Hiii, this is a problem that I have encountered and I need help ASAP.
This is the figure:
Thanks a lot!!
Hello,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