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

#### Video Explanation

#### Explanatory Answer

Step 1: Frame equations based on information about areas and perimeter of the two squares.

Let the side of one square be a m

Let the side of second square be b m

Perimeter of the 1st square = 4a

Perimeter of the 2nd square = 4b

Given Information: Difference between their perimeters is 24.

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

4(a - b) = 24

a - b = 6

a = b + 6 ------(1)

Area of the 1^{st} square = a^{2}

Area of the 2^{nd} square = b^{2}

Given Information: Sum of the areas of two squares is 468 m^{2}.

Sum of their areas = a^{2} + b^{2} = 468 ------------(2)

Substitute a = b + 6 in (2)

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

b^{2} + 12b + 36 + b^{2} = 468

2b^{2} + 12b - 432 = 0

Divide the equation by 2

b^{2} + 6b - 216 = 0

##### Step 2: Factorize the quadratic equation to find the value of â€˜bâ€™

b^{2} + 18b - 12b - 216 = 0

b(b + 18) - 12(b + 18) = 0

(b - 12) (b + 18) = 0

b = 12 or b = - 18

b cannot be negative

âˆ´ b = 12 m

a = b + 6 = 12 + 6 = 18 m

The sides of the two squares measure 12 m and 18 m.