Unit 59: Creative Geometry

It is likely that you will encounter a problem with a figure, usually near the end of section 4, that falls under the umbrella of “creative geometry”.  There’s no way of knowing exactly how the problem will appear, hence the need to get creative.  In general, though, there are a few things to look for.

Here’s an example:

Point P is the center of the circle in the figure above.  What is the value of x ?


  1. Always try and make triangles.  You can often add in your own lines to figures to create triangles, and then use interior angles or special right triangle rules to solve.
  2. Use interior angles of a polygon formula: (mathbf{180}boldsymbol(boldsymbol nboldsymbol-mathbf2boldsymbol)) where n is the number of sides.
  3. Identify 90-degree angles that aren’t stated.  For example, a tangent to a circle always makes a 90-degree angle with a radius of the circle.
  4. Know your circle rules – degrees to radians, fractions of a circle, circumference of a circle, etc.  A lot of these problems incorporate circles.
  5. Use similar triangle properties – same angles, same ratio of sides.

Back to our example:

By adding in our own line, we’ve now made two isosceles triangles.  We know they are isosceles because two of the legs are radii.  This means two of the angles are 20°, and the remaining one must be 140°.  

If  (angle APB=140) and (angle APC=140), then (angle BPC=360-280), or 80°. 

This is a “creative” problem because there are at least 2 other ways to get to the correct answer besides this explanation.  

Here’s another example:

The semicircle above has radius of r inches, and chord (overline{CD}) is parallel to diameter (overline{AB}) .  If the length of (overline{CD}) is (frac23) the length of (overline{AB}), what is the distance between the chord and the diameter in terms of ?

A) (frac13pi r)

B) (frac23pi r)

C) (frac{sqrt2}2r)

D) (frac{sqrt5}3r)

At first glance, it might be hard to see where to go with this problem.  When in doubt, try making triangles!

Now we have a triangle with our distance between the chord and diameter as one of the sides.  As it turns out, we know the other two sides.  

From C to the center of the circle is another radius, r.  

From C to the center of the chord is (frac23r).  

If the full length of the chord is (frac23) the length of the diameter, then half of the chord must be (frac23) the length of half of the diameter.

When we know 2 sides of a right triangle, we can get the third by Pythagorean.


Finally taking the square root of both sides gives us x, the distance in terms of r.


We’ve finally come around to our answer, D).  

Both of the examples given here are tough problems.  Keep these methods in the back of your mind for when you come to another geometry problem that might require some creative thinking.