Chapter 6, Problem 36CR

To determine

To calculate:

Area of triangle by using Heron’s area formula.

Answer to Problem 36CR

Area of triangle by using Heron’s area formula is $\Delta ABC=15.44sq.units$ .

Explanation of Solution

Given information:

  $a=8.55,b=5.14,c=12.73$

Calculation:

Let’s consider,

  $a=8.55,b=5.14,c=12.73$

Heron’s area formula forgiven triangle with sides $a,b,c$ then area 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 $a=8.55,b=5.14,c=12.73$

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

At first, find the values of

  $\begin{array}{l}s=\frac{a+b+c}{2}\\ s=\frac{8.55+5.14+12.73}{2}\\ s=\frac{26.42}{2}\\ s=13.21\end{array}$

Put the values in above equation,

  $\begin{array}{l}\Delta ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\\ \Delta ABC=\sqrt{13.21\left(13.21-8.55\right)\left(13.21-5.14\right)\left(13.21-12.73\right)}\\ \Delta ABC=\sqrt{\left(13.21\right)\left(4.66\right)\left(8.07\right)\left(0.48\right)}\\ \Delta ABC=\sqrt{238.45}\\ \Delta ABC=14.44\end{array}$

Therefore, value of Area of triangle by using Heron’s area formula is $\Delta ABC=15.44sq.units$