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

Answer 1

given a circle of radius $12cm$.

$OA=OB=12cm=$ radius of the circle and .Let $AB$ be the chord of $12cm$. hence we get an equilateral triangle $OAB$ inside the circle that means $\angle O=\angle A=\angle B={60}^o$ to find the length of the arcs $\left(APBandADB\right)$ of circle.

circumference of the circle $=2\pi r=2\pi \times 12$ length of the arc $APB$ of the circle $=\frac{2\pi \times 12\times 60}{360}$

$=\frac{2\pi \times 12\times 60}{360}=4\pi =12.56cm$

Now the length of the arc $AQB=\frac{2\pi \times 12\times (360-60)}{360}$

$=\frac{2\pi \times 12\times 300}{360}=20\pi$

$=62.80cm$

Now to find the Area of the minor segment

= Area of the sector $ABCA$= Area of the sector $AOBCA$

= Area of the triangle $OAB$

Area of the sector $AOBCA=\pi \times (12{)}^2\times \frac{60}{360}$

$=\pi \times 12\times 2=24\pi =75.36c{m}^2$

Area of the triangle OAB $=\frac{\sqrt{3}}{4}\times 12\times 12=36\sqrt{3}$

$=62.28c{m}^2$

hence area of the minor segment is $=(75.36-62.28)c{m}^2$

$=13.08c{m}^2$