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Circle in a Semicircle - Sept. 16-20, 1996

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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?

- Annie Fetter

Solutions

Annie says:

This isn't a real tough problem, but I think it's neat that it works out to half. Mathematically that makes sense, but you might not guess it from looking at a picture - there are too many curves to be sure.

A comment: leave your answer in terms of pi until the very end of the problem. This not only lets you avoid any rounding errors that you get by multiplying by pi, but it makes the numbers easier to handle, and helps people think of circles the whole way.

Sarah Townsend made a good observation about why you should care about area - it makes things like rearranging furniture a lot easier if you know it will all fit. Patrick Beebe included an excellent opening paragraph: he tells the reader exactly how he's going to solve the problem, which makes reading the rest of the problem a lot easier! Katie Quinn-Kerins noted that the area of a circle inscribed in a semi-circle would always be equal to one quarter of the area of the whole big circle, but she didn't say why - look at the formula for areas and see what happens when you take the square of half as much (so 4^2 vs 2^2, since the pi's cancel each other out).

We got our first very POW response from India, which is exciting!

A list of all the people who got this problem right and most of the solutions are also available.

---

Sarah Townsend
Grade 9, Period 1
School: College Park High School, Pleasant Hill, California

Problem - The diameter of the circle and the radius of the 
semicircle is 12 units.  What is the area of the region of the 
semicircle that is outside the circle.  A + B = ?

Steps to solving the problem :

First find the area of the small circle with the diameter of 12.
      A = Pi r^2
      A = (3.14)(6^2)
      A = (3.14)(36)
      A = 113.04 units 
      Area of the small circle is 113.04 units.

  - Then find the area of the semicircle using 12 units as a radius.
      A = Pi r^2
      A = (3.14)(12^2)
      A = (3.14)(144)
      A = 452.16 units = 226.08
      Area of the semicircle is 226.08 units.

  - Next subtract the area of the small circle from the area of 
    the semicircle.
      226.08 units
     -113.04 units
      113.04 units

Conclusion - My conclusion is that the area of the region around
the circle (A+B) is equal to the area of the circle inside the
semicircle, which is 113.04 units.

Importance - I think this problem is important to know because it
helps you in the real world. For example, if I wanted to 
rearrange my room or add more furniture, then it would be
important to know how much room (area) I would have left. 
To find this out I could find the area of my room and the area
of my furniture and by using math I could solve my problem the 
same way we solved this problem.



Patrick Beebe
Grade: 12
School: Sammamish High School, Bellevue, Washington

The area inside the semicircle but not inside the inscribed
circle is found by finding the total area inside the semicircle
and then subtracting from it the area of the inscribed circle.
I am using the formula (Pi)*(Radius^2) = Area for the
following steps, with some modifications.

Area of semicircle: (1/2)*(Pi)*(12^2)= 72*(Pi)
The 1/2 was added to account for the fact that we are dealing
with a semi or half circle.

Area of inscribed circle: (Pi)*(6^2) = 36*(Pi)

The radius of six is found by taking the given diameter of 12 and
simply dividing by two. Or you could plug the diameter in the
equation 2*Diameter=Radius.

Area of the region of the semicircle that is outside the circle:
72*(Pi) - 36*(Pi) = 36*(Pi)

Therefore, half of the semicircle's area is located outside of the
inscribed circle.



Katie Quinn-Kerins
Grade: 10
School: Germantown Academy, Fort Washington, Pennsylvania

First I found the area of the large circle and divided it in half, 
to find the area of the semicircle in which the smaller circle was 
inscribed.

  Large circle:  A = pi x r x r
                 A = 3.14 x 12 x 12
                 A = 452.389 square units
               A/2 = 226.195 square units

Next I found the area of the small circle.

  Small circle:  A = pi x r x r
                 A = 3.14 x 6 x 6
                 A = 113.097 square units

To find the area of the semicircle not filled in by the smaller 
circle, I subtracted the area of the small circle from the area of 
the semicircle.

        Area = (226.195 square units) - (113.097 square units)
        Area = 113.097 square units

Furthermore, if the radius of one circle is double the radius of 
another circle, the area of the greater circle is four times 
greater than the area of the smaller circle.  



N. Siva
Grade: 5
School: Padma Seshadri, Madras, South India

Here's the solution (I hope) to your problem of a circle within a 
semicircle.

Area left out of circle = ar(semi-circle) - ar(circle)
                        = 0.5*(pi*(r1^2)) -(pi*d2^2)/4
                        = 0.5 *pi * 12 *12 -  pi * 12 *12 * 0.25 
                        = pi * 12 *12( 0.5 - 0.25)
                        = pi * 12 *12( 0.25)
                        = pi * 36
                        = 36pi units

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