Chapter 0, Problem 43PRT

To determine

Find the area of triangle.

Answer to Problem 43PRT

  $\text{Area of trinagle}=96{\text{ cm}}^2$

Explanation of Solution

Given information:

  

Formula used:

Area of triangle= $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

Where a, b, and c are sides of triangle and $s=\frac{a+b+c}{2}$

As we know that 12,16 and 20 are sides of triangle, now by using $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ we will find area of triangle:

  $\begin{array}{l}s=\frac{a+b+c}{2}\\ s=\frac{12+16+20}{2}=\frac{48}{2}=24\end{array}$

  $\begin{array}{l}\text{Area of trinagle=}\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\\ \\ \text{Area of trinagle}=\sqrt{24\left(24-20\right)\left(24-16\right)\left(24-12\right)}\\ \\ \text{Area of trinagle}=\sqrt{24\left(4\right)\left(8\right)\left(12\right)}\\ \\ \text{Area of trinagle=}\sqrt{9216}\end{array}$

  $\text{Area of trinagle}=96{\text{ cm}}^2$