Areas Related to Circles

• Sector: The portion (or part) of the circular region enclosed by two radii and the corresponding arc of a circle is called a sector of the circle.
  

Here region OAPB is a sector of the circle with centre $0.\angle AOB$ is called the angle of the sector. Here unshaded region OAPB is called minor sector and shaded region OAQB is called major sector. Angle of the major sector $=360^{\circ }-\angle \mathrm{AOB}$.

• Segment: The portion (or part) of the circular region enclosed between a chord and the corresponding arc of a circle is called a segment of the circle. Shaded region APB is a segment of the circle. The unshaded region AQB is another segment of the circle formed by the chord $AB$.
  
  APB is called the minor segment and AQB is called the major segment.

1.0Method to calculate area of sector and segment

Let OAPB be a sector of a circle with centre 0 and radius $r$. Let $\angle AOB=\theta$. Now area of the circle (in fact a circular region of disc) $=\pi r^2$. We can regard this circular region on disc as a sector forming an angle of $360^{\circ }$ (i.e. of degree measure 360 ) at the centre 0 . By unitary method, area of the sector OAPB can be calculated as follows When the degree measure of the angle at the centre is 360 , area of the sector $=\pi r^2$

When the degree measure is 1 , area of sector $=\frac{1}{360}\times \pi r^2$ When the degree measure is $\theta$, area of sector $=\frac{\theta }{360}\times \pi r^2$ Thus, area of the sector OAPB (i.e., sector of angle $\theta$ ) $=\frac{\theta }{360}\times \pi r^2$ where $r$ is the radius of the circle and $\theta$ is the angle of the sector. Next, we shall find the length of the arc APB corresponding to this sector by unitary method.

If angle subtended at the centre is $360^{\circ }$, then length of the $\mathrm{arc}=2\pi r$

Now, area of segment APB. $=$ area of sector OAPB - area of $△\mathrm{OAB}$

Note: Area of $△\mathrm{OAB}$ with $\angle \mathrm{AOB}=\theta$ is $\frac{1}{2}{\mathrm{r}}^2\sin \theta$ or ${\mathrm{r}}^2\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)$

• If $\theta =90^{\circ }$, then $△AOB$ is right triangle. So, area of $△AOB=\frac{1}{2}\times AO\times BO=\frac{1}{2}\times r^2$
• If $\theta =60^{\circ }$, then $△AOB$ is equilateral triangles. So, of $△AOB=\frac{\sqrt{3}}{4}\times r^2$
• If $\theta =120^{\circ }$, then $△AOB$ is isosceles triangle. So, area of $△AOB=\frac{1}{2}{r}^2\sin 120^{\circ }$ $$\begin{gathered} =\frac{1}{2}{r}^2\times \frac{\sqrt{3}}{2} \\ =\frac{\sqrt{3}}{4}{r}^2 \end{gathered}$$

2.0Area of combinations of plane figures

In this section, we shall calculate the areas of some combinations of plane figures. In our daily life, we see figures like flowerbeds, drain covers, window designs, designs on table covers etc. These are combinations of plane figures.

3.0Numerical Ability

• The perimeter of a semi-circular protractor is $32.4\mathbf{cm}$. Calculate: (i) The radius of the protractor in cm (ii) The area of the protractor in ${\mathrm{cm}}^2$. Solution: (i) Let the radius of the protractor be rcm . Then, its perimeter $=(\pi r+2r)cm$. $\therefore \pi r+2r=32.4$ $\Rightarrow (\pi +2)r=32.4$ $\Rightarrow \left(\frac{22}{7}+2\right)r=32.4$ $\Rightarrow \frac{36}{7}r=32.4$ $\Rightarrow \mathrm{r}=\left(32.4\times \frac{7}{36}\right)\mathrm{cm}$ $=6.3\mathrm{cm}$. Radius of the protractor $=6.3\mathrm{cm}$. (ii) Area of the protractor $$\begin{gathered} =\frac{1}{2}\pi r^2=\left(\frac{1}{2}\times \frac{22}{7}\times 6.3\times 6.3\right){\mathrm{cm}}^2 \\ =62.37{\mathrm{cm}}^2. \end{gathered}$$ $\therefore$ Area of the protractor $=62.37{\mathrm{cm}}^2$.
• Two circles touch externally. The sum of their areas is $130\pi \mathrm{sq}$. cm and the distance between their centres is $14\mathbf{cm}$. Determine the radii of the circles. Solution: Let the radii of the given circles be Rcm and rcm respectively as shown in figure. As the circles touch externally, distance between their centres $=(R+r)cm$.
  
  $\therefore \mathrm{R}+\mathrm{r}=14$ Sum of their areas $=\left(\pi R^2+\pi r^2\right){\mathrm{cm}}^2=\pi \left({\mathrm{R}}^2+{\mathrm{r}}^2\right){\mathrm{cm}}^2$. $\therefore \pi \left({\mathrm{R}}^2+{\mathrm{r}}^2\right)=130\pi$ $\Rightarrow {\mathrm{R}}^2+{\mathrm{r}}^2=130$ We have the identity, $(R+r{)}^2+(R-r{)}^2=2\left(R^2+r^2\right)$ $\Rightarrow (14{)}^2+(R-r{)}^2=2\times 130$ [From (i) and (ii)] $\Rightarrow (\mathrm{R}-\mathrm{r}{)}^2=64$ $\Rightarrow \mathrm{R}-\mathrm{r}=8$ On solving (i) and (iii), we get $\mathrm{R}=11\mathrm{cm}$ and $\mathrm{r}=3\mathrm{cm}$. Hence, the radii of the given circles are 11 cm and 3 cm .
• A chord of a circle of radius 14 cm makes a right angle at the centre. Calculate: (i) The area of the minor segment of the circle. (ii) The area of the major segment of the circle. Solution: Let AB be the chord of a circle with centre 0 and radius 14 cm such that $\angle \mathrm{AOB}=90^{\circ }$. Thus, $\mathrm{r}=14\mathrm{cm}$ and $\theta =90^{\circ }$.
  
  (i) Area of sector $\mathrm{OACB}=\frac{\pi {\mathrm{r}}^2\theta }{360}=\left(\frac{22}{7}\times 14\times 14\times \frac{90}{360}\right){\mathrm{cm}}^2=154{\mathrm{cm}}^2$. $\text{ Area of }△\mathrm{OAB}=\frac{1}{2}{\mathrm{r}}^2\sin \theta =\left(\frac{1}{2}\times 14\times 14\times \sin 90^{\circ }\right){\mathrm{cm}}^2=98{\mathrm{cm}}^2.$ $\therefore$ Area of minor segment $ACBA=($ Area of sector $OACB)-($ Area of $△OAB)$ $=(154-98){\mathrm{cm}}^2=56{\mathrm{cm}}^2.$ (ii) Area of major segment $\mathrm{BDAB}=($ Area of the circle $)-($ Area of minor segment ACBA$)$

$=\left[\left(\frac{22}{7}\times 14\times 14\right)-56\right]{\mathrm{cm}}^2=(616-56){\mathrm{cm}}^2=560{\mathrm{cm}}^2.$

• The minute hand of a clock is 10.5 cm long. Find the area swept by it in 15 minutes. Solution Angle described by minute hand in 60 minutes $=360^{\circ }$. Angle described by minute hand in 15 minutes $={\left(\frac{360}{60}\times 15\right)}^{\circ }=90^{\circ }$. Thus, the required area is the area of a sector of a circle with central angle, $\theta =90^{\circ }$ and radius, $\mathrm{r}=10.5\mathrm{cm}$. $\therefore$ Required area $=\left(\frac{\pi r^2\theta }{360}\right)$ $=\left(\frac{22}{7}\times 10.5\times 10.5\times \frac{90}{360}\right){\mathrm{cm}}^2=86.63{\mathrm{cm}}^2.$
• The figure shows a running track surrounding a grass enclosure PQRSTU. The enclosure consists of a rectangle $PQST$ with a semi-circular region at each end. Given, $PQ=200\mathrm{m}$ and $PT=70\mathrm{m}$. (i) Calculate the area of the grassed enclosure in ${\mathrm{m}}^2$. (ii) Given that the track is of constant width 7 m , calculate the outer perimeter ABCDEF of the track.
  
  Solution: (i) Diameter of each semi-circular region of grassed enclosure $=\mathrm{PT}=70\mathrm{m}$, $\therefore$ Radius of each one of them $=35\mathrm{m}$. Area of grassed enclosure $\Rightarrow$ (Area of rect. PQST) +2 (Area of semi-circular region with radius 35 m ) $\Rightarrow (\mathrm{PQ}\times \mathrm{PT})+2\times \frac{1}{2}\pi {\mathrm{r}}^2=\left[(200\times 70)+\frac{22}{7}\times 35\times 35\right]{\mathrm{m}}^2=17850{\mathrm{m}}^2$. (ii) Diameter of each outer semi-circle of the track $\Rightarrow \mathrm{AE}=(\mathrm{PT}+7+7)\mathrm{m}=84\mathrm{m}$. $\therefore$ Radius of each one of them $=42\mathrm{m}$. Outer perimeter ABCDEF $=(\mathrm{AB}+\mathrm{DE}+$ semi-circle BCD + semi-circle EFA $)$ $\Rightarrow (2\mathrm{PQ}+2\times$ circumference of semi-circle with radius 42 m$)$ $\Rightarrow (2\times 200+2\times \pi \times 42)\mathrm{m}=\left[2\times 200+2\times \frac{22}{7}\times 42\right]\mathrm{m}=664\mathrm{m}$.

• In an equilateral $\Delta$ of side a units, Height $=\frac{\sqrt{3}}{2}$ a, Inradius $=\frac{a}{2\sqrt{3}}$, Circumradius $=\frac{a}{\sqrt{3}}$

• In an equilateral triangle of side $24\mathbf{cm}$, a circle is inscribed, touching its sides. Find the area of the remaining portion of the triangle. Take $\sqrt{3}=1.73$ and $\pi =3.14$. Solution: Let $△ABC$ be the given equilateral triangle in which a circle is inscribed. Side of the triangle, $a=24\mathrm{cm}$.
  
  Height of the triangle, $\mathrm{h}=\left(\frac{\sqrt{3}}{2}\times \mathrm{a}\right)\mathrm{cm}=\left(\frac{\sqrt{3}}{2}\times 24\right)\mathrm{cm}=12\sqrt{3}\mathrm{cm}$. Radius of the incircle, $\mathrm{r}=\frac{1}{3}\mathrm{h}=\left(\frac{1}{3}\times 12\sqrt{3}\right)\mathrm{cm}=4\sqrt{3}\mathrm{cm}$. $\therefore$ Required Area $=$ Area of the shaded region $\Rightarrow$ (Area of $△ABC$ ) - (Area of incircle) $\Rightarrow \left(\frac{\sqrt{3}}{4}\times 24\times 24-\pi \times 4\sqrt{3}\times 4\sqrt{3}\right){\mathrm{cm}}^2$ $\Rightarrow (144\sqrt{3}-3.14\times 48){\mathrm{cm}}^2=(144\times 1.73-3.14\times 48){\mathrm{cm}}^2$ $\Rightarrow [48\times (3\times 1.73-3.14)]{\mathrm{cm}}^2=(48\times 2.05){\mathrm{cm}}^2=98.4{\mathrm{cm}}^2$
• In the given figure, ABCPA is a quadrant of a circle of radius 22 cm . With AC as diameter, a semi-circle is drawn. Find the area of the shaded region. Solution: Since, ABCPA is a quadrant of a circle of radius 22 cm .
  
  we have, $\mathrm{AB}=\mathrm{BC}=22\mathrm{cm}$ and $\angle \mathrm{ABC}=90^{\circ }$. $\therefore \mathrm{AC}=\sqrt{{\mathrm{AB}}^2+{\mathrm{BC}}^2}$ [by Pythagoras theorem] $=\sqrt{(22{)}^2+(22{)}^2}\mathrm{cm}=\sqrt{2\times 484}\mathrm{cm}=22\sqrt{2}\mathrm{cm}$ $\therefore$ Radius of the semicircle $=\frac{1}{2}\mathrm{AC}=11\sqrt{2}\mathrm{cm}$ $\therefore$ Required area $=($ area of the semicircle with $AC$ as diameter $)+($ area of $△ABC)-($ area of the quadrant with $r=22\mathrm{cm}$ ) $\Rightarrow \left[\frac{1}{2}\times \frac{22}{7}\times (11\sqrt{2}{)}^2+\frac{1}{2}\times 22\times 22-\frac{1}{4}\times \frac{22}{7}\times 22\times 22\right]{\mathrm{cm}}^2$ $\Rightarrow \frac{2662}{7}+242-\frac{2662}{7}=242{\mathrm{cm}}^2$

4.0Memory map