Chapter 11 QuaDraTiC appliCaTionS 421

3. The length is 7 inches more than the width, so L = W + 7. The diagonal
   is 17 inches. The formula DL^22 =  + W^2 becomes 17^2 = (W + 7)^2 + W^2.

17 7

289 77

289 14 49

2

222

2

22

=++

=+ ++

=+ ++

()

()()

WW

WW W

WWW

889 21449

02 14 240

1

2

0 1

2

214 240

2

2

2

=++

=+−

=+−

WW

WW

() ()WW)

()()

07120

0815

=+^2 −

=− +

WW

WW

W − 8 = 0 W + 15 = 0 (This does not lead to a solution.)

W = 8

The rectangle’s width is 8 inches and its length is 8 + 7 = 15 inches.

4. The width is three-fourths its length, so WL= 43. The diagonal is

10 inches, so the formula DL^22 =  + W^2 becomes (^102234)
2
=  + LL().
10 3
4
100 9
16
100 1 9
16
22
2
2
22
2
=+



=+
=+



LL
LL
L
1100 16
16
9
16
100 25
16
16
25
100
64
2
2
2
=+





=

L
L
L
L
()
22
64
8

=
L
L
The rectangle’s length is 8 inches and its width is (^8) ()^34  =  6 inches.