# Area Problem #14

Find the area of the semi-circle.

## 6 thoughts on “Area Problem #14”

1. Rob says:

Nice problem

2. My first thought was “How do I know it’s a semicircle?”
Well, it is, and the radius was 15, but it wouldn’t be if the 12 had been 10.
So I considered the problem “Are there other pairs of whole numbers in place of the 6 and the 12 which would give a semicircle, and a whole number for the radius.” That was fun. Simple proportion stuff but it would keep your math whiz kids busy for quite a while.

3. If the numbers were 6 and 10, it would still be a semicircle, and the radius would be 34/3.

To find integer pairs besides 6 and 12 that will result in radii that are also integers seems related to Pythagorean triples.

Let’s use m in place of 6, and n in place of the 12 for the general case.

If you start with any Pythagorean triple triangle, put one acute angle at the centre of the semicircle, one side along the diameter, the hypotenuse is the radius, and the hypotenuse minus the length of the side along the base becomes m, and the height becomes n.

• flyingcoloursmaths says:

The rest of the diameter is n^2 / m (using similar triangles), so the radius is an integer if n(n+m)/(2m) is.

• Another use for the diagram: Find an equation for the circle. If you go at it the best way it’s a two liner!