Question

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

Answer 1

Hint: Here, we need to find the area of the minor segment enclosed by the arc and the chord. To solve this question, we will find the area of the sector \[OPQ\] and the area of the triangle \[OPQ\]. Then, we will subtract these to get the required area of the minor segment.

Formula Used: The area of a sector of a circle is given by the formula \[\dfrac{\theta }{{360^\circ }}\pi {r^2}\], where \[r\] is the radius of the circle and \[\theta \] is the angle between the two radii forming the sector.
The area of a triangle can be calculated using the length of any two of its sides \[a\] and \[b\], and the angle between them \[C\] using the formula \[\dfrac{1}{2}ab\sin C\].

Complete step-by-step answer:
We will find the areas of sector \[OPQ\] and triangle \[OPQ\], and subtract them to get the required area enclosed by the arc \[PQ\] and the chord \[PQ\].
First, we will draw the diagram using the given information.

Here, \[OP\] and \[OQ\] are radii of the circle of 15 cm length.
We have to find the area enclosed by the arc \[PQ\] and the chord \[PQ\], that is the area of the minor segment .
First, we will find the area of the sector \[OPQ\].
We know that the area of a sector of a circle is given by the formula \[\dfrac{\theta }{{360^\circ }}\pi {r^2}\], where \[r\] is the radius of the circle and \[\theta \] is the angle between the two radii forming the sector.
Substituting \[\theta = 30^\circ \] and \[r = 15{\rm{ cm}}\], we get
Area of sector \[OPQ\] \[ = \dfrac{{30^\circ }}{{360^\circ }}\pi {\left( {15} \right)^2}\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow \dfrac{{30^\circ }}{{360^\circ }}\pi {\left( {15} \right)^2} = \dfrac{1}{{12}}\pi \left( {225} \right)\\ \Rightarrow \dfrac{{30^\circ }}{{360^\circ }}\pi {\left( {15} \right)^2} = \dfrac{{75}}{4}\pi \end{array}\]
Substituting \[\pi \] as \[\dfrac{{22}}{7}\], we get
\[\begin{array}{l} \Rightarrow \dfrac{{30^\circ }}{{360^\circ }}\pi {\left( {15} \right)^2} = \dfrac{{75}}{{14}} \times \dfrac{{22}}{7}\\ \Rightarrow \dfrac{{30^\circ }}{{360^\circ }}\pi {\left( {15} \right)^2} = \dfrac{{825}}{{14}}\end{array}\]
Therefore, the area of the sector \[OPQ\] is \[\dfrac{{825}}{{14}}{\rm{ c}}{{\rm{m}}^2}\].
Now, we will find the area of the triangle \[OPQ\].
The area of a triangle can be calculated using the length of any two of its sides \[a\] and \[b\], and the angle between them \[C\] using the formula \[\dfrac{1}{2}ab\sin C\].
Therefore, we get the area of the triangle \[OPQ\] as \[\dfrac{1}{2}\left( {OP} \right)\left( {OQ} \right)\sin \angle POQ\].
Substituting \[OP = 15{\rm{ cm}}\], \[OQ = 15{\rm{ cm}}\], and \[\angle POQ = 30^\circ \], we get
Area of triangle \[OPQ\] \[ = \dfrac{1}{2}\left( {15} \right)\left( {15} \right)\sin 30^\circ \]
Simplifying the expression, we get
\[ \Rightarrow \dfrac{1}{2}\left( {15} \right)\left( {15} \right)\sin 30^\circ = \dfrac{{225}}{2}\sin 30^\circ \]
We know that the sine of the angle measuring \[30^\circ \] is \[\dfrac{1}{2}\].
Substituting \[\sin 30^\circ = \dfrac{1}{2}\] in the expression, we get
\[\begin{array}{l} \Rightarrow \dfrac{1}{2}\left( {15} \right)\left( {15} \right)\sin 30^\circ = \dfrac{{225}}{2} \times \dfrac{1}{2}\\ \Rightarrow \dfrac{1}{2}\left( {15} \right)\left( {15} \right)\sin 30^\circ = \dfrac{{225}}{4}\end{array}\]
Therefore, the area of the triangle \[OPQ\] is \[\dfrac{{225}}{4}{\rm{ c}}{{\rm{m}}^2}\].
Finally, we will subtract the area of the triangle\[OPQ\] from the area of sector \[OPQ\] to get the area enclosed by the arc \[PQ\] and the chord \[PQ\].
Therefore, we get
Area enclosed by arc \[PQ\] and the chord \[PQ\] \[ = \left( {\dfrac{{825}}{{14}} - \dfrac{{225}}{4}} \right){\rm{ c}}{{\rm{m}}^2}\]
Taking the L.C.M., we get
\[ \Rightarrow \left( {\dfrac{{825}}{{14}} - \dfrac{{225}}{4}} \right){\rm{ c}}{{\rm{m}}^2} = \dfrac{{1650 - 1575}}{{28}}{\rm{ c}}{{\rm{m}}^2}\]
Subtracting the terms in the numerator, we get
\[ \Rightarrow \left( {\dfrac{{825}}{{14}} - \dfrac{{225}}{4}} \right){\rm{ c}}{{\rm{m}}^2} = \dfrac{{75}}{{28}}{\rm{ c}}{{\rm{m}}^2}\]
Dividing 75 by 28, we get
\[ \Rightarrow \left( {\dfrac{{825}}{{14}} - \dfrac{{225}}{4}} \right){\rm{ c}}{{\rm{m}}^2} = 2.68{\rm{ c}}{{\rm{m}}^2}\]
\[\therefore\] We get the area enclosed by arc \[PQ\] and the chord \[PQ\]as \[2.68{\rm{ c}}{{\rm{m}}^2}\].

Note: If we don’t remember the formula of a triangle \[\dfrac{1}{2}ab\sin C\], we also find the area of the triangle \[OPQ\] using the formula \[\dfrac{1}{2} \times {\rm{Base}} \times {\rm{Height}}\]. Let as assume \[AD\] be the height. Using trigonometric ratios, we can find the base and height of the isosceles triangle \[OPQ\] as \[\dfrac{{15\left( {\sqrt 3 - 1} \right)}}{{\sqrt 2 }}\] and \[\dfrac{{15\left( {\sqrt 3 + 1} \right)}}{{2\sqrt 2 }}\] respectively.