The sides of triangle XYZ are XY = XZ = 25 and YZ = 40. A semicircle is inscribed in triangle XYZ so that its diameter lies on line YZ, and is tangent to the
other two sides. Find the area of the semicircle.
I'm not sure what to do. I dropped an altitude to split it in half but it didn't really help. Could someone please help? Thank you, and I hope you have a nice day and stay safe!
+128473 Draw altitude XD . . .(as you did)
We can find this altitude as
XD = sqrt [ 25^2 - 20^2 ]= sqrt [ 625 - 400 ] = sqrt (225) = 15
So...
sin XZY = 15 / 25 = 3/5
From D, draw DE such that E is the point of tangency between the circle and the side XZ of the triangle
Then DE = the radius of the semi-circle
And angle DEZ is right
So...by the Law of Sines
DE / sin XYZ = DZ /sin 90
DE / (3/5) = 20
DE = 20 (3/5) = 12
This is the radius of the semi-circle.....can you take it from there ???
