Area of Isosceles Trapezoid

**The Fonz** · May 3rd 2009, 12:24 PM

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?

**Soroban** · May 3rd 2009, 04:44 PM

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 $ABCD$ with diagonal $AC = 25$
. . and altitudes $AE = BF = 15.$

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

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

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

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

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

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

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