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Topic: POW Solution, Dec. 6-10
Replies: 1   Last Post: Dec 13, 1993 10:08 AM

 Messages: [ Previous | Next ]
 Annie Fetter Posts: 524 Registered: 12/3/04
POW Solution, Dec. 6-10
Posted: Dec 13, 1993 10:03 AM

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Problem of the Week for December 6-10:

A circle is inscribed in a semicircle. The diameter of the circle and the
radius of the semicircle are 12 units. What is the area of the region of
the semicircle that is outside the circle?

**********************

Correct solutions to the problem were submitted by

Dan Hirschhorn
Jonathan Jacobs
Jason Jenkins, Boston U.
Kim Chatha, Chris Bernardi, and Tim Luff of Dover-Sherborn High School

Dan and Jonathan's solutions are included below, as they are quite
different from one another.

**********************

Solutions:

Since ratio of diameters from semicircle to circle is 2:1, by Fund.
thm of Similarity, area of Full circle to inscribed circle is 4:1 so area
of semicircle to inscribed circle is 2:1. Thus inscribed circle takes up
1/2 semicircle! Area of outside half = area of inscribed circle = 36 .
---
Daniel B. Hirschhorn | ISU Mathematics
danh@math.ilstu.edu | (309) 438-7849

-------------------

Explanation, eh? Well, let's see ...
One-half of a circle at radius 12 has area:
(1/2)pi(12)^2
or 72pi
The complete circle at diameter 12, radius 6 has area:
pi(6)^2
or 36pi
So, subtracting the smaller, inscribed circle from the larger semicircle
gives us:
72pi - 36pi, or _36 pi_.
Now do I get full credit, or is it too late? (Honest ... I had the
workthere originally, but I erased it ...)
-Jon (pigpen@hardy.u.washington.edu)

Date Subject Author
12/13/93 Annie Fetter
12/13/93 Problem of the Week