Chapter 11.3, Problem 23WE

To determine

To Find: The height and perimeter.

Answer to Problem 23WE

  $h=15unit$

  $P=54units$

Explanation of Solution

Given: The isosceles trapezoid has bases are $12and28$ and area is $300$

Draw the figure using the given information.

  

Observe the figure.

  $\begin{array}{l}AB=EF=12\\ DE=FC\\ DC=DE+EF+FC\\ DC=2DE+12\\ 28=2DE+12\\ 2DE=28-12\\ DE=8units\end{array}$

Now finding height.

  $\begin{array}{l}A=\frac{h}{2}(b_1+b_2)\\ 300=\frac{h}{2}(12+28)\\ h=15unit\end{array}$

Now using Pythagoras theorem.

  $\begin{array}{l}{a}^2=8^2+{15}^2\\ a^2=64+225\\ a^2=289\\ a=17\end{array}$

Thus, perimeter is

  $\begin{array}{l}P=17+17+12+28\\ P=54units\\ \end{array}$

Hence, height of the trapezoid is $h=15unit$ and perimeter is $P=54units$ .