7.6: Area of a Circle

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In chapter VI we defined the area of a closed figure to be the number of square units contained in the figure. To apply this definition to the circle, we will again assume a circle is a regular polygon with a large number of sides. The following formula is then obtained:

Theorem $\PageIndex{1}$

The area of circle is $\pi$ times the square of its radius.

\[A = \pi r^2\]

Proof: The area of a circle with radius $r$ is approximately equal to the area of a regular polygon with apothem $a =r$ circumscribed about the circle (Figure $\PageIndex{1}$). The approximation becomes more exact as the number of sides of the polygon becomes larger. At the same time the perimeter of the polygon approximates the circumference of the circle (= $2\pi r$).

$Screen Shot 2021-01-08 at 4.24.34 PM.png$
Figure $\PageIndex{1}$: Regular polygon with apothem $a = r$ circumscribed about circle with radius $r$.

Example $\PageIndex{1}$

Find the area of the circle:

$Screen Shot 2021-01-08 at 4.21.46 PM.png$

Solution

$A = \pi r^2 = \pi (3)^2 = 9\pi = 9(3.14) = 28.26$

Answer: 28.26

Using the formula for the area of a regular polygon (Theorem 7.1.4, section 7.1) we have

\[\text{area of circle} = \text{area of polygon} = \dfrac{1}{2}aP = \dfrac{1}{2} r(2\pi r) = \pi r^2.\]

Example $\PageIndex{2}$

Find the shaded area:

$Screen Shot 2021-01-08 at 4.30.36 PM.png$

Solution

Screen Shot 2021-01-08 at 4.32.51 PM.png — Figure $\PageIndex{2}$: A circle divided into six equal parts.

The shaded area $OAB$ is $\dfrac{60}{360} = \dfrac{1}{6}$ of the total area (see Figure $\PageIndex{2}$). The area of the whole circle = $\pi r^2 = \pi (3)^2 = 9\pi = 9(3.14) = 28.26$. Therefore the area of $OAB = \dfrac{1}{6} (28.26) = 4.71$.

Answer: 4.71

The shaded area in Example $\PageIndex{2}$ is called a sector of the circle. Example $\PageIndex{2}$ suggests the following formula for the area of a sector:

\[\text{Area of sector} = \dfrac{\text{Degrees in arc of sector}}{360} \cdot \text{Area of circle}\]

or simply

\[A = \dfrac{D}{360} \pi r^2\]

Using this formula, the solution of Example $\PageIndex{2}$ would be

\[A = \dfrac{D}{360} \pi r^2 = \dfrac{60}{360} (3.14)(3)^2 = \dfrac{1}{6} (3.14)(9) = \dfrac{1}{6} (28.26) = 4.71\]

Example $\PageIndex{3}$

Find the shaded area:

$Screen Shot 2021-01-08 at 4.36.45 PM.png$

Solution

Let us first find the area of triangle $OAB$ (Figure $\PageIndex{3}$).

Screen Shot 2021-01-08 at 4.37.44 PM.png — Figure $\PageIndex{3}$: Triangle $OAB$ with base $b$ and height $h$.

$\triangle OAB$ is equilateral with base $b = AB = 10$. Drawing in height $h = OC$ we have that $\triangle AOC$ is a $30^{\circ} - 60^{\circ} -90^{\circ}$ triangle with $AC = 5$ and $h = 5\sqrt{3}$. Therefore area of $\triangle OAB = \dfrac{1}{2} bh = \dfrac{1}{2} (10) (5\sqrt{3}) = 25 \sqrt{3}$. Therefore

\[\begin{array} {rcl} {\text{shaded area}} & = & {\text{area of sector } OAB - \text{area of triangle } OAB} \\ {} & = & {\dfrac{D}{360} \pi r^2 - \dfrac{1}{2} bh} \\ {} & = & {\dfrac{60}{360} \pi (10)^2 - \dfrac{1}{2} (10) (5\sqrt{3})} \\ {} & = & {\dfrac{1}{6} (100\pi) - \dfrac{1}{2} (50\sqrt{3})} \\ {} & = & {\dfrac{50\pi}{3} - 25\sqrt{3}} \\ {} & = & {\dfrac{50(3.14)}{3} - 25(1.732)} \\ {} & = & {52.33 - 43.30 = 9.03} \end{array}\]

Answer: $\dfrac{50\pi}{3} - 25\sqrt{3}$ or 9.03.

The shaded area in Example $\PageIndex{3}$ is called a segment of the circle. The area of a segment is obtained by subtracting the area of the triangle from the area of the sector.

Example $\PageIndex{4}$

Find the shaded area:

$Screen Shot 2021-01-08 at 4.47.59 PM.png$

Solution

The area of the large semicircle = $\dfrac{1}{2} \pi r^2 = \dfrac{1}{2} \pi (20)^2 = \dfrac{1}{2} (400) \pi = 200 \pi$. The area of each of the smaller semicircles = $\dfrac{1}{2} \pi r^2 = \dfrac{1}{2} \pi (10)^2 = \dfrac{1}{2} (100) \pi = 50\pi$. Therefore

\[\begin{array} {rcl} {\text{shaded area}} & = & {\text{area of large semicircle - (2)(area of small semicircles)}} \\ {} & = & {200 \pi - 2(50\pi)} \\ {} & = & {200\pi - 100\pi} \\ {} & = & {100 \pi = 100 (3.14) = 314} \end{array}\]

Answer: $100 \pi$ or 314.

Historical Note

Problem 50 of the Rhind Papyrus, a mathematical treatise written by an Egyptian scribe in about 1650 B.C., states that the area of a circular field with a diameter of 9 units is the same as the area of a square with a side of 8 units. This is equivalent to using the formula $A = (\dfrac{8}{9} d)^2$ to find the area of a circle. If we let $d = 2r$ this becomes $A = (\dfrac{8}{9} d)^2 = (\dfrac{8}{9} \cdot 2r)^2 = (\dfrac{16}{9} r)^2 = \dfrac{256}{81} r^2$ or about $3.16 r^2$. Comparing this with our modern formula $A = \pi r^2$ we find that the ancient Egyptains had a remarkably good approximation, 3.16, for the value of $\pi$.

In the same work in which he calculated the value of $\pi$, Archimedes gives a formula for the area of a circle (see Historical Note, section 7.5). He states that the area of a circle is equal to the area of a right triangle whose base $b$ is as long as the circumference and whose altitude $h$ equals the radius. Letting $b = C$ and $h = r$ in the formula for the area of a triangle, we obtain $A = \dfrac{1}{2} bh = \dfrac{1}{2} Cr = \dfrac{1}{2} (2\pi r) = \pi r^2$, the modern formula.