Q.OAB is a sector of the circle having centre at O and radius 12 cm. If $m\angle AOB = 45°$, find the difference between the area of sector OAB and triangle AOB.

Step 1: $\theta = 45°\times\frac{\pi}{180}=\frac{\pi}{4}$ radian, $r=12$ cm.

Step 2: Area of sector $OAB = \frac12 r^2\theta = \frac12(144)\left(\frac{\pi}{4}\right)=18\pi$ sq.cm.

Step 3: Area of $\triangle AOB = \frac12 r^2\sin\theta = \frac12(144)\sin45° = 72\times\frac{\sqrt2}{2}=36\sqrt2$ sq.cm. …

Compute the sector area (½r²θ) and the enclosed triangle's area (½r²si …

• Using sinθ with θ still in degree inside a formula that expects a numeric radian angle (though sin45° is fine either way here, it is easy to mix conventions on other angles). …