Question

The sides of a triangle are in the ratio of 1:4:6 and its perimeter is 132.Then find the length of its sides

Answer 1

Answer :

12 , 48 , 72 units respectively

Given :

• The sides of a triangle are in the ratio of 1:4:6 and its perimeter is 132

To Find :

• The length of its sides

Solution :

Let the sides of the triangle be , " x " , " 4x " , " 6x "

We know that , " Perimeter of the triangle is the sum of all of its sides "

⇒ x + 4x + 6x = 132

⇒ 11x = 132

⇒ x = 12

So ,The length of the sides of triangle are ,

1st side = x = 12 units

2nd side = 4x = 48 units

3rd side = 6x = 72 units

Answer 2

SOLUTION :

$\underline{\bf{Given\::}}}}$

The sides of a triangle are in the ratio of 1:4:6 & it's perimeter is 132.

$\underline{\bf{To\:find\::}}}}$

The length of it's sides.

$\underline{\bf{Explanation\::}}}}$

Attach figure :

$\setlength{\unitlength}{0.8cm}\begin{picture}(6,5)\thicklines\put(1,0.5){\line(2,1){3}}\put(4,2){\line(-2,1){2}}\put(2,3){\line(-2,-5){1}}\put(0.7,0.3){ A }\put(4.05,1.9){ B }\put(1.7,2.95){ C }\put(3.1,2.5){ \sf 6r }\put(1.3,1.7){ \sf\:\:\:\:4r }\put(2.5,1.05){ \:\:\:\sf r }\end{picture}$

We know that formula of the perimeter of triangle :

$\boxed{\bf{Perimeter\:of\:\triangle =Side+Side+Side}}}}$

A/q

$\longrightarrow\sf{r+4r+6r=132}\\\\\longrightarrow\sf{11r=132}\\\\\longrightarrow\sf{r=\cancel{132/11}}\\\\\longrightarrow\sf{r=12\:unit}$

Thus;

• 1st side of Δ = r = 12 unit
• 2nd side of Δ = 4r = 4 × 12 = 48 unit
• 3rd side of Δ = 6r = 6 × 12 = 72 unit