# Another Challenging Shaded Region Problem

We’ve written about shaded region problems before (here and here) and recently offered up a challenging problem (here) on this topic. Here’s another problem that will put your math brain to work. Give it a try!

In the diagram below, segment AB is tangent to the circle at Point A. The measure of ∠OBA is 30 degrees and the circumference of the circle is 8π units. What is the area, to the nearest tenth of a square unit, of the shaded region? A.  4.8
B.  5.5
C.  5.8
D.  6.0
E.  6.1

Scroll down for the solution.
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This problem is hard, in part, because there are a number of different things that you need to remember from Geometry class in order to get the correct answer. For instance, you were taught that if a segment is tangent to a circle, it is perpendicular to the radius of the circle at that point. We can, therefore, conclude that ΔOAB is a right triangle. Because one of the angles is given to us as 30 degrees, we also know that this is a 30-60-90 triangle and we can label its side lengths once we find the measure of one of them (make sure you know the side relationships for the “special” 30-60-90 and 45-45-90 triangles!).

Because we know the circumference of the circle, we can figure out the length of the radius, which is also one of the sides of the triangle.

C = 2πr
8π = 2πr
4 = r

Now we can label the sides of the triangle. Next, we need to figure out the area of that shaded region. Oftentimes, the shaded region can be found indirectly by finding the difference between areas that we can compute. In this case, the area of the shaded region is equal to the area of the sector subtracted from the area of the triangle. Let’s find each of those.

Triangle:

A = ½bh
A = ½(4√3)(4)
A = 8√3

Sector:
There’s a bit of work involved here since we’ll need to first find the area of the entire circle and then figure out what part of that circle the sector is.

A = πr²
A = π(4)² = 16π

The central angle for the sector is 60 degrees, so the sector occupies 60/360 or 1/6 of the area of the whole circle. It’s area will be (1/6)(16π) = 8π/3 square units.