Question

The perimeter of a rectangle is 24 ft. Its length is five times its width Let x be the length and y be the width. What is the area of the rectangle?

Answer 1

Here ,

Given perimeter = 24 ft.

Then , given breadth = y

Length= x

Forming the linear equation in two variables :

x = 5y

Write X = 5 y :

2(length + breadth) = 24

2(5y+y) = 24

12 y = 24

y = 2 ft.

X = 5 y = 5 × 2 = 10 ft

Then area of the rectangle = xy = 10 × 2 = 20 ft²

Answer 2

To find : We have to find the area of the rectangule whose length is five times it's width and the perimeter is 24 ft.

Answer : The length of the rectangle is 10 ft. and width is 2 ft.

Solution :

Let, length = x ft. and width = y ft.

Given that, length = 5 × width

⇒ x = 5y .....(i)

Given that, perimeter of the rectangle = 24 ft.

⇒ 2 (length + width) = 24

⇒ 2 (x + y) = 24

⇒ x + y = 12

⇒ 5y + y = 12 , by substituting x = 5y from (i)

⇒ 6y = 12

⇒ y = 2

From (i), we get

x = 5 × 2 = 10

Thus, the length of the rectangle is 10 ft. and width is 2 ft.

Therefore, the area of the rectangle

    = length × width

    = 10 × 2 ft.²

    = 20 ft.²