Chapter 6, Problem 35CR

To determine

To calculate:

Area of triangle by using Heron’s area formula.

Answer to Problem 35CR

Area of triangle by using Heron’s area formula is $\Delta ABC=511.71sq.units$ .

Explanation of Solution

Given information:

  $a=64.8,b=49.2,c=24.1$

Calculation:

Let’s consider,

  $a=64.8,b=49.2,c=24.1$

Heron’s area formula forgiven triangle with sides $a,b,c$ then aea of triangle is,

  $\Delta ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

Where

  $s=\frac{a+b+c}{2}$

Here, given values are $a=64.8,b=49.2,c=24.1$

  $\Delta ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

At first, find the values of

  $\begin{array}{l}s=\frac{a+b+c}{2}\\ s=\frac{64.8+49.2+24.1}{2}\\ s=\frac{138.1}{2}\\ s=69.05\end{array}$

Put the values in above equation,

  $\begin{array}{l}\Delta ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\\ \Delta ABC=\sqrt{69.05\left(69.05-64.8\right)\left(69.05-49.2\right)\left(69.05-24.1\right)}\\ \Delta ABC=\sqrt{\left(69.05\right)\left(4.28\right)\left(19.85\right)\left(44.95\right)}\\ \Delta ABC=\sqrt{263692.42}\\ \Delta ABC=513.5\end{array}$

Therefore, value of Area of triangle by using Heron’s area formula is $\Delta ABC=511.71sq.units$