Question

OPQ is the sector of a circle having
centre at 0 and radius 15 cm. If
mZPOQ 30°, find the area enclosed
by arc PQ and chord PQ.​

Answer 1

Answer:

ANSWER

Area of ΔOPQ=

2

1

r

2

sin(θ)=

2

1

⋅225⋅

2

1

=

4

225

Area of sector = π

360

o

θ

⋅r

2

=

12

225

π=

4

75

π

Area between the chord and arc = Area of sector - Area of triangle =

4

75

π−

4

225

=

4

75

(π−3)

Answer 2

Step-by-step explanation:

$\tt Area\:of\:\Delta OPQ = \frac{1}{2}×{r}^{2}sin\theta \\ \\\ \tt \longrightarrow Area\:of\:\Delta OPQ = \frac{1}{2} × 15 × 15 × sin30 \\ \\ \tt \longrightarrow Area\:of\:\Delta OPQ = \frac{225}{4} \\ \\ \tt Area\:of\:sector = \pi \frac{\theta}{360} {r}^{2} \\ \\ \longrightarrow Area\:of\:sector = \pi × \frac{30}{360} × 225 \\ \\ \tt \longrightarrow Area\:of\:sector = \frac{75}{4} \pi \\ \\ \tt Area\:enclosed\:by\:arc\:PQ\:and\:chord\:PQ \\ \\ \tt \longrightarrow A(sector)-A(∆OPQ) \\ \\ \tt \longrightarrow \frac{75}{4}\pi - \frac{225}{4} \\ \\ \tt \longrightarrow \frac{75}{4}(\pi - 3 ) \\ \\ \boxed{\bold{\underline{\red{\tt Area\:enclosed\:by\:arc\:PQ\:and\:chord\:PQ \: is \: \frac{75}{4}(\pi - 3 )}}}}$