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.

ChiIIBill Jan 13, 2024

#1**+1 **

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

DS2011 Jan 13, 2024

#2**+1 **

Let O be the center of the semi-circle

Let P be the tangent point of the semi-circle with the triangle

Connect OP = the radius of the semi-circle

We have triangle OPZ

The triangle is equilateral so angle PZO = 60

Angle OPZ = 90 (a radius meeting a tangent forms a 90 degree angle )

So angle POZ = 30

We have a 30-60 -90 right triangle

OZ = (1/2)YZ = (1/2) (10) = 5

PZ = 5/2

So OP = (5 /2)sqrt (3) = the radius of the semi-circle

Area of semi-circle = (1/2 pi * [ ( 5/2)sqrt 3 ) ] ^2 = (75 / 8) pi

CPhill Jan 13, 2024