Question

The sum of the areas of two squares is `640m^2` . If the difference in their perimeter be 64m, find the sides of the two square 

Answer 1

Let the length of the side of the first and the second square be x and y. respectively. According to the question: 

`x^2+y^2=640`                                 ................(1)

Also, 

`4x-4y=64` 

⇒`x-y=16` 

⇒`x=16+y`  

Putting the value of x in (1), we get:  

`x^2+y^2=640` 

⇒`(16+y)^2+y^2=640` 

⇒`256+32y+y^2+y^2=640` 

⇒`2y^2+32y-384=0` 

⇒`y^2+16y-192=0` 

⇒`y^2+(24-8)y-192=0` 

⇒`y^2+24y-8y-192=0` 

⇒`y(y+24)-8(y+24)=0` 

⇒`(y+24)(y-8)=0` 

⇒`y=-24  or  y=8` 

∴ y=8                                    (∵ Side cannot be negativ) 

∴` x=16+y=16+8=24m` 

Thus, the sides of the squares are 8 m and 24 m.