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

SOLUTION

Let, Sides of the first and second square be S₁ and S₂

• Area of the first square = (S₁)²
• Area of the second square = (S₂)²

According to Question

(S₁)² - (S₂)² = 468 m² _(1)

• Perimeter of first square = 4 S₁
• Perimeter of second square = 4 S₂

According to question

4 S₁ - 4 S₂ = 24 _(2)

From equation (2) we get,

4 S₁ - 4 S₂ = 24

S₁ - S₂ = 6

S₁ = 6 + S₂

Putting the value of S₁ in equation (1)

• (S₁)² + (S₂)² = 468, (6+S₂)² + (S₂)² = 468
• (6)² + (S₂)² + 2 × 6 × S₂ + (S₂)² = 468
• 36 + S₂² + 12 S₂ + S₂ ² = 468
• 2S₂ ² + 12 S₂ - 432 = 0
• S₂ ² + 6 S₂ – 216 = 0
• S₂ ² + 18 S₂ – 12 S₂ - 216 = 0
• S₂(S₂ + 18) – 12(S₂ + 18) = 0
• (S₂ + 18) (S₂ - 12) = 0
• S₂ = -18 or 12

Putting S₂ = 12 in Eqn (3)

S₁ = 6 + S₂ = 18

Side of first square = S₁ = 18 m

Side of second square = S₂ = 12 m