Chapter 1.1, Problem 40PPS

To determine

To find: the area of the given triangle.

Answer to Problem 40PPS

  $A=31.5{\text{in}}^2$

Explanation of Solution

Given information:

A triangle is given as shown:

  

Area of a triangle is

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

Where s is semi perimeter of the triangle.

  $s=\frac{a+b+c}{2}$

a , b , c are sides of the triangle

Calculation:

Consider the given triangle.

  

Here, base is $a=6,b=11,c=14$

So, semi perimeter is calculated as

  $\begin{array}{c}s=\frac{a+b+c}{2}\\ =\frac{6+11+14}{2}\\ =\frac{31}{2}\text{in}\end{array}$

So, the Area of the triangle can be evaluated as

  $\begin{array}{c}A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\\ =\sqrt{\frac{31}{2}\left(\frac{31}{2}-6\right)\left(\frac{31}{2}-11\right)\left(\frac{31}{2}-14\right)}\\ =\sqrt{\frac{31}{2}\times \frac{19}{2}\times \frac{9}{2}\times \frac{3}{2}}\\ =31.5{\text{in}}^{\text{2}}\end{array}$

Hence, the area is $31.5{\text{in}}^{\text{2}}$.