Question

if breadth of a rectangle is one third of its length and its area is 27 square CM then find its perimeter​

Answer 1

Let the length be x cm.

and, the breadth be 1x

3

Area of rectangle = 27 sq.cm

l x b = 27 sq.cm

x* 1x = 27 sq.cm

3

x sq. = 27 sq.cm

3

x sq. = 27 sq. cm x 3

x sq. = 81 sq. cm

x = √81

x = 9

Hence, length = 9cm

breadth = 3cm

according to question,

perimeter of rectangle= 2( l + b )

= 2 ( 9 + 3)

= 2 x 12

= 24 cm

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Answer 2

Answer :

Perimeter of the rectangle is 24 cm.

Given :

• If the breadth of a rectangle is 1/3 of its length and its area is 27 cm².

To find :

• Perimeter of the rectangle.

Solution :

Consider,

• Length of the rectangle = x cm

★ The breadth of the rectangle is 1/3 of its length.

Then,

• Breadth of the rectangle = x ×(1/3) = x/3 cm

★ Area of the rectangle is 27 cm².

According to the question :-

$\to\sf{x\times\dfrac{x}{3}=27}$

$\to\sf{\dfrac{x^2}{3}=27}$

$\to\sf{x^2=27\times\:3}$

$\to\sf{x^2=81}$

$\to\sf{x=\sqrt{81}}$

$\to\sf{x=9}$

★ Length of the rectangle = 9 cm.

★ Breadth of the rectangle = 9×(1/3) = 3 cm

Formula used :-

${\boxed{\sf{Perimeter\:of\: rectangle=2(Length+Breadth)}}}$

Perimeter of the rectangle,

= {2(9+3) } cm

= 2 × 12 cm

= 24 cm.

Therefore, the perimeter of the rectangle is 24 cm.