Question

Sum of the areas of two squares is 468 m square if the difference of their perimeter is 24 m find the sides of the two square​

Answer 1

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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.

Answer 2

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Let the side of the first square be 'a'm and that of the second be `A` m.

Area of the first square = a²sqm

Area of the second square = A² sqm

Their perimeters would be 4a and 4A respectively.

Given ,

4A - 4a = 24

A - a = 6 -- (1)

A² + a² = 468 --(2)

From (1) a = a + 6

Substituting for A in (2), we get

→ (a +6)² + a² = 468

→ a² + 12a + 36 + a² = 468

→ 2a² + 12a + 36 = 468

→ a² + 6a + 18 = 234

→ a² + 6a - 216 = 0

→ a² + 18a - 12a - 216 = 0

→ a(a+18) − 12(a+18) = 0

→ (a−12) (a+18) = 0

→ a = 12, -18

So, the side of the first square is 12 m.

The side of the second square is 18 m.