Question

Sum of the areas of the squares is 468m2. It
the difference of their perimeters is 24m, find
the sides of the two squares​

Answer 1

Let the sides of first and second square be X and Y .

Area of first square = (X)²

And,

Area of second square = (Y)²

According to question,

(X)² + (Y)² = 468 m² ------------(1).

Perimeter of first square = 4 × X

and,

Perimeter of second square = 4 × Y

According to question,

4X - 4Y = 24 -----------(2)

From equation (2) we get,

4X - 4Y = 24

4(X-Y) = 24

X - Y = 24/4

X - Y = 6

X = 6+Y ---------(3)

Putting the value of X in equation (1)

(X)² + (Y)² = 468

(6+Y)² + (Y)² = 468

(6)² + (Y)² + 2 × 6 × Y + (Y)² = 468

36 + Y² + 12Y + Y² = 468

2Y² + 12Y - 468 +36 = 0

2Y² + 12Y -432 = 0

2( Y² + 6Y - 216) = 0

Y² + 6Y - 216 = 0

Y² + 18Y - 12Y -216 = 0

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

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

(Y+18) = 0 Or (Y-12) = 0

Y = -18 OR Y = 12

Putting Y = 12 in EQUATION (3)

X = 6+Y = 6+12 = 18

Side of first square = X = 18 m

and,

Side of second square = Y = 12 m