A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths 3,4,and 5. What is the area of the triangle?
Not sure how to start the problem...
Could I get some hints please?
You must draw a diagram. Inscribe a triangle in a circle.
Draw three radial segments, one from each vertex to the center,
Now you have three sub-triangles. The sum of their areas is the area you want.
The radius of the circle is $\displaystyle r=\frac{6}{\pi}$. How and why?
The angle subtending the arc of length 5 measures $\displaystyle \theta_5=\frac{5\pi}{6}$. Again how and why?.
If $\displaystyle a~\&~b$ are the length of two sides of a triangle and $\displaystyle \theta$ is the measure of the angle between then the area to that triangle is $\displaystyle \frac{1}{2}a\cdot b\sin(\theta)$.
I think I got it. It took me a while though....
I understand the radius is 6/pi because 2pir = 12 r= 6/pi.
Then I understood the angle subtending the arc of length 5 is 5pi/6 because 2pi X 5/12 = 5pi/6
arc of length 4 2pi X 4/12 = 2pi/3
arc of length 3 2pi X 3/12 = pi/2
I fully understand why 1/2absin(theta) gets the area therefore,
1/2 X 6/pi X 6/pi X sin 5pi/6 + 1/2 X 6/pi X 6/pi X sin 2pi/3 + 1/2 X 6/pi X 6/pi X sin pi/2 = 9/pi squared X (3 + root3)
Thanks!!!