+0  
 
+1
2
2
avatar+113 

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.

 

 Jan 13, 2024
 #1
avatar+287 
+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

 Jan 13, 2024
 #2
avatar+126663 
+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

 

cool cool cool

 Jan 13, 2024

5 Online Users

avatar
avatar
avatar