the chord PQ of a circle with radius 15 CM subtends an angle of 60° with the centre of the circle find the area of the minor as well as major segment
Answers
It requires to be solved more but buddy I did this much
sry for inconvinence
Steps:
1) Given :
Angle of sector: \theta=60\degreeθ=60°
Radius of circle,r = 15 cm
Area of minor segment ;
\begin{lgathered}A_{minor}= \frac{r^{2}}{2} ( \frac{\pi*\theta}{180\degree}- sin(\theta) ) \\ \\ =\ \textgreater \ \frac{15^{2}}{2} ( \frac{\pi *60\degree}{180\degree}-sin(60\degree) ) \\ \\ =\ \textgreater \ \frac{225}{2}( \frac{\pi}{3}- \frac{ \sqrt{3} }{2} ) \\ \\ =\ \textgreater \ \frac{225}{2} ( \frac{3.14}{3}- \frac{1.73}{2} ) \\ \\ =\ \textgreater \ 20.475cm^{2}\end{lgathered}
A
minor
=
2
r
2
(
180°
π∗θ
−sin(θ))
= >
2
15
2
(
180°
π∗60°
−sin(60°))
= >
2
225
(
3
π
−
2
3
)
= >
2
225
(
3
3.14
−
2
1.73
)
= > 20.475cm
2
2) Area of major segment :
\begin{lgathered}A_{major}=A_{circle} -A_{minor} \\ \\ =\ \textgreater \ \pi*r^{2} - 20.475 \\ \\ =\ \textgreater \ 3.14*15^{2}-20.475 \\ \\ =\ \textgreater \ 686.025\:cm^{2}\end{lgathered}
A
major
=A
circle
−A
minor
= > π∗r
2
−20.475
= > 3.14∗15
2
−20.475
= > 686.025cm
2