Ex 4.3 Q11. CBSE 10th Maths NCERT exercise solution

#### Video Explanation

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 â€˜bâ€™

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
âˆ´ b = 12 m
a = b + 6 = 12 + 6 = 18 m
The sides of the two squares measure 12 m and 18 m.