use the block as your unit of length. )

the same two corners?

9.

*lo.

Vertices W, X and Z of rectangle WXYZ (0,0), ( a , ~ ) and ( 0 , b ) respectively.

have coordinates

Y?

a.

What are the coordinates of

b.

Prove, using coordinates, that

a.

Using 3-dimensional coordinates (see Problem 12 of Problem Set 17-31, compute the distance between (0,0,0)

and b.

WY

=

XZ.

(2,3,6).

Write a formula for the distance between

(0,0,0) and

(x,Y,z) c.

Write a formula for the distance between and

p2(x2,yp. z 2 )

.

~ ~ ( x ~ , ~ ~ ,

17-7. The Mid-Point Formula. In S e c t i o n 17-8 we wlll be proving geometric theorems by t h e u s e of coordinate systems. In some of these proofs, we will need to f i n d the coordinates of t h e mid-point o f a segment PIP2 in terms of the coordinates o f PI and P2. F i r s t let us t a k e the case where x - a x i s , with xl < x2, l i k e t h i s :

PI

and

P2

and P is the mid-point, with coordinate x. S i n c e we h o w that PIP = x - x1 and PP2 = x2 - x . Since

P

are on the

xl


xl.)

*9.

a.

prove: and if PP1

If PI = (xl,yl), P is between PI

YP; = C* b.

then

x =

P2 and

mr 2 + SXl r + s

=

(x2,y2 ) and P = (x,Y) P2 such t h a t

and

y =

rY2

"

SY1

r + s

'

Use the result of p a r t (a) to find a point P on the aegment j o i n i n g p1(5,11) and ~ ~ ( 2 5 , 3 6 1such that

17-8. Proofs of Geometric Theorems. L e t us now put our c o o r d i n a t e systems to work in provlng a f e w geometric theorems. We start w i t h a theorem that we have already proved by other methods.

Theorem A . The s e p e n t between the mid-points o f two s l d e s of a triangle is parallel to the third s i d e and half as long. -r

Restatement: In h ABC and Then 11

z.

D and E be the mid-points o f and DE = %c. let

Proof : The flrst step In using coordinates to prove a theorem l i k e t h i s l a to introduce a suitable coordinate system. That I s , we must decide which l i n e is to be the x-axis, which the y-axis, and which d i r e c t i o n to t a k e as p o s i t i v e along each axis. We have many choices, and sometimes a c l e v e r choice can greatly s h p l i f y o u r work. In the present case it seems reasonably t3 + simple t o take BC as our x-axis, with BC as the p o s i t i v e + d i r e c t i o n . The y-axis we take to pass through A , with OA as the p o s i t i v e direction, llke t h i s :

The next s t e p is to determine the coordinates of the v a r i o u s p o i n t s of the figure. The x-coordinate of A is zero; the y-coordinate could be any p o s i t i v e number, so we write A = ( ~ , p ) , with the only restriction on p being p > 0. Similarly, B = ( q , ~ ) and C = ( r , ~ ) ,with r q. ( ~ o t et h a t we might

>

have m y of the cases q < r ( O, q < r = 0, g < 0 < r, 0 = q < r, 0 < q < r. Our figure i l l u s t r a t e a the t h i r d case.) The coordinates o f D and E can now be found by the mid-point formula. We get

Therefore the slope of

is

5-%

0

==r;9=

0,

( s i n c e q # r t h e denominator I s n o t z e r o ) . Likewise, the slope of BC is 0

and so

DE

f-4 1 1 E.

and so that

DE

-

=

BC I

0 = 0;

Finally, by t h e distance formula,

=

PC.

J(r

-

q18+

(O

- o)* = r -

q,

The algebra in t h i s proof can be made even e a s i e r by a simple device. Instead of setting A = ( ~ , p ) , B = ( q , ~ , ) C = ( r , ~ )we could just as well have put A = (0,2p), B = (2q,0), C = (2r,0); that is, take p, q and r as half the coordinates of t h e p o i n t s A, B and C. If we do It t h i s way, then no fractions a r i s e when we divide by 2 in the mid-point formula. This s o r t of t h i n g happens f a i r l y o f t e n ; f o r e s i g h t at the beginning can t a k e the place of patience later on.

Theorem B. If t h e diagonals o f a parallelogram a r e congruent, the parallelogram is a rectangle.

Restatement: L e t ABCD Then ABCD i s a rectangle.

be a parallelogram, and let

AC = BD.

P r o o f : Let us take the axes as shown i n t h e figure. Then A = (0,0), and B = ( p , ~ ) with p > 0. I f we assume nothing about the f i g u r e except t h a t ABCD is a parallelogram D could be anywhere in the upper half-plane, so that D = (q,r) w i t h r > 0, but no o t h e r r e s t r i c t f o n on q or r, However, C is now determined by the f a c t that ABCD is a parallelogram. It is fairly obvious (see the preceding proof f o r d e t a i l s ) that f o r DC to be parallel to AB we must have C = (s,r). s can be determined by t h e condition BC 11 like t h i s :

AD,

slope of

BC =

q = s

s l o p e of

-

p,

AD,

(since

r # 0)

ÿÿ he coordinates (p + q,r) f o r C oan be written down by inspection if one is willing to asaume earlier theorems about parallelog~ams, f o r example, that ABCD is a parallelogram if % 11 and AB CD.) Now we finally put in the condition that AC = BD. Using t h e diatance formula, we get

-

Squaring gives (p

+

q12

+

r2

= (9

-

P)

2

+

2

r

,

Now 4 0 and p # 0; hence, q = 0. This means that D lies on the y-axis, so that L BAD is a right angle and ABCD is a rectangle.

Problem Set 17-8

Prove the following theorems uafng coordinate geometm:

.

1.

The diagonals of a rectangle have equal lengtha (~int: Place the axes &a shown.)

2.

The mid-point of the hypotenuse of a right triangle is equidistant from its three vertices

3.

Every point equidistant the axie In putation as

.

.

on t h e perpendicular bisector of a segment is from the ends of the s e p e n t . (~int: S e l e c t a posftion which will make the algebraic comsimple as possible. )

4.

Every point equidistant from the ends of a segment lies on the perpendicular* bisector of the s e p e n t F

5.

6,

The diagonals of a p a r a l l e l o gram b i s e c t each o t h e r . ( ~ i n:t Give the vertices of p a r a l l e l o gram ABCD the coordinates shown in the diagram. Show that both diagonala have t h e same mid-point. )

.

4

I

The line segment joining t h e mid-points of t h e diagonala of

a trapezoid is parallel to the bases and equal in length to half the difference of their lengths. In t h e figure R and S are mid-points of the diagonals AC and of trapezoid ABCD. F 4

The segments joining midp o i n t s of oppoalte sides of any quadrilateral bisect each o t h e r . h he 4's in the diagram are suggested by the f a c t the mid-points of segments joining midp o i n t s m u t be found. )

8.

I

2

( a , r ) , B = (b,8) and C = ( c , t ) . (~lnt: Find three trapezoids in

where

crc,t 1

A =

A ( qr), 1

the figure.)

9.

*p e(b,s 1

The area of b ABC 1 s a(t - s ) + b ( r - t) + c [ s - r )

1

I

m

I

Y

2

X

%

In A XYZ, L X l a acute and is an altitude. Given: Prove:

ZY~=XZ*+XY~-SY*XR.

,X

lo.

-

is any quadrilateral with diagonals, AC and E, &nd If M and N are the mid-points of these diagonals,

If ABCD

then 11.

In A Am, Prove:

CM

+

is amedian to aide

A C ~ B C = ~

$ + m2.

z.

of-a Condition. 17-9. The Graph By a graph we mean simply a figure In the plane, that l a , a s e t of p o i n t a . For example, triangles, rays, lines and h a l f planes are graph&. We can describe a graph by atatlng a condition which is satisfied by all points of the graph, and by no o t h e r points. Here are some examples showing a condition, a deacrfption of the graph, and the figure f o r each:

Condition 1.

Graph

Both of the coordinates of t h e point P are pos it ive

1.

The f i r s t quadrant.

The distance

2.

Themcircle with center at the origin, and radfus 2.

.

2.

OP

is

2.

The i n t e r i o r of the c i r c l e w i t h center at the origin and radius 1.

4.

X = 0.

The y-axis.

5.

y = 0.

The x - a x i s .

6.

x > O

and y = O .

The ray

7.

x = O

and

ys0.

The ray

-

The seven graphs look l i k e this:

+ OA,

+ OB,

where where

You should check carefully, in each of these cases, that the graph is really accurately described by the condition in the left-hand column above. Notice t h a t we use diagonal cross-hatching t o indicate a region. If a graph is described by a certafn condition, then the graph is called the graph of that conditiod. For example, the first quadrant is the graph of the condition x > 0 and y > 0; the c i r c l e in Figure 2 is the graph of the condition OP = 2; the y-axis is the graph of the condition x = 0; the x-axis is the graph of t h e condition y = 0; and so on. Very often the condition describing a graph will be s t a t e d in the form of an equation. In these casea we naturally speak of the graph of the given equation. If you remember Chapter 14, you have probably noticed that we are doing the same t h i n g here that we d l d in Sections 14-1 and 14-2, namely, characterizing a set by a property of i t s points. The fact that here we use the word 'graph" instead of " s e t " i~ n o t important; it I s almply customary to use the word "graph" when working with coordinate systems.

--

Problem Set 17-9 Sketch and deacribe the graphs of the conditions s t a t e d below:

4.

-1 L x l 5 .

5.

-2 3

8.

a.

x

is a p o s i t i v e integer.

b.

y

Is a positive i n t e g e r .

c

9.

lo.

.

and

Both

y 0, y > 0, and y > x . 1< x c 3 and 1 2 ~ 5 5 .

x

-.

*i1.

1x1