1. Using Properties of Circles
A Belt That Fits
Finding the length of the tangent segments and the length of arcs (or circumference) using is featured in this highlighted problem for Math Practice 8.
Using Properties of Circles

2. A certain machine is to contain two wheels, one of radius 3 centimeters and one of radius 5 centimeters, whose centers are attached to points 14 centimeters apart. The manufacturer of this machine needs to produce a belt that will fit snugly around the two wheels, as shown in the diagram below. How long should the belt be?
Initiate student discussion about appropriate strategies to solve this length problem.

3. Let’s take a closer look.
The length of the belt consists of four parts:
two segments tangent to the two circles, and two circular arcs.
Have students draw a label a picture (next slide). Be sure to label the distance between the two circle centers.

4. To determine the length of the tangent segments, we draw a line from the center of the small circle perpendicular to the radius of the large circle.
The quadrilateral that is formed by the 2 radii, the tangent segment, and the segment connecting the centers of the circle is a trapezoid since the two radii are parallel. This trapezoid can be decomposed into a rectangle and right triangle by sketching an auxiliary line. Have students pay special attention to the position of the right angles (at the points of tangency).
Since the radius of each circle is perpendicular to the tangent segment at the
point of tangency, this forms a rectangle.
The length of the tangent segment is equal to the length of the opposite side
of the rectangle, which is a leg of a right triangle with a hypotenuse of 14 cm
and another leg of length 2 cm.

5. Use Pythagorean theorem to find the length of each tangent segment.
a2 + b2 = c2
22 + b2 = 142
b2 = 142 – 22
b =
Since the trapezoid is now decomposed into a right triangle, use Pythagorean theorem to find the longer leg of the right triangle recognizing the following givens: Hypotenuse = 14 cm, Leg = 2 cm. Since the remaining figure is a rectangle, the tangent segment is also equal to longer leg, 8√3. The two tangent segments are equal to 16√3.
For students not familiar with trigonometry, the remaining distances around the circles can be found using the circumference formula, C=2πr. Since the distance is only half of the circumference, distance = ½ (2π(3)) + ½ (2π(5)) = 8π ≈ cm.

6. angle θ in our right triangle:
To compute the lengths of the two circular arcs, we start by computing the
angle θ in our right triangle:
This is also the angle between the radius of the small wheel and the horizontal line joining the two centers in the diagram, and twice this angle is the central angle that stretches across the circular arc on the small wheel. (Hint: Create auxiliary lines to show parallel lines and the transversal. The angles are corresponding.)
So the length of the part of the belt that is wrapped around the small wheel is approximately
Using trigonometry, the length of the arcs can also be found.
First, find angle θ in the right triangle by cos-1 (2/14). This angle is congruent to the angle created by the small circle’s radius and the extended segment through the two circle centers. This central angle is congruent to the arc it intercepts (half of the distance around the semicircle).
Using the arc length formula, θ/360 x 2πr, the distance around the small semicircle would be twice the distance or ≈ 8.56.

7. So the total length of the belt should be
Meanwhile, the central angle that stretches across the circular arc on the large wheel is 2[180 - arccos(2/14)] radians, and thus the length of the part of the belt that is wrapped around the large wheel is
So the total length of the belt should be
Using trigonometry to find the distance around the larger half circle, first extend the segment that connects the two circle’s centers.
Since Angle θ = cos-1(2/14), the central angle associated with the larger circle would be cos-1(2/14). To find the length of the arc, use the arc length formula, θ/360 x 2πr, again multiplying by two to find the entire length around the larger half circle.
Add 2 tangent segments and two arc lengths together.