Area enclosed by a full circle or parts of a circle

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Area Enclosed by a Circle

... as a word equation ...

$$\mathrm{Area}=\mathrm{\pi }\times {\text{radius}}^2$$

or
... the algebra ...

$$A=\mathrm{\pi }{r}^2$$

• where
• $A$ is the area enclosed;
• and
• $r$ is the radius of the circle.

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Video and Questions:

Proof

Practice Questions

Textbook Exercise

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Area Enclosed by a semi-circle and a Diameter

... as a word equation ...

$\text{Area of a semicircle}=\text{half the area of a circle}$

or
... the algebra ...

$$A=\frac{\mathrm{\pi }{r}^2}{2}$$

Practice Questions

Textbook Exercise

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Quarter-Circle Area

... as a word equation ...

$\text{Area of a semicircle}=\text{a quarter the area of a circle}$

or
... the algebra ...

$$A=\frac{\mathrm{\pi }{r}^2}{4}$$

Practice Questions

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Sector Area

$$A=\frac{\mathrm{\pi }{r}^2\theta }{360}$$

Practice Questions

Textbook Exercise

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Segment Area

$$A=\frac{1}{2}{r}^2(\frac{\pi \theta }{180}-sin\theta )$$

Practice Questions

Textbook Exercise

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Proof

Each sector of the circle can be approximated by a triangle. Each triangle has an area of $\frac{1}{2}rl$.

The areas of all the triangles approximates to the area of the triangle, $A$

$$A=\frac{1}{2}r{l}_1+\frac{1}{2}r{l}_2+\frac{1}{2}r{l}_3+...+\frac{1}{2}r{l}_{n-2}+\frac{1}{2}r{l}_{n-1}+\frac{1}{2}r{l}_n$$ $$A=\frac{1}{2}r(l_1+l_2+l_3+...+l_{n-2}+l_{n-1}+l_n)$$

The length of the arc for each sector, $c_n$ is approximately equal to $l_n$, so,

$$l_1+l_2+l_3+...+l_{n-2}+l_{n-1}+l_n$$ $$\approx c_1+c_2+c_3+...+c_{n-2}+c_{n-1}+c_n$$ $$=2\mathrm{\pi }r$$

This approximation becomes more true as $n\to \mathrm{\infty }$. Therefore, substituting this into the equation for the area,

$$A=\frac{1}{2}r\times 2\mathrm{\pi }r$$ $$=\mathrm{\pi }{r}^2$$

The first person identified as using this method is Johannes Kepler

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