+625 A semicircle is inscribed in triangle XYZ so that its diameter lies on YZ, and is tangent to the other two sides. If XY = 10, XZ = 10, and YZ = 10*sqrt(2), then find the area of the semicircle.
+1628 By special right triangls, triangle XYZ is a 45-45-90 with legs XY and XZ = 10.
Let the radius of the semicircle = r. Also, let the point of tangency between the semicircle and XY, XZ be P and Q respectively. Let the center of the semicircle be O (which also happens to be the midpoint of YZ).
OP = OQ = r, and since P and Q are points of tangency, they are perpendicular to the tangent line XY and XZ.
[XYZ] = 10*10/2 = 50 units squared
[XYZ] = XY*PO/2 + YZ*QO/2 (because you can divide the triangle in two, but still obtain the same area).
[XYZ] = 50 = 10r/2 + 10r/2 = 10r, so r = 5.
If r = 5, then the area of the semicirlce is pi*r^2 / 2 = 25pi/2
