# Thread: Area of Isosceles Trapezoid

1. ## Area of Isosceles Trapezoid

Hi there,

How do you find the area of an Isosceles trapezoid if all you have is that the altitude is 15 inches and the diagonal is 25 inches?

2. Aaaaay, The Fonz!

Find the area of an Isosceles trapezoid with altitude 15 and diagonal 25.
Code:
              A       x       B
*---------------*
/:  *            :\
/ :     *         : \
/  :        *      :  \
/   :        25 *   :   \
/  15:              *:    \
/     :               : *   \
/      :               :    * \
D *-------*---------------*-------* C
: 20-x  E       x       F 20-x  :
: - - - -  20 - - - - - :

We have isosceles trapezoid $\displaystyle ABCD$ with diagonal $\displaystyle AC = 25$
. . and altitudes $\displaystyle AE = BF = 15.$

In right triangle $\displaystyle AEC\!:\;\;AE^2 + EC^2 \:=\:AC^2 \quad\Rightarrow\quad 15^2 + EC^2 \:=\:25^2$

. . Hence: .$\displaystyle EC^2 \:=\:400 \quad\Rightarrow\quad EC \:=\:20$

Let $\displaystyle AB \,=\,EF \,=\,x$
. . Then: .$\displaystyle FC\,=\,DE \,=\,20-x$

Hence: .$\displaystyle DC \:=\:(20-x) + x + (20-x) \:=\:40-x$

The area of a trapezoid is: .$\displaystyle A \;=\;\frac{h}{2}(b_1 + b_2)$

We have: .$\displaystyle A \;=\;\frac{AE}{2}(AB + DC) \;=\;\frac{15}{2}\bigg[x + (40-x)\bigg] \;=\;\frac{15}{2}\cdot40 \;=\;\boxed{300}$