Hiii, this is a problem that I have encountered and I need help ASAP.

This is the figure:

http://img404.imageshack.us/img404/1...helpppp6yk.gif

Thanks a lot!!

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- Apr 8th 2006, 06:45 PMYumiGeometry: Urgent Help: Area of Trapezoid Circum. Abt Circle
Hiii, this is a problem that I have encountered and I need help ASAP.

This is the figure:

http://img404.imageshack.us/img404/1...helpppp6yk.gif

Thanks a lot!! - Apr 8th 2006, 10:57 PMearbothQuote:

Originally Posted by**Yumi**

I've attached a diagram to demonstrate what I calculated.

Let r be the radius of the circle.

Let angle(CBA)= alpha. Then angle(DCB)=180°-alpha.

The triangle (BMO) is a right triangle. The triangle (OSC) is a right triangle.

Now use the tangens:

$\displaystyle \frac{r}{9}=\tan \left( \frac{\alpha}{2}}\right) $

$\displaystyle \frac{r}{4}=\tan \left(90^\circ- \frac{\alpha}{2}}\right) $

with $\displaystyle \tan \left(90^\circ- \frac{\alpha}{2}} \right) =\frac{1}{\tan \left( \frac{\alpha}{2}}\right) }$

So you get:

$\displaystyle \frac{r}{9}=\frac{1}{ \frac{r}{4}}$

Solve for r and you'll get r = 6.

That means the height of the trapezoid is 12. Therefore the area is 156 (square units).

Greetings

EB - Apr 9th 2006, 06:46 AMThePerfectHacker
There is something easier which you can do. If I remember correctly the height of a trapezoid where a circle can be inscribed is the geometric mean of its bases.

- Apr 9th 2006, 06:32 PMYumi
I see! Thank you so much for the detailed responses and thanks to ThePerfectHacker also :D Both gets to the same answer :D Once again, thank you so much!