Question

if the sum of the areas of two squares is 468 m2 and the difference of their perimeters is24m .find the meserments of their sides ​

Answer 1

Answer:

Let the Sides of Square 1 be n, and Sides of Square 2 be m respectively.

☯ Difference of the Perimeter :

⇾ Square 1 – Square 2 = 24

⇾ 4n – 4m = 24

⇾ 4(n – m) = 24

• Dividing both term by 4

⇾ n – m = 6

⇾ n = (6 + m) ⠀— eq. ( I )

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☯ According to the Question :

⇴ Sum of Areas of Squares = 468 m²

⇴ (Square 1)² + (Square 2)² = 468 m²

⇴ (n)² + (m)² = 468

• Putting value of n from eq. ( I )

⇴ (6 + m)² + m² = 468

⇴ 36 + m² + 12m + m² = 468

⇴ 2m² + 12m + 36 = 468

⇴ 2(m² + 6m + 18) = 468

• Dividing both term by 2

⇴ m² + 6m + 18 = 234

⇴ m² + 6m + 18 – 234 = 0

⇴ m² + 6m – 216 = 0

⇴ m² + 18m – 12m – 216 = 0

⇴ m(m + 18) – 12(m + 18) = 0

⇴ (m - 12)(m + 18) = 0

⇴ m = 12 m⠀or,⠀m ≠ - 18 m [ Impossible ]

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☯ Putting value of m as 12 in eq. ( I ) :

⟶ n = (6 + m)

⟶ n = (6 + 12)

⟶ n = 18 m

∴ Hence, Sides of Squares are 18 metres and 12 metres respectively.

Answer 2

Answer:

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Let us say that the sides of the two squares are 'a' and 'b'

Sum of their areas = a^2 + b^2 = 468

Difference of their perimeters = 4a - 4b = 24

=> a - b = 6

=> a = b + 6

So, we get the equation

$= (b + 6)^2 + b^2 = 468 \\ </p><p></p><p>= \: 2b^2 + 12b + 36 = 468 \\ </p><p></p><p>= \: b^2 + 6b - 216 = 0 \\ </p><p></p><p>= \: b = 12 \\ </p><p></p><p>= \: a = 18$

The sides of the two squares are 12 and 18.