Question #9532

Sub 24
Ques. 18
Find the dimensions of a rectangle with an area of 60 ft^2 and a diagonal of 13 ft.

Expert's answer

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

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

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