# Find the Sides of the Two Squares

## Question: The sum of the areas of two squares is 468 m^{2}. If the difference of their perimeters is 24 m, find the sides of the two squares

A perimeter is defined as the total length of the boundaries of a triangle

## Answer: The side of the first square and the second square whose sum of the areas is 468 m^{2} and the difference of their perimeters is 24m is 18m and 12m respectively

The Area of the square is defined as the surface enclosed by a square given by the square of it's sides and the Perimeter of a square is defined as the total length of all four sides of a square

## Explanation:

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

Area of the first square = a^{2}

Area of the second square = A^{2}

Perimeter of the first square = 4a

Perimeter of the second square = 4A

Given, difference of their perimeters is 24 m

4a − 4A = 24

a - A = 6 -------- (1)

Also given, sum of the areas of two squares is 468 m^{2}

a^{2} + A^{2} = 468 ------------ (2)

From (1), a = A + 6

Substituting the value of 'a' in (2), we get

(A + 6)^{2} + A^{2} = 468

=> A^{2} + 12A + 36 + A^{2 }= 468 (By using the algebraic identity (a + b)^{2} = a^{2} + 2ab + b^{2})

=> 2A^{2} + 12A + 36 = 468

=> A^{2} + 6A + 18 = 234 (By taking 2 as a common factor)

=> A^{2} + 6A − 216 = 0

=> A^{2 }+ 18A − 12A − 216 = 0 (By splitting the middle term)

=> A (A + 18) - 12 (A + 18) = 0

=> (A + 18) (A - 12) = 0

Thus, A = 12, -18

Since, length cannot be negative, neglect A = -18

Hence, A = 12

Substitute A = 12 in (1)

=> a - A = 6

=> a - 12 = 6

=> a = 18

### Therefore, the side of the first square and the second square whose sum of the areas is 468 m^{2} and the difference of their perimeters is 24m is 18m and 12m respectively

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