# Right trapeziod area problem

• Oct 13th 2009, 09:17 AM
jlefholtz
Right trapeziod area problem
A right trapezoid has parallel sides AB and CD (AB > CD), a perpendicular side AD. The diagnoal AC equals to the side AB and the side AD equals to the side DC. A line through C, perpendicular to side BC intersects side AD in the point F. Find the area of the trapezoid if we know that DF=1.

Can someone please tell me how to set this problem up? I can solve it from there...

Thank you.
• Oct 13th 2009, 09:43 AM
red_dog
$\widehat{ACD}=\widehat{CAD}=\frac{\pi}{4}$

Let $\widehat{ACB}=\widehat{ABC}=x$. Then $2x+\frac{\pi}{4}=\pi\Rightarrow x=\frac{3\pi}{8}$

$\widehat{FCA}=\frac{\pi}{2}-x=\frac{\pi}{2}-\frac{3\pi}{8}=\frac{\pi}{8}$

Then $\widehat{DCF}=\frac{\pi}{4}-\frac{\pi}{8}=\frac{\pi}{8}$.

In the triangle CDF: $\tan\frac{\pi}{8}=\frac{DF}{CD}=\frac{1}{CD}$

Then $CD=\frac{1}{\tan\frac{\pi}{8}}=\cot\frac{\pi}{8}=\ sqrt{\frac{1+\cos\frac{\pi}{4}}{1-\cos\frac{\pi}{4}}}=\sqrt{2}+1=AD$.

$AB=AC=AD\sqrt{2}=2+\sqrt{2}$

Now you have all the elements to find the area.
• Oct 13th 2009, 09:45 AM
Soroban
Let's see if I can make this clear enough for you . . .

If you can't show a little initiative on your part,
show some work, evidence of some effort,
STOP WASTING OUR TIME!

Mod: Is THAT the message I should give all visitors?
.
• Oct 13th 2009, 11:03 AM
jlefholtz
not homework
This is not for homework. I am doing this problem for fun. Thanks.
• Oct 13th 2009, 11:07 AM
jlefholtz
Thanks, red_dog. I wasn't sure whether or not I was correct in using tan...I am out of practice.