Area of a major sector (1 Viewer)

Demise · Oct 18, 2012

How would you do c ii)?

Timothy6340 · Oct 18, 2012

area of circle minus area of small sector

Sy123 · Oct 18, 2012

So from c i

We find the radius to be:

$l=r \theta \\ 15=r \times \frac{\pi}{5} \\ r=\frac{75}{\pi}$

Now since pi/5 is less than pi, the sector formed is a minor sector, so to find the major sector, we can either find the reflex angle at the centre and use the area of sector formula, or we can find area of whole cirlce minus area of minor sector.

FIRST WAY:

$A_{maj}=A_{circle}-A_{min} \\ \\ A=\pi r^2-\frac{1}{2}r^2 \theta=r^2(\pi-\theta /2) \\ \\ \frac{75^2}{\pi^2}(\pi-\pi/10)=\frac{75^2 \cdot 9}{10 \pi}$

SECOND WAY:

$\theta_{major}=2\pi-\frac{\pi}{5}=\frac{9\pi}{5} \\ A=\frac{1}{2}\frac{75^2}{\pi^2} \cdot \frac{9\pi}{5}=\frac{75^2 \cdot 9}{10\pi}$

Demise · Oct 18, 2012

Sy123 said:
So from c i

We find the radius to be:

$l=r \theta \\ 15=r \times \frac{\pi}{5} \\ r=\frac{75}{\pi}$

Now since pi/5 is less than pi, the sector formed is a minor sector, so to find the major sector, we can either find the reflex angle at the centre and use the area of sector formula, or we can find area of whole cirlce minus area of minor sector.

FIRST WAY:

$A_{maj}=A_{circle}-A_{min} \\ \\ A=\pi r^2-\frac{1}{2}r^2 \theta=r^2(\pi-\theta /2) \\ \\ \frac{75^2}{\pi^2}(\pi-\pi/10)=\frac{75^2 \cdot 9}{10 \pi}$

SECOND WAY:

$\theta_{major}=2\pi-\frac{\pi}{5}=\frac{9\pi}{5} \\ A=\frac{1}{2}\frac{75^2}{\pi^2} \cdot \frac{9\pi}{5}=\frac{75^2 \cdot 9}{10\pi}$

I never learned about the major sector, ffs thank you so much!

Area of a major sector (1 Viewer)

Demise

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Timothy6340

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Sy123

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Demise

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