Question

The perimeter of a rectangle is 52 cm. If its width is 2 cm more than one–third of its length,

find the dimensions of the rectangle​

Answer 1

Answer:

let length be 3x.

width = x+2

perimeter=2(3x+x+2)

52=2(4x+2)

52=8x+4

8x=48

x=6

length=3×x=18

width=x+2=8

Answer 2

Answer :-

Dimensions of rectangle are 18 cm and 8 cm.

Solution :-

Let the length of the rectangle be x

Breadth of the rectangle = 2 cm more than one third of its length = 1/3 (x) + 3 = x/3 + 2

We know that

Perimeter of a rectangle = 2(Length + Breadth)

Perimeter of the given rectangle = 52 cm

⇒ 2(x + x/3 + 2) = 52

Transpose 2 to RHS

⇒ x + x/3 + 2 = 52/2

⇒ x + x/3 + 2 = 26

Transpose 2 to RHS

⇒ x + x/3 = 26 - 2

⇒ x + x/3 = 24

Taking LCM

⇒ x(3)/1(3) + x/3 = 24

⇒ 3x/3 + x/3 = 24

⇒ (3x + x)/3 = 24

⇒ 4x/3 = 24

⇒ x = 24 * 3/4

⇒ x = 6 * 3

⇒ x = 18

Length of the rectangle = x = 18 cm

Breadth of the rectangle =x/3 + 2 = 18/3 + 2 = 6 + 2 = 8 cm

Therefore dimensions of rectangle are 18 cm and 8 cm.