Chapter 9.4, Problem 29PSC

To determine

To calculate: The perimeter of atrapezoidAFED .

Answer to Problem 29PSC

The perimeter of a trapezoidAFEDis 50.

Explanation of Solution

Given information:

Side $FB=8,$

Side $FE=9,$

Side $FD=17.$

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Calculation:

  

Apply Pythagoras Theorem in triangle FBD and ECD .

In right angled triangle FBD , we get

  $\begin{array}{l}\left(BD\right)+{\left(8\right)}^2={\left(17\right)}^2\\ \therefore BD=15\\ SinceBC=FE=9,\\ CD=AB=6\end{array}$

In right angled triangle ECD , we get

  $\begin{array}{l}{\left(ED\right)}^2={\left(8\right)}^2+{\left(6\right)}^2\\ {\left(ED\right)}^2=64+36\\ ED=\sqrt{100}\\ ED=10\\ \overline{ED}\cong \overline{FA}\end{array}$

Perimeter is sum of all sides.

  $\begin{array}{l}P=AF+FE+ED+DC+CB+BA\\ P=10+9+10+6+9+6\\ P=50\end{array}$