ACT Math: A Challenging Circle Problem

Here’s another challenging problem for you, this one involving circles. You’ll need to use concepts you’ve learned in both Algebra and Geometry to get this one right. Good luck!

In the circle below with center at Point O, the length of chord AB is 30 units. The radius in the diagram is perpendicular to the chord AB, which cuts the radius into two pieces with lengths of x units and x + 1 units as labeled in the diagram. What is the length, in units, of the radius of the circle?

Challenge 4-15A

A.  13
B.  15
C.  17
D.  18
E.  20

Scroll down for the solution.
One the facts about circles that you learned in Geometry is that if a radius is perpendicular to a chord of the circle, it bisects that chord. Therefore, we can conclude that both segment AC and segment CB have lengths of 15 units.

Now draw in the radius OB of the circle. An expression for this radius can be found by adding the expressions for the two pieces of the labeled radius. Let’s label OB as 2x + 1.

Challenge 4-15B

You can now use the Pythagorean Theorem to find the value of x.

a² + b² = c²
x² + 15² = (2x + 1)²
x² + 225 = (2x + 1)(2x + 1)
x² + 225 = 4x² + 4x + 1
0 = 3x² + 4x – 224
0 = (3x + 28)(x – 8)
x = -28/3 or x = 8

(Note: If you had a hard time with the factoring, remember that you can get the values for x by using the Quadratic Formula as well.)

Because the radius can’t be negative, you can conclude that x has to equal 8. That makes the radius 2x + 1 = 2(8) + 1 = 17 units long.  The correct answer is Choice C.

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