Question

Prove the formula \(A = \frac{1}{2}{r^2}\theta \) for the area of a sector of a circle with radius \(r\) and central angle \(\)\(\theta \) . (Hint: Assume \(\)\(0 < \theta < {\raise0.7ex\hbox{\(\pi \)} \!\mathord{\left/

{\vphantom {\pi 2}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{\(2\)}}\) and place the center of the circle at the origin so it has the equation \({x^2} + {y^2} = {r^2}\) . Then \(A\) is the sum of the area of the triangle \(POQ\) and the area of the region \(\)\(PQR\) in the figure.)

Short answer

We proved that area of the sector is \(\frac{1}{2}{r^2}\theta \)

Step-by-step solution

Step-1: Given

Given curve \({x^2} + {y^2} = {r^2}\)

To prove the area of sector = \(\frac{1}{2}{r^2}\theta \)

\(r\)=Radius

Step-2: Calculation of Area

The total area of sector = area of right-angle triangle \(\left( {PO\theta } \right)\) + area of under curve \(\left( {P\theta RP} \right)\)

\(\begin{aligned}{l}{A_1} + {A_2} = \frac{1}{2}O\theta \times OP + \int\limits_{r\cos \theta }^r {\sqrt {{r^2} - {x^2}} dx\left( {y > 0} \right)} \\\frac{1}{2} \times r\cos \theta \times r\sin \theta + \left( {\frac{x}{2}\sqrt {{r^2} - {x^2}} + \frac{{{r^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{x}{r}} \right)} \right)_{r\cos \theta }^r\end{aligned}\)

Step-3: Putting an upper limit-lower limit

Total Area =

\(\begin{aligned}{l}\frac{1}{2}{r^2}\cos \theta \sin \theta + 0 + \frac{{{r^2}}}{2}\left( {\frac{\pi }{2}} \right) - \frac{{r\cos \theta }}{2}\sqrt {{r^2} - {r^2}{{\cos }^2}\theta } - \frac{{{r^2}}}{2}{\sin ^{ - 1}}\left( {\frac{{r\cos \theta }}{r}} \right)\\ = \frac{1}{2}{r^2}\cos \theta \sin \theta + \frac{{\pi {r^2}}}{4} - \frac{{{r^2}}}{2}\cos \theta \sin \theta - \frac{{{r^2}}}{2}{\sin ^{ - 1}}\left( {\sin \left( {\frac{\pi }{2} - \theta } \right)} \right)\\ = \frac{1}{2}{r^2}\cos \theta \sin \theta + \frac{{\pi {r^2}}}{4} - \frac{{{r^2}}}{2}\cos \theta \sin \theta - \frac{{{r^2}}}{2}\left( {\frac{\pi }{2} - \theta } \right)\\ = \frac{{{r^2}}}{2}\cos \theta \sin \theta + \frac{{\pi {r^2}}}{4} - \frac{{{r^2}}}{2}\cos \theta \sin \theta - \frac{{\pi {r^2}}}{4} + \frac{1}{2}{r^2}\theta \\ = \frac{1}{2}{r^2}\theta \end{aligned}\)

Hence proved that area of the sector is \(\frac{1}{2}{r^2}\theta \)