Question: Sum of the areas of two squares is 468 m2. 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 1st square = a2
Area of the 2nd square = b2
Given Information: Sum of the areas of two squares is 468 m2.
Sum of their areas = a2 + b2 = 468 ------------(2)
Substitute a = b + 6 in (2)
(b + 6)2 + b2 = 468
b2 + 12b + 36 + b2 = 468
2b2 + 12b - 432 = 0
Divide the equation by 2
b2 + 6b - 216 = 0
Step 2: Factorize the quadratic equation to find the value of ‘b’
b2 + 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
∴ b = 12 m
a = b + 6 = 12 + 6 = 18 m
The sides of the two squares measure 12 m and 18 m.