Area Proof

[Clifford]

Problem: C is any point on a circle with diameter AB. On sides AC and BC of triangle ABC, semicricles are drawn outside the triangle. Prove that the total area of the two shaded lunes is equal to the area of triangle ABC.



First of all I have calculated the area of the semi circle around the triangle which is pi1/2AB^2.
I also got the area of the other two semi circles which are: pi1/2AC^2 and pi1/2CB^2
I am not sure how to get the area of the triangle in this situation.

After I get all the areas, would I do:
area of triangle = area of semi circle (with radius 1/2AC) + area of semi circle (with radius 1/2CB) - (area of semi circle with radius 1/2AB - area of the triangle)?

Can somebody help me out with the area of the triangle and specify if I am doing it right or not. Lastly, is this the right approach at this question, or is there a better / easier way to tackle it?

 

[skeeter]

since triangle ABC is inscribed in a semi-circle, it is a right triangle.

let a = triangle side opposite angle A
b = triangle side opposite angle B
c = triangle side opposite angle C = circle diameter

area of triangle ABC = ab/2

area of two smaller semicircles = (pi/2)(a²/4 + b²/4)

area of region between the circle and triangle ABC = (pi/2)(c²/4) - ab/2

area of two lunes = (pi/2)(a²/4 + b²/4) - [(pi/2)(c²/4) - ab/2]

now, remember ... a² + b² = c²
should be a cake-walk.

 

[Clifford]

Thanks for the help! I didn't know that the triangle would be a right angle which made my ponder how I could calculate the area of it. Thanks again.