Question

OPQ is the sector of a circle having a center at O and radius 15cm. If \[m\angle POQ={{30}^{\circ }}\], find the area enclosed by arc PQ and chord PQ.

Answer 1

Hint: In this problem, we are given that OPQ is the sector of a circle having a center at O and radius 15cm. We have to find the area enclosed by arc PQ and chord PQ if \[m\angle POQ={{30}^{\circ }}\]. We can first find the area of the triangle OPQ with the given angle and radius using the formula \[\dfrac{1}{2}{{r}^{2}}\sin \theta \]. We can then find the area of sector with the formula \[\pi {{r}^{2}}\dfrac{\theta }{{{360}^{\circ }}}\]. We can then subtract them to get the area enclosed by the arc PQ and the chord PQ.

Complete step by step answer:
Here we are given that OPQ is the sector of a circle having a center at O and radius 15cm.
We have to find the area enclosed by arc PQ and chord PQ if \[m\angle POQ={{30}^{\circ }}\].

We can now find the area of the triangle OPQ with the given angle and radius using the formula \[\dfrac{1}{2}{{r}^{2}}\sin \theta \], we get
Area of the triangle OPQ = \[\dfrac{1}{2}{{\left( 15 \right)}^{2}}\sin {{30}^{\circ }}=\dfrac{225}{5}c{{m}^{2}}\]
We can now find the area of sector with the formula \[\pi {{r}^{2}}\dfrac{\theta }{{{360}^{\circ }}}\], we get
Area of sector = \[\pi {{\left( 15 \right)}^{2}}\dfrac{{{30}^{\circ }}}{{{360}^{\circ }}}=\dfrac{225\pi }{12}=\dfrac{75}{4}\pi \]
From the diagram, we can see that,
the area enclosed by arc PQ and chord PQ = Area of sector – Area of triangle OPQ.
the area enclosed by arc PQ and chord PQ = \[\dfrac{75}{4}\pi -\dfrac{225}{4}=\dfrac{75}{4}\left( \pi -3 \right)c{{m}^{2}}\].
Therefore, the area enclosed by arc PQ and chord PQ = \[\dfrac{75}{4}\left( \pi -3 \right)c{{m}^{2}}\]

Note: We should always remember that the formula to find the area with the given angle and radius is \[\dfrac{1}{2}{{r}^{2}}\sin \theta \] and the formula to find the area of the sector with the given angle and radius is \[\pi {{r}^{2}}\dfrac{\theta }{{{360}^{\circ }}}\]. Here we can find the required area from the proper diagram.