Hi everyone,
This is a question from a June 1984 cambridge past paper. I'm getting stuck with the part (c) and the 'hence show...'
Please someone can help, I'd be very grateful.
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base $AD = 2r\cos t$ and the height is $r\sin t.$ you can compute the area. – abel Apr 18 '15 at 09:35
2 Answers
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$OA=OD$ and so $\angle ADO=\theta$, so $\angle AOD=\pi-\theta$. So, area of triangle $AOD$ is $\frac{1}{2}\sin(\pi-2\theta)r\times r=\frac{1}{2}\sin(2\theta)r^2$.
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part (a)
$\triangle AOD$ is isosceles triangle with similar sides $= r$ (also called as legs), let $OO'$ be the altitude, since altitude bisects the base,
$l(AD)=2\sqrt{(\text{leg})^2-(\text{altitude})^2}=2\sqrt{r^2-r^2\sin^2 \theta}=2r\cos \theta$
part (b)
Area of the sector formed by $AB, AD$ and arc $BCD$ with center at $A$ is
(fraction of the angle forming the sector) $\times \pi \times (\text {radius})^2$ (why?)
$$\frac{2\theta}{2\pi}\times \pi \times (2r\cos\theta)^2 = 4 r^2 \theta\cos^2 \theta$$
Vikram
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part (c), area of triangle=1/2 * base * height, Area of $OAD=\frac{1}{2}\times 2r\cos\theta \times r \sin \theta$ – Vikram Apr 18 '15 at 10:14