Thread: triangle

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 . How and why?
The angle subtending the arc of length 5 measures . Again how and why?.

If are the length of two sides of a triangle and is the measure of the angle between then the area to that triangle is .

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!!!

Originally Posted by Veronica1999

....... = 9/pi squared X (3 + root3)

Agree.
AS a general case:
r = radius; a,b,c = 3 given arc lengths with u,v,w = corresponding central angles

Area = r^2[SIN(u) + SIN(v) + SIN(w)] / 2, where:

r = (a + b + c) / (2pi)
u = 360a / (a + b + c)
v = 360b / (a + b + c)
w = 360c / (a + b + c)