Find the Sides of the Two Squares


Question: The sum of the areas of two squares is 468 m2. 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 m2 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 = a2 

Area of the second square = A2

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 m2

a2 + A2 = 468 ------------ (2)

From (1), a = A + 6

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

(A + 6)2 + A2 = 468

=> A2 + 12A + 36 + A= 468              (By using the algebraic identity (a + b)2 = a2 + 2ab + b2)

=> 2A2 + 12A + 36 = 468

=> A2 + 6A + 18 = 234                       (By taking 2 as a common factor)

=> A2 + 6A − 216 = 0

=> A+ 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 m2 and the difference of their perimeters is 24m is 18m and 12m respectively