Chapter 13.7, Problem 17PSB

To determine

To find: The area of the isosceles trapezoid with the vertices at $\left(4,8\right),\left(2,3\right),\left(14,3\right)\left(12,8\right)$ .

Answer to Problem 17PSB

The area of the trapezoid is $50$ .

Explanation of Solution

Given:

The given vertices are $\left(4,8\right),\left(2,3\right),\left(14,3\right)\left(12,8\right)$ .

Calculation:

Consider the base width of the trapezoid is,

  $\begin{array}{c}{b}_1=12-4\\ =8\end{array}$

Also,

  $b_2=14-2=12$

Consider the height of the vertices is,

  $\begin{array}{c}A=8-3\\ =5\end{array}$

Then, the area of the trapezoid is,

  $\begin{array}{c}A=\left(\frac{b_1+b_2}{2}\right)h\\ =\left(\frac{8+12}{2}\right)5\\ =50\end{array}$