Question

find the dimensions of rectangle whose perimeter is 112cm and whose breadth is 4 cm less than half its length​

Answer 1

Answer:

40 x 16 ( in the format l X b)

Step-by-step explanation:

Let 'b' be the breadth and 'l' be the length of the rectangle.

b=l/2 - 4

perimeter,p = 2(l+b)

i.e p = (l + l/2 - 4)*2

p = 3l - 8 = 112

i.e 3l = 120

Therefore, l = 40cm and b = 16cm..

Answer 2

The length of rectangle = 40 cm and

The breadth of rectangle = 16 cm

Step-by-step explanation:

Given,

The perimeter of rectangle = 112 cm

Let the length of rectangle = x and

The breadth of rectangle = $\dfrac{1}{2} x-4$

To find, the length and breadth of rectangle = ?

We know that,

The perimeter of rectangle = 2(l + b)

∴ 2(x + $\dfrac{1}{2} x-4$) = 112

⇒ x + $\dfrac{1}{2} x-4$ = $\dfrac{112}{2} =56$

⇒ $\dfrac{2x+x}{2} =56+4=60$

⇒ $\dfrac{3x}{2}=60$

⇒ 3x = 60 × 2 = 120

⇒ x = $\dfrac{120}{3} =40$

∴ The length of rectangle = 40 cm and

The breadth of rectangle = $\dfrac{1}{2}(40)-4$ = 16 cm