Question

Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, determine the sides of the two squares.

Answer 1

Let x and y be the side length of squares.

 

x2 + y2 = 468 ----(1)

 

4x - 4y = 24 

 

x - y = 6

 

x = 6 + y ----(2)

 

(6+y)2 + y2 = 468 

 

36+y2+12y+y2-468 = 0 

 

2y2 + 12y - 432 = 0

 

y2 + 6y - 216 = 0

 

(y - 12)(y + 18) = 0

 

y = 12 and y = -18 (not admissible)

 

If y = 12, then x = 18

 

So, the side length of required squares are 12 and 18 respectively.

the side length of required squares are 12 and 18 respectively.