Advertisement Remove all ads

The sum of the areas of two squares is 400 sq.m. If the difference between their perimeters is 16 m, find the sides of two squares. - Algebra

Advertisement Remove all ads
Advertisement Remove all ads
Sum

The sum of the areas of two squares is 400 sq.m. If the difference between their perimeters is 16 m, find the sides of two squares.

Advertisement Remove all ads

Solution

Let the sides of the two squares be x cm and y cm(x > y).

Then, their areas are x 2 and y 2 and their perimeters are 4x and 4y.

According to the first condition,

sum of the areas of two squares is 400 sq.m

∴ x2 + y2 = 400 ........(i)

According to the second condition,

difference between the perimeters is 16 m

∴ 4x − 4y = 16

∴ 4(x − y) = 16

∴ x − y = 4

∴ x = y + 4

Substituting the value of x in equation (i), we get

(y + 4)2 + y2 = 400

∴ y2 + 8y + 16 + y2 = 400

∴ 2y2 + 8y – 384 = 0

∴ y2 + 4y – 192 = 0

∴ y2 + 16y – 12y – 192 = 0

∴ y(y + 16) – 12(y + 16) = 0

∴ (y + 16) (y – 12) = 0

∴ y + 16 = 0 or y – 12 = 0

∴ y = – 16 or y = 12

But, y ≠ – 16 as the side of a square cannot be negative.

∴ y = 12

∴ x = y + 4 = 12 + 4 = 16

∴ The sides of the two squares are 16 cm and 12 cm.

Concept: Solutions of Quadratic Equations by Completing the Square
  Is there an error in this question or solution?
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×