+359 A semicircle is inscribed in triangle $XYZ$ so that its diameter lies on $\overline{YZ}$, 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.
+289 My idea: Reflect triangle XYZ across line segment YZ, to form a square, which is possible because this is a 45 45 90 triangle where YZ is the hypotenuse. We can find the area of the semicircle by finding the area of the circle and then dividing it by 2.
The diameter of the circle will be 10 because it is inscriibed in the square of length 10, so the radius is 5.
using A = pir^2, we get 25 pi, or 12.5 pi for the semicircle.
We can approximate 12.5pi as 36.27
Answer: 12.5pi or 36.27
+289 My idea: Reflect triangle XYZ across line segment YZ, to form a square, which is possible because this is a 45 45 90 triangle where YZ is the hypotenuse. We can find the area of the semicircle by finding the area of the circle and then dividing it by 2.
The diameter of the circle will be 10 because it is inscriibed in the square of length 10, so the radius is 5.
using A = pir^2, we get 25 pi, or 12.5 pi for the semicircle.
We can approximate 12.5pi as 36.27
Answer: 12.5pi or 36.27
