Question

The length of a rectangle is thrice its breadth. If the perimeter is 80m, find the length and the breadth​

Answer 1

Let length of rectangle=lcm

breadth of rectangle= bcm

Given length is 3 times its breadth

i.e. l= 3×b

Perimeter of the rectangle= 120 cm

i.e.2×( l+ b)=120

2×(3×b+ b)=120

2×(4×b)=120

8×b=120

b= 15 cm

...

Answer 2

Answer:-

$\bigstar$ Length of the rectangle = 15cm.

$\bigstar$ Breadth of the rectangle = 5cm.

  SOLUTION  

Let us assume that the length of the rectangle is 'l' cm.

Let us assume that the breadth of the rectangle is 'b' cm .

Then according to the question , the length of a rectangle is thrice its breadth.

$\implies$

l = 3b

Here,

Perimeter of the rectangle = 80cm .

Formula of perimeter of a rectangle = 2×( l + b)

$\implies$  2×( l + b) = 80

$\sf ( l+ b) = \dfrac{80}{2} \longrightarrow (1)$

Substituting , l = 3b in (1) , we get

2×( 3 × b + b) = 40

$\implies$    2 × 4b = 40

$\implies$ 8b = 40

Therefore ,

$\sf \red\bigstar b = \dfrac{40}{8} = 5cm$

We know that l = 3b , now let us substitute b = 5cm in it to get the value of l.

$\blue\bigstar$ l = 3 × 5 = 15 cm.

The required rectangle is  

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 15 cm}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 5 cm}\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\end{picture}$