NCERT Solutions Class 10 Maths Chapter 12

Q.1 The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Ans.

$\begin{array}{l}\text{It is given that the radii of first and second circles are}\\ \text{19 cm and 9 cm respectively.}\\ \text{Let the radius of the third circle is r.}\\ \text{According to question,}\\ \text{Circumference of the third circle}=\text{Sum of the circumferences}\\ \text{of the first and second circles}\\ \text{or 2}\mathrm{\pi }\text{r}=2\mathrm{\pi }\times 19+2\mathrm{\pi }\times 9=2\mathrm{\pi }(19+9)\\ \text{or 2}\mathrm{\pi }\text{r}=2\times 28\times \mathrm{\pi }\\ \text{or r}=\frac{2\times 28\times \mathrm{\pi }}{2\times \mathrm{\pi }}=28\text{cm}\\ \text{Therefore, the radius of the circle which has circumference}\\ \text{equal to the sum of the circumferences of the given circles}\\ \text{is 28 cm.}\end{array}$

Q.2 The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

Ans.

$\begin{array}{l}\text{It is given that the radii of first and second circles are}\\ \text{8 cm and 6 cm respectively.}\\ \text{Let the radius of the third circle is r.}\\ \text{According to question,}\\ \text{Area of the third circle}=\text{Sum of the areas of the}\\ \text{first and second circles}\\ \text{or}\mathrm{\pi }{\text{r}}^2=\mathrm{\pi }\times 8^2+\mathrm{\pi }\times 6^2=\mathrm{\pi }(64+36)\\ \text{or}\mathrm{\pi }{\text{r}}^2=100\mathrm{\pi }\\ \text{or r}=\sqrt{100}=10\text{cm}\\ \text{Therefore, the radius of the circle which has area}\\ \text{equal to the sum of the areas of the given circles}\\ \text{is 10 cm.}\end{array}$

Q.3 The following figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.


Ans.

$\begin{array}{l}\text{Radius of the region representing Gold score}=\mathrm{r}=\frac{21}{2}=10.5\text{cm}\\ \text{It is given that each band}\left(\text{except gold}\right)\text{is 10.5 cm wide.}\\ \text{Therefore,}\\ \text{Radius of the outer circle of the Red score}\\ ={\mathrm{r}}_1=10.5+10.5=21\text{cm\hspace{0.17em}}\\ \text{Radius of the outer circle of the Blue score}\\ ={\mathrm{r}}_2=21+10.5=31.5\text{cm}\\ \text{Radius of the outer circle of the Black score}\\ ={\mathrm{r}}_3\\ =31.5+10.5\\ =42\text{cm}\\ \text{Radius of the outer circle of the White score}\\ ={\mathrm{r}}_4\\ =42+10.5\\ =52.5\text{cm}\\ \text{Now,}\\ \text{Area of the Gold score}={\mathrm{\pi r}}^2\\ =\frac{22}{7}\times \frac{21}{2}\times \frac{21}{2}\\ =346.5{\text{cm}}^2\\ \text{Area of the Red score}={{\mathrm{\pi r}}_1}^2-\text{Area of the Gold score}\\ =\frac{22}{7}\times {21}^2-346.5\\ =1039.5{\text{cm}}^2\\ \text{Area of the Blue score}={{\mathrm{\pi r}}_2}^2-{{\mathrm{\pi r}}_1}^2\\ =\frac{22}{7}\times \{{(31.5)}^2-{21}^2\}\\ =1732.5{\text{cm}}^2\\ \text{Area of the Black score}={{\mathrm{\pi r}}_3}^2-{{\mathrm{\pi r}}_2}^2\\ =\frac{22}{7}\times \{{\left(42\right)}^2-{(31.5)}^2\}\\ =2425.5{\text{cm}}^2\\ \text{Area of the White score}={{\mathrm{\pi r}}_4}^2-{{\mathrm{\pi r}}_3}^2\\ =\frac{22}{7}\times \{{(52.5)}^2-{\left(42\right)}^2\}\\ =3118.5{\text{cm}}^2\end{array}$

Q.4 The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

Ans.

$\begin{array}{l}\text{It is given that the wheels of a car are of diameter 80 cm each.}\\ \text{Therefore, circumference of each wheel}=\mathrm{\pi d}=\frac{22}{7}\times 80\\ \\ \text{Speed of the car}=66\text{km/h}=\frac{66\times 1000\times 100}{60}\\ =110000\text{cm/minute}\\ \text{Distance covered by the car in 10 minutes}=11,00,000\text{cm}\\ \text{Number of revolutions made by each wheel to}\\ \text{cover}11,00,000\text{cm}=\frac{11,00,000}{\frac{22}{7}\times 80}=4375\end{array}$

Q.5 Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units (B) π units (C) 4 units (D) 7 units

Ans.

$\begin{array}{l}\text{Let the radius of the circle be}\mathrm{r}.\\ \text{Given that,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Circumference of the circle}=\text{Area of the circle}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{\pi r}={\mathrm{\pi r}}^2\\ \text{or 2}=\mathrm{r}\\ \text{Therefore, radius of the circle is 2 units.}\end{array}$

Q.6 Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Ans.

$\begin{array}{l}\text{It is given that radius of the circle is 6 cm and angle of the}\\ \text{sector is 60°.}\\ \text{Area of a sector of angle}\mathrm{\theta }=\frac{\mathrm{\theta }}{360\mathrm{°}}\times {\mathrm{\pi r}}^2\\ \therefore \text{Area of a sector of angle}60\mathrm{°}=\frac{60\mathrm{°}}{360\mathrm{°}}\times \frac{22}{7}\times 6^2=\frac{132}{7}{\text{cm}}^2\end{array}$

Q.7 Find the area of a quadrant of a circle whose circumference is 22 cm.

Ans.

$\begin{array}{l}\text{Let the radius of the circle be}\mathrm{r}.\\ \text{Given that,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Circumference of the circle}=\text{22 cm}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{\pi r}=\text{22 cm}\\ \text{or}\mathrm{r}=\frac{\text{22}}{2}\times \frac{7}{22}=\frac{7}{2}\\ \text{Quadrant of a circle subtends an angle of 90° at the centre.}\\ \therefore \text{Area of a quadrant of the given circle}=\frac{90\mathrm{°}}{360\mathrm{°}}\times \frac{22}{7}\times {\left(\frac{7}{2}\right)}^2\\ =\frac{77}{8}{\text{cm}}^2\end{array}$

Q.8 The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Ans.

\begin{array}{l}The\; length\; of\; minute\; hand,r=14\text{cm}\\ \text{In 60 minutes, a minute hand subtends an angle of 360°}\\ \text{at the centre}\text{.}\\ \text{So, in 5 minutes a minute hand subtends an angle}\\ \text{of}\frac{\text{360°}\times 5}{\text{60°}}=\text{30°}\\ \therefore \text{Area swept by the minute hand in 5 minutes}\\ =\frac{30°}{360°}\times \frac{22}{7}\times {\left(14\right)}^2\\ =\frac{154}{3}{\text{cm}}^2\end{array}

Q.9 A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major sector. (Use π = 3.14)

Ans.

$\begin{array}{l}\text{Let AB be the chord of the circle subtending 90° angle at}\\ \text{centre O of the circle.}\\ \left((\mathrm{i}\right)\text{) Area of minor sector OACB}=\frac{90°}{360°}\times {\mathrm{\pi r}}^2\\ =\frac{1}{4}\times 3.14\times 10\times 10\\ =78.5{\text{cm}}^2\\ \text{Area of}\mathrm{\Delta }\text{OAB}=\frac{1}{2}\times \text{OA}\times \text{OB}=\frac{1}{2}\times 10\times 10=50{\text{cm}}^2\\ \text{Area of minor segment ACB}\\ =\text{Area of minor sector OACB}-\text{Area of}\mathrm{\Delta }\text{OAB}\\ =78.5{\text{cm}}^2-50{\text{cm}}^2=28.5{\text{cm}}^2\\ \left(\left(\mathrm{ii}\right)\right)\text{Area of major sector OADB}=\frac{360°-90°}{360°}\times {\mathrm{\pi r}}^2\\ =\frac{270°}{360°}\times 3.14\times 10\times 10\\ =235.5{\text{cm}}^2\end{array}$

Q.10 In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

1. 1. the length of the arc
   2. area of the sector formed by the arc
   3. area of the segment formed by the corresponding chord

Ans.

$\begin{array}{l}\text{Radius of the given circle}=\mathrm{r}=21\text{cm}\\ \text{Angle subtended by the given arc}=60\mathrm{°}\\ \text{Length of an arc of a sector of angle}\mathrm{\theta }=\frac{\mathrm{\theta }}{360\mathrm{°}}\times 2\mathrm{\pi r}\\ \therefore \text{Length of arc ACB}=\frac{60\mathrm{°}}{360\mathrm{°}}\times 2\times \frac{22}{7}\times 21=22\text{cm}\\ \text{Area of sector OACB}=\frac{60\mathrm{°}}{360\mathrm{°}}\times \frac{22}{7}\times 21\times 21=231{\text{cm}}^2\\ \text{In}\mathrm{\Delta }\text{OAB,}\\ \angle \text{OAB}=\angle \text{OBA [As OA}=\text{OB]}\\ \angle \text{OAB}+\angle \text{OBA}+\angle \text{AOB}=180\mathrm{°}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\angle \text{OAB}+60\mathrm{°}=180\mathrm{°}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{OAB}=\frac{180\mathrm{°}-60\mathrm{°}}{2}=\frac{120\mathrm{°}}{2}=60\mathrm{°}\\ \text{Therefore,}\mathrm{\Delta }\text{OAB is an equilateral triangle.}\\ \text{Area of}\mathrm{\Delta }\text{OAB}=\frac{\sqrt{3}}{4}\times {\left(\text{side}\right)}^2=\frac{\sqrt{3}}{4}\times {\left(\text{21}\right)}^2=\frac{441\sqrt{3}}{4}{\text{cm}}^2\\ \text{Area of segment ACB}=\text{Area of sector OACB}-\text{Area of}\mathrm{\Delta }\text{OAB}\\ =(231-\frac{441\sqrt{3}}{4}){\text{cm}}^2\end{array}$

Q.11

\begin{array}{l}Achordofacircleofradius15cmsubtendsanangleof\\ 60°atthecentre.Findtheareasofthecorresponding\\ minorandmajorsegmentsofthecircle.\\ \left(Use\pi =3.14and\sqrt{3}=1.73\right)\end{array}

Ans.

$\begin{array}{l}\text{Radius of the given circle}=\mathrm{r}=15\text{cm}\\ \text{Angle subtended by the given chord}=60\mathrm{°}\\ \text{Area of sector OACB}=\frac{60\mathrm{°}}{360\mathrm{°}}\times 3.14\times 15\times 15=117.75{\text{cm}}^2\\ \text{In}\mathrm{\Delta }\text{OAB,}\\ \angle \text{OAB}=\angle \text{OBA [As OA}=\text{OB]}\\ \angle \text{OAB}+\angle \text{OBA}+\angle \text{AOB}=180\mathrm{°}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\angle \text{OAB}+60\mathrm{°}=180\mathrm{°}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{OAB}=\frac{180\mathrm{°}-60\mathrm{°}}{2}=\frac{120\mathrm{°}}{2}=60\mathrm{°}\\ \text{Therefore,}\mathrm{\Delta }\text{OAB is an equilateral triangle.}\\ \text{Area of}\mathrm{\Delta }\text{OAB}=\frac{\sqrt{3}}{4}\times {\left(\text{side}\right)}^2=\frac{\sqrt{3}}{4}\times {\left(\text{15}\right)}^2\\ =\frac{225\times 1.73}{4}{\text{cm}}^2=97.3125{\text{cm}}^2\\ \text{Area of segment ACB}=\text{Area of sector OACB}-\text{Area of}\mathrm{\Delta }\text{OAB}\\ =117.75-97.3125=20.4375{\text{cm}}^2\\ \text{Area of major segment ADB}\\ =\text{Area of circle}-\text{Area of segment ACB}\\ =\mathrm{\pi }{\left(15\right)}^2-20.4375\\ =3.14\times 225-20.4375=706.5-20.4375\\ =686.0625{\text{cm}}^2\end{array}$

Q.12

$\begin{array}{l}\text{A chord of a circle of radius 12 cm subtends an}\\ \text{angle of 120° at the centre. Find the area of the}\\ \text{corresponding segment of the circle.}\\ (\mathbf{Use}\mathrm{\pi }=\mathbf{3}.\mathbf{14}\mathbf{and}\sqrt{3}=\mathbf{1}.\mathbf{73})\end{array}$

Ans.

$\begin{array}{l}\text{Let AB is a chord of a circle of radius 12 cm which subtends an}\\ \text{angle of 120° at the centre}\mathrm{O}\text{.}\\ \text{Radius of the given circle}=\mathrm{r}=\text{OA}=\text{OB}=12\text{cm}\\ \text{Angle subtended by the given chord}=120\mathrm{°}\\ \text{Area of sector OACB}=\frac{120\mathrm{°}}{360\mathrm{°}}\times 3.14\times 12\times 12=150.72{\text{cm}}^2\\ \text{Let us draw a perpendicular OD on chord AB. It bisects the}\\ \text{chord AB.}\\ \therefore \text{AD}=\text{DB}\\ \text{In}\mathrm{\Delta }\text{ODA,}\\ \frac{\text{OD}}{\mathrm{OA}}=\cos 60\mathrm{°}\\ \mathrm{or}\frac{\mathrm{OD}}{12}=\frac{1}{2}\\ \mathrm{or}\text{OD}=6\text{cm}\\ \text{Also,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}sin60°}=\frac{\mathrm{AD}}{\mathrm{OA}}=\frac{\mathrm{AD}}{12}\\ \mathrm{or}\frac{\sqrt{3}}{2}=\frac{\mathrm{AD}}{12}\\ \mathrm{or}\mathrm{AD}=6\sqrt{3}\text{cm}\\ \mathrm{Now},\\ \mathrm{AB}=2\mathrm{AD}=2\times 6\sqrt{3}=12\sqrt{3}\text{cm}\\ \text{Area of}\mathrm{\Delta }\text{OAB}=\frac{1}{2}\times \mathrm{AB}\times \mathrm{OD}=\frac{1}{2}\times 12\sqrt{3}\times 6=62.28{\text{cm}}^2\\ \text{Area of segment ACB}=\text{Area of sector OACB}-\text{Area of}\mathrm{\Delta }\text{OAB}\\ =150.72-62.28=88.44{\text{cm}}^2\end{array}$

Q.13

$\begin{array}{l}\text{A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m}\\ \text{long rope}\left(\text{see the following figure}\right)\text{. Find}\\ \left(\text{i}\right)\text{the area of that part of the field in which the horse can graze.}\\ \left(\text{ii}\right)\text{the increase in the grazing area if the rope were 10 m long instead of 5 m.}(\mathbf{Use}\mathrm{\pi }=\mathbf{3}.\mathbf{14})\end{array}$

Ans.

$\begin{array}{l}\left(\left(\mathrm{i}\right)\right)\text{The horse can graze a sector of 90° in a circle of radius 5 m.}\\ \text{Area that can be grazed by horse}\\ =\text{Area of sector of 90° in a circle of radius 5 m}\\ =\frac{90°}{360°}\times 3.14\times 25=19.625{\text{m}}^2\\ \left(\left(\mathrm{ii}\right)\right)\text{When length of rope is 10 m,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}area that can be grazed by horse}\\ =\frac{90°}{360°}\times 3.14\times 100=78.5{\text{m}}^2\\ \mathrm{Increase}\mathrm{in}\text{grazing area}=\left(78.5-19.625\right){\text{m}}^2=58.875{\text{m}}^2\end{array}$

Q.14 A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the following figure. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.


Ans.

$\begin{array}{l}\text{Total length of wire required will be the length of 5 diameters}\\ \text{and the circumference of the brooch.}\\ \text{Radius of circle}=\frac{35}{2}\text{mm}\\ \text{Circumference of brooch}=2\mathrm{\pi r}=2\times \frac{22}{7}\times \frac{35}{2}=110\text{mm}\\ \text{Length of wire required}=110+5\times 35=110+175=285\text{mm}\\ \text{From the given figure, we observe that each of 10 sectors of the}\\ \text{circle subtends 36° at the centre of the circle.}\\ \text{Therefore, area of each sector}=\frac{36\mathrm{°}}{360\mathrm{°}}\times {\mathrm{\pi r}}^2\\ =\frac{1}{10}\times \frac{22}{7}\times {\left(\frac{35}{2}\right)}^2\\ =\frac{385}{4}{\text{mm}}^2\end{array}$

Q.15 An umbrella has 8 ribs which are equally spaced (see the following figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.


Ans.

$\begin{array}{l}\mathrm{There}\text{are 8 ribs in the given umbrella. The arc between two}\\ \text{consecutive ribs subtends}\frac{360\mathrm{°}}{8}=45\mathrm{°}\text{at the centre of the}\\ \text{assumed flat circle.}\end{array}$

$\begin{array}{l}\text{Area between two consecutive ribs of circle}\\ =\frac{45\mathrm{°}}{360\mathrm{°}}\times {\mathrm{\pi r}}^2\\ =\frac{1}{8}\times \frac{22}{7}\times {\left(45\right)}^2=\frac{22275}{28}{\text{cm}}^2\end{array}$

Q.16 A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Ans.

$\begin{array}{l}\mathrm{It}\text{is obvious that each blade of wiper will sweep an area of}\\ \text{a sector of 115° in a circle of 25 cm radius.}\\ \\ \text{Area of a sector of 115° in a circle of 25 cm radius}\\ =\frac{115\mathrm{°}}{360\mathrm{°}}\times {\mathrm{\pi r}}^2\\ =\frac{115\mathrm{°}}{360\mathrm{°}}\times \frac{22}{7}\times {\left(25\right)}^2=\frac{158125}{252}{\text{cm}}^2\\ \therefore \text{Area swept by two blades}=2\times \frac{158125}{252}{\text{cm}}^2\\ =\frac{158125}{126}{\text{cm}}^2\end{array}$

Q.17 To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)

Ans.

$\begin{array}{l}\mathrm{It}\text{is obvious that the lighthouse spreads light across a sector}\\ \text{of 80° in a circle of 16.5 km radius.}\\ \therefore \text{Required area}\\ =\frac{80\mathrm{°}}{360\mathrm{°}}\times {\mathrm{\pi r}}^2\\ =\frac{80\mathrm{°}}{360\mathrm{°}}\times 3.14\times {(16.5)}^2=189.97{\text{km}}^2\end{array}$

Q.18

$\begin{array}{l}\text{A round table cover has six equal designs as shown}\\ \text{in the following figure. If the radius of the cover is}\\ \text{28 cm, find the cost of making the designs at the}\\ {\text{rate of ₹ 0.35 per cm}}^{\text{2}}\text{.}(\text{Use}\sqrt{\text{3}}=\text{1.7})\end{array}$

Ans.

$\begin{array}{l}\mathrm{It}\text{is obvious from the above figure that the designs are}\\ \text{segments of the circle.}\\ \text{Let us consider the segment APB. Chord AB is a side of}\\ \text{the hexagon. Each chord subtends}\frac{360°}{6}=60°\text{at the centre}\\ \text{of the circle.}\\ \text{In}\mathrm{\Delta }\text{OAB,}\\ \angle \text{OAB}=\angle \text{OBA [As OA}=\mathrm{OB}]\\ \angle \text{AOB}=60°\\ \mathrm{Also},\\ \angle \text{OAB}+\angle \text{OBA}+\angle \text{AOB}=180°\\ \mathrm{or}\text{2}\angle \text{OAB}=180°-60°=120°\\ \mathrm{or}\angle \text{OAB}=60°\\ \text{Therefore,}\mathrm{\Delta }\text{OAB is an equilateral triangle.}\\ \text{Area of}\mathrm{\Delta }\text{OAB}=\frac{\sqrt{3}}{4}\times {\left(\text{side}\right)}^2=\frac{\sqrt{3}}{4}\times {\left(\text{28}\right)}^2\\ =333.2{\text{cm}}^2\\ \therefore \text{Area of sector OAPB}\\ =\frac{60°}{360°}\times {\mathrm{\pi r}}^2\\ =\frac{1}{6}\times \frac{22}{7}\times {\left(28\right)}^2=\frac{1232}{3}{\text{cm}}^2\\ \text{Area of segment APB}=\text{Area of sector OAPB}-\text{Area of}\mathrm{\Delta }\text{OAB}\\ =\left(\frac{1232}{3}-333.2\right){\text{cm}}^2\\ \therefore \text{Area of designs}=6\times \left(\frac{1232}{3}-333.2\right){\text{cm}}^2=464.8{\text{cm}}^2\\ \text{Cost of making}464.8{\text{cm}}^2\text{design}=464.8\times 0.35=₹162.68\end{array}$

Q.19

$\begin{array}{l}\text{Tick the correct answer in the following:}\\ \text{Area of a sector of angle p}\left(\text{in degrees}\right)\text{of a circle}\\ \text{with radius R is}\\ \text{\hspace{0.17em}\hspace{0.17em}(A)\hspace{0.17em}\hspace{0.17em}}\frac{\text{P}}{180}\times 2\mathrm{\pi }\text{R \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(B)\hspace{0.17em}\hspace{0.17em}}\frac{\text{P}}{180}\times \mathrm{\pi }{\text{R}}^2\\ \text{\hspace{0.17em}\hspace{0.17em}(C)}\frac{\text{P}}{360}\times 2\mathrm{\pi }\text{R \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(D)}\frac{\text{P}}{720}\times 2\mathrm{\pi }{\text{R}}^2\end{array}$

Ans.

$\begin{array}{l}\text{We know that area of a sector of angle}\mathrm{\theta }=\frac{\mathrm{\theta }}{360\mathrm{°}}\times {\mathrm{\pi R}}^2\\ \therefore \text{Area of a sector of angle}\mathrm{P}=\frac{\mathrm{P}}{360\mathrm{°}}\times {\mathrm{\pi R}}^2=\frac{\mathrm{P}}{360\mathrm{°}}\times \frac{2}{2}{\mathrm{\pi R}}^2\\ =\frac{\mathrm{P}}{720\mathrm{°}}\times 2{\mathrm{\pi R}}^2\\ \mathrm{Hence},\text{Option}\left(\text{D}\right)\text{is the correct answer.}\end{array}$

Q.20 Find the area of the shaded region in the following figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.


Ans.

$\begin{array}{l}\text{In the given figure, RQ is diameter. Therefore,}\angle \text{RPQ}=90\mathrm{°}\\ \text{By applying Pythagoras Theorem in}\mathrm{\Delta }\text{PQR, we get}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}QR}}^2={\mathrm{PQ}}^2+{\mathrm{PR}}^2={24}^2+7^2=625\\ \mathrm{or}\text{QR}=\sqrt{625}=25\\ \mathrm{Therefore},\text{radius of the circle}=\mathrm{OR}=\frac{\mathrm{QR}}{2}=\frac{25}{2}\\ \mathrm{Area}\text{of the semi-circle RPQOR}=\frac{1}{2}{\mathrm{\pi r}}^2=\frac{1}{2}\times \frac{22}{7}\times {\left(\frac{25}{2}\right)}^2=\frac{6875}{28}{\text{cm}}^2\\ \mathrm{Area}\text{of}\mathrm{\Delta }\text{PQR}=\frac{1}{2}\times \mathrm{PQ}\times \mathrm{PR}=\frac{1}{2}\times 24\times 7=84{\text{cm}}^2\\ \mathrm{Area}\text{of the shaded region}\\ =\mathrm{Area}\text{of semi-circle RPQOR}-\mathrm{Area}\text{of}\mathrm{\Delta }\text{PQR}\\ =\frac{6875}{28}-84=\frac{4523}{28}{\text{cm}}^2\end{array}$

Q.21

Ans.

$\begin{array}{l}\mathrm{Radius}\text{of inner circle}=7\mathrm{cm}\\ \mathrm{Radius}\text{of outer circle}=14\mathrm{cm}\\ \mathrm{Area}\text{of the shaded region}\\ =\mathrm{Area}\text{of sector OAFC}-\mathrm{Area}\text{of sector OBED}\\ =\frac{40\mathrm{°}}{360\mathrm{°}}\times \mathrm{\pi }{\left(14\right)}^2-\frac{40\mathrm{°}}{360\mathrm{°}}\times \mathrm{\pi }{\left(7\right)}^2\\ =\frac{1}{9}\times \frac{22}{7}\times 196-\frac{1}{9}\times \frac{22}{7}\times 49\\ =\frac{616}{9}-\frac{154}{9}\\ =\frac{462}{9}\\ =\frac{154}{3}{\text{cm}}^2\end{array}$

Q.22 Find the area of the shaded region in the following figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.


Ans.

$\begin{array}{l}\text{Radius of each semicircle}=7\text{cm}\\ \mathrm{Area}\text{of each semicircle}=\frac{1}{2}{\mathrm{\pi r}}^2=\frac{1}{2}\times \frac{22}{7}\times 7\times 7=77{\text{cm}}^2\\ \mathrm{Area}\text{of square ABCD}={\left(\mathrm{side}\right)}^2={14}^2=196{\text{cm}}^2\\ \mathrm{Area}\text{of shaded region}\\ =\mathrm{Area}\text{of square ABCD}-\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{two}\text{semicircles}\\ =196-2\times 77=42{\text{cm}}^2\end{array}$

Q.23 Find the area of the shaded region in the following figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.


Ans.

$\begin{array}{l}\text{Area of the circle}={\mathrm{\pi r}}^2=\frac{22}{7}\times 6^2=\frac{22}{7}\times 36=\frac{792}{7}{\text{cm}}^2\\ \text{Area of the sector}=\frac{60\mathrm{°}}{360\mathrm{°}}\times \frac{22}{7}\times 6^2=\frac{792}{42}{\text{cm}}^2=\frac{132}{7}{\text{cm}}^2\\ \text{Area of the given equilateral triangle}=\frac{\sqrt{3}}{4}{\left(\mathrm{side}\right)}^2\\ =\frac{\sqrt{3}}{4}\times {12}^2\\ =36\sqrt{3}{\text{cm}}^2\\ \text{Area of the shaded region}=\text{Area of the circle}+\text{Area of the triangle}\\ -\text{Area of the sector}\\ =\frac{792}{7}+36\sqrt{3}-\frac{132}{7}=(\frac{660}{7}+36\sqrt{3}){\text{cm}}^2\end{array}$

Q.24 From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the following figure. Find the area of the remaining portion of the square.


Ans.

$\begin{array}{l}\text{Area of the square ABCD}={\left(\mathrm{side}\right)}^2=4^2=16{\text{cm}}^2\\ \text{Area of each sector}=\frac{90\mathrm{°}}{360\mathrm{°}}\times \frac{22}{7}\times 1^2=\frac{11}{14}{\text{cm}}^2\\ \text{Area of the circle}={\mathrm{\pi r}}^2=\frac{22}{7}\times 1^2=\frac{22}{7}{\text{cm}}^2\\ \text{Area of the shaded region}=\text{Area of the square ABCD}\\ -\text{Area of the circle}\\ -4\times \text{Area of the sector}\\ =16-\frac{22}{7}-4\times \frac{11}{14}=\frac{68}{7}{\text{cm}}^2\end{array}$

Q.25 In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the following figure. Find the area of the design (shaded region).


Ans.

$\begin{array}{l}\text{Radius of the circle}=32\text{cm}\\ \text{AD is the median of the triangle ABC.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AO}=\frac{2}{3}\mathrm{AD}=32\\ \text{or \hspace{0.17em}\hspace{0.17em}AD}=48\text{cm}\\ \text{In}\mathrm{\Delta }\text{\hspace{0.17em}ABD,}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}}^2={\mathrm{AD}}^2+{\mathrm{BD}}^2={48}^2+{\left(\frac{\mathrm{AB}}{2}\right)}^2\\ \text{or \hspace{0.17em}\hspace{0.17em}AB}=32\sqrt{3}\text{cm}\\ \mathrm{Area}\text{of equilateral}\mathrm{\Delta }\text{\hspace{0.17em}ABC}=\frac{\sqrt{3}}{4}{\left(\mathrm{side}\right)}^2=\frac{\sqrt{3}}{4}{\left(32\sqrt{3}\right)}^2=768\sqrt{3}{\text{cm}}^2\\ \text{Area of the circle}={\mathrm{\pi r}}^2=\frac{22}{7}\times {\left(32\right)}^2=\frac{22528}{7}{\text{cm}}^2\\ \text{Area of the design}=\text{Area of the circle}-\mathrm{Area}\text{of \hspace{0.17em}}\mathrm{\Delta }\text{\hspace{0.17em}ABC}\\ =(\frac{22528}{7}-768\sqrt{3}){\text{cm}}^2\end{array}$

Q.26 In the following figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded region.


Ans.

$\begin{array}{l}\text{Area of the square ABCD}={\left(\mathrm{side}\right)}^2={14}^2=196{\text{cm}}^2\\ \text{Distance between two centres of circles}=\text{AB}=14\text{cm}\\ \text{Radius of each circle}=\frac{\mathrm{AB}}{2}=\frac{14}{2}=7\text{cm}\\ \text{Area of each sector}=\frac{90\mathrm{°}}{360\mathrm{°}}\times \frac{22}{7}\times 7^2=\frac{77}{2}{\text{cm}}^2\\ \text{Area of the shaded region}=\text{Area of the square ABCD}\\ -4\times \text{Area of the sector}\\ =196-4\times \frac{77}{2}=42{\text{cm}}^2\end{array}$

Q.27 The following figure depicts a racing track whose left and right ends are semicircular.

The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:

1. 1. the distance around the track along its inner edge
   2. the area of the track.

Ans.

$\begin{array}{l}\text{Distance around the track along its inner edge}\\ =\mathrm{AB}+\mathrm{arc}\text{BEC}+\mathrm{CD}+\mathrm{arc}\text{DFA}\\ =106+\frac{1}{2}\times 2\mathrm{\pi r}+106+\frac{1}{2}\times 2\mathrm{\pi r}=\frac{2804}{7}\text{m}\\ \text{Area of track}\\ =\mathrm{area}\mathrm{of}\text{rectangle GHIJ}-\mathrm{area}\mathrm{of}\text{rectangle ABCD}\\ +\text{area of semicircle HKI}-\text{area of semicircle BEC}\\ +\text{area of semicircle GLJ}-\text{area of semicircle AFD}\\ =106\times 80-106\times 60+\frac{1}{2}\times \frac{22}{7}\times {\left(40\right)}^2-\frac{1}{2}\times \frac{22}{7}\times {\left(30\right)}^2\\ +\frac{1}{2}\times \frac{22}{7}\times {\left(40\right)}^2-\frac{1}{2}\times \frac{22}{7}\times {\left(30\right)}^2\\ =4320{\text{m}}^2\end{array}$

Q.28 In the following figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.


Ans.

$\begin{array}{l}\text{Area of smaller circle}={\mathrm{\pi r}}^2=\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}=\frac{77}{2}{\text{cm}}^2\\ \text{Area of a quadrant of the bigger circle}\\ \text{=}\frac{1}{4}\mathrm{\pi }{\text{(7)}}^2\\ \text{=}\frac{1}{4}\times \frac{22}{7}\times {\text{(7)}}^2\\ =\frac{77}{2}{\text{cm}}^2\\ \mathrm{Area}\text{of triangle OBC =}\frac{1}{2}\times 7\times 7=\frac{49}{2}{\text{cm}}^2\\ \mathrm{Area}\text{of a shaded segment of the bigger circle}\\ \text{=Area of a quadrant}-\mathrm{Area}\text{of triangle OBC}\\ =\frac{77}{2}-\frac{49}{2}=\frac{28}{2}{\text{= 14 cm}}^2\\ \mathrm{Total}\text{area of the shaded region}\\ \text{=Area of smaller circle}+2\left(\mathrm{Area}\text{of a shaded segment}\right)\\ \text{=}\frac{77}{2}\text{+2}\times 14=\frac{133}{2}{\text{=66.5 cm}}^2\end{array}$

Q.29

$\begin{array}{l}{\text{The area of an equilateral triangle ABC is 17320.5 cm}}^2\text{.}\\ \text{With each vertex of the triangle as centre, a circle is}\\ \text{drawn with radius equal to half the length of the side}\\ \text{of the triangle}\left(\text{see the following figure}\right)\text{. Find the}\\ \text{area of the shaded region.}(\text{Use}\mathrm{\pi }=\text{3.14 and}\sqrt{\text{3}}=\text{1.73205})\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{side of equilateral triangle be}\mathrm{a}\text{.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Area of equilateral triangle}=17320.5{\text{cm}}^2\\ \mathrm{or}\frac{\sqrt{3}}{4}{\mathrm{a}}^2=17320.5\\ \mathrm{or}{\mathrm{a}}^2=\frac{4\times 17320.5}{\sqrt{3}}\\ \mathrm{or}\mathrm{a}=200\text{cm}\\ \text{Each sector is of 60°.}\\ \text{So area of sector ADEF}=\frac{60\mathrm{°}}{360\mathrm{°}}{\mathrm{\pi r}}^2=\frac{1}{6}\times 3.14\times {\left(100\right)}^2\\ =\frac{15700}{3}\\ \text{Area of}\mathrm{shaded}\text{region}=\mathrm{Area}\text{of equilateral triangle}\\ -3\times \text{area of each sector}\\ =17320.5-3\times \frac{15700}{3}\\ =1620.5{\text{cm}}^2\end{array}$

Q.30 On a square handkerchief, nine circular designs each of radius 7 cm are made (see the following figure). Find the area of the remaining portion of the handkerchief.


Ans.

$\begin{array}{l}\mathrm{Side}\text{of the square}=42\text{cm}\\ \therefore \text{Area of square}=1764{\text{cm}}^2\\ \mathrm{Area}\text{of each circle}=\frac{22}{7}\times 7^2=154{\text{cm}}^2\\ \mathrm{Area}\text{of 9 circles}=9\times 154=1386{\text{cm}}^2\\ \mathrm{Required}\text{area}=1764-1386=378{\text{cm}}^2\end{array}$

Q.31 In the following figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the
(i) quadrant OACB, (ii) shaded region.


Ans.

$\begin{array}{l}\mathrm{Given}\text{that, r}=3.5\text{cm and OD}=2\text{cm}\\ \mathrm{Area}\text{of the quadrant OACB}=\frac{90\mathrm{°}}{360\mathrm{°}}\times \frac{22}{7}\times {(3.5)}^2=\frac{77}{8}{\text{cm}}^2\\ \mathrm{Area}\text{of}\mathrm{\Delta }\text{\hspace{0.17em}OBD}=\frac{1}{2}\times \mathrm{OB}\times \mathrm{OD}=\frac{1}{2}\times 3.5\times 2=3.5{\text{cm}}^2\\ \text{Area of}\mathrm{shaded}\text{region}=\mathrm{Area}\text{of the quadrant OACB}\\ -\mathrm{Area}\text{of}\mathrm{\Delta }\text{\hspace{0.17em}OBD}\\ =\frac{77}{8}-3.5=\frac{49}{8}{\text{cm}}^2\end{array}$

Q.32 In the following figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)


Ans.

$\begin{array}{l}\mathrm{Given}\text{that, OA}=20\text{cm}\\ \text{Area of square OABC}={20}^2=400{\text{cm}}^2\\ \text{Radius of the quadrant\hspace{0.17em}\hspace{0.17em}OPBQ}=\mathrm{Diagonal}\text{of the square OABC}\\ =20\sqrt{2}\text{cm}\\ \mathrm{Area}\text{of the quadrant OPBQ}=\frac{90°}{360°}\times 3.14\times {\left(20\sqrt{2}\right)}^2\\ =628{\text{cm}}^2\\ \text{Area of}\mathrm{shaded}\text{region}=\mathrm{Area}\text{of the quadrant OPBQ}\\ -\text{Area of square OABC}\\ =628-400=228{\text{cm}}^2\end{array}$

Q.33 AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see the following figure). If ∠AOB = 30°, find the area of the shaded region


Ans.

$\begin{array}{l}\text{Area of}\mathrm{shaded}\text{region}=\mathrm{Area}\text{of the sector ABO}\\ -\mathrm{Area}\text{of the sector CDO}\\ =\frac{30\mathrm{°}}{360\mathrm{°}}\times \frac{22}{7}\times {\left(21\right)}^2-\frac{30\mathrm{°}}{360\mathrm{°}}\times \frac{22}{7}\times {\left(7\right)}^2\\ =\frac{308}{3}{\text{cm}}^2\end{array}$

Q.34 In the following figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.


Ans.

$\begin{array}{l}\text{Area of}\mathrm{shaded}\text{region}\\ =\mathrm{Area}\text{of the semicircle}+\mathrm{Area}\mathrm{of}\mathrm{\Delta ABC}\\ -\mathrm{Area}\text{of the quadrant ABC}\\ =\frac{1}{2}\times \frac{22}{7}\times {\left(\frac{\sqrt{{14}^2+{14}^2}}{2}\right)}^2+\frac{1}{2}\times {\left(14\right)}^2-\frac{90°}{360°}\times \frac{22}{7}\times {\left(14\right)}^2\\ =154+98-154\\ =98{\text{cm}}^2\end{array}$

Q.35 Calculate the area of the designed region in the following figure common between the two quadrants of circles of radius 8 cm each.


Ans.

$\begin{array}{l}\text{The designed area is the common region between two}\\ \text{equal sectors BAEC and DAFC.}\\ \mathrm{Area}\text{of the sector BAEC}=\mathrm{Area}\text{of the sector DAFC}\\ =\frac{90\mathrm{°}}{360\mathrm{°}}\times \frac{22}{7}\times {\left(8\right)}^2=\frac{352}{7}{\text{cm}}^2\\ \mathrm{Area}\text{of}\mathrm{\Delta }\text{ADC}=\mathrm{Area}\text{of}\mathrm{\Delta }\text{ABC}=\frac{1}{2}\times 8\times 8=32{\text{cm}}^2\\ \mathrm{Area}\text{of the designed portion}\\ =2\times \left(\mathrm{Area}\text{of segment AEC}\right)\\ =2\times (\mathrm{Area}\text{of sector BAEC}-\mathrm{Area}\text{of}\mathrm{\Delta }\text{ABC})\\ =2\times (\frac{352}{7}-32)=\frac{256}{7}{\text{cm}}^2\end{array}$