Question

Find the lengths of the arcs cut off from a circle of radius 12 cm by a chord 12 cm long. Also, find the area of the minor segment.

Answer 1

Let AB be the chord. Joining A and B to O, we get an equilateral triangle OAB.
Thus, we have:
$\angle O=\angle A=\angle B=60°$

Length of the arc ACB:
$$\begin{gathered} 2\mathrm{\pi }\times 12\times \frac{60}{360} \\ =4\mathrm{\pi } \\ =12.56\mathrm{cm} \end{gathered}$$

Length of the arc ADB:

$$\begin{gathered} \mathrm{Circumference}\mathrm{of}\mathrm{the}\mathrm{circle}-\mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{arc}\mathrm{ACB} \\ =2\mathrm{\pi }\times 12-4\mathrm{\pi } \\ =20\mathrm{\pi }\mathrm{cm} \\ =62.80\mathrm{cm} \end{gathered}$$

Now,
Area of the minor segment:

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{sector}-\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{triangle} \\ =\left[\mathrm{\pi }\times {\left(12\right)}^2\times \frac{60}{360}-\frac{\sqrt{3}}{4}\times {\left(12\right)}^2\right] \\ =13.08{\mathrm{cm}}^2 \end{gathered}$$