+0

0
127
4

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

#2
0

#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
0

#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
0

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