Geometry: Urgent Help: Area of Trapezoid Circum. Abt Circle

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April 8th 2006, 06:45 PM
Yumi

Geometry: 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!!
April 8th 2006, 10:57 PM
earboth

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Quote:

Originally Posted by Yumi

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!!

Hello,

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:

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

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

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

So you get:

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

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
April 9th 2006, 06:46 AM
ThePerfectHacker

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.
April 9th 2006, 06:32 PM
Yumi

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!