Chapter 5.6, Problem 25E

Solution

To find: If a triangle can be formed using the given lengths and to determine the area of a triangle using the Heron’s formula.

The triangle is possible and the area of the triangle is $216.15{\text{unit}}^2$ .

Given information:

The length of the sides of the triangle are $a=19.3\text{units}$ , $b=22.5\text{units}$ , and $c=31\text{units}$ .

Formula used:

The conditions to check if the triangle is possible or not are as follows.

  $\begin{array}{l}a+b>c\\ b+c>a\\ a+c>b\end{array}$

The Heron’s formula to determine the area of triangle is

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

Here, $s$ is the semi-perimeter and $s=\frac{a+b+c}{2}$ .

Calculation:

Substitute $19.3$ for $a$ , $22.5$ for $b$ and $31$ for $c$ in the three conditions to check whether triangle is possible.

  $\begin{array}{l}a+b>c\\ 19.3+22.5>31\end{array}$

This condition is satisfied.

  $\begin{array}{l}b+c>a\\ 22.5+31>19.3\end{array}$

This condition is satisfied.

  $\begin{array}{l}a+c>b\\ 19.3+31>22.5\end{array}$

This condition is satisfied.

Since all the three conditions are satisfied, so the trainable is possible.

Substitute $19.3$ for $a$ , $22.5$ for $b$ and $31$ for $c$ in the formula $s=\frac{a+b+c}{2}$ , to calculate the semi perimeter.

  $\begin{array}{c}s=\frac{19.3+22.5+31}{2}\\ =36.4\end{array}$

Substitute $19.3$ for $a$ , $22.5$ for $b$ and $31$ for $c$ , and $36.4$ for $s$ into the Heron’s formula to calculate the area of triangle.

  $\begin{array}{c}\text{Area}=\sqrt{36.4\left(36.4-19\right)\left(36.4-22.5\right)\left(36.4-31\right)}\\ =\sqrt{64720.34}\\ =216.15\end{array}$

Therefore, the area of the triangle is $216.15{\text{unit}}^2$ .