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

Answer:

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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.

Answer 2

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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.