New way to derive sectors of a circle (easy)

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So for starters the area of an entire circle has 360º,right?

So we can say that: ##1∏r^2## is ##\equiv## to ##360º##

So by that logic ##0.5∏r^2## is ##\equiv## to ##180º##

And finally ##0.25∏r^2## is ##\equiv## to ##90º##

Divide both sides by 9, and you get : ##0.25∏r^2/9## is ##\equiv## to ##10º##

From that it's much simpler to multiply both sides by some variable.

Simple right?
 
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shadowboy13 said:
So for starters the area of an entire circle has 360º,right?
For starters, the area of a circle is not 360°. That's the measure of the angle of a sector.
shadowboy13 said:
So we can say that: ##1∏r^2## is ##\equiv## to ##360º##

So by that logic ##0.5∏r^2## is ##\equiv## to ##180º##

And finally ##0.25∏r^2## is ##\equiv## to ##90º##

Divide both sides by 9, and you get : ##0.25∏r^2/9## is ##\equiv## to ##10º##

From that it's much simpler to multiply both sides by some variable.

Simple right?
 
Try using \pi in your latex code to produce ##\pi## instead of using the product symbol.

If you want to find the area of a sector of a circle that has angle ##\theta## then multiply the area of a circle by ##\theta/2\pi## so

[tex]A=\pi r^2\frac{\theta}{2\pi}=\frac{r^2\theta}{2}[/tex]

However, this assumes that the angle is in radians, but if you want to use degrees instead then just use the conversion

[tex]\text{angle in radians}=\text{angle in degrees}\times \frac{\pi}{180^o}[/tex]

So the formula is then

[tex]A=\pi r^2\cdot\frac{\phi}{360}[/tex]

Where ##\phi## is in degrees. So if ##\phi=360## which would be the entire circle, then as expected, you get ##A=\pi r^2##