Question

the perimeter of a rectangle is 52 cm . If its width is 2 cm more than one-third of its length, find the length and breadth of the rectangle

Answer 1

Perimeter of rectangle= 2(l+b)

FIven perimeter= 52 cm

Given b= 2cm+1/3(l)

So 2(l+b) = 52

   2(l+ (2+(l/3)) = 52

   l+((6+l)/3)=52

  (3l+6+l)/3=52

   4l+6= 52*3

   4l=156-6

  4l=150

  l= 150/4

 l=39 cm

so b= 2+ ( 39/3)

      = 2+ 13

     =15

So l=39 cm and b=15 cm

Area = l*b

       = 39*15

       =585 sq cm

Answer 2

Given,

• Perimeter of a Rectangle is 52 cm.
• If breadth is 2 cm more than one third of its length .

To Find,

• Length of Rectangle = ?
• Breadth of Rectangle = ?

Solution,

☯$\sf{Suppose \ the \ length \ of \ the \ Rectangle be \ "x"}$

☯$\sf{Therefore, \ Breadth \ of \ the \ Rectangle \ be = \dfrac{1}{3}x + 2}$

Now According to the Question :

$\implies \boxed{\sf{Perimeter \ of \ Rectangle = 2(Length + Breadth)}}$

$\implies \sf{2(x + \dfrac{1}{3}x + 2 ) = 52}$

$\implies \sf{2(\dfrac{3x + x }{3} + 2) = 52}$

$\implies \sf{2(\dfrac{4x}{3} + 2) = 52}$

$\implies \sf{2(\dfrac{4x + 6}{3} ) = 52}$

$\implies \sf{\dfrac{8x + 12}{3} = 52}$

$\implies \sf{8x + 12 = 52 \times 3}$

$\implies \sf{8x + 12 = 156}$

$\implies \sf{8x = 156 - 12}$

$\implies \sf{8x = 144}$

$\implies \sf{x = \dfrac{144}{8}}$

$\implies \boxed{\sf{x = 18}}$

Now Find Breadth of Rectangle :-

$\implies \sf{Breadth \ of \ Rectangle = \dfrac{1}{3}x + 2 }$

$\implies \sf{Breadth \ of \ Rectangle = \dfrac{1}{3} \times 18 + 2}$

$\implies \sf{Breadth \ of \ Rectangle = \dfrac{18}{3} + 2}$

$\implies \sf{Breadth \ of \ Rectangle = \dfrac{18 + 6}{3} + 2}$

$\implies \sf{Breadth \ of \ Rectangle = \dfrac{24}{3}}$

$\implies \sf{Breadth \ of \ Rectangle = 8 \ cm}$

$\large{\sf{Therefore,}}$

$\boxed{\bold{Length \ of \ Rectangle = 18 \ cm}}$

$\boxed{\bold{Breadth \ of \ Rectangle = 8 \ cm}}$

$\rule{200}2$