# [SOLVED] Incribed Rectangle and Triangle

• Mar 27th 2009, 05:46 AM
amla
[SOLVED] Incribed Rectangle and Triangle
A rectangle of base b and height h and an isosceles triangle of base b are inscribed in a circle of radius one as shown. For what values of b and h will the rectangle and triangle have the same area?

http://www.mcs.alma.edu/hsmathchalle...march/fig1.gif
How would you solve this? I got an expression for the height of the triangle, but it led to h=0, or b=0. Thanks in advance.
• Mar 27th 2009, 07:38 AM
red_dog
The height of the triangle is $\displaystyle 1-\frac{h}{2}$.

The area of the rectangle is $\displaystyle S_1=bh$

The area of the triangle is $\displaystyle S_2=\frac{b\left(1-\frac{h}{2}\right)}{2}=\frac{b(2-h)}{4}$

$\displaystyle S_1=S_2\Rightarrow bh=\frac{b(2-h)}{4}\Rightarrow h=\frac{2}{5}$

We have $\displaystyle b^2+h^2=4$ Replacing h we get $\displaystyle b=\frac{4\sqrt{6}}{5}$
• Mar 27th 2009, 07:08 PM
amla
Thanks for replying, but I how did you know the center of the rectangle was on the center of the circle (i.e. why is the height of the triangle $\displaystyle 1-h/2$)? Also, where did you get $\displaystyle b^2+h^2=4$?

Thanks again.
• Mar 28th 2009, 01:24 AM
red_dog
Quote:

Originally Posted by amla
how did you know the center of the rectangle was on the center of the circle

The angles of the rectangle are of $\displaystyle 90^{\circ}$, so each diagonal is diameter of the circle.

Quote:

Also, where did you get $\displaystyle b^2+h^2=4$?
In a triangle formed by two sides of the rectangle and a diameter apply Pitagora.
• Mar 28th 2009, 04:12 AM
amla
Thanks!