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1. A quarter-circle with radius \(5\) is drawn. A circle is drawn inside the sector, which is tangent to the sides of the sector, as shown. Find the radius of the inscribed circle.

 

2. 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 \(\overline{YZ}\), and is tangent to the other two sides. Find the area of the semicircle.

Thank you very much for your help!

 May 25, 2020
 #1
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Anyone??

 May 25, 2020
 #2
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#2)  Drop a perpendicular from X to YZ. Call this point O.

       O will be the midpoint of YZ.

       Since YZ = 40, OZ = 20.

       Use the Pythagorean Theorem on right trianle(XOZ)

       --->   XO2 + OZ2  =  XZ2   --->   XO2 + 202  =  252   --->   XO  =  15

 

       Draw a perpendicular from O to XZ. Call the point of intersection P.

       Since this is perpendicular to XZ, this is the point of tangency to the semicircle.

       Consider triangle(OPZ); this right triangle is similar to triangle((XOZ) by AA.

       Therefore:  OP / OZ  =  XO / XZ.

       OP / 20  =  15 / 25   --->   OP  =  20 · 15 / 25   --->   OP  =  12.

 

       OP is the radius of the semicircle; now use the area formula:

            area semicircle  =  ½ · pi · radius2   to find the area of the semicircle.

 May 25, 2020
 #3
avatar+21953 
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#1)  Draw a line segment from O to the point of tangency of te quarter-circle (call that point T).

       The center of the circle will be on OT.

       Call the cdnter of the circle C.

       Drop perpendiculars from C to AO (call that point X) and from C to OB (call tht point Y).

       The distances from C to T, from C to X, and from C to Y are all equal and they are radii of

              the smaller circle.

       Let the radius of the smaller circle be x.

       All the distances:  XO, YO, CX, CY, and OT are equal to x.

       Since OXCY is a square with side = x, the distance from O to C  =  x·sqrt(2).

       The distance from  OT  =  OC + CT

                                      OT  =  x·sqrt(2) + x  =  x(sqrt(2) + 1)

       But, OT is a radius of the quarter-circle  =  5.

                                        5 =  x(sqrt(2) + 1)

                                        x  =  5 / (sqrt(2) + 1)

 May 25, 2020
 #4
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Thank you. #1 was wrong though. Any ideas on where you went wrong? It's fine if you don't I might be able to figure it out.

 May 25, 2020

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