Question

The side of a square exceeds the side of another square by 4cm and the sum of the areas of the squares is 400cm². Find the dimensions of the squares.

Answer 1

Let the side of the smaller square = x
∴ the side of the larger square = x + 4
We know, The area of a square with side s = s²
∴ The area of a square with side x = x²
and, The area of a square with side x + 4 = (x + 4)²
Now, the sum of the two area = 400
⇒ x² + (x+ 4)² = 400
⇒ x² + x² + 16 + 8x = 400
⇒ 2x² + 8x + 16 = 400
⇒ 2(x² + 4x + 8) = 2(200)
⇒ x² + 4x + 8 = 200
⇒ x² + 4x - 192 = 0
Splittting the middle term, we have
x² + 16x - 12x - 192 = 0
⇒ x(x + 16) - 12(x + 16) = 0
⇒ (x + 16)(x - 12) = 0
⇒ x = -16, x = 12
But x is the length of the side of a square,
∴ x ≠ -16
∴ x = 12
⇒ the side ofthe smaller square = 12cm
∴ the side of the larger square
= 12 + 4
= 16cm.