You need the area of the sector created by the radii and the intercepted arc minus the area of the triangle created by the central angle of 120 and the segment BG. The area of the sector is, according to our info: [tex]A= \frac{120}{360}*(3.14)(100) ^{2} [/tex] which is 10466.6666. Now we need the area of the triangle which is not as simple. If the central angle of the whole triangle is 120, then if we split it in half and deal with right triangles only, the angle measure is now 60, and we have a 30-60-90 special right triangle. If the hypotenuse (radius) is 100, in our pythagorean triple for that right triangle means that 100=2x and x = 50, which is going to be the side length for the side opposite the 30. That happens to be the height of the triangle, which we need for the area. We also need the base, which according to our Pythagorean triple, being the side across from the 60 degree angle, is 50 sqrt 3. but that's only half the side (we are still dealing with half of the whole triangle) and that means that the length of the whole side is 100 sqrt 3. Now we can find the area of the triangle: [tex]A= \frac{1}{2}(100 \sqrt{3})(50) [/tex] which is [tex]2500 \sqrt{3} [/tex]. Now subtract the area of the triangle from the area of the sector: [tex]10466.666-2500 \sqrt{3} =6136.5[/tex]
