# Semicircle and isosceles problem

• Mar 11th 2010, 12:44 PM
josephk
Semicircle and isosceles problem
A semicircle and an isosceles triangle have the same base and the same area. Find the two identical angles of the isosceles triangle.
• Mar 11th 2010, 01:33 PM
TKHunny
What have you done? What do you know about circles and triangles?

Given the Base = 2B and the Area = A

1) What is the radius of the semi-circle? B

2) What is the area of the semi circle? $\displaystyle A = \frac{1}{2}\pi\cdot B^{2}$

3) What is the area of the triangle? A = ½*B*(height)

4) How do you find the height, given the Base and the Area?

5) How do you find the angle, given the Base and the Height?

You just have to think it through. A solution should present itself.
• Mar 11th 2010, 01:49 PM
Quote:

Originally Posted by josephk
A semicircle and an isosceles triangle have the same base and the same area. Find the two identical angles of the isosceles triangle.

Semicircle area = $\displaystyle \frac{{\pi}R^2}{2}$

With angle B at the lower left, and the 2 equal angles opposite that one,
then the length of both sides making angle A is 2R.

$\displaystyle \frac{{\pi}R^2}{2}=\frac{1}{2}(2R)(2R)SinB$

$\displaystyle \frac{\pi}{2}=2SinB$

$\displaystyle B=Sin^{-1}\left(\frac{\pi}{4}\right)=51.76^o$

The equal angles A are $\displaystyle A=\frac{180^o-51.76^0}{2}$

Alternatively, the equal angles could be the lower left and lower right angles.

$\displaystyle \frac{{\pi}R^2}{2}=\frac{1}{2}2Rh=Rh$

$\displaystyle TanA=\frac{h}{R}\ \Rightarrow\ h=RTanA$

$\displaystyle \frac{{\pi}R^2}{2}=Rh=R^2TanA\ \Rightarrow\ \frac{\pi}{2}=TanA$

$\displaystyle A=Tan^{-1}\left(\frac{\pi}{2}\right)=57.52^o$