Question

if sum of are of two squares is 468 and difference in perimeter is 24​

Answer 1

Step-by-step explanation:

Let the sides of the first and second square be X and Y . Area of the first square = (X)²

Area of the 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

Side of second square = Y = 12 m.