Table of Contents
Exercise Ex. 16A
Question 1
Solution 1
Question 2
The circumference of a circle is 22 cm. Find the area of its quadrant.Solution 2
Question 3
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?Solution 3
Question 4
If the area of a circle is numerically equal to twice its circumference, then what is the diameter of the circle?Solution 4
Question 5
What is the perimeter of a square which circumscribes a circle of radius a cm?Solution 5
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 cmQuestion 6
Find the length of the arc of a circle of diameter 42 cm which subtends an angle of 60° at the centre.Solution 6
Question 7
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.Solution 7
Question 8
Find the area of a circle whose circumference is 8π.Solution 8
Question 9
Find the perimeter of a semicircular protractor whose diameter is 14 cm.Solution 9
Question 10
Find the radius of a circle whose perimeter and area are numerically equal.Solution 10
Question 11
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.Solution 11
Question 12
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.Solution 12
Question 13
Find the area of the sector of a circle having radius 6 cm and of angle 30°. [Take π = 3.14.]Solution 13
Question 14
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the length of the arc.Solution 14
Question 15
The circumferences of two circles are in the ratio 2 : 3. What is the ratio between their areas?Solution 15
Question 16
The areas of two circles are in the ratio 4 : 9. What is the ratio between their circumferences?Solution 16
Question 17
A square is inscribed in a circle. Find the ratio of the areas of the circle and the square.Solution 17
Question 18
The circumference of a circle is 8 cm. Find the area of the sector whose central angle is 72°.Solution 18
Question 19
A pendulum swings through an angle of 30 and describes an arc 8.8cm in length. Find the length of the pendulumSolution 19
Length of the pendulum = radius of sector = r cm
Question 20
The minute hand of a clock is 15cm long. Calculate the area swept by it in 20minutes. Take = 3.14Solution 20
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°)
Question 21
A sector of 56o, cut out from a circle, contains . Find the radius of the circle.Solution 21
= 56o and let radius is r cm
Area of sector =
Hence radius= 6cmQuestion 22
The area of the sector of a circle of radius 10.5cm is . Find the central angle of the sector.Solution 22
Area of the sector of circle =
Radius = 10.5 cm
Question 23
The perimeter of a certain sector of circle of radius 6.5cm is 31 cm. Find the area of the sector.Solution 23
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
Question 24
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.Solution 24
Length of arc of circle = 44 cm
Radius of circle = 17.5 cm
Area of sector =
Question 25
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.Solution 25
Question 26
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.]
Solution 26
Question 27
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.Solution 27
Question 28
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.
Solution 28
Question 29
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.
Solution 29
Question 30
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.
Solution 30
Question 31
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.
Solution 31
Question 32
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.
Solution 32
Question 33
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.
Solution 33
Question 34
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.
Solution 34
Shaded area = (area of quadrant) – (area of DAOD)
Question 35
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.
Solution 35
Question 36
In a circle of radius 7 cm, a square ABCD is inscribed. Find the area of the circle which is outside the square.Solution 36
Question 37
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.
Solution 37
Question 38
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.]
Solution 38
Question 39
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.]
Solution 39
Question 40
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.
Solution 40
Question 41
Find the area of a quadrant of a circle whose circumference is 44 cm.Solution 41
Question 42
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.
Solution 42
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 = 42Question 43
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.
Solution 43
Question 44
A wire is bent to form a square enclosing an area of 484 m2. Using the same wire, a circle is formed. Find the area of the circle.Solution 44
Question 45
A square ABCD is inscribed in a circle of radius r. Find the area of the square.Solution 45
Question 46
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 m2. Find the cost of ploughing the field.
Solution 46
Question 47
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)
Solution 47
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 mQuestion 48
In the given figure, PQSR represents a flower bed. If OP = 21 m and OR = 14 m, find the area of the flower bed.
Solution 48
Area of flower bed = (area of quadrant OPQ)
-(area of the quadrant ORS)
Question 49
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.
Solution 49
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
Question 50
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.
Solution 50
Question 51
Find the area of the major segment APB of a circle of radius 35 cm and ∠AOB = 90°, as shown in the given figure.
Solution 51
Exercise Ex. 16B
Question 1
The circumference of a circle is 39.6 cm. find its area.Solution 1
Circumference of circle = 2r = 39.6 cm
Question 2
The area of a circle is 98.56 cm2. Find its circumference.Solution 2
Question 3
The circumference of a circle exceeds its diameter by 45 cm. Find the circumference of the circle.Solution 3
Question 4
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.Solution 4
Area of square =
Perimeter of square = 4 side = 4 22 = 88 cm
Circumference of circle = Perimeter of square
Question 5
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.Solution 5
Area of equilateral =
Perimeter of equilateral triangle = 3a = (3 22) cm
= 66 cm
Circumference of circle = Perimeter of circle
2r = 66 r =
Area of circle =
= Question 6
The length of a chain used as the boundary of a semicircular park is 108 m. Find the area of the park.Solution 6
Question 7
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.Solution 7
Let the radii of circles be x cm and (7 – x) cm
Circumference of the circles are 26 cm and 18 cmQuestion 8
Find the area of a ring whose outer and inner radii are respectively 23 cm and 12 cm.Solution 8
Area of outer circle =
= 1662.5
Area of ring = Outer area – inner area
= (1662.5 – 452.5) Question 9
A path of 8m width runs around the outside of a circular park whose radius is 17m. Find the area of the path.Solution 9
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 =
Question 10
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.Solution 10
Let r m and R m be the radii of inner circle and outer boundaries respectively.
Then, 2r = 352 and 2R = 396
Width of the track = (R – r) m
Area the track =
Question 11
A sector is cut from a circle of radius 21cm. The angle of the sector is 150o. Find the length of the arc and the area of the sector.Solution 11
Length of the arc
Length of arc =
Area of the sector =
Question 12
The area of the sector of a circle of radius 10.5 cm is 69.3 cm2. Find the central angle of the sector.Solution 12
Question 13
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.Solution 13
Length of arc =
Circumference of circle = 2 r
Area of circle =
Question 14
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.Solution 14
Question 15
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, Solution 15
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
Question 16
A chord 10cm long is drawn in a circle whose radius is . Find the areas of both the segments. Take = 3.13Solution 16
Let OA = , OB =
And AB = 10 cm
Area of AOB =
Area of minor segment = (area of sector OACBO) – (area of OAB)
=
Question 17
Find the areas of both the segments of a circle of radius 42 cm with central angle 120o. Solution 17
Area of sector OACBO
Area of minor segment ACBA
Area of major segment BADB
Question 18
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, Solution 18
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)] = 2744.25Question 19
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.Solution 19
Let the major arc be x cm long
Then, length of the minor arc =
Circumference =
Question 20
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 = 3.14Solution 20
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 cmQuestion 21
Find the area of a quadrant of a circle whose circumference is 88 cm.Solution 21
Question 22
A rope by which a cow is tethered is increased from 16m to 23m. How much additional ground does it have now to graze?Solution 22
Area of plot which cow can graze when r = 16 m is
= 804.5 m2
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 Question 23
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?Solution 23
Area which the horse can graze = Area of the quadrant of radius 21 m
Area ungrazed =
Question 24
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.Solution 24
Each angle of equilateral triangle is 60
Area that the horse cannot graze is 36.68 m2Question 25
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 = 3.14Solution 25
Ungrazed area
Question 26
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.
Solution 26
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
Question 27
The side of a square is 10 cm. Find (i) the area of the inscribed circle, and (ii) area of the circumscribed circle. Take = 3.14Solution 27
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 Question 28
If a square is inscribed in a circle, find the ratio of the areas of the circle and the square.Solution 28
Let the radius of circle be r cm
Then diagonal of square = diameter of circle = 2r cm
Area of the circle =
Question 29
The area of a circle inscribed in an equilateral triangle is 154 . Find the perimeter of the triangle. Take Solution 29
Let the radius of circle be r cm
Let each side of the triangle be a cm
And height be h cm
Question 30
The radius of the wheel of a vehicle is 42 cm. How many revolutions will it complete in 19.8 km long journey?Solution 30
Radius of the wheel = 42 cm
Circumference of wheel =2r =
Distance travelled = 19.8 km = 1980000 cm
Number of revolutions = Question 31
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.Solution 31
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 = Question 32
The wheels of a car make 2500 revolutions in covering a distance of 4.95 km. Find the diameter of a wheel.Solution 32
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 cmQuestion 33
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.Solution 33
Radius of the wheel
Circumference of the wheel = 2r =
Distance covered in 140 revolution
Distance covered in one hour = Question 34
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?Solution 34
Distance covered by a wheel in 1minute
Circumference of a wheel =
Number of revolution in 1 min = Question 35
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.Solution 35
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 = Question 36
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.
Solution 36
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 Question 37
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.
Solution 37
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 Question 38
Four equal circles, each of radius a units, touch each other. Show that the area between them is sq. units.Solution 38
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
Question 39
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
Solution 39
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 cm2Question 40
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 .
Solution 40
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°]
Question 41
In the given figure, ABCD is a trapezium of area 24.5 cm2. 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.
Solution 41
Question 42
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.
Solution 42
Question 43
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.
Solution 43
Question 44
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.
Solution 44
Question 45
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.]
Solution 45
Question 46
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.]
Solution 46
Question 47
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.
Solution 47
Question 48
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.
Solution 48
Question 49
The perimeter of the quadrant of a circle is 25 cm. Find its area.Solution 49
Question 50
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.]Solution 50
Question 51
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 m2. [Use π = 3.14.]Solution 51
Question 52
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.Solution 52
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
Question 53
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. Solution 53
Question 54
A circular disc of radius 6cm is divided into three sectors with central angles 90o, 120o and 150o. What part of the whole circle is the sector with central angle 150o? Also, calculate the ratio of the areas of the three sectors.Solution 54
Question 55
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
Solution 55
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 Question 56
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
Solution 56
In PQR, P = 90, PQ = 24 cm, PR = 7 cm
Area of semicircle
Area of PQR =
Shaded area = 245.31 – 84 = 161.31 Question 57
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.
Solution 57
In ABC, A = 90°, AB = 6cm, BC = 10 cm
Area of ABC =
Let r be the radius of circle of centre O
Question 58
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.
Solution 58
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)
Question 59
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
Solution 59
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)
Question 60
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.
Solution 60
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
Exercise MCQ
Question 1
The area of a circle is 38.5 cm2. The circumference of the circle is
- 6.2 cm
- 12.2 cm
- 11 cm
- 22 cm
Solution 1
Question 2
The area of a circle is 49πcm2. Its circumference is
- 7π cm
- 14π cm
- 21π cm
- 28π cm
Solution 2
Question 3
The difference between the circumference and radius of a circle is 37 cm. The area of the circle is
a. 111 cm2
b. 184 cm2
c. 154 cm2
d. 259 cm2Solution 3
Question 4
The perimeter of a circular field is 242 m. The area of the field is
- 9317 m2
- 18634 m2
- 4658.5 m2
- none of these
Solution 4
Question 5
On increasing the diameter of a circle by 40%, its area will be increased by
- 40%
- 80%
- 96%
- 82%
Solution 5
Question 6
On decreasing the radius of circle by 30%, its area is decreased by
- 30%
- 60%
- 45%
- none of these
Solution 6
Question 7
The area of a square is the same as the area of a circle. Their perimeters are in the ratio
Solution 7
Question 8
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
Solution 8
Question 9
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
Solution 9
Question 10
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
Solution 10
Question 11
If the sum of the areas of two circles with radii R1 and R2 is equal to the area of a circle of radius R then
- R1 + R2 = R
- R1 + R2< R
- R12 + R22< R2
- R12 + R22 = R2
Solution 11
Question 12
If the sum of the circumferences of two circles with radii R1 and R2 is equal to the circumference of a radius R then
- R1 + R2 = R
- R1 + R2> R
- R1 + R2< R2
- none of these
Solution 12
Question 13
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
Solution 13
Question 14
The radii of two concentric circles are 19 cm and 16 cm respectively. The area of the ring enclosed by these circle is
- 320 cm2
- 330 cm2
- 332 cm2
- 340 cm2
Solution 14
Question 15
The areas of two concentric circles are 1386 cm2 and 962.5 cm2. The width of the ring is
- 2.8 cm
- 3.5 cm
- 4.2 cm
- 3.8 cm
Solution 15
Question 16
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
Solution 16
Question 17
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
Solution 17
Question 18
The radius of a wheel is 0.25 m. How many revolutions will it make in covering 11 km?
- 2800
- 4000
- 5500
- 7000
Solution 18
Question 19
The diameter of a wheel is 40 cm. How many revolutions will it make in covering 176 m?
- 140
- 150
- 160
- 166
Solution 19
Question 20
In making 1000 revolutions, a wheel covers 88 km. The diameter of the wheel is
- 14 m
- 24 m
- 28 m
- 40 m
Solution 20
Question 21
The area of a sector of angle θ° of a circle with radius R is
Solution 21
Question 22
The length of an arc of a sector of angle θ° of a circle with radius R is
Solution 22
Question 23
The length of the minute hand of a clocks is 21 cm. The area swept by the minute hand in 10 minutes is
- 231 cm2
- 210 cm2
- 126 cm2
- 252 cm2
Solution 23
Question 24
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 cm2
- 34.5 cm2
- 28.5 cm2
- 30.5 cm2
Solution 24
Question 25
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
Solution 25
Question 26
- 120.56 cm2
- 124.63 cm2
- 118.24 cm2
- 130.57 cm2
Solution 26
Exercise FA
Question 1
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 cm2
- 228 cm2
- 242 cm2
- 248 cm2
Solution 1
Question 2
The diameter of a wheel is 84 cm. How many revolutions will it make to cover 792 m?
- 200
- 250
- 300
- 350
Solution 2
Question 3
The area of a sector of a circle with radius r, making an angle of x° at the centre is
Solution 3
Question 4
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 cm2
- 266 cm2
- 272 cm2
- 254 cm2
Solution 4
Question 5
The circumference of a circle is 22 cm. Find its area. Solution 5
Question 6
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the length of the arc.Solution 6
Question 7
The minute hand of a clock is 12 cm long. Find the area swept by it in 35 minutes.Solution 7
Question 8
The perimeter of a sector of a circle radius 5.6 cm is 27.2 cm. Find the area of the sector.Solution 8
Question 9
A chord of a circle of radius 14 cm makes right angle at the centre. Find the area of the sector.Solution 9
Question 10
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.
Solution 10
Question 11
Solution 11
Question 12
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. Solution 12
Question 13
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.
Solution 13
Question 14
In the given, ABCD is a square each of whose side measures 28 cm. Find the area of the shaded region.
Solution 14
Question 15
In the given figure, an equilateral triangle has been inscribed in a circle of radius 4 cm. Find the area of the shaded region.
Solution 15
Question 16
The minute hand of a clock is 7.5 cm long. Find the areaSolution 16
Question 17
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.Solution 17
Question 18
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. Solution 18
Question 19
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]Solution 19
Question 20
A square tank has an area of 1600 m2. There are four semicircular plots around it. Find the cost of turfing the plots at Rs. 12.50 per m2.[Take π = 3.14]Solution 20
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