NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle

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 circumference of the two circles.

Answer :

Let r be the radius of the circle whose circumference is equal to the sum of the circumference of two circles of radii 19 cm and 9 cm.

? $2?r=2?(19)+2?(9)$

$=>r=19+9=28$

Hence, the radius of the new circle is 28 cm.

Q2. 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 area of the two circles.

Answer :

Let r be the radius of the circle whose area is equal to the sum of the area of the circles of radii 8 cm and 6 cm.

$?r^2=?(8{)}^2+?(6{)}^2$

=> $r^2=8^2+6^2$

$=>r^2=64+36=>r=10$

Hence, the radius of the new circle is 10 cm.

Q3. 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 reach of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

Answer :

The area of each of the five scoring regions are as under :

For Gold : $?(10.5{)}^2$ $=\frac{22}{7}\times 110.25$

$=\frac{2425.5}{7}=346.5c{m}^2$

For Red : $?[(21{)}^2?(10.5{)}^2]=\frac{22}{7}(441?110.25)=\frac{22}{7}\times 330.75$

$=\frac{7276.5}{7}=1039.5c{m}^2$

For Blue : $?[(31.5{)}^2?(21{)}^2]=\frac{22}{7}(992.25?441)$

$=\frac{22}{7}\times 551.25=\frac{12127.5}{7}=1732.5c{m}^2$

For Black : $?[(42{)}^2?(31.5{)}^2]=\frac{22}{7}(1764?992.25)$

$=\frac{22}{7}\times 771.75=\frac{16978.5}{7}=2425.5c{m}^2$

For White : $?[(52.5{)}^2?(42{)}^2]=\frac{22}{7}(2756.25?1764)$

$=\frac{22}{7}\times 992.25=\frac{21829.5}{7}=3118.5c{m}^2$

Q4. 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?

Answer :

Distance covered by the car in 10 minutes

$=\frac{66\times 1000\times 100\times 10}{60}$ $=\frac{6600000}{6}$ cm

Circumference of the wheel of the car

$=(2\times \frac{22}{7}\times 40)$cm

? Number of revolutions in 10 minutes when the car is travelling at a speed of 66 km/hr.

$=\frac{6600000}{6}\times \frac{7}{2\times 22\times 40}=4375$

Hence, the wheel of car makes 4375 revolutions in 10 minutes.

Q5. 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) $?$ units
(c) $4$ units
(d) $7$ units

Answer :

Here, $2?r=?r^2$, where r is the radius

$=>r^2?2r=0$

$=>r(r?2)=0$

$=>r=0orr=2$

But, $r?0$ ? $r=2$ units

Hence the correct option is (a).

Q1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°

Answer :

We know that the area A of a sector of angle $?$ in a circle of radius r is given by $A=\frac{?}{360}\times ?r^2$.

Here $r=6cm$, $?=60°$

? $A=\frac{60}{360}\times \frac{22}{7}\times 36=(\frac{132}{7})c{m}^2$

Q.2 Find the area of a quadrant of a circle whose circumference is 22 cm.

Answer :

Let r be the radius of the circle. Then,

Circumference $=22$ cm

$=>2?r=22=>2\times \frac{22}{7}\times r=22$ $=>r=\frac{7}{2}$cm

Area of the quadrant of a circle

$=\frac{1}{4}?r^2=(\frac{1}{4}\times \frac{22}{7}\times \frac{49}{4})$

$=\frac{539}{56}=\frac{77}{8}c{m}^2$

Q3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Answer :

Clearly, minute hand of a clock describes a circle of radius equal to its length i.e., 14 cm. Since the minutes hand rotates through 6° in one minute, therefore, area swept by the minute hand in one minute is the area of sector of angle 6° in a circle of radius 14 cm. Hence, the required area i.e., the area swept in 5 minutes.

$=(\frac{?}{360}\times ?r^2\times 5)=\frac{6}{360}\times \frac{22}{7}\times (14{)}^2\times 5$

$=\frac{1}{60}\times \frac{22}{7}\times 196\times 5=\frac{154}{3}c{m}^2$

Q4. 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 $?=3.14$)

Answer :

Here, $r=10$ cm, $?=90°$

(i) Area of the minor segment

$=r^2[\frac{??}{360}?\frac{1}{2}sin?]$

$=(10{)}^2[\frac{3.14\times 90}{360}?\frac{1}{2}sin90°]$

$=100(0.785?0.5)=100\times 0.285=28.5c{m}^2$

(ii) Area of the major sector

$=\frac{?}{360}\times ?r^2$, where $r=10$ and $?=270°$

$=\frac{270}{360}\times 3.14\times {10}^2$
$=(\frac{3}{4}\times 3.14\times 100)=235.5c{m}^2$

Q5. In a circle of a radius 21 cm, an arc substends an angle of 60º at the centre. Find :
(i) the length of the arc

(ii) area of the sector formed by the arc

(iii) area of the segment formed by the corresponding chord.

Answer :

Here $r=21$cm $?=60°$
(i) Length of the arc, $l=\frac{?}{360}\times 2?r$

$=\frac{60}{180}\times \frac{22}{7}\times 21=22$cm

(ii) Area of the sector, $A=\frac{?}{360}\times ?r^2$

$=\frac{60}{360}\times \frac{22}{7}\times 21\times 21=231$ cm²

(iii) Area of the segment

$=r^2[\frac{??}{360}?\frac{1}{2}sin?]$

$=(21{)}^2[\frac{22}{7}\times \frac{60}{360}?\frac{1}{2}sin60°]=441(\frac{11}{21}?\frac{1}{2}\times \frac{\sqrt{3}}{2})$

$=21\times 11?\frac{441\sqrt{3}}{4}=(231?\frac{441\sqrt{3}}{4})$cm²

Q6. A chord of a circle of radius 15 cm subtends an angle of 60º at the centre. Find the areas of the corresponding minor and major segment of the circle. (Use $?=3.14$and$\sqrt{3}=1.73$)

Answer :

Here $r=15$ cm and $?=60°$

? Area of the minor segment

$=r^2[\frac{??}{360}?\frac{1}{2}sin?]$

$=(15{)}^2[\frac{3.14\times 60}{360}?\frac{1}{2}\times sin60]$

$=225[\frac{3.14}{6}?\frac{1}{2}\times \frac{\sqrt{3}}{2}]=225(\frac{3.14\times 2?3\sqrt{3}}{12})$

$=225(\frac{6.28?3\times 1.73}{12})=\frac{225\times (6.28?5.19)}{12}$

$=\frac{225\times 1.09}{12}=20.4375$cm²

Area of the major segment

= Area of the circle - Area of the minor segment

$=3.14\times 225?20.4375=706.5?20.4375$

$=686.0625$ cm²

Q.7 A chord of a circle of radius 12 cm subtends an angle of 120º at the centre. Find the area of the corresponding segment of the circle.
(Use $?=3.14$ and $\sqrt{3}=1.73$)

Answer :

Here$r=12$ cm and $?=120°$

Area of the corresponding segment of the circle

= Area of the minor segment

$=r^2[\frac{??}{360}?\frac{1}{2}sin?]$

$=(12{)}^2[\frac{3.14\times 120}{360}?\frac{1}{2}\times sin120]$

$=144[\frac{3.14}{3}?\frac{1}{2}\times \frac{\sqrt{3}}{2}]=225(\frac{3.14\times 2?3\sqrt{3}}{12})$ [? $sin120°=sin(180°?60°)=sin60°=\frac{\sqrt{3}}{2}$]

$=144\times \frac{3.14\times 4?3\sqrt{3}}{12}=12\times (12.56?3\times 1.73)$

$=12\times (12.56?5.19)=12\times 7.37=88.44$cm²

Q8. 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 long rope (see figure). Find


(i) The area of that part of the field in which the horse can graze.

(ii) The increase in the grazing area if the rope were 10 m long instead of 5 m. (Use $?=3.14$ )

Answer :

(i) The horse will graze over a quadrant of a circle with centre at the corner A of the field and radius AF = 5 m.

Then the area of the quadrant of this circle

$=\frac{?\times 5^2}{4}=\frac{3.14\times 25}{4}=19.625$m²

(ii) In the 2nd case, radius = 10 m.

The area of the quadrant of this circle

$=\frac{?\times {10}^2}{4}=\frac{3.14\times 100}{4}=78.5$ m²

? Increase in the grazing area

$=78.5?19.625=58.875$m²

Q.9 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 sector as shown in figure. Find :
(i) The total length of the silver wire required.

(ii) The area of each sector of the brooch.

Answer :

(i) Length of silver wire used to make a brooch

$=2?r$, where $r=\frac{35}{2}$mm

$=2\times \frac{22}{7}\times \frac{35}{2}=110$ mm

Wire used in 5 diameters $=5\times 35=175$ mm

? Total wire used $=(110+175)=285$ mm

(ii) The area of each sector of the brooch

$=\frac{1}{10}\times$ Area of the circle

$=\frac{1}{10}\times ?r^2$, where $r=\frac{35}{2}$mm

$=\frac{1}{10}\times \frac{22}{7}\times (\frac{35}{2}{)}^2=\frac{385}{4}$ mm²

Q.10 An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Answer :

Area between two consecutive ribs

= ${\frac{1}{8}}^{th}$ times the Total Area of the umbrella

$=\frac{1}{8}\times ?r^2$ ,where $r=45$cm

$=\frac{1}{8}\times \frac{22}{7}\times (45{)}^2=\frac{22275}{28}$ cm²

Q.11 A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115º. Find the total area cleaned at each sweep of the blades.

Answer :

Here $r=25$ cm, $?=115º$

Total area cleaned at each sweep of the blades

$=2\times$ Area of the sector (having $r=25$ and $?=115º$ )

$=2\times \frac{?}{360}\times ?r^2$

$=2\times \frac{115}{360}\times \frac{22}{7}\times (25{)}^2$

$=\frac{23\times 11\times 625}{18\times 7}=\frac{158125}{126}$ cm²

Q.12 To warn ships for underwater rocks a lighthouse spreads a red coloured light over a sector of angle 80º to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use $?=3.14$ )

Answer :

Area of the sea over which the ships are warned

= Area of the sector (having $r=16.5$ km and $?=80°$)

$=\frac{?}{360}\times ?r^2$

$=\frac{80}{360}\times 3.14\times (16.5{)}^2$

$=\frac{68389.2}{360}$

$=189.97$ km²

Q.13 A round table cover has six equal designs as shown in figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs. 0.35 per cm2. (Use $\sqrt{3}=1.7$)

Answer :

Clearly from the figure,



Area of one design

= Area of the sector AOB - Area ( $?AOB$ )

$=\frac{?}{360}\times ?r^2?\frac{1}{2}\times (r{)}^2\times sin60°$, where $r=28$ cm

$=\frac{60}{360}\times$ $\frac{22}{7}\times (28{)}^2$ $?\frac{1}{2}\times (28{)}^2\times \frac{\sqrt{3}}{2}$

$=\frac{1232}{3}?333.2$

$=\frac{1232?999.6}{3}$

$=\frac{232.4}{3}$cm²

Area of 6 such design $=6\times \frac{232.4}{3}=464.8$cm²

Cost of making such designs @ Rs. 0.35 per cm²

= Rs $0.35\times 464.8$

= Rs 162.68

Q.14 Tick the correct answer in the following : Area of a sector of angle p (in degrees) of a circle with radius R is :
(a) $\frac{p}{180}\times 2?R$
(b) $\frac{p}{180}\times ?R^2$
(c) $\frac{p}{360}\times 2?R$
(d) $\frac{p}{720}\times 2?R^2$

Answer :

We know that area A of a sector of angle $?$ in a circle of radius r is given by

$A=\frac{?}{360}\times ?r^2$

But here , $r=R$ and $?=p$

? $A=\frac{p}{360}\times ?R^2=\frac{p}{720}\times 2?R^2$

? (D) is the correct answer.

Q.1 Find the area of the shaded region in figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

Answer :

? ROQ is a diameter

? $?RPQ=90°$

In rt $?PRQ$ , $R{Q}^2=R{P}^2+P{Q}^2$

$=>R{Q}^2=7^2+{24}^2=49+576=625$

$=>RQ=25$ cm

? Radius $r=\frac{1}{2}RQ=\frac{25}{2}$cm

Area of the semi circle $=\frac{1}{2}?r^2=\frac{1}{2}\times \frac{22}{7}\times (\frac{25}{2}{)}^2=\frac{6875}{28}c{m}^2$

and area of $?RPQ=\frac{1}{2}\times RP\times PQ$

$=12\times 7\times 24=84$ cm²

Area of the shaded region = Area of the semi circle - Area ( $?RPQ$ )

$=\frac{6875}{28}?84=\frac{4523}{28}$ cm²

Q.2 Find the area of the shaded region in figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and $?AOC=40°$

Answer :

Area of the shaded region

= Area of sector AOC - Area of sector OBD

$=\frac{40}{360}\times ?\times {14}^2?\frac{40}{360}\times ?\times 7^2$

$=\frac{1}{9}\times \frac{22}{7}(196?49)$ $=\frac{22}{63}\times 147=\frac{154}{3}$cm²

Q.3 Find the area of the shaded region in figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Answer :

Area of the square ABCD $=(14{)}^2=196$ cm²

Diameter of the semicircles = AD or BC = 14 cm

? Radius of each semicircle $=7$ cm

Area of the semicircular regions $=2\times \frac{1}{2}?r^2=?r^2$

$=\frac{22}{7}\times 49=154$cm²
? Area of the shaded portion

= Area of the square ABCD - Area of the semicircular regions

$=196?154=42$cm²

Q.4 Find the area of the shaded region in 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.

Answer :

Area of the circular portion

= Area of the circle - Area of the sector

$=?r^2?\frac{60}{360}?r^2=?r^2(1?\frac{1}{6})$

$=\frac{5}{6}?r^2$, where r = 6

$=\frac{5}{6}\times \frac{22}{7}\times 36=\frac{660}{7}$ cm²

Area of the equilateral $?OAB$

$=\frac{\sqrt{3}}{4}(side{)}^2=(\frac{\sqrt{3}}{4}\times 144)$

$=36\sqrt{3}$cm²

? Area of the shaded region

$=(\frac{660}{7}+36\sqrt{3})$ cm²

Q.5 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 figure. Find the area of the remaining portion of the square.

Answer :

The area of the whole square ABCD $=4^2=16$cm²

The sum of the area of the four quadrants at the four corners of the square

= The area of a circle of radius 1 cm