On Area percentage -2

In this discussion I will prove the following statement:

In above diagram area of black region is $100(\frac{1}{2}-\frac{\sin(\angle COB)}{\pi})%$ of the circle

Proof:

Let $a=\angle COB$ and $\overline{AO}=r$

$\sin(a)=\frac{\overline{CD}}{r}$ $\Rightarrow \overline{CD}=r\sin(a)$ $Area_{\triangle}=\frac{1}{\cancel{2}}\cancel{2}r^2\sin(a)=r^2\sin(a)$ $Area_{semi-circle}=\frac{1}{2}\pi r^2$ $Area_{black}=Area_{semi-circle}-Area_{\triangle}=\frac{1}{2}\pi r^2-r^2\sin(a)$ $Area_{black}\%=\frac{100}{\pi r^2}(\frac{1}{2}\pi r^2-r^2\sin(a))=100(\frac{1}{2}-\frac{\sin(a)}{\pi})$

#Geometry

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Comments

Wait, in line three, shouldn’t it be $Area_{\triangle}=\frac{1}{2}2r(base)r\sin (a)(height)=\frac{1}{2}2r^2\sin (a),$ further implying the percentage should be $100(\frac{1}{2}-\frac{\sin (a)}{\pi})$ ?

Jeff Giff - 11 months, 3 weeks ago

Sure.

Zakir Husain - 11 months, 3 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$