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.

$[Take\pi =3.14and\sqrt{3}=1.73]$

Answer 1

ANSWER:

Let AB be the chord. Joining A and B to O, we get an equilateral triangle OAB.
Thus, we have:
$O=\angle A=\angle B=\angle {120}^{\circ }$.

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

Length of the arc ADB:

Circumference of the circle - Length of the arc ACB=$2\pi \times 12-4\pi =20\pi cm=62.80cm$

Now,
Area of the minor segment:

Area of the sector - Area of the triangle=$\pi \times (12{)}^2\times \frac{60}{360}-\sqrt{3}4\times (12{)}^2=13.08c{m}^2$