# 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.

**Solution:**

Let the side of the first square be x and the side of the second square be y.

Area of a square = side × side

Perimeter of a square = 4 × side

Therefore, the area of the first and second square is x^{2} and y^{2} respectively.

Also, the perimeters of the first and second square are 4x and 4y respectively.

Applying the known conditions:

(i) x^{2} + y^{2} = 468 ...(1)

(ii) 4x - 4y = 24 ....(2) considering x > y

4(x - y) = 24

x - y = 6

x = 6 + y

Substitute x = y + 6 in equation (1)

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

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

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

2(y^{2} + 6y + 18) = 468

y^{2} + 6y + 18 = 234

y^{2} + 6y + 18 - 234 = 0

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

Solving by factorization method,

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

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

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

y + 18 = 0 and y - 12 = 0

y = -18 and y = 12

y cannot be a negative value as it represents the side of the square.

Side of the first square x = y + 6 = 12 + 6 = 18 m

Side of the second square = 12 m.

**Video Solution:**

## Sum of the areas of two squares 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

### Class 10 Maths NCERT Solutions - Chapter 4 Exercise 4.3 Question 11:

**Summary:**

Sum of the areas of two squares 468 m², if the difference of perimeters is 24 m, then the side of the first square is 18 m and the side of the second square is 12 m.