Question

In a circle with a 12 inch radius, find the legnth of a segment joining the midpoint of a 20 inch chord and the center of the circle

Answer 1

Conditions

In a circle with a 12 inch radius, find the length of a segment joining the midpoint of a 20 inch chord and the center of the circle

Solution

Let's consider a graph:

As we can see, BC is a chord and its length is 20 inches. OA is a radius, which is intersect BC in a point D with an angle 90 degrees (we can build radius in this way). Its length is 12 inches. It's obvious to notice, that BD=DC, OB=OC, AC=AB. But OD doesn't equal to AD.

As BC is a chord, than we know a formula which links the radius, the chord length and the angle BOC:

$$B C = 2 \cdot O A \cdot \sin \frac {\angle B O C}{2}$$

So we can find this angle now:

$$2 0 = 2 \cdot 1 2 \cdot \sin \frac {\angle B O C}{2}$$ $$\sin \frac {\angle B O C}{2} = \frac {2 0}{2 4} = \frac {5}{6}$$ $$\angle B O C = 2 \sin^ {- 1} \frac {5}{6}$$

Then, we can find the answer on the question in the task. Consider, for example, triangle ODB. Angle O is a half of angle BOC and it's $\sin^{-1}\frac{5}{6}$. As this is a right triangle, we know that:

$$OD = \frac{BD}{\tan\left(\sin^{-1}\frac{5}{6}\right)} = \frac{10}{\tan\left(\sin^{-1}\frac{5}{6}\right)}$$