# Thread: [SOLVED] Incribed Rectangle and Triangle

1. ## [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?

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.

2. The height of the triangle is $1-\frac{h}{2}$.

The area of the rectangle is $S_1=bh$

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

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

We have $b^2+h^2=4$ Replacing h we get $b=\frac{4\sqrt{6}}{5}$

3. 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 $1-h/2$)? Also, where did you get $b^2+h^2=4$?

Thanks again.

4. 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 $90^{\circ}$, so each diagonal is diameter of the circle.

Also, where did you get $b^2+h^2=4$?
In a triangle formed by two sides of the rectangle and a diameter apply Pitagora.

5. Thanks!