Question

find the primeter of a triangle whose sides are 5 1\2cm,7 3/5cm and 3 1/4cm respectively

Answer 1

• The perimeter of the triangle = $\sf 16 \dfrac{7}{20} \: cm$

Given :

• Sides of the triangle = $\sf 5\dfrac{1}{2} \: cm, \: \: 7\dfrac{3}{5} \: cm, \: \: 3 \dfrac{1}{4} \: cm$

To Find :

• The perimeter of a triangle.

Solution :

We know that,

Perimeter of a triangle = a + b + c

a, b and c are sides of the triangle.

We have,

• a = $\sf 5\dfrac{1}{2}$ cm.
• b = $\sf 7\dfrac{3}{5}$ cm.
• c = $\sf 3\dfrac{1}{4}$ cm.

First, we have to convert this (mixed fraction) into simple frac kotion.

For converting,

The whole number is multiply by the denominator and then we get the multiplication then we add the numerator to the multiplication result and the denominator remains same.

So,

• a = $\sf 5\dfrac{1}{2}$ cm.

$\implies \dfrac{5 \times 2 + 1}{ 2} \: cm$

$\implies \dfrac{10 + 1}{2} \: cm$

$\implies \dfrac{11}{2} \: cm$

Hence, the first side is $\sf \dfrac{11}{2} \: cm$

• b = $\sf 7\dfrac{3}{5}$ cm.

$\implies \dfrac{7 \times 5 + 3}{5} \: cm$

$\implies \dfrac{35 + 3}{5} \: cm$

$\implies \dfrac{38}{5} \: cm$

Hence, the second side is $\sf \dfrac{38}{5} \: cm$

• c = $\sf 3\dfrac{1}{4}$ cm.

$\implies \dfrac{3 \times 4 + 1}{4} \: cm$

$\implies \dfrac{12 + 1}{4} \: cm$

$\implies \dfrac{13}{4} \: cm$

Hence, the third side is $\sf \dfrac{13}{4} \: cm$

Now we have,

• a = $\sf \dfrac{11}{2} \: cm$
• b = $\sf \dfrac{38}{5} \: cm$
• c = $\sf \dfrac{13}{4} \: cm$

Now, substitute this values in the formula of the perimeter of the triangle.

$\implies \dfrac{11}{2} \: cm + \dfrac{38}{5} \: cm + \dfrac{13}{4} \: cm$

Now, takin L.C.M. of the denominators (2, 5 and 4).

L.C.M. of 2, 5 & 4 = 20

So,

$\implies \dfrac{(11 \times 10) + (38 \times 4) + (13 \times 5)}{20} \: cm$

$\implies \dfrac{110 + 152 + 65}{20} \: cm$

$\implies \dfrac{327}{20} \: cm$

Now, we need to convert this simple fraction into mixed fraction.

For converting we have to divide this.

$\sf\Large\qquad\quad16\\ \begin{array}{cc} \cline{2 - 2}\sf 20 )&\sf \ \ 327\\&\sf - 20 \downarrow\\ \cline{2-2}& \sf \ \ \ \ 127\ \ \\ &\sf \ - 120 \\ \cline{2-2} & \sf \ \ 007 \\ \cline{2-2} \end{array}\\\\\\ divisor = 20 \\ \\ Quotient = 16\\\\Remainder = 7$

After dividing, we get 7 as a remainder, 16 as a quotient and 20 is the divisor.

For mixed fraction, the remainder becomes numerator, the divisor becomes denominator and the quotient becomes whole number.

So, the mixed fraction :

$\implies 16 \dfrac{7}{20} \: cm$

Hence,

The perimeter of the triangle is $\sf 16 \dfrac{7}{20} \: cm$