## Chapter 18 – Areas of Circle, Sector and Segment Exercise Ex. 18A

The

circumference of a circle is 22 cm. Find the area of its quadrant.

What

is the diameter of a circle whose area is equal to the sum of the areas of

two circles of diameters 10 cm and 24 cm?

If

the area of a circle is numerically equal to twice its circumference, then

what is the diameter of the circle?

What

is the perimeter of a square which circumscribes a circle of radius a cm?

Since square circumscribes a circle of radius a cm, we have

Side of the square = 2 ⨯

radius of circle = 2a cm

Then, Perimeter of the square = (4 ⨯

2a) = 8a cm

Find

the length of the arc of a circle of diameter 42 cm which subtends an angle

of 60° at the centre.

Find

the diameter of the circle whose area is equal to the sum of the areas of two

circles having radii 4 cm and 3 cm.

Find

the area of a circle whose circumference is 8π.

Find

the perimeter of a semicircular protractor whose diameter is 14 cm.

Find

the radius of a circle whose perimeter and area are numerically equal.

The

radii of two circles are 19 cm and 9 cm. Find the radius of the circle which

has circumference equal to the sum of the circumferences of the two circles.

The

radii of two circles are 8 cm and 6 cm. Find the radius of the circle having

area equal to the sum of the areas of the two circles.

Find

the area of the sector of a circle having radius 6 cm and of angle 30°. [Take

π

= 3.14.]

In

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

the length of the arc.

The

circumferences of two circles are in the ratio 2 :

3. What is the ratio between their areas?

The

areas of two circles are in the ratio 4 : 9. What is

the ratio between their circumferences?

A

square is inscribed in a circle. Find the ratio of the areas of the circle

and the square.

The

circumference of a circle is 8 cm. Find the area of the sector whose central

angle is 72°.

A pendulum swings through an angle of 30 and describes an arc 8.8cm in length. Find the length of the pendulum

Length of the pendulum = radius of sector = r cm

The minute hand of a clock is 15cm long. Calculate the area swept by it in 20minutes. Take = 3.14

Angle described by the minute hand in 60 minutes = 360°

Angle described by minute hand in 20 minutes

Required area swept by the minute hand in 20 minutes

=Area of the sector(with r = 15 cm and = 120°)

A sector of 56^{o}, cut out from a circle, contains . Find the radius of the circle.

= 56^{o} and let radius is r cm

Area of sector =

Hence radius= 6cm

The area of the sector of a circle of radius 10.5cm is . Find the central angle of the sector.

Area of the sector of circle =

Radius = 10.5 cm

The perimeter of a certain sector of circle of radius 6.5cm is 31 cm. Find the area of the sector.

Let sector of circle is OAB

Perimeter of a sector of circle =31 cm

OA + OB + length of arc AB = 31 cm

6.5 + 6.5 + arc AB = 31 cm

arc AB = 31 – 13

= 18 cm

The radius of a circle is 17.5 cm. Find the area of the sector enclosed by two radii and an arc 44cm in length.

Length of arc of circle = 44 cm

Radius of circle = 17.5 cm

Area of sector =

Two

circular pieces of equal radii and maximum area, touching each other are cut

out from a rectangular cardboard of dimensions 14 cm x 7 cm. Find the area of

the remaining cardboard.

In

the given figure, ABCD is a square of side 4 cm. A quadrant of a circle of

radius 1 cm is drawn at each vertex of the square and a circle of diameter 2

cm is also drawn. Find the area of the shaded region. [Use π

= 3.14.]

From

a rectangular sheet of paper ABCD with AB = 40 cm and AD = 28 cm, a

semicircular portion with BC as diameter is cut off. Find the area of the remaining paper.

In

the given figure, OABC is a square of side 7 cm. If COPB is a quadrant of a

circle with centre C find the area of the shaded region.

In

the given figure, three sectors of a circle of radius 7 cm, making angles of

60°, 80° and 40° at the centre are shaded. Find the area of the shaded

region.

In

the given figure, PQ and AB are respectively the arcs of two concentric

circles of radii 7 cm and 3.5 cm with centre O. If ∠POQ

= 30°, find the area of the shaded region.

In

the given figure, find the area of the shaded region, if ABCD is a square of

side 14 cm and APD and BPC are semicircles.

In

the given figure, the shape of the top of a table is that of a sector of a

circle with centre 0 and ∠A0B =90°. If AO =0B

= 42 cm, then find the perimeter of the top of the table.

In

the given figure, ABCD is a square of side 7 cm, DPBA and DQBC are quadrants

of circles each of the radius 7 cm. Find the area of shaded region.

In the given figure, AOBCA represents a quadrant of a circle of radius 3.5 cm with centre O. Calculate the area of the shaded portion.

Shaded area = (area of quadrant) – (area of DAOD)

Find

the perimeter of the shaded region in the figure, if ABCD is a square of side

14 cm and APB and CPD are semicircles.

In

a circle of radius 7 cm, a square ABCD is inscribed. Find the area of the

circle which is outside the square.

In

the given figure, APB and CQD are semicircles of diameter 7 cm each, while

ARC and BSD are semicircles of diameter 14 cm each. Find the

(i) perimeter,

(ii)

area of the shaded region.

In

the given figure, PSR, RTQ and PAQ are three semicircles of diameter 10 cm, 3

cm and 7 cm respectively. Find the perimeter of shaded region. [Use π

= 3.14.]

In

the given figure, a square OABC is inscribed in a quadrant OPBQ of a circle.

If OA = 20 cm, find the area of the shaded region. [Use π

= 3.14.]

In

the given figure, APB and AQO are semicircles and AO = OB. If the perimeter

of the figure is 40 cm, find the area of the shaded region.

Find

the area of a quadrant of a circle whose circumference is 44 cm.

In the given figure, find area of the shaded region, where ABCD is a square of side 14cm and all circles are of the same diameter.

Side of the square ABCD = 14 cm

Area of square ABCD = 14 14 = 196

Radius of each circle =

Area of the circles = 4 area of one circle

Area of shaded region = Area of square – area of 4 circles

= 196 – 154 = 42

Find

the area of the shaded region in the given figure, if ABCD is a rectangle

with sides 8 cm and 6 cm and O is the centre of the circle.

A

wire is bent to form a square enclosing an area of 484 m^{2}. Using

the same wire, a circle is formed. Find the area of the circle.

A

square ABCD is inscribed in a circle of radius r. Find the area of the

square.

The

cost of fencing a circular field at the rate of Rs.25 per metre is Rs.5500.

The field is to be ploughed at the rate of 50 paise

per m^{2}. Find the cost of ploughing the

field.

A park is in the form of a rectangle 120m by 90m. At the centre of the park there is a circular lawn as shown in the figure. The area of the park excluding the lawn is 2950_{}. Find the radius of the circular lawn. (given: _{}=3.14)

Area of rectangle = (120 × 90) _{}

= 10800 _{}

Area of circular lawn = [Area of rectangle – Area of park excluding circular lawn]

= [10800 – 2950] _{}= 7850 _{}

Area of circular lawn = 7850 _{}_{}

_{}

Hence, radius of the circular lawn = 50 m

In the given figure, PQSR represents a flower bed. If OP = 21 m and OR = 14 m, find the area of the flower bed.

Area of flower bed = (area of quadrant OPQ)

-(area of the quadrant ORS)

In the given figure O is the centre of the bigger circle, and AC is its diameter. Another circle with AB as diameter is draw. If AC = 54cm and BC = 10 cm, find the area of the shaded region.

Diameter of bigger circle = AC = 54 cm

Radius of bigger circle = _{}

_{}

Diameter AB of smaller circle

_{}

Radius of smaller circle = _{}

Area of bigger circle = _{}

= 2291. 14 _{}

Area of smaller circle = _{}

= 1521. 11 _{}

Area of shaded region = area of bigger circle – area of smaller circle

_{}

From

a thin metallic piece in the shape of a trapezium ABCD in which AB ‖⃦

CD and ∠BCD

= 90°, a quarter circle BFEC is removed. Given, AB = BC = 3.5 cm and DE = 2

cm, calculate the area of remaining (shaded) part of metal sheet.

Find

the area of the major segment APB of a circle of radius 35 cm and ∠AOB

= 90°, as shown in the given figure.

## Chapter 18 – Areas of Circle, Sector and Segment Exercise Ex. 18B

The circumference of a circle is 39.6 cm. find its area.

Circumference of circle = 2_{}r = 39.6 cm

_{}

The

area of a circle is 98.56 cm^{2}. Find its circumference.

The

circumference of a circle exceeds its diameter by 45 cm. Find the

circumference of the circle.

A copper wire when bent in the form of a square encloses an area of 484 _{}. The same wire is now bent in the form of a circle. Find the area enclosed by the circle.

Area of square =

Perimeter of square = 4 side = 4 22 = 88 cm

Circumference of circle = Perimeter of square

_{}

A wire when bent in the form of an equilateral triangle encloses an area of 121_{}. The same wire is bent to form a circle. Find the area enclosed by the circle.

Area of equilateral = _{}

_{}

Perimeter of equilateral triangle = 3a = (3 22) cm

= 66 cm

Circumference of circle = Perimeter of circle

2_{}r = 66 _{}r = _{}

Area of circle = _{}

= _{}

The

length of a chain used as the boundary of a semicircular park is 108 m. Find

the area of the park.

The sum of the radii of two circles is 7 cm, and the difference of their circumferences is 8 cm. Find the circumferences of the circles.

Let the radii of circles be x cm and (7 – x) cm

Circumference of the circles are 26 cm and 18 cm

Find the area of a ring whose outer and inner radii are respectively 23 cm and 12 cm.

Area of outer circle =

= 1662.5 _{}

Area of ring = Outer area – inner area

= (1662.5 – 452.5) _{}

A path of 8m width runs around the outside of a circular park whose radius is 17m. Find the area of the path.

Inner radius of the circular park = 17 m

Width of the path = 8 m

Outer radius of the circular park = (17 + 8)m = 25 m

Area of path =

A racetrack is in the form of a ring whose inner circumference is 352m and outer circumference is 396m. Find the width and the area of the track.

Let r m and R m be the radii of inner circle and outer boundaries respectively.

Then, 2_{}r = 352 and 2_{}R = 396

_{}

Width of the track = (R – r) m

_{}

Area the track =

A sector is cut from a circle of radius 21cm. The angle of the sector is 150^{o}. Find the length of the arc and the area of the sector.

Length of the arc

Length of arc =

Area of the sector =

The

area of the sector of a circle of radius 10.5 cm is 69.3 cm^{2}. Find

the central angle of the sector.

The length of an arc of a circle, subtending an angle of 54° at the centre, is 16.5 cm. Calculate the radius, circumference and area of the circle.

Length of arc =

Circumference of circle = 2 r

Area of circle =

The

radius of a circle with centre O is 7 cm. Two radii OA and OB are drawn at

right angles to each other. Find the areas of minor and major segments.

Find the lengths of the arcs cut off from a circle of radius 12 cm by a chord 12 cm long. Also, find the area of the minor segment. Take _{}= 3.14, _{}

_{}OAB is equilateral.

So, _{}AOB = 60

Length of arc BDA = (2_{} 12 – arc ACB) cm

= (24_{} – 4_{}) cm = (20_{}) cm

= (20 3.14) cm = 62.8 cm

Area of the minor segment ACBA

_{}

A chord 10cm long is drawn in a circle whose radius is . Find the areas of both the segments. Take = 3.13

Let OA = , OB =

And AB = 10 cm

Area of AOB =

Area of minor segment = (area of sector OACBO) – (area of OAB)

=

Find the areas of both the segments of a circle of radius 42 cm with central angle 120^{o}.

Area of sector OACBO

Area of minor segment ACBA

Area of major segment BADB

A chord of a circle of radius 30cm makes an angle of 60° at the centre of the circle. Find the areas of the minor and major segments. Take = 3.14,

Let AB be the chord of circle of centre O and radius = 30 cm such that AOB = 60°

Area of the sector OACBO

Area of OAB =

Area of the minor segment ACBA

= (area of the sector OACBO) – (area of the OAB)

=(471 – 389.25) = 81.75

Area of the major segment BADB

= (area of circle) – (area of the minor segment)

= [(3.14 × 30 × 30) – 81.75)] = 2744.25

In a circle of radius 10.5cm, the minor arc is one-fifth of the major arc. Find the area of the sector corresponding to the major arc.

Let the major arc be x cm long

Then, length of the minor arc =

Circumference =

The short and long hands of a clock are 4cm and 6cm long respectively. Find the sum of distances travelled by their tips in 2days. Take = 3.14

In 2 days, the short hand will complete 4 rounds

Distance travelled by its tip in 2 days

=4(circumference of the circle with r = 4 cm)

= (4 × 2 × 4) cm = 32 cm

In 2 days, the long hand will complete 48 rounds

length moved by its tip

= 48(circumference of the circle with r = 6cm)

= (48 × 2 × 6) cm = 576 cm

Sum of the lengths moved

= (32 + 576) = 608 cm

= (608 × 3.14) cm = 1909.12 cm

Find

the area of a quadrant of a circle whose circumference is 88 cm.

A rope by which a cow is tethered is increased from 16m to 23m. How much additional ground does it have now to graze?

Area of plot which cow can graze when r = 16 m is _{}

_{}

= 804.5 m^{2}

Area of plot which cow can graze when radius is increased to 23 m

_{}

Additional ground = Area covered by increased rope – old area

= (1662.57 – 804.5) _{}= 858 _{}

A horse is placed for grazing inside a rectangular field 70m by 52m. It is tethered to one corner by a rope 21 m long. On how much area can it graze? How much area is left ungrazed?

Area which the horse can graze = Area of the quadrant of radius 21 m

_{}

Area ungrazed =

_{}

A horse is tethered to one corner of a field which is in the shape of an equilateral triangle of side 12m. If the length of the rope is 7m, find the area of the field which the horse cannot graze. Take _{}. Write the answer correct to 2 places of decimal.

Each angle of equilateral triangle is 60

Area that the horse cannot graze is 36.68 m^{2}

Four cows are tethered at the four corners of a square field of side 50 m such that each can graze the maximum unshared area. What area will be left ungrazed? Take _{}= 3.14

Ungrazed area

_{}

In the given figure, OPQR is a rhombus, three of whose vertices lie on a circle with centre O. If the area of the rhombus is _{}, find the radius of the circle.

OP = OR = OQ = r

Let OQ and PR intersect at S

We know the diagonals of a rhombus bisect each other at right angle.

Therefore we have

_{}

_{}

The side of a square is 10 cm. Find (i) the area of the inscribed circle, and (ii) area of the circumscribed circle. Take = 3.14

Diameter of the inscribed circle = Side of the square = 10 cm

_{}Radius of the inscribed circle = 5 cm

Diameter of the circumscribed circle

= Diagonal of the square

_{}

Radius of circumscribed circle = _{}

(i)Area of inscribed circle = _{}

(ii)Area of the circumscribed circle _{}

If a square is inscribed in a circle, find the ratio of the areas of the circle and the square.

Let the radius of circle be r cm

Then diagonal of square = diameter of circle = 2r cm

Area of the circle =

The area of a circle inscribed in an equilateral triangle is 154 _{}. Find the perimeter of the triangle. Take _{}

Let the radius of circle be r cm

_{}

Let each side of the triangle be a cm

And height be h cm

_{}

The radius of the wheel of a vehicle is 42 cm. How many revolutions will it complete in 19.8 km long journey?

Radius of the wheel = 42 cm

Circumference of wheel =2r = _{}

Distance travelled = 19.8 km = 1980000 cm

Number of revolutions = _{}

The wheels of the locomotive of a train are 2.1m in radius. They make 75 revolutions in one minute. Find the speed of the train in km per hour.

Radius of wheel = 2.1 m

Circumference of wheel = _{}

Distance covered in one revolution = 13.2 m

Distance covered in 75 revolutions = (13.2 75) m = 990 m

= _{}

Distance a covered in 1 minute = _{}

Distance covered in 1 hour = _{}

The wheels of a car make 2500 revolutions in covering a distance of 4.95 km. Find the diameter of a wheel.

Distance covered by the wheel in 1 revolution

_{}

_{}The circumference of the wheel = 198 cm

Let the diameter of the wheel be d cm

_{}

Hence diameter of the wheel is 63 cm

A boy is cycling in such a way that the wheels of his bicycle are making 140 revolutions per minute. If the diameter of a wheel is 60 cm, calculate the speed(in km/h) at which the boy is cycling.

Radius of the wheel _{}

Circumference of the wheel = 2_{}r = _{}

_{}

Distance covered in 140 revolution

_{}

Distance covered in one hour = _{}

The diameter of the wheels of a bus is 140cm, How many revolutions per minute do the wheels make when the bus is moving at a speed of 72.6 km per hour?

Distance covered by a wheel in 1minute

_{}

Circumference of a wheel = _{}

Number of revolution in 1 min = _{}

The diameter of the front and rear wheels of a tractor are 80cm and 2 m respectively. Find the number of revolutions that a rear wheel makes to cover the distance which the front wheel covers in 800 revolutions.

Radius of the front wheel = 40 cm =

Circumference of the front wheel=

Distance moved by it in 800 revolution

Circumference of rear wheel = (2 1)m = (2) m

Required number of revolutions =

Four equal circles are described about the four corners of a square so that each touches two of the others, as shown in the figure. Find the area of the shaded region, if each of the square measures 14cm.

Each side of the square is 14 cm

Then, area of square = (14 × 14)

= 196

Thus, radius of each circle 7 cm

Required area = area of square ABCD

-4 (area of sector with r = 7 cm, = 90°)

Area of the shaded region = 42

Four equal circles, each of radius 5 cm, touch each other, as shown in the figure. Find the area included between them. Take = 3.14.

Let A, B, C, D be the centres of these circles

Join AB, BC, CD and DA

Side of square = 10 cm

Area of square ABCD

Area of each sector =

= 19.625

Required area = [area of sq. ABCD – 4(area of each sector)]

= (100 – 4 19.625)

= (100 – 78.5) = 21.5

Four equal circles, each of radius a units, touch each other. Show that the area between them is _{}sq. units.

Required area = [area of square – areas of quadrants of circles]

Let the side = 2a unit and radius = a units

Area of square = (side side) = (2a 2a) sq. units

_{}

Three equal circles, each of radius of 6cm, touchone another as shown in the figure. Find the area enclosed between them. Take = 3.14 and

Let A, B, C be the centres of these circles. Joint AB, BC, CA

Required area=(area of ABC with each side a = 12 cm)

-3(area of sector with r = 6, = 60°)

The area enclosed = 5.76 cm^{2}

If three circles of radius a each, are drawn such that each touches the other two, prove thatthe area included between them is equal to . Take .

Let A, B, C be the centers of these circles. Join AB, BC, CA

Required area= (area of ABC with each side 2)

-3[area of sector with r = a cm, = 60°]

In

the given figure, ABCD is a trapezium of area 24.5 cm^{2}. If AD ∥

BC, ∠DAB

= 90°, AD =10 cm, BC = 4 cm and ABE is quadrant of a circle then find the

area of the shaded region.

ABCD

is a field in the shape of a trapezium, AD ∥

BC, ∠ABC

= 90° and ∠ADC

= 60°. Four sectors are formed with centres A, B, C and D, as shown in the

figure. The radius of each sector is 14 m. Find the following:

i. total area of the four sectors,

ii. area of

the remaining portion, given that AD = 55 m, BC = 45 m and AB = 30 m.

Find

the area of the shaded region in the given figure, where a circular arc of

radius 6 cm has been drawn with vertex of an equilateral triangle of side 12

cm as centre and a sector of circle of radius 6 cm with centre B is made.

In

the given figure, ABCD is a rectangle with AB = 80 cm and BC = 70 cm, ∠AED

= 90° and DE = 42 cm. A semicircle is drawn, taking BC as diameter. Find the

area of the shaded region.

In

the given figure, from a rectangular region ABCD with AB = 20 cm, a right

triangle AED with AE = 9 cm and DE =12 cm, is cut off. On the other end,

taking BC as diameter, a semicircle is added on outside the region: Find the

area of the shaded region. [Use π = 3.14.]

In

the given figure, 0 is the centre of the circle with AC = 24 cm, AB = 7 cm

and ∠BOD

= 90°. Find the area of shaded region. [Use π = 3.14.]

In

the given figure, a circle is inscribed in an equilateral triangle ABC of

side 12 cm. Find the radius of inscribed circle and the area of the shaded

region.

On

a circular table cover of radius 42 cm, a design is formed by a girl leaving

an equilateral triangle ABC in the middle, as shown in the figure. Find the

covered area of the design.

The

perimeter of the quadrant of a circle is 25 cm. Find its area.

A

chord of a circle of radius 10 cm subtends a right angle at the centre. Find

the area of the minor segment. [Use π = 3.14.]

The

radius of a circular garden is 100 m. There is a road 10 m wide, running all

around it. Find the area of the road and the cost of levelling it at Rs.20 per m^{2}.

[Use π

= 3.14.]

The area of an equilateral triangle is .Taking each angular point as centre, circles are drawn with radius equal to half the length of the side of the triangle. Find the area of the triangle not included in the circles. Take.

Area of equilateral triangle ABC = 49

Let a be its side

Area of sector BDF =

Area of sector BDF = Area of sector CDE = Area of sector AEF

Sum of area of all the sectors

Shaded area = Area of ABC – sum of area of all sectors

A child draws the figure of an aeroplane as shown. Here, the wings ABCD and FGHI are parallelograms, the tail DEF is an isosceles triangle, the cockpit CKI is a semicircle and CDFI is a square. In the given figure, BP CD, HQ FI and EL DF. IF CD = 8cm, BP = HQ = 4 cm and DE = EF = 5cm, find the area of the whole figure. Take = 3.14.

A circular disc of radius 6cm is divided into three sectors with central angles 90^{o}, 120^{o} and 150^{o}. What part of the whole circle is the sector with central angle 150^{o}? Also, calculate the ratio of the areas of the three sectors.

A round table cover has six equal designs as shown in the given figure. If the radius of the cover is 35cm, then find the total area of the design. Use

ABCDEF is a hexagon

AOB = 60, Radius = 35 cm

Area of sector AOB

Area of AOB =

Area of segment APB = (641.083 = 530.425)= 110.658

Area of design (shaded area) = 6 110.658= 663.948

= 663.95

In the given figure, PQ = 24cm, PR = 7cm and O is the centre of the circle. Find the area of the shaded region. Take = 3.14

In PQR, P = 90, PQ = 24 cm, PR = 7 cm

Area of semicircle

Area of PQR =

Shaded area = 245.31 – 84 = 161.31

In the given figure, ABC is right angled at A. find the area of the shaded region if AB = 6cm, BC = 10 cm and O is the centre of thein circle of ABC. Take = 3.14.

In ABC, A = 90°, AB = 6cm, BC = 10 cm

Area of ABC =

Let r be the radius of circle of centre O

In the given figure, ABC is right-angled at A. Semicircles are drawn on AB, AC and BC as diameters. It is given that AB = 3cm and AC = 4cm. Find the area of the shaded region.

Area of shaded region = Area of ABC + Area of semi-circle APB

+ Area of semi circle AQC – Area of semicircle BAC

Further in ABC,A = 90

Adding (1), (2), (3) and subtracting (4)

PQRS is a diameter of a circle of radius 6cm. The lengths PQ, QR, and RS are equal. Semicircles are drawn with PQ andQS as diameters, as shown in the given figure. If PS = 12cm, find the perimeter and area of the shaded region. Take = 3.14

PS = 12 cm

PQ = QR = RS = 4 cm, QS = 8 cm

Perimeter = arc PTS + arc PBQ + arc QES

_{}

Area of shaded region = (area of the semicircle PBQ)

+ (area of semicircle PTS)-(Area of semicircle QES)

_{}

The inside perimeter of a running track shown in the figure is 400 m. The length of each of the straight portions is 90m, and the ends are semicircles. If the track is 14 m wide everywhere, find the area of the track. Also, find the length of the outer boundary of the track.

Length of the inner curved portion

= (400 – 2 90) m

= 220 m

Let the radius of each inner curved part be r

Inner radius = 35 m, outer radius = (35 + 14) = 49 m

_{}Area of the track = (area of 2 rectangles each 90 m 14 m)

+ (area of circular ring with R = 49 m, r = 35 m

_{}

Length of outer boundary of the track

_{}

## Chapter 18 – Areas of Circle, Sector and Segment Exercise FA

In the given figure, a square OABC

has been inscribed in the quadrant OPBQ. If OA = 20cm then the area of the

shaded region is [take π=3.14]

- 214 cm
^{2} - 228 cm
^{2} - 242 cm
^{2} - 248 cm
^{2}

The diameter of a wheel is 84 cm.

How many revolutions will it make to cover 792 m?

- 200
- 250
- 300
- 350

The area of a sector of a circle

with radius r, making an angle of x° at the centre is

In the given figure, ABCD is a rectangle inscribed in a

circle having length 8 cm and breadth 6 cm. If π

= 3.14 then the area of the shaded region is

- 264 cm
^{2} - 266 cm
^{2} - 272 cm
^{2} - 254 cm
^{2}

The circumference of a circle is 22 cm. Find its area.

In

a circle of radius 21 cm, an arc subtends an angle of 60°

at the centre. Find the length of the arc.

The minute hand of a clock is 12 cm

long. Find the area swept by it in 35 minutes.

The

perimeter of a sector of a circle radius 5.6 cm is 27.2 cm. Find the area of

the sector.

A

chord of a circle of radius 14 cm makes right angle at the centre. Find the

area of the sector.

In the given figure, the sectors of two concentric circles

of radii 7 cm and 3.5 cm are shown. Find the area of the shaded region.

The wheel of a cart is making 5 revolutions per second. If

the diameter of the wheel is 84 cm. find its speed in km per hour.

OACB

is a quadrant of a circle with centre O and its radius is 3.5 cm. If OD = 2

cm, find the area of (i) the quadrant OACB (ii) the shaded region.

In

the given, ABCD is a square each of whose side measures 28 cm. Find the area

of the shaded region.

In

the given figure, an equilateral triangle has been inscribed in a circle of

radius 4 cm. Find the area of the shaded region.

The

minute hand of a clock is 7.5 cm long. Find the area

A racetrack is in the form of a ring whose inner

circumference is 352 m and outer circumference is 396 m. Find the width and

the area of the track.

A chord of circle of radius 30 cm makes an angle of 60°

at the centre of the circle. Find the area of the minor and major segment.

Four

cows are tethered at the four corners of a square field of side 50 m such

that each can graze the maximum unshared area. What area will be left ungrazed?

[Take π

= 3.14]

A

square tank has an area of 1600 m^{2}. There are four semicircular

plots around it. Find the cost of turfing the plots at Rs. 12.50 per m^{2}.[Take π = 3.14]

## Chapter 18 – Areas of Circle, Sector and Segment Exercise MCQ

The area of a circle is 38.5 cm^{2}. The circumference of the circle is

- 6.2 cm
- 12.2 cm
- 11 cm
- 22 cm

The area of a circle is 49πcm^{2}. Its circumference is

- 7π cm
- 14π cm
- 21π cm
- 28π cm

The difference between the circumference and radius of a circle is 37 cm. The area of the circle is

a. 111 cm^{2}

b. 184 cm^{2}

c. 154 cm^{2}

d. 259 cm^{2}

The perimeter of a circular field is 242 m. The area of the field is

- 9317 m
^{2} - 18634 m
^{2} - 4658.5 m
^{2} - none of these

On increasing the diameter of a circle by 40%, its area will be increased by

- 40%
- 80%
- 96%
- 82%

On decreasing the radius of circle by 30%, its area is decreased by

- 30%
- 60%
- 45%
- none of these

The area of a square is the same as the area of a circle. Their perimeters are in the ratio

The circumference of a circle is equal to the circumferences of two circle having diameters 36 cm and 20 cm. The radius of the new circle is

- 16 cm
- 28 cm
- 42 cm
- 56 cm

The area of a circle is equal to the sum of the areas of two circles of radii 24 cm and 7 cm. The diameter of the new circle is

- 25 cm
- 31 cm
- 50 cm
- 62 cm

If the perimeter of a square is equal to the circumference of a circle then the ratio of their areas is

- 4:π
- π:4
- π:7
- 7: π

If the sum of the areas of two circles with radii R_{1} and R_{2} is equal to the area of a circle of radius R then

- R
_{1}+ R_{2}= R - R
_{1}+ R_{2}< R - R
_{1}^{2}+ R_{2}^{2}< R^{2} - R
_{1}^{2}+ R_{2}^{2}= R^{2}

If the sum of the circumferences of two circles with radii R_{1} and R_{2} is equal to the circumference of a radius R then

- R
_{1}+ R_{2}= R - R
_{1}+ R_{2}> R - R
_{1}+ R_{2}< R^{2} - none of these

If the circumference of a circle and the perimeter of a square are equal then

- area of the circle = area of the square
- (area of the circle) > (area of the square)
- (area of the circle) < (area of the square)
- none of these

The radii of two concentric circles are 19 cm and 16 cm respectively. The area of the ring enclosed by these circle is

- 320 cm
^{2} - 330 cm
^{2} - 332 cm
^{2} - 340 cm
^{2}

The areas of two concentric circles are 1386 cm^{2} and 962.5 cm^{2}. The width of the ring is

- 2.8 cm
- 3.5 cm
- 4.2 cm
- 3.8 cm

The circumferences of two circles are in the ratio 3:4. The ratio of their areas is

- 3:4
- 4:3
- 9:16
- 16:9

The area of two circles is in the ratio 9:4. The ratio of their circumference is

- 3:2
- 4:9
- 2:3
- 81:16

The radius of a wheel is 0.25 m. How many revolutions will it make in covering 11 km?

- 2800
- 4000
- 5500
- 7000

The diameter of a wheel is 40 cm. How many revolutions will it make in covering 176 m?

- 140
- 150
- 160
- 166

In making 1000 revolutions, a wheel covers 88 km. The diameter of the wheel is

- 14 m
- 24 m
- 28 m
- 40 m

The area of a sector of angle θ° of a circle with radius R is

The length of an arc of a sector of angle θ° of a circle with radius R is

The length of the minute hand of a clocks is 21 cm. The area swept by the minute hand in 10 minutes is

- 231 cm
^{2} - 210 cm
^{2} - 126 cm
^{2} - 252 cm
^{2}

A chord of a circle of radius 10 cm subtends right angles at the centre. The area of the minor segment

( given, π=3.14) is

- 32.5 cm
^{2} - 34.5 cm
^{2} - 28.5 cm
^{2} - 30.5 cm
^{2}

In a circle of radius 21 cm, an arc subtends an angles of 60° at the centre. The length of the arc is

- 21 cm
- 22 cm
- 18.16 cm
- 23.5 cm

- 120.56 cm
^{2} - 124.63 cm
^{2} - 118.24 cm
^{2} - 130.57 cm
^{2}