Answer to Question #9532 in Geometry for Candice

Question

Sub 24
Ques. 18

Find the dimensions of a rectangle with an area of 60 ft^2 and a diagonal of 13 ft.

Answer 1

Let a,b be the sides of the rectangle.
Then
the area
ab = 60
and
the diagonal
d = sqrt( a^2+b^2 ) = 13

Thus we obtain the
following system:

ab=60
a^2 + b^2 = 13^2=169

Hence

(a-b)^2 = a^2+b^2-2ab = 169 - 2*60 = 169-120=49 = 7^2
therefore

a-b=7


On the other hand
(a+b)^2 = a^2+b^2+2ab = 169 + 2*60 =
169+120=289 = 17^2,
so
a+b=17


Thus
a-b=7

a+b=17

adding both equation we get
2a = 7+17 = 24 =>
a=12
Therefore
b = 17-a=17-12=5


Answer: the sides are 12ft
and 5ft