Rs Aggarwal 2019 Solutions for Class 10 Math Chapter 16 Area Of Circle, Sector And Segment are provided here with simple step-by-step explanations. These solutions for Area Of Circle, Sector And Segment are extremely popular among Class 10 students for Math Area Of Circle, Sector And Segment Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 10 Math Chapter 16 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 705:

Question 1:

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

Answer:

Circumference = 39.6 cm
We know:
Circumference of a circle = 2πr
39.6 =2×227×r39.6×72×22=rr=6.3 cm

Also,
Area of the circle = πr2
                             =227×6.3×6.3=124.74 cm2

Page No 705:

Question 2:

The area of a circle is 98.56 cm2. Find its circumference.

Answer:

Let the radius of the circle be r.
​Now,
Area=98.56πr2=98.56227×r2=98.56r=5.6
Now,
Circumference = 2πr=2×227×5.6=35.2 cm
Hence, the circumference of the circle is 35.2 cm.

Page No 705:

Question 3:

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

Answer:

Let the radius of the circle be r.
Now,
Circumference = Diameter + 45
2πr=2r+452πr-2r=452r227-1=452r×157=45r=10.5 cm
∴ Circumference = Diameter + 45 = 2(10.5) + 45 = 66 cm
Hence, the circumference of the circle is 66 cm.

Page No 705:

Question 4:

A copper wire when bent in the form of a square encloses an area of 484 cm2. The same wire is not bent in the form of a circle. Find the area enclosed by the circle.

Answer:

Area of the circle = 484 cm2
Area of the square = Side2
484=Side2222=Side2Side=22 cm
Perimeter of the square = 4×Side
Perimeter of the square = 4×22
                                       = 88 cm
Length of the wire = 88 cm
Circumference of the circle = Length of the wire = 88 cm
Now, let the radius of the circle be r cm.
​Thus, we have:
​ 2πr=882×227×r=88r=14 

Area of the circle = πr2
                             =227×14×14=616 cm2

Thus, the area enclosed by the circle is 616 cm2.

Page No 705:

Question 5:

A wire when bent in the form of an equilateral triangle encloses an area of 1213cm2. The same wire is bent to form a circle. Find the area enclosed by the circle.

Answer:

Area of an equilateral triangle=34×Side21213=34×Side2121×4=Side2Side=22 cm

Perimeter of an equilateral triangle = 3×Side
                                                =3×22=66 cm

Length of the wire=66 cm
Now, let the radius of the circle be r cm.
We know:
Circumference of the circle = Length of the wire
2πr =662×227×r=66r=66×72×22r=212r =10.5 
Thus, we have:
Area of the circle =πr2
                             =227×10.5×10.5=346.5 sq.cm
Area enclosed by the circle = 346.5 cm2

Page No 705:

Question 6:

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

Answer:

Let the radius of the park be r.
Length of chain = Perimeter of the semicircular park
⇒ 108 = Length of the arc + Diameter
108=12×2πr+2r108=r227+2108=367rr=21 m
Now, Area of park=12πr2=12×227×212=693 m2
Hence, the area of the park is 693 m2 .

Page No 705:

Question 7:

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

Answer:

Let the radii of the two circles be r1 cm and r2cm.
Now,
Sum of the radii of the two circles = 7 cm
r1 + r2=7      ...i 

Difference of the circumferences of the two circles = 88 cm
2πr1-2πr2 =82π(r1-r2)=8(r1-r2)=82πr1-r2=82×227r1- r2 =8×744

r1-r2=5644r1-r2=1411       ...(ii)

Adding (i) and (ii), we get:
2r1=9111r1=9122

∴ Circumference of the first circle = 2πr1
                                                    =2×227×9122=26 cm
Also,
r1-r2=14119122-r2=14119122-1411=r2r2=6322

∴ Circumference of the second circle = 2πr2

                                                             =2×227×6322=18 cm

Therefore, circumferences of the first and second circles are 18 cm and 26 cm, respectively.

Page No 705:

Question 8:

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

Answer:

Let r1 cm and r2 cm be the radii of the outer and inner boundaries of the ring, respectively.
We have:

r1=23 cmr2=12 cm

Now,
Area of the outer ring = πr12

                                    =227×23×23=1662.57 cm2
 
Area of the inner ring = πr22

                                    =227×12×12=452.57 cm2

Area of the ring = Area of the outer ring - Area of the inner ring
                           = 1662.57 - 452.57
                           = 1210 cm2 

Page No 705:

Question 9:

(i) A path of 8 m width runs around the outside of a circular park whose radius is 17 m. Find the area of the path.
(ii) A park is of the shape of a circle of diameter 7 m. It is surrounded by a path of width of 0.7 m. Find the expenditure of cementing the path, if its cost is ₹ 110 per sq m.

Answer:

(i) The radius (r) of the inner circle is 17 m.
The radius (R) of the outer circle is 25 m.          [Includes path, i.e., (17 + 8)]

Area of the path = πR2-πr2
                           =πR2-r2=227252-172=227×(25 - 17) (25 + 17)=227×8×42=1056 m2

∴ Area of the path = 1056 m2                

(ii)


Diameter of the circular park = 7 m
∴ Radius of the circular park, 72 = 3.5 m
Width of the path = 0.7 m
∴  Radius of the park including the path, R = 3.5 + 0.7 = 4.2 m
Area of the path
=πR2-πr2=πR2-r2
=2274.22-3.52=227×4.2-3.5×4.2+3.5=227×0.7×7.7=16.94 m2
Rate of cementing the path = Rs 110/m2      (Given)
∴ Total cost of cementing the path
= 16.94 × 110
= Rs 1863.40
Thus, the expenditure of cementing the path is Rs 1863.40.

Page No 705:

Question 10:

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

Answer:

Let r m and R m be the radii of the inner and outer tracks.
Now,
Circumference of the outer track = 2πR
396=2×227×RR=396×744R=63

Circumference of the inner track = 2πr
352=2×227×rr=352×744r=56

Width of the track = Radius of the outer track - Radius of the inner track
                              =63-56=7 m
   
Area of the outer circle = πR2
                                      =227×63×63=12474 m2

Area of the inner circle = πR2
                                      =227×56×56=9856 m2

Area of the track = 12474 - 9856
                             = 2618 m2

Page No 705:

Question 11:

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

Answer:

Given:
Radius = 2 cm
Angle of sector = 150
Now,
Length of the arc=2πrθ360                              =2×227×21×150360                              = 55 cm

Area of the sector =πr2θ360

                              =227×21×21×150360=577.5 cm



Page No 706:

Question 12:

A chord PQ of a circle of radius 10 cm subtends an angle of 60° at the centre of the circle. Find the area of major and minor segments of the circle.

Answer:


Radius of the circle, r = 10 cm

Area of sector OPRQ
=60°360°×πr2=16×3.14×102=52.33 cm2

In ΔOPQ,
∠OPQ = ∠OQP     (As OP = OQ)
∠OPQ + ∠OQP + ∠POQ = 180°
2∠OPQ = 120°
∠OPQ = 60°
ΔOPQ is an equilateral triangle.
So, area of ΔOPQ
=34×Side2=34×102=10034cm2=43.30 cm2

Area of minor segment PRQ
= Area of sector OPRQ − Area of ΔOPQ
= 52.33 − 43.30
= 9.03 cm2

Area of major segment PSQ
= Area of circle − Area of minor segment PRQ
=π102-9.03=314-9.03=304.97 cm2

Page No 706:

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.

Answer:

Length of the arc = 16.5 cm
θ=54
Radius = ?
Circumference=?
We know:
Length of the arc =2πrθ360
16.5=2×227×r×54360r=16.5×360×744×54r=17.5 cm

c=2πr=2×227×17.5=110 cm

Circumference = 110 cm

Now,
Area of the circle =πr2
                             =227×17.5×17.5=962.5 cm2

Page No 706:

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.

Answer:


Area of minor segment = Area of sector AOBC − Area of right triangle AOB
=90°360°πOA2-12×OA×OB=14×227×72-12×7×7=14×227×72-12×7×7=38.5-24.5=14 cm2
Area of major segment APB = Area of circle − Area of minor segment
=πOA2-14=227×72-14=154-14=140 cm2
Hence, the area of major segment is 140 cm2 .

Page No 706:

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.

Answer:


Let AB be the chord. Joining A and B to O, we get an equilateral triangle OAB.
Thus, we have:
O=A=B=60°

Length of the arc ACB:
2π×12×60360=4π =12.56 cm

Length of the arc ADB:

Circumference of the circle - Length of the arc ACB=2π×12-4π=20π cm = 62.80 cm

Now,
Area of the minor segment:

Area of the sector - Area of the triangle=π×122×60360-34×122=13.08 cm2

Page No 706:

Question 16:

A chord 10 cm long is drawn in a circle whose radius is 52 cm. Find the areas of both the segments.

Answer:

Let O be the centre of the circle and AB be the chord.

Consider OAB.

OA=OB=52 cm

OA2+OB2=50+50=100

Now,
100 = 10 cm = AB

Thus, OAB is a right isosceles triangle.

Thus, we have:
Area of OAB = 12×52×52=25 cm2

Area of the minor segment = Area of the sector - Area of the triangle
                                            =90360×π×522-25=14.25 cm2

Area of the major segment = Area of the circle - Area of the minor segment
                                            =π×522-14.25=142.75 cm2

Page No 706:

Question 17:

Find the areas of both the segments of a circle of radius 42 cm with central angle 120°.

Answer:

Areaoftheminorsector=120360×π×42×42=13×π×42×42=π×14×42=1848 cm2

Area of the triangle = 12R2sinθ
Here, R is the measure of the equal sides of the isosceles triangle and θ is the angle enclosed by the equal sides.
Thus, we have:
12×42×42×sin120°=762.93 cm2

Area of the minor segment = Area of the sector - Area of the triangle
                                            =1848-762.93=1085.07 cm2

Area of the major segment = Area of the circle - Area of the minor segment
                                            =π×42×42-1085.07=5544-1085.07=4458.93 cm2

Page No 706:

Question 18:

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

Answer:


Let the chord be AB. The ends of the chord are connected to the centre of the circle O to give the triangle OAB.

OAB is an isosceles triangle. The angle at the centre is 60°

Area of the triangle = 12302 sin 60°=450 × 32=389.25 cm2

Area of the sector OACBO = 60360×π×30×30=150π=471 cm2

Area of the minor segment = Area of the sector - Area of the triangle
                                            =471-389.25=81.75 cm2

Area of the major segment = Area of the circle - Area of the minor segment
                                            =π×30×30-81.29=2744.71 cm2

Page No 706:

Question 19:

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

Answer:

Let the length of the major arc be x cm
Radius of the circle = 10.5 cm
∴ Length of the minor arc = x5 cm

Circumference =  x+x5=6x5 cm

Using the given data, we get:

6x5=2×227×2126x5=66Or, x = 55 

∴ Area of the sector corresponding to the major arc = 12×55×212=288.75 cm2

Page No 706:

Question 20:

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

Answer:

In 2 days, the short hand will complete 4 rounds.
Length of the short hand = 4 cm
 
Distance covered by the short hand = 4×2π×4=32π cm

In the same 2 days, the long hand will complete 48 rounds.
Length of the long hand = 6 cm
Distance covered by the long hand = 48×2π×6=576π cm

∴ Total distance covered by both the hands = Distance covered by the short hand + Distance covered by the long hand

                                                                       =32π + 576π = 608π =608 × 3.14= 1909.12 cm

Page No 706:

Question 21:

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

Answer:

Let the radius of the circle be r.
​Now,
Circumference=882πr=88r=14 cm
Now,
Area of quadrant = 14πr2=14×227×142=154 cm2
Hence, the area of the quadrant of the circle is 154 cm2.

Page No 706:

Question 22:

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

Answer:


r1 = 16 m
r2 = 23 m

Amount of additional ground available = Area of the bigger circle - Area of the smaller circle
                                                               =πr12-r22=π232-162=π23+1623-16=858 m2

Page No 706:

Question 23:

A horse is placed for grazing inside a rectangular field 70 m by 52 m. 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?

Answer:

Radius of the quadrant of the circle = 21 m
The shaded portion shows the part of the field the horse can graze.

Area of the grazed field = Area of the quadrant OPQ
                                       =14 of the circle having radius OP
                                       =14πr2=14×227×21×21=346.5 m2

Total area of the field = 70×52=3640 m2

Area left ungrazed = Area of the field - Area of the grazed field
                               = 3640-346.5=3293.5 m2

Page No 706:

Question 24:

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

Answer:

Side of the equilateral triangle = 12 m
Area of the equilateral triangle =34×(Side)2
                                                   =34×12×12=62.28 m2
Length of the rope = 7 m
Area of the field the horse can graze is the area of the sector of radius 7 m .Also, the angle subtended at the centre is 60°

=θ360×πr2
 
=60360×227×(7)2=25.67 m2

Area of the field the horse cannot graze = Area of the equilateral triangle - Area of the field the horse can graze
                                                                 =62.28-25.67=36.61 m2
                                                       

Page No 706:

Question 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?

Answer:


Each cow can graze a region that cannot be accessed by other cows.
∴ Radius of the region grazed by each cow = 502=25 m

Area that each cow grazes =  14×π×r2
                                            =14×3.14×25×25=490.625 cm2

Total area grazed = 4×490.625=1963.49 m2
Area of the square=(Side)2=502=2500 cm2

Now,
Area left ungrazed = Area of the square - Grazed area
                               = 2500 - 1963.49 = 536.51 m 2



Page No 707:

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 323, find the radius of the circle.

Answer:

In a rhombus, all sides are congruent to each other.

Thus, we have:
OP=PQ=QR=RO

Now, consider QOP.

OQ=OP (Both are radii.)

Therefore, QOP is equilateral.

Similarly, QOR is also equilateral and QOP  QOR.

Ar.QROP = Ar.QOP+AQOR=2Ar.QOP

Ar.QOP=12×323=163Or,163=34s2  (where s is the side of the rhombus)Or, s2=16×4=64s=8 cm

OQ = 8 cm

Hence, the radius of the circle is 8 cm.

Page No 707:

Question 27:

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

Answer:

(​i)​ If a circle is inscribed in a square, then the side of the square is equal to the diameter of the circle.
Side of the square = 10 cm
Side = Diameter = 10
∴ Radius = 5 cm
Area of the inscribed circle = πr2
                                            =3.14×5×5=78.5 cm2

(ii) If a circle is circumscribed in a square, then the diagonal of the square is equal to the diameter of the circle.
Diagonal of the square = 2×Side of the square=2×10=102 cm

Diagonal = Diameter = 102 cm
r=52 cm

Now,
Area of the circumscribed circle = πr2
                                                     =3.14×522=3.14×50=157 cm2
                          

Page No 707:

Question 28:

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

Answer:

If a square is inscribed in a circle, then the diagonals of the square are diameters of the circle.
Let the diagonal of the square be d cm.
Thus, we have:
Radius, r = d2 cm

Area of the circle=πr2
                       = πd24 cm2

We know: d=2×SideSide=d2 cm

 Area of the square=(Side)2=d22=d22 cm2

Ratio of the area of the circle to that of the square:
=πd24d22=π2
Thus, the ratio of the area of the circle to that of the square is π:2.

Page No 707:

Question 29:

The area of a circle inscribed in an equilateral triangle is 154 cm2. Find the perimeter of the triangle.

Answer:

Let the radius of the inscribed circle be r cm.
Given:
Area of the circle = 154 cm2
We know:
Area of the circle =πr2
154=227r2154×722=r2r2=49r=7
In a triangle, the centre of the inscribed circle is the point of intersection of the medians and altitudes of the triangle. The centroid divides the median of a triangle in the ratio 2:1.
Here,
AO:OD = 2:1  

Now,
Let the altitude be h cm.
We have:
ADB =90OD = 13ADOD=h3
h=3rh=21

Let each side of the triangle be a cm.
In the right-angled ADB, we have:AB2=AD2+DB2a2=h2+a224a2=4h2+a23a2=4h2a2=4h23a=2h3a=423
∴ Perimeter of the triangle = 3a
​                                            =3×423=3×42=72.66 cm

Page No 707:

Question 30:

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

Answer:

Radius of the wheel = 42 cm
​Circumference of the wheel = 2πr
                                             =2×227×42=264 cm=264100 m=2.64 m

Distance covered by the wheel in 1 revolution = 2.64 m

Total distance = 19.8 km or 19800 m

∴ Number of revolutions taken by the wheel = 198002.64=7500

Page No 707:

Question 31:

The wheels of the locomotive of a train are 2.1 m in radius. They make 75 revolutions in one minute.

Answer:

Radius of the wheel = 2.1 m
Circumference of the wheel = 2πr
                                             =2×227×2.1=13.2 m
Distance covered by the wheel in 1 revolution = 13.2 m
Distance covered by the wheel in 75 revolutions = (13.2×75)=990×1100 m
                                                                                                =990×11000 km
Distance covered by the wheel in 1 minute = Distance covered by the wheel in 75 revolutions =9901000 km

∴ Distance covered by the wheel in 1 hour = 9901000×60
                                                                      = 59.4 km/h

                                                                                              

Page No 707:

Question 32:

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

Answer:

Distance = 4.95 km = 4.95×1000×100 cm
∴ Distance covered by the wheel in 1 revolution =Total distance covered Number of revolutions

                                                                               =4.95×1000×1002500=198 cm

Now,
Circumference of the wheel = 198 cm
2πr=1982×227×r=198r=198×744r=31.5cm

∴ Diameter of the wheel = 2r
                                         = 2(31.5)
                                         = 63 cm

Page No 707:

Question 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.

Answer:

Diameter of the wheel = 60 cm
​∴ Radius of the wheel = 30 cm
​Circumference of the wheel = 2πr
                                             =2×227×30=13207 cm
Distance covered by the wheel in 1 revolution = 13207 cm
∴ Distance covered by the wheel in 140 revolutions = 13207×140×1100 m
                                                                                    =1320×1407×100×11000 km=2641000 km

Now,
Distance covered by the wheel in 1 minute = Distance covered by the wheel in 140 revolutions = 2641000 km

∴ Distance covered by the wheel in 1 hour = 2641000×60=15.84 km/h

Hence, the speed at which the boy is cycling is 15.84 km/h.

Page No 707:

Question 34:

The wheel of a motorcycle is of radius 35 cm. How many revolutions per minute must the wheel make so as to keep a speed of 66 km/hr?

Answer:

The radius of wheel of a motorcycle = 35 cm = 0.35 m.
So, the distance covered by this wheel in 1 revolution will be equal to perimeter of wheel i.e. 2πr=2×227×0.35=2.2 m.
Since speed is given to be 66 km/hr=66×1000m60min=1100 m/min.
As we know speed=distancetime
1100=number of revolutions×perimeter of wheeltime1100=number of revolutions×2.21number of revolutions=11002.2=500 revolutions.

Page No 707:

Question 35:

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

Answer:

Radius of the front wheel = 40 cm = 25 m

Circumference of the front wheel = 2π×25 m=4π5 m

Distance covered by the front wheel in 800 revolutions = 4π5×800 m=640π m
Radius of the rear wheel = 1 m
Circumference of the rear wheel = 2π×1=2π m

∴ Required number of revolutions = Distance covered by the front wheel in 800 revolutionsCircumference of the rear wheel
                                                        =640π2π=320

Page No 707:

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 side of the square measures 14 cm.

Answer:

Side of the square = 14 cm
Radius of the circle =142= 7 cm
Area of the quadrant of one circle = 14πr2
                                                        =14×227×7×7=38.5 cm2
Area of the quadrants of four circles = 38.5×4=154 cm2

Now,
Area of the square = Side2
                               =142=196 cm2

Area of the shaded region = Area of the square - Area of the quadrants of four circles
                                           = 196 - 154
                                           = 42 cm2

Page No 707:

Question 37:

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

Answer:


Radius = 5 cm 
AB = BC = CD = AD = 10 cm
All sides are equal, so it is a square.
Area of a square = Side2
Area of the square = 102=100 cm2
Area of the quadrant of one circle = 14πr2
  
                                                        =14×227×5×5=19.64 cm2
Area of the quadrants of four circles = 19.64×4=78.57 cm2
Area of the shaded portion = Area of the square - Area of the quadrants of four circles
                                            =100-78.57=21.43 cm2

Page No 707:

Question 38:

Four equal circles, each of radius a units, touch each other. Show that the area between them is 67a2 sq units.

Answer:


When four circles touch each other, their centres form the vertices of a square. The sides of the square are 2a units.

Area of the square = 2a2=4a2 sq. units

Area occupied by the four sectors
=4×90360×π×a2=πa2 sq. units

Area between the circles = Area of the square - Area of the four sectors
=4-227a2=67a2 sq. units



Page No 708:

Question 39:

Three equal circles, each of radius 6 cm, touch one another as shown in the figure. Find the area of enclosed between them.

Answer:


Join ABC. All sides are equal, so it is an equilateral triangle.
Now,
Area of the equilateral triangle =34×Side2

                                                   =1.734×12×12=62.28 cm2


Area of the arc of the circle=60360πr2=16πr2=16×227×6×6=18.86 cm2Area of the three sectors=3×18.86=56.57 cm2

                                      
Area of the shaded portion = Area of the triangle - Area of the three quadrants

=62.28-56.57 = 5.71 cm2
                                

Page No 708:

Question 40:

If three circles of radius a each, are drawn such that each touches the other two, prove that the area included between them is equal to 425a2.

Answer:



When three circles touch each other, their centres form an equilateral triangle, with each side being 2a.

Area of the triangle = 34×2a×2a=3a2

Total area of the three sectors of circles = 3×60360×227×a2=12×227×a2=117a2

Area of the region between the circles = Area of the triangle - Area of the three sectors
 =3-117a2=1.73-1.57a2=0.16a2=425a2

Page No 708:

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. [CBSE 2014]

Answer:

Area of trapezium = 12AD+BC×AB
24.5=1210+4×ABAB=3.5 cm
Area of shaded region = Area of trapezium ABCD − Area of quadrant ABE
=24.5-14πAB2=24.5-14×2273.52=24.5-9.625=14.875 cm2
Hence, the area of shaded region is 14.875 cm2

Page No 708:

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:            [CBSE 2013C]
(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.

Answer:

(i) Area of fours sector = Area of sector having central angle 60° + Area of sector having central angle 90° + Area of sector having central angle 90° + Area of sector having central angle 120°
=60°360°π142+90°360°π142+90°360°π142+120°360°π142=60°360°+90°360°+90°360°+120°360°π142=360°360°π142=616 m2

(ii) Area of the remaining portion = Area of trapezium ABCD − Area of four quadrants
=12AD+BC×AB-616=1255+45×30-616=1500-616=884 m2

Page No 708:

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. [Use 3=1.73, π=3.14]
[CBSE 2014]

Answer:

In equilateral traingle all the angles are of  60°
∴ ∠ABO = ∠AOB = 60°
 Area of the shaded region = (Area of triangle  AOB − Area of sector having central angle 60°) + Area of sector having central angle (360° − 60°)
=34AB2-60°360°π62+300°360°π62=1.734122-16×3.1462+56×3.1462=62.28-18.84+94.2=137.64 cm2
Hence, the area of shaded region is 137.64 cm2

Page No 708:

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. [CBSE 2014]

Answer:

We know that the opposite sides of a rectangle are equal
AD = BC =  70 cm
In right triangle AED
AE2 = AD2 − DE2  
= (70)2 − (42)2 
= 4900 − 1764
= 3136
∴ AE2 = 3136
⇒ AE = 56
= Area of the shaded region = Area of rectangle − (Area of triangle  AED + Area of semicircle)
=AB×BC-12×AE×DE+12πBC22=80×70-12×56×42+12×2277022=5600-3101=2499 cm2
Hence, the area of shaded region is 2499 cm2



Page No 709:

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] [CBSE 2014]

Answer:

In right triangle AED
AD2 = AE2 + DE2  
= (9)2 + (12)2 
= 81 + 144
= 225
∴ AD2 = 225
⇒ AD = 15 cm
We know that the opposite sides of a rectangle are equal
AD = BC =  15 cm
= Area of the shaded region = Area of rectangle − Area of triangle  AED + Area of semicircle
=AB×BC-12×AE×DE+12πBC22=20×15-12×9×12+12×3.141522=300-54+88.31=334.31 cm2
Hence, the area of shaded region is 334.31 cm2

Page No 709:

Question 46:

In the given figure, O 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] [CBSE 2012]

Answer:

In right triangle ABC
BC2 = AB2 + AC2  
= (7)2 + (24)2 
= 49 + 576
= 625
∴ BC2 = 625
⇒ BC = 25
Now, ∠COD + ∠BOD = 180°            (Linear pair angles)
⇒∠COD = 180° − 90° = 90°
 Now, Area of the shaded region = Area of sector having central angle (360° − 90°) −  Area of triangle  ABC
=270°360°πBC22-12AB×AC=34×3.142522-12×7×24=367.97-84=283.97 cm2
Hence, the area of shaded region is 283.97 cm2

Page No 709:

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.
[Use 3=1.73, π=3.14]     [CBSE 2014]

Answer:

We can find the radius of the incircle by using the formula
r=2×Area of trianglePerimeter of triangle=2×34×1223×12=23 cm
Now, area of shaded region = Area of triangle − Area of circle
=34×122-3.14×232=62.28-37.68=24.6 cm2
Hence, the area of shaded region is 24.6 cm2

Page No 709:

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.  [Use 3=1.73, π=227] [CBSE 2013C]

Answer:

Construction:  Join AO and extend it to D on BC.


Radius of the circle, r = 42 cm
∠OCD= 30°

cos30°=DCOC32=DC42DC=213BC=2×DC=423 =72.66 cmsin30°=ODOC12=OD42OD=21 cmNow, AD=AO+OD=42+21=63 cm

Area of shaded region = Area of circle − Area of triangle ABC
=πOA2-12×AD×AB=227422-12×63×72.66=5544-2288.79=3255.21 cm2

Page No 709:

Question 49:

The perimeter of the quadrant of a circle is 25 cm. Find its area.          [CBSE 2012]

Answer:

Let the radius of the circle be r
Now, Perimeter of quadrant = 142πr+2r
25=12×227×r+2r25=25r7r=7 cm
Area of quadrant = 14πr2=14×227×7×7=38.5 cm2
Hence, the area of quadrant is 38.5 cm2

Page No 709:

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.] [CBSE 2012]

Answer:



Area of minor segment = Area of sector AOBC − Area of right triangle AOB
=θ360°πOA2-12×OA×OB=90°360°×3.14102-12×10×10=78.5-50=28.5 cm2
Hence, the area of minor segment is 28.5 cm2

Page No 709:

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]        [CBSE 2011]

Answer:

Area of the road = Area of outer circle − Area of inner circle
=πR2-πr2=π1102-1002=3.14×2100=6594 m2
Cost of levelling the road = Area of the road ⨯ Rate
= 6594 ⨯ 20
= Rs 131880

Page No 709:

Question 52:

The area of an equilateral triangle is 493 cm2 . 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 π = 1.73]      [CBSE 2009]

Answer:

Area of equilateral triangle=49334×Side2=493Side2=72×22Side=14 cm

Radius of the circle = Half of the side of the triangle = 7 cm
Area of triangle not included in the circle = Area of triangle − Area of 3 sectors having central angle 60∘
=493-3×60°360°×227×7×7=84.77-77=7.77 cm2
Hence, the required area is 7.77 cm2



Page No 710:

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, HQFI and ELDF. If CD = 8 cm, BP = HQ = 4 cm and DE = EF = 5 cm, find the area of the whole figure.

Answer:

CD = 8 cm
BP = HQ = 4 cm
DE = EF = 5 cm
Area of the parallelogram ABCD = B×H
                                                     =8×4=32 sq.cm

Area of parallelogram FGHI = B×H
                                               =8×4=32 sq.cm
Area of the square = Side2
                               =82 =64 sq.cm
    
In ELF, we have:
EL2=52-42EL2=9EL=3 cm

Area of DEF = 12×B×H
                        =12×8×3=12 sq.cm

Area of the semicircle =12πr2

                                    =12×227×16=25.12 sq.cm
∴ Total Area =  Area of the parallelogram ABCD + Area of the parallelogram FGHI + Area of the triangle DEF + Area of the semicircle CKI + Area of the square
Total Area = 165.12 cm2

Page No 710:

Question 54:

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

Answer:

Area of sector having central angle 150° =  =150°360°π62=512×Area of circular disc

Now,  Area of sector having central angle 90° : Area of sector having central angle 120° : Area of sector having central angle 150°
=90°360°π62:120°360°π62:150°360°π62=14:13:512=3:4:5

Page No 710:

Question 55:

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

Answer:


Join each vertex of the hexagon to the centre of the circle.

The hexagon is made up of six triangles.

Total area of the design=Area of the circle-Area of six triangles= 227×35×35-6×34×35×35=35×35227-6×1.7324=1225227-2.598=1225×0.542=663.95 cm2

Page No 710:

Question 56:

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

Answer:

In the right RPQ, we have:
RQ=RP2+PQ2      =72+242      =49 + 576      = 25 cm
​
OR = OQ = 12.5 cm
Now,
Area of the circle = πr2
                             =3.14×12.5×12.5=490.625 sq.cm 

Area of the semicircle = 490.6252=245.31 sq.cm
Area of the triangle =12×b×h=12×7×24=84 sq.cm
Thus, we have:

Area of the shaded part =Area of the semicircle- Area of the triangle= 245.31 - 84 = 161.31 cm2

Page No 710:

Question 57:

In the given figure, ∆ABC is right-angled at A. Find the area of the shaded region if AB = 6 cm, BC = 10 cm and O is the centre of the incircle of ∆ABC.

Answer:

Using Pythagoras' theorem for triangle ABC, we have:

CA2+AB2=BC2

 CA=BC2-AB2=100-36=64= 8 cm

Now, we must find the radius of the incircle. Draw OE, OD and OF perpendicular to AC, AB and BC, respectively.



Consider quadrilateral AEOD.
Here,
EO=OD (Both are radii.)

Because the circle is an incircle, AE and AD are tangents to the circle.
AEO=ADO =90°

Also,
A=90°
Therefore, AEOD is a square.
Thus, we can say that AE=EO=OD=AD=r.

CE=CF=8-rBF=BD=6-rCF+BF=108-r+6-r=1014-2r=10r=2 cm

Area of the shaded part = Area of the triangle - Area of the circle
                                       =12×6×8-π×2×2=24-12.56=11.44 cm2

Page No 710:

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 = 3 cm and AC = 4 cm. Find the area of the shaded region.

Answer:

In triangle ABC, we have:
BC=AB2+AC2=9+16=25=5 cm

Ar(shaded part) = ArABC+Arsemicircle APB+Arsemicircle AQC-Arsemicircle BAC=12×3×4+12π×1.5×1.5+12π×2×2-12π×2.5×2.5=6+12π4+94-254=6+0=6 cm2



Page No 711:

Question 59:

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

Answer:

Perimeter (circumference of the circle) = 2πr
We know:
Perimeter of a semicircular arc = πr
Now,
For the arc PTS, radius is 6 cm.
∴ Circumference of the semicircle PTS = πr=6π cm

For the arc QES, radius is 4 cm.
​∴ Circumference of the semicircle QES = πr=4π cm

For the arc PBQ, radius is 2 cm.
∴ Circumference of the semicircle PBQ = πr=2π cm

Now,
Perimeter of the shaded region = 6π+4π+2π
                                                   =12πcm
                                                   =12×3.14=37.68 cm

Area of the semicircle PBQ = 12πr2

                                            =12×3.14×2×2=6.28 cm2

Area of the semicircle PTS = 12πr2

                                            =12×3.14×6×6=56.52 cm2

Area of the semicircle QES = 12πr2
  
                                            =12×3.14×4×4=25.12 cm2

Area of the shaded region = Area of the semicircle PBQ + Area of the semicircle PTS - Area of the semicircle QES
                                           =6.28 + 56.52 - 25.12 = 37.68 cm2

Page No 711:

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 90 m, 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.

Answer:

Length of the inner curved portion (400-2×90)=220 m
∴ Length of each inner curved path = 2202 = 110 m
​Thus, we have:
πr=110227r=110r=110×722r=35 m

Inner radius = 35 m
Outer radius = (35 + 14) = 49 m
Area of track = {Area of the two rectangles [each (90×14) m] + Area of the circular ring with R = 49 m and r = 35 m)}
=(2×90×14)+227×492-352=2520+227×2401-1225=2520+227×1176=2520+3696=6216 m2

​Length of the outer boundary of the track
=2×90+2×227×49=488 m

Therefore, the length of the outer boundary of the track is 488 m and the area of the track is 6216 sq. m.

Page No 711:

Question 61:

In the given figure, ABCD is a rectangle of dimensions 21 cm × 14 cm. A semicircle is drawn with BC as diameter. Find the area and the perimeter of the shaded region in the figure.          [CBSE 2017]

Answer:

Area of the shaded region

= Area of the rectangle − Area of the semicircle 

=21×14-12×π×1422=294-12×227×7×7=294-77=217 cm2

Therefore, area of shaded region is 217  cm2.

Length of the boundary (or perimeter) of the shaded region

=AB+AD+DC+arc BC=AB+AD+DC+12×2π142=21+14+21+227×7=56+22=78 cm

Therefore, the perimeter of the shaded region is 78 cm.

Page No 711:

Question 62:

In the given figure, two concentric circles with centre O have radii 21 cm and 42 cm. If ∠AOB = 60°, find  the area of the shaded region. Use π=227      [CBSE 2017]

Answer:

Given: Radius of the inner circle with radius OC, r = 21 cm
Radius of the inner circle with radius OA, R = 42 cm
∠AOB = 60°

Area of the circular ring

=πR2-πr2=πR2-r2=π422-212 cm2

Area of ACDB = area of sector AOB − area of COD
=60360×π×R2-60360×π×r2=60360×πR2-r2=60360×π422-212                 

Area of shaded region = area of circular ring − area of ACDB
=π422-212-60360π422-212=π422-2121-60360=22742-2142+21×300360=3465 cm2

Page No 711:

Question 63:

Three semicircles each of diameter 3 cm, a circle of diameter 4.5 cm and a semicircle of radius 4.5 cm are drawn in the given figure. Find the area of the shaded region.

Answer:


Area of the shaded region

= Area of the semi-circle with diameter of 9 cm − Areas of two semi-circles with diameter 3 cm − Area of the circle with diameter 4.5 cm + Area of semi-circle with diameter 3 cm

= Area of the semi-circle with radius of 4.5 cm − 2 × Area of semi-circle with radius 1.5 cm − Area of the circle with radius 2.25 cm + Area of semi-circle with radius 1.5 cm



3.9375×227

= 12.375 cm2

Thus, the area of the shaded region is 12.375 cm2.

Page No 711:

Question 64:

In the given figure, the side of square is 28 cm and radius of each circle is half of the length of the side of the square, where O and O' are centres of the circles. Find the area of shaded region.            [CBSE 2017]

Answer:

We have,
Side of square = 28 cm and radius of each circle = 282 cm


Area of the shaded region

= Area of the square + Area of the two circles − Area of the two quadrants

=282 + 2×π×2822 - 2×14×π×2822=282 + 32×π×2822=2821+32×227×12×12=2821+3328=282×6128=28×61=1708 cm2

Therefore, the area of the shaded region is 1708 cm2.



Page No 712:

Question 65:

With the vertices A, B and C of a triangle ABC as centres, arcs are drawn with radii 5 cm each as shown in the given figure. If AB = 14 cm, BC = 48 cm and CA = 50 cm then find the area of the shaded region. [Use π = 3.14]

Answer:

Given that ABC is a triangle with sides AB = 14 cm, BC = 48 cm, CA = 50 cm.
Clearly it is a right angled triangle.
Area of ABC=12×base×height=12×48×14=336 cm2.

Now we need to remove the area of 3 arcs each of radius 5 cm from this area of triangle.
Here, B=90°, A+C=90°.
The area of these 3 arcs would exactly be equal to a sector of circle with radius 5 cm and angle = A+B+C=180° i.e. a semi circle.
Since we know that area of semi circle is 12πr2=12×3.14×5×5=12×314100×25=3148=1574 cm2.
Hence, the area of shaded region is = area of ABC-combined area of 3 arcs=336-1574=336-39.25=296.75 cm2.

Page No 712:

Question 66:

If the diameters of the concentric circles shown in the figure below are in the ratio 1 : 2 : 3 then find the ratio of the areas of three regions.

Answer:

The diameters of concentric circles are in ratio 1 : 2 : 3.
let the diameters be d1 = 1xd2 = 2xd3 = 3x.
So, the radii are r1=x2, r2=x, r3=3x2.
The areas of three regions are given be 
A1=πr12=πx22=πx24, A2=πr22-πr12=πx2-πx24=3πx24, A3=πr32-πr22=π3x22-πx2=5πx24
So, their ratio is A1:A2:A3=πx24:3πx24:5πx24=1:3:5.



Page No 722:

Question 1:

The difference between the circumference and radius of a circle is 37cm. Using π=227 , find the circumference of the circle.   [CBSE 2013]

Answer:

Let the radius of the circle be r and circumference C.
​
Now,
C-r=372πr-r=37r2×227-1=37r=37×737=7 cm
Now, C=2πr=2×227×7=44 cm
Hence, the circumference of the circle is 44 cm.

Page No 722:

Question 2:

The circumference of a circle is 22 cm. Find the area of its quadrant. [CBSE 2012]

Answer:

Let the radius of the circle be r.
​Now,
Circumference=222πr=22r=22×744=72 cm
Now, Area of quadrant = 14πr2=14×227×722=778 cm2
Hence, the area of the quadrant of the circle is 778 cm2.

Page No 722:

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?     [CBSE 2012]

Answer:

Let the diameter of the required circle be d.
Now, Area of required circle = Area of circle having diameter 10 cm + Area of circle having diameter 24 cm

πd22=π1022+π2422d22=25+144d22=132d2=13d=26 cm
Hence, the diameter of the of the circle is 26 cm.

Page No 722:

Question 4:

If the area of the circle is numerically equal to twice its circumference then what is the diameter of the circle?   [CBSE 2011]

Answer:

Let the diameter of the required circle be d.
Now, Area of circle = 2 ⨯ Circumference of the circle
πd22=22π×d2d22=2dd2=8dd2-8d=0dd-8=0d=8 cm          d0
Hence, the diameter of the of the circle is 8 cm.

Page No 722:

Question 5:

What is the perimeter of a square which circumscribes a circle of radius a cm?     [CBSE 2011]

Answer:

We know that if a square circumscribes a circle, then the side of the square is equal to the diameter of the circle.
∴ Side of Square = 2a
Now, Perimeter of the square = 4 ⨯ Side of square = 4 ⨯ 2a = 8a cm
Hence, the perimeter of the square is 8a cm.

Page No 722:

Question 6:

Find the length of the arc of a circle of diameter 42 cm which subtends an angle of 60° at the centre.  [CBSE 2012]

Answer:

We have r=diameter2=422=21 cm and θ=60°
Length of arc = θ360°×2πr=60°360°×2×227×21=22 cm
Hence, the length of the arc of the circle is 22 cm.

Page No 722:

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.   [CBSE 2011]

Answer:

Let the diameter of the required circle be d.
Now, Area of the required circle = Area of circle having radius 4 cm + Area of circle having radius 3 cm

πd22=π42+π32d22=16+9d22=25=52d2=5d=10 cm

Hence, the diameter of the circle is 10 cm.

Page No 722:

Question 8:

Find the area of a circle whose circumference is 8π.    [CBSE 2014]

Answer:

Let the radius of the circle be r.
​Now,
Circumference=8π2πr=8πr=4 cm
Now, Area of circle= πr2=π×42=16π cm2
Hence, the area of the circle is 16π cm2.

Page No 722:

Question 9:

Find the perimeter of a semicircular protractor whose diameter is 14 cm.                    [CBSE 2009]

Answer:

Perimeter of a semicircular protractor = Circumference of semicircular protractor + diameter of semicircular protractor
=122πr+d=πr+d=πd2+d=227×7+14=22+14=36 cm
Hence, the perimeter of a semicircular protractor is 36 cm.

Page No 722:

Question 10:

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

Answer:

Let the radius of the required circle be r.
Now, Area of circle = Perimeter of the circle
πr2=2π×rr2=2rr2-2r=0rr-2=0r-2=0       r0r=2 units
Hence, the radius of the circle is 2 units.

Page No 722:

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.

Answer:

Let the radius of the required circle be r.
Now, Circumference of the required circle = Circumference of circle having radius 19 cm + Circumference of circle having radius 9 cm

2πr=2π×19+2π×9r=19+9r=28 cm
Hence, the radius of the required circle is 28 cm.

Page No 722:

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.

Answer:

Let the radius of the required circle be r.
Now, Area of the required circle = Area of circle having radius 8 cm + Area of circle having radius 6 cm

πr2=π82+π62r2=64+36r2=102r=10 cm
Hence, the radius of the circle is 10 cm.

Page No 722:

Question 13:

Find the area of the sector of a circle having radius 6 cm and of angle 30°.      [Take π = 3.14]

Answer:

We have r=6 cm and θ=30°
Now, Area of sector = θ360°×πr2=30°360°×3.14×36=9.42 cm2
Hence, the area of the sector of the circle is 9.42 cm2.

Page No 722:

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.

Answer:

We have r=21 cm and θ=60°
Length of arc = θ360°×2πr=60°360°×2×227×21=22 cm
Hence, the length of the arc of the circle is 22 cm.

Page No 722:

Question 15:

The circumferences of two circles are in the ratio 2: 3. What is the ratio between their areas?

Answer:

Let the the radii of the two circles be r and R, the circumferences of the circles be c and C and the areas of the two circles be a and A.
Now,
cC=232πr2πR=23rR=23
Now, the ratio between their areas is given by
aA=πr2πR2=rR2=232=49
Hence, the ratio between their areas is 4 : 9.



Page No 723:

Question 16:

The areas of two circles are in the ratio 4: 9. What is the ratio between their circumferences?

Answer:

Let the radii of the two circles be r and R, the circumferences of the circles be c and C and the areas of the two circles be a and A.
Now,
aA=49πr2πR2=232rR=23
Now, the ratio between their circumferences is given by
cC=2πr2πR=rR=23
Hence, the ratio between their circumferences is 2 : 3.

Page No 723:

Question 17:

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

Answer:

Let the side of the square be a and radius of the circle be r
We know that if a circle circumscribes a square, then the diameter of the circle is equal to the diagonal of the square.
2a=2ra=2r
Now,
Area of circleArea of square=πr2a2=πr22r2=πr22r2=π2
Hence, the ratio of the areas of the circle and the square is π : 2

Page No 723:

Question 18:

The circumference of a circle is 8 cm. Find the area of the sector whose central angle is 72°.

Answer:

Let the radius of the circle be r.
​Now,
Circumference=82πr=8r=1411 cm
We have r=1411 cm and θ=72°
Area of sector = θ360°×πr2=72°360°×227×14112=1.02 cm2
Hence, the area of the sector of the circle is 1.02 cm2.

Disclaimer : If we take the circumference of the circle is 8 cm then the area of the sector will be 1.02 cm2. But if we take the circumference of the circle is 88 cm then the area of the sector will be 123.2 cm2

Page No 723:

Question 19:

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

Answer:

Given:
Length of the arc = 8.8 cm
And,
θ=30

Now,
Length of the arc =2πrθ360
8.8=2×227×r×30360r=8.8×360×744×30r=16.8 cm

∴ Length of the pendulum = 16.8 cm

Page No 723:

Question 20:

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

Answer:

Angle inscribed by the minute hand in 60 minutes = 360
Angle inscribed by the minute hand in 20 minutes = 36060×20=120

We have:
θ=120 and r= 15 cm

∴ Required area swept by the minute hand in 20 minutes = Area of the sector with r = 15 cm and θ=120
                                                                                             =πr2θ360

                                                                                             =3.14×15×15×120360=235.5 cm2

Page No 723:

Question 21:

A sector of 56°, cut out from a circle, contains 17.6 cm2. Find the radius of the circle.

Answer:

Area of the sector =17.6 cm2
Area of the sector=πr2θ360

17.6=227×r2×56360r2=17.6×7×36022×56r2=36r=6 cm

∴ Radius of the circle = 6 cm

Page No 723:

Question 22:

The area of the sector of a circle of radius 10.5 cm is 69.3 cm2. Find the central angle of the sector.

Answer:

Given:
Area of the sector = 63 cm2
Radius = 10.5 cm

Now,
Area of the sector =πr2θ360
69.3=227×10.5×10.5×θ360θ=69.3×7×36022×10.5×10.5θ=72

∴ Central angle of the sector = 72

Page No 723:

Question 23:

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

Answer:

Given:
Radius = 6.5 cm


Let O be the centre of the circle with radius 6.5 cm and OACBO be its sector with perimeter 31 cm.
Thus, we have:
OA + OB + arc AB = 31 cm

6.5+6.5+arc AB=31arc AB =31 - 13arc AB = 18 cm

Now,
Area of the sector OACBO = 12×Radius×Arc
                                           =12×6.5×18=58.5 cm2
​                

Page No 723:

Question 24:

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

Answer:

Given:
Radius = 17.5 cm
Length of the arc = 44 cm

Now,
Length of the arc =2πrθ360

44=2×227×17.5×θ360θ=44×7×36044×17.5θ=144
Also,
Area of the sector =πr2θ360

                              =227×17.5×17.5×144=385 cm2
                     

Page No 723:

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 ⨯ 7 cm. Find the area of the remaining cardboard.   [CBSE 2013]

Answer:

We know that we can cut two circular pieces of equal radii and maximum area from the rectangular cardboard whose diameter is equal to the width of the rectangular cardboard.
∴ Radii of two circuar pieces = Half of the width of the rectangular cardboard = 3.5 cm
Now,
Area of remaining cardboard = Area of rectangular cardboard − 2 ⨯ Area of circular piece having radius 3.5 cm
=14×7-2227×3.5×3.5=98-77=21 cm2
Hence, the area of the remaining cardboard is 21 cm2

Page No 723:

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.]

Answer:

Area of the square ABCD = (Side)2 
                                      =42=16 cm2

Area of the circle = πr2
Radius = 1 cm
Area = 3.14×(1)2=3.14 cm2

Area of the quadrant of one circle = 14πr2
                                                   =14×3.14×12=0.785 cm2

Area of the quadrants of four circles = 0.785×4 = 3.14 cm2
Area of the shaded region = Area of the square - Area of the circle - Area of the quadrants of four circles
                                           =16-3.14-3.14=9.72 cm2

Page No 723:

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.    [CBSE 2012]

Answer:

We know that the opposite sides of rectangle are equal
∴ AD = BC = 28 cm
Now, Radius of semicircular portion = 12BC=14 cm
∴ Area of remaining paper = Area of rectangular sheet −Area of semicircular portion
=40×28-12227×14×14=1120-308=812 cm2
Hence, the area of the remaining paper is 812 cm2

Page No 723:

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.

Answer:

Area of shaded region = Area of square OABC − Area of quadrant COPB having radius OC
=Side2-14π×r2=72-14227×72=49-38.5=10.5 cm2
Hence, the area of the shaded region is 10.5 cm2



Page No 724:

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.    [CBSE 2012]

Answer:

Area of the shaded region = Area of sector having central angle 60∘ + Area of sector having central angle 80∘ + Area of sector having central angle 40∘
=60°360°×π72+80°360°×π72+40°360°×π72=π7260°360°+80°360°+40°360°=π72180°360°=2277212=77 cm2
Hence, the area of the shaded region is 77 cm2.

Page No 724:

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.  [CBSE 2012]

Answer:

Area of the shaded portion = Area of sector OPQ − Area of sector OAB
=30°360°×π72-30°360°×π3.52=227×11272-3.52=227×11272-722=227×112×1474=778 cm2
Hence, the area of the shaded portion is 778 cm2.

Page No 724:

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.     [CBSE 2012]
 

Answer:

Area of the shaded region = Area of Square ABCD − (Area of semicircle APD + Area of semicircle BPC)
=142-12π1422+12π1422=142-22772=42 cm2
Hence, the area of the shaded region is 42 cm2.

Page No 724:

Question 32:

In the given figure, the shape of the top of a table is that of a sector of a circle with centre O and ∠AOB = 90°. If AO = OB = 42 cm,· then find the perimeter of the top of the table.    [CBSE 2012]

Answer:

We have r=42 cm and θ=360°-90°=270°
Perimeter of the top of the table = Length of the major arc AB + Length of OA + Length of OB
=θ360°×2πr+42+42=270°360°×2×227×42+84=198+84=282 cm

Hence, the perimeter of the top of the table is 282 cm.

Page No 724:

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.

Answer:

Area of the shaded portion = (Area of quadrant DPBA + Area of quadrant DQBC) − Area of Square ABCD
=14π72+14π72-72=12×22772-49=28 cm2
Hence, the area of the shaded portion is 28 cm2.

Page No 724:

Question 34:

In the given figure, OABC is a quadrant of a circle of radius 3.5 cm with centre O. If OD = 2 cm, find the area of the shaded portion.

Answer:

Area of the right-angled COD = 12×b×h
                                                   =12×3.5×2=3.5 cm2

Area of the sector AOC  =θ360×π×r2
                                       =90360×227×3.52= 9.625 cm2

Area of the shaded region = Area of the COD - Area of the sector AOC

                                           =9.625-3.50 = 6.125 cm2



Page No 725:

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.    [CBSE 2011]

Answer:

Permieter of shaded region = Length of the arc APB + Length of the arc CPD + Length of AD + Length of BC
=12×2πr+12×2πr+14+14=2πr+28=2×227×142+28=72 cm

Hence, the perimeter of the shaded region is 72 cm.

Page No 725:

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.   [CBSE 2011]

Answer:

Let the diagonal of the square be d.
We know that if a circle circumscribes a square, then the diameter of the circle is equal to the diagonal of the square.
∴ d = 2 ⨯ 7 = 14 cm
Now,
Area of required region = Area of circle − Area of square
=πr2-12d2=227×72-12×142=56 cm2

Hence, the required area is 56 cm2 .

Page No 725:

Question 37:

In the given figure, APB and CQD are semicircles of diameter 7 cm each, while ARC an BSD are semicircles of diameter 14 cm each. Find the   
(i) perimeter, (ii) area of the shaded region.                                [CBSE 2011]
 

Answer:

(i) Perimeter of shaded region = Length of the arc APB + Length of the arc ARC + Length of the arc BSD + Length of the arc CQD
=12×2πr1+12×2πr2+12×2πr3+12×2πr4=12×2π72+12×2π7+12×2π7+12×2π72=2π72+2π7=2π72+7=2×227×212=66 cm

(ii) Area of shaded region = Area of the arc ARC + Area of the arc BSD − (Area of the arc APB + Area of the arc CQD)
=12×πr12+12×πr22-12×πr32+12×πr42=12×π72+12×π72-12×π722+12×π722=π72-π722=π49-494=227×1474=115.5 cm2

Page No 725:

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]           [CBSE 2014]

Answer:

Perimeter of shaded region = Length of the arc PAQ + Length of the arc PSR + Length of the arc RTQ
=12×2πr1+12×2πr2+12×2πr3=12×2π72+12×2π102+12×2π32=72π+5π+32π=72π+32π+5π=5π+5π=10π=31.4 cm
Hence, the perimeter of shaded region is 31.4 cm.

Page No 725:

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.]        [CBSE 2014]

Answer:

Construction: Join OB

In right triangle AOB
OB2 = OA2 + AB2
= 202 + 202
= 400 + 400
= 800
∴ OB2 = 800
Area of the shaded region = Area of quadrant OPBQ − Area of Square OABC
=14πOB2-OA2=14×3.14×800-400=628-400=228 cm2
Hence, the area of the shaded region is 228 cm2.



Page No 726:

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.     [CBSE 2015]

Answer:

Permieter of shaded region = Length of the arc AQO + Length of the arc APB + Length of OB
40=12×2πAO2+12×2πOB+OB40=117AO+227OB+OB40=117OB+227OB+OB        AO=OB40=407OBOB=7 cm        
Area of the shaded portion = Area of semicircle AQO + Area of semicircle APB
=12π722+12π72=12×227×722+12×227×72=96.25 cm2
Hence, the area of the shaded portion is 96.25 cm2.

Page No 726:

Question 41:

Find the area of a quadrant of a circle whose circumference is 44 cm.  [CBSE 2011]

Answer:

Let the radius of the circle be r.
​Now,
Circumference=442πr=44r=7 cm
Now,
Area of quadrant = 14πr2=14×227×72=38.5 cm2
Hence, the area of the quadrant of the circle is 38.5 cm2.

Page No 726:

Question 42:

In the given figure, find the area of the shaded region, where ABCD is a square of side 14 cm and all circles are of the same diameter.  [CBSE 2014]

Answer:


Area of the square = Side2=142=196 sq. cm

Area of the circles = 4×π×3.5×3.5=154 sq. cm

Area of the shaded region = Area of the square - Area of four circles
                                           =196-154=42 cm2

Page No 726:

Question 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.  [CBSE 2014]

Answer:

In right triangle ABC
AC2 = AB2 + BC2
= 82 +62
= 64 + 36
= 100
∴ AC2 = 100
⇒ AC = 10 cm
Now, Radius of circle(OA)=12AC=5 cm
Area of the shaded region = Area of circle − Area of rectangle OABC
=πOA2-AB×BC=227×52-8×6=78.57-48=30.57 cm2
Hence, the area of the shaded region is 30.57 cm2.

Page No 726:

Question 44:

A wire is bent to form a square enclosing an area of 484 cm2.Using the same wire, a circle is formed. Find the area of the circle.

Answer:

Area of the circle = 484 cm2
Area of the square = Side2 
484=Side2222=Side2Side=22 cm 
Perimeter of the square = 4×Side
Perimeter of the square = 4×22
                                       = 88 cm
Length of the wire = 88 cm
Circumference of the circle = Length of the wire = 88 cm
Now, let the radius of the circle be r cm.
​Thus, we have:
​ 2πr=882×227×r=88r=14 

Area of the circle = πr2
                             =227×14×14=616 cm2

Thus, the area enclosed by the circle is 616 cm2.

Page No 726:

Question 45:

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

Answer:

Let the diameter of the square be d and having circumscribed circle of radius r.
We know that if a circle circumscribes a square, then the diameter of the circle is equal to the diagonal of the square.
d = 2r
Now,
Area of square=12d2=122r2=2r2 sq units
Hence, the area of the square ABCD is 2r2 sq units.

Page No 726:

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. [Take π=227].

Answer:

Circumference=Total cost of fencingRate of fencing=550025=220
Let the radius of the circle be r.
​Now,
Circumference=2202πr=220r=35 cm
Now,
Area of field = πr2=227×352=3850 cm2
Cost of ploughing = Rate ⨯ Area of field = 0.5 ⨯ 3850 = Rs 1925
Hence, the cost of ploughing the field is Rs 1925.

Page No 726:

Question 47:

A park is in the form of a rectangle 120 m by 90 m. 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 m2. Find the radius of the circular lawn.

Answer:

Area of the rectangle = l×b
                                   =120×90=10800 sq. m

Area of the park excluding the lawn = 2950 m2

Area of the circular lawn = Area of the park - Area of the park excluding the lawn
                                          = 10800 - 2950
                                          = 7850 m2
Area of the circular lawn = πr2
7850 = 227×r27850×722=r2r2=2497.72r=49.97Or,r50 m

Thus, the radius of the circular lawn is 50 m.
                      



Page No 727:

Question 48:

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

Answer:

Area of the flower bed is the difference between the areas of sectors OPQ and ORS.

Area of the flower bed= θ360×π(PO2-OR2)=90360×227212-142=14×227×35×7=192.5 m2

Page No 727:

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 drawn. If AC = 54 cm and BC = 10, find the area of the shaded region.

Answer:

We have:
OA = OC = 27 cm
AB = AC - BC
      = 54 - 10
      = 44
AB is the diameter of the smaller circle.
Thus, we have:
Radius of the smaller circle = AB2=442= 22 cm
Area of the smaller circle = πr2
                                          =227×22×22=1521.14 cm2

Radius of the larger circle = AC2=542= 27 cm
Area of the larger circle = πr2
                                        =227×27×27=2291.14 cm2
∴ Area of the shaded region = Area of the larger circle - Area of the smaller circle
                                               = 2291.14 - 1521.14
                                               = 770 cm2

Page No 727:

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.                                                                                                                         [CBSE 2011]

Answer:

Since, BFEC is a quarter of a circle.
Hence, BC = EC = 3.5 cm
Now, DC = DE + EC = 2 + 3.5 = 5.5 cm
Area of shaded region = Area of the trapezium ABCD − Area of the quadrant BFEC
=12×AB+DC×BC-14×πEC2=12×3.5+5.5×3.5-14×227×3.52=6.125 cm2
Hence, the area of the shaded region is 6.125 cm2 .

Page No 727:

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.

Answer:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB
=90°360°×πOA2-12×OA×OB=14×227×352-12×35×35=14×227×352-12×35×35=962.5-612.5=350 cm2
Area of major segment APB = Area of circle − Area of minor segment
=πOA2-350=227×352-350=3850-350=3500 cm2
Hence, the area of major segment is 3500 cm2



Page No 733:

Question 1:

The area of a circle is 38.5 cm2. The circumference of the circle is
(a) 6.2 cm
(b) 12.1 cm
(c) 11 cm
(d) 22 cm

Answer:

(d) 22 cm
Let the radius be r cm.
We know:
Area of a circle=πr2 cm2
Thus, we have:
πr2=38.5
227×r2=38.5r2=38.5×722r2=38510×722r2=494r=72
Now,
Circumference of the circle=2πr
                                            =2×227×r=2×227×72=22 cm

Page No 733:

Question 2:

The area of a circle is 49 π cm2. Its circumference is
(a) 7 π cm
(b) 14 π cm
(c) 21 π cm
(d) 28 π cm

Answer:

(b) 14π cm
Let the radius be r cm.
We know:
Area of a circle=πr2
Thus, we have:
πr2=49πr2=49r=49r=7

Now,
Circumference of the circle=2πr
                                            =2×227×7 cm=14π cm

Page No 733:

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 cm2

Answer:

(c) 154 cm2
Let the radius be r cm.
We know:
Circumference of the circle=2πr
Thus, we have:
2πr-r=37r2π-1=37r2×227-1=37r377=37r=37×737r=7 cm

Radius = 7 cm
Now,
Area of the circle=πr2
                            =227×7×7 cm2=154 cm2



Page No 734:

Question 4:

The perimeter of a circular field is 242 m. The area of the field is
(a) 9317 m2
(b) 18634 m2
(c) 4658.5 m2
(d) none of these

Answer:

(c) 4658.5 m2
Let the radius be r m.
We know:
Perimeter of a circle=2πr 
Thus, we have:
2πr=242
2×227×r=242447×r=242r=242×744r=772

∴ Area of the circle=πr2
                                =227×772×772m2=4658.5 m2

Page No 734:

Question 5:

On increasing the diameter of a circle by 40%, its area will be increased by
(a) 40%
(b) 80%
(c) 96%
(d) 82%

Answer:

(c) 96%
Let d be the original diameter.
Radius=d2
Thus, we have:
Original area=π×d22
                      =πd24
New diameter=140% of d
                       =140100×d=7d5
Now,
New radius=7d5×2
                   =7d10
New area=π×7d102
                =49πd210
Increase in the area=49πd210-πd24
                                =24πd2100=6πd225
We have:
Increase in the area=6πd225×4πd2×100%
                                = 96%

Page No 734:

Question 6:

On decreasing the radius of a circle by 30%, its area is decreased by
(a) 30%
(b) 60%
(c) 45%
(d) none of these

Answer:

(d) None of these
Let r be the original radius.
Thus, we have:
Original area=πr2
Also,
New radius=70% of r
                   =70100×r=7r10
New area=π×7r102
                =49πr2100
Decrease in the area=πr2-49πr2100
                                 =59πr2100
Thus, we have:
Decrease in the area=59πr2100×1πr2×100%
                                  =51%

Page No 734:

Question 7:

The area of a square is the same as the area of a square. Their perimeters are in the ratio
(a) 1 : 1
(b) 2 : π
(c) π : 2
(d) π:2

Answer:

(d) π:2
Let a be the side of the square.
We know:
Area of a square=a2
Let r be the radius of the circle.
We know:
Area of a circle=πr2
Because the area of the square is the same as the area of the circle, we have:
a2=πr2r2a2=1πra=1π
∴ Ratio of their perimeters=2πr4a   Because perimeter of the circle is 2πr and perimeter of the square is 4a
                                                        
                                            =πr2a=π2×ra=π2×1π   Since ra=1π=π2=π:2

Page No 734:

Question 8:

The circumference of a circle is equal to the sum of the circumference of two circles having diameters 36 cm and 20 cm. The radius of the new circle is
(a) 16 cm
(b) 28 cm
(c) 42 cm
(d) 56 cm

Answer:

(b) 28 cm
Let r cm be the radius of the new circle.
We know:
Circumference of the new circle = Circumference of the circle with diameter 36 cm + Circumference of the circle with diameter 20 cm
Thus, we have:
2πr=2πr1+2πr2
2πr=2π×18+2π×102πr=2π×18+102πr=2π×282πr=2×227×28
2πr=1762×227×r=176r=176×744r=28 cm

Page No 734:

Question 9:

The area of 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
(a) 25 cm
(b) 31 cm
(c) 50 cm
(d) 62 cm

Answer:

(c) 50 cm
Let r cm be the radius of the new circle.
Now,
Area of the new circle = Area of the circle with radius 24 cm + Area of the circle with radius 7 cm
Thus, we have:
πr2=πr12+πr22
πr2=π×242+π×72 cm2πr2=π×576+π×49 cm2πr2=π×576+49 cm2r2=625π cm2r2=625r=25 

∴ Diameter of the new circle=25×2 cm
                                               = 50 cm

Page No 734:

Question 10:

If the perimeter of a square is equal to the circumference of a circle then the ratio of their areas is
(a) 4 : π
(b) π : 4
(c) π : 7
(d) 7 : π

Answer:

Let the side of the square be a and the radius of the circle be r.
Now, Perimeter of circle = Circumference of the circle
4a=2πrar=π2Now,Area of squareArea of circle=a2πr2=1π×ar2=1π×π24=π4
Hence, the correct answer is option (b).

Page No 734:

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
(a) R1+R2=R
(b) R1+R2<R
(c) R12+R22<R2
(d) R12+R22=R2

Answer:

(d) R12+R22=R2
Because the sum of the areas of two circles with radii R1 and R2 is equal to the area of a circle with radius R, we have:
πR12+πR22=πR2π(R12+R22)=πR2R12+R22=R2

Page No 734:

Question 12:

If the sum of the circumferences of two circles with radii R1 and R2 is equal to the circumference of a circle of radius R, then
(a) R1+R2=R
(b) R1+R2>R
(c) R1+R2<R
(d) none of these

Answer:

(a) R1+R2=R
Because the sum of the circumferences of two circles with radii R1 and R2 is equal to the circumference of a circle with radius R, we have:

2πR1+2πR2=2πR2π(R1+R2)=2πRR1+R2=R

Page No 734:

Question 13:

If the circumference of a circle and the perimeter of a square are equal, then
(a) area of the circle = area of the square
(b) (area of the circle) > (area of the square)
(c) (area of the circle) < (area of the square)
(d) none of these

Answer:

(b) Area of the circle > Area of the square
Let r be the radius of the circle.
We know:
Circumference of the circle=2πr
Now,
Let a be the side of the square.
We know:
Perimeter of the square = 4a
Now,
2πr=4ar=4a2π
∴ Area of the circle=πr2
                                =π×4a2π2=π×16a24π2=4a2π=4×7a222=14a211

Also,
Area of the square=a2
Clearly, 14a211>a2.
∴ Area of the circle > Area of the square

Page No 734:

Question 14:

The radii of two concentric circles are 19 cm and 16 cm respectively. The area of the ring enclosed by these circles is
(a) 320 cm2
(b) 330 cm2
(c) 332 cm2
(d) 340 cm2

Answer:

(b) 330 cm2
Let:
R = 19 cm and r = 16 cm
Thus, we have:
Area of the ring=πR2-r2
                          =πR+rR-r=227×19+16×19-16 cm2=227×35×3 cm2=330 cm2



Page No 735:

Question 15:

The areas of two concentric circles are 1386 cm2 and 962.5 cm2. The width of the ring is
(a) 2.8 cm
(b) 3.5 cm
(c) 4.2 cm
(d) 3.8 cm

Answer:

 (b) 3.5 cm
Let r cm and R cm be the radii of two concentric circles.
Thus, we have:
πR2=1386
227×R2=1386R2=1386×722 cm2R2=441 cm2R=21 cm

Also,
πr2=962.2227×r2=962.2r2=962.2×722 cm2r2=962.210×722 cm2r2=12254 cm2r=352 cm
∴ Width of the ring=R-r
                               =21-352 cm=72 cm=3.5 cm

Page No 735:

Question 16:

The circumferences of two circles are in the ratio 3 : 4. The ratio of their areas is
(a) 3: 4
(b) 4 : 3
(c) 9 : 16
(d) 16: 9

Answer:

Let the the radii of the two circles be r and R, the circumferences of the circles be c and C and the areas of the two circles be a and A.
Now,
cC=342πr2πR=34rR=34
Now, the ratio between their areas is given by
aA=πr2πR2=rR2=342=916
Hence, the correct answer is option (c).

Page No 735:

Question 17:

The areas of two circles are in the ratio 9 : 4. The ratio of their circumferences is
(a) 3 : 2
(b) 4 : 9
(c) 2 : 3
(d) 81 : 16

Answer:

Let the the radii of the two circles be r and R, the circumferences of the circles be c and C and the areas of the two circles be a and A.
Now,
aA=94πr2πR2=322rR=32
Now, the ratio between their circumferences is given by
cC=2πr2πR=rR=32
Hence, the correct answer is option (a)

Page No 735:

Question 18:

The radius of a wheel is 0.25 m. How many revolutions will it make in covering 11 km?
(a) 2800
(b) 4000
(c) 5500
(d) 7000

Answer:

(d) 7000
Distance covered in 1 revolution=2πr
                                                    =2×227×0.25 m=2×227×25100 m=117 m
Number of revolutions taken to cover 11 km=11×1000×711
                                                                        = 7000

Page No 735:

Question 19:

The diameter of a wheel is 40 cm. How many revolutions will it make in covering 176 m?
(a) 140
(b) 150
(c) 160
(d) 166

Answer:

(a) 140
Distance covered by the wheel in 1 revolution=πd
                                                                         =227×40 cm=8807 cm=8807×100 m

Number of revolutions required to cover 176 m =1768807×100
                                                                             =176×100×7880
                                                                              =140

Page No 735:

Question 20:

In making 1000 revolutions, a wheel covers 88 km. The diameter of the wheel is
(a) 14 m
(b) 24 m
(c) 28 m
(d) 40 m

Answer:

(c) 28 m
Distance covered by the wheel in 1 revolution=88×10001000 m
                                                                          = 88 m
We have:
Circumference of the wheel = 88 m
Now, let the diameter of the wheel be d m.
Thus, we have:

πd=88227×d=88d=88×722d=28 m

Page No 735:

Question 21:

The area of the sector of angle θ° of a circle with radius R is
(a) 2πRθ180
(b) πR2θ180
(c) 2πRθ360
(d) πR2θ360

Answer:

(d) πR2θ360

Page No 735:

Question 22:

The length of an arc of the sector of angle θ° of a circle with radius R is
(a) 2πRθ180
(b) 2πRθ360
(c) πR2θ180
(d) πR2θ360

Answer:

(b) 2πRθ360

Page No 735:

Question 23:

The length of the minute hand of a clock is 21 cm. The area swept by the minute hand in 10 minutes is [CBSE 2012]
(a) 231 cm2
(b) 210 cm2
(c) 126 cm2
(d) 252 cm2

Answer:

Angle subtends by the minute hand in 1 minute = 6∘
∴ Angle subtends by the minute hand in 10 minutes = 60∘
Now,
Area of the sector=θ360°πr2=60°360°×227212=231 cm2
Hence, the correct answer is option (a).

Page No 735:

Question 24:

A chord of a circle of radius 10 cm subtends a right angle at the centre. The area of the minor segments (given, π = 3.14) is
(a) 32.5 cm2
(b) 34.5 cm2
(c) 28.5 cm2
(d) 30.5 cm2

Answer:


Area of minor segment = Area of sector AOBC − Area of right triangle AOB
=90°360°πOA2-12×OA×OB=14×3.14×102-12×10×10=78.5-50=28.5 cm2
Hence, the correct answer is option (c).

Page No 735:

Question 25:

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. The length of the arc is
(a) 21 cm
(b) 22 cm
(c) 18.16 cm
(d) 23.5 cm

Answer:

We have r=21 cm and θ=60°
Length of arc = θ360°×2πr=60°360°×2×227×21=22 cm
Hence, the correct answer is option (b)

Page No 735:

Question 26:

In a circle of radius 14 cm, an arc subtends an angle of 120° at the centre. If 3= 1.73 then the area of the segment of the circle is
(a) 120.56 cm2
(b) 124.63 cm2
(c) 118.24 cm2
(d) 130.57 cm2

Answer:



Radius of the circle, r = 14 cm
Draw a perpendicular OD to chord AB. It will bisect AB.
∠A = 180° − (90° + 60°) = 30°

cos30°=ADOA32=AD14AD=73AB=2×AD=143 cmsin30°=ODOA12=OD14OD=7 cm

Area of minor segment = Area of sector OAPB − Area of triangle AOB
=θ360°πOA2-12×OD×AB=120°360°×227142-12×7×143=205.33-84.77=120.56 cm2
Hence, the correct answer is option (a).



Page No 739:

Question 1:

In the given figure, a square OABC has been inscribed in the quadrant OPBQ. If OA = 20 cm, then the area of the shaded region is


(a) 214 cm2
(b) 228 cm2
(c) 242 cm2
(d) 248 cm2

Answer:


(b) 228 cm2
Join OB
Now, OB is the radius of the circle.
We have:OB2=OA2+AB2   By Pythagoras' theoremOB2=202+202 cm2

OB2=400+400 cm2OB2=800 cm2OB=202 cm

Hence, the radius of the circle is 202 cm.
Now,
Area of the shaded region = Area of the quadrant - Area of the square OABC
                                          =14×3.14×202×202-20×20 cm2=14×314100×800-400 cm2=628-400 cm2=228 cm2

Page No 739:

Question 2:

The diameter of a wheel is 84 cm. How many revolutions will it make to cover 792 m?
(a) 200
(b) 250
(c) 300
(d) 350

Answer:

(c) 300

Let d cm be the diameter of the wheel.
We know:
Circumference of the wheel=π×d
                                            =227×84 cm=264 cm

Now,
Number of revolutions to cover 792 m=792×1000264
                                                              =300

Page No 739:

Question 3:

The area of a sector of a circle with radius r, making an angle of at the centre is
(a) x180×2πr
(b) x180×πr2
(c) x360×2πr
(d) x360×πr2

Answer:

(d) x360×πr2

The area of a sector of a circle with radius r making an angle of x° at the centre is x360×πr2.

Page No 739:

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


(a) 264 cm2
(b) 266 cm2
(c) 272 cm2
(d) 254 cm2

Answer:

All options are incorrect; the correct answer is 30.5 cm.

Join AC.
Now, AC is the diameter of the circle.
We have:AC2=AB2+BC2   By Pythagoras' theoremAC2=82+62 cm2AC2=64+36 cm2AC2=100 cm2AC=10 cm
∴ Radius of the circle=102  cm
                                   =5 cm
Now,
Area of the shaded region = Area of the circle with radius 5 cm - Area of the rectangle ABCD
                                          =3.14×5×5-8×6 cm2=314100×25-48 cm2=1572-48 cm2=612 cm2=30.5 cm2

Page No 739:

Question 5:

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

Answer:

Let r cm be the radius of the circle.
Now,
Circumference of the circle:
2πr=222×227×r=22r=22×744 cmr=72 

Also,
Area of the circle=πr2
                            =227×72×72 cm2=772 cm2=38.5 cm2

Page No 739:

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.

Answer:


Let ACB be the given arc subtending at an angle of 60° at the centre.
Now, we have:
r=21 cm and θ=60°
∴ Length of the arc ACB=2πr360
                                        =2×227×21×60360=22 cm

Page No 739:

Question 7:

The minute hand of a clock is 12 cm long. Find the area swept by in it 35 minutes.

Answer:

Angle described by the minute hand in 60 minutes=360°
Angle described by the minute hand in 35 minutes=36060×35°
                                                                                =210°
Now,
r=12 cm and θ=210° 
∴ Required area swept by the minute hand in 35 minutes = Area of the sector with r=12 cm and θ=210° 
                                                                                             =πr2θ360=227×12×12×210360 cm2=264 cm2

Page No 739:

Question 8:

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

Answer:

Let O be the centre of the circle with radius 5.6 cm and OACB be its sector with perimeter 27.2 cm.
Thus, we have:
OA+OB+arc AB=27.25.6+5.6+arc AB=27.2arc AB=16 cm
Now,
Area of the sector OACBO=12×Radius×l square units
                                          =125.6×16 cm2=44.8 cm2

Page No 739:

Question 9:

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

Answer:

Let r cm be the radius of the circle and θ be the angle.
We have:
r=14 cm and θ=90°
Area of the sector=πr2θ360
                             =227×14×14×90360 cm2=154 cm2

Page No 739:

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.

Answer:

Area of the shaded region = (Area of the sector with r=7 cm and θ=30°- (Area of the sector with r=3.5 cm and θ=30°)
                                          =227×7×7×30360-227×3.5×3.5×30360 cm2=776-7724 cm2=778 cm2=9.625 cm2



Page No 740:

Question 11:

A wire when bent in the form of an equilateral triangle encloses an area of 1213cm2. If the same wire is bent into the form of a circle, what will be the area of the circle?

Answer:

Let a cm be the side of the equilateral triangle.
Now,
Area of the equilateral triangle=34a2
We have:
34a2=1213a24=121a2=484 a=22 

Perimeter of the triangle = Circumference of the circle
Perimeter of the triangle = (22 + 22 + 22) cm
                                        = 66 cm
Now, let r cm be the radius of the circle.
We know:
Circumference of the circle=2πr
2×227×r=66r=66×744 cmr=212 cm
Also,
Area of the circle=πr2
                            =227×212×212 cm2=6932 cm2=346.5 cm2

Page No 740:

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.

Answer:

Distance covered in 1 revolution=π×d
                                                    =227×84 cm=264 cm
Distance covered in 1 second=5×264 cm
                                               = 1320 cm
Distance covered in 1 hour=60×60×1320 cm

                                           =4752000 cm=47520001000×100 km=47.52 km

Page No 740:

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) quadrant OACB (ii) the shaded region.

Answer:

(i) Area of the quadrant OACB =14×227×3.5×3.5 cm2
                                                  =14×227×3510×3510 cm2=778 cm2=9.625 cm2

(ii) Area of the shaded region = Area of the quadrant OACB - Area of AOD
                                                =778-12×3.5×2 cm2=778-3510 cm2=498 cm2=6.125 cm2

Page No 740:

Question 14:

In the given figure, ABCD is a square each of whose sides measures 28 cm. Find the area of the shaded region.

Answer:

Let r be the radius of the circle.
Thus, we have:
 r=282 cm
   =14 cm
Now,
Area of the shaded region = (Area of the square ABCD- 4(Area of the sector where r=14 cm and θ=90°)
                                          =28×28-4227×14×14×90360 cm2=784-4154 cm2=784-616 cm2=168 cm2 

Page No 740:

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.

Answer:


Draw ODBC.
Because ABC is equilateral, A=B=C=60°.
Thus, we have:

OBD=30°ODOB=sin 30°ODOB=12OD=12×4 cm    OB=radiusOD=2 cm

BD2=OB2-OD2   By Pythagoras' theoremBD2=42-22 cm2BD2=16-4 cm2BD2=12 cm2BD=23 cm

Also,
BC=2×BD    =2×23 cm    =43 cm
∴ Area of the shaded region = (Area of the circle) - (Area of  ABC)
                                              =3.14×4×4-34×43×43 cm2=50.24-12×1.73 cm2=50.24-20.76 cm2 =29.48  cm2

Page No 740:

Question 16:

The minute hand of a clock is 7.5 cm long. Find the area of the face of the clock described by the minute hand in 56 minutes.

Answer:

The length of minute hand of clock = 7.5 cm.
The angle made by minute hand in 60 minutes = 360°.
The angle made by minute hand in 1 minute = 6°.
The angle made by minute hand in 56 minutes = 56×6°=336°.
So, the area of clock described by minute hand in 56 minutes = area of sector with angle 336°=πr2×θ360°=227×7.5×7.5×336°360°=165 cm2.

Page No 740:

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.

Answer:

Let r m and R m be the inner and outer boundaries, respectively.
Thus, we have:

2πr=352 r=3522π Also,2πR=396R=3962π

Width of the track=R-r
                             =3962π-3522π m=12π396-352 m=12×722×44 m=7 m

Area of the track=πR2-r2
                            =πR+rR-r=π3962π+3522π×3962π-3522π m2=π×7482π×7 m2=7482×7 m2=2618 m2

Page No 740:

Question 18:

A chord of a circle of radius 30 cm makes an angle of 60° at the centre of the circle. Find the area of the minor and major segments.

Answer:


Let AB be the chord of a circle with centre O and radius 30 cm such that AOB=60°.
Area of the sector OACBO =πr2θ360
                                           =3.14×30×30×60360 cm2=471 cm2
Area of OAB=12r2 sin θ
                     =12×30×30×sin 60° cm2=225×1.732 cm2=389.7 cm2

Area of the minor segment = (Area of the sector OACBO- (Area of OAB)
                                            =471-389.7 cm2=81.3 cm2

Area of the major segment = (Area of the circle) - (Area of the minor segment)
                                            =3.14×30×30-81.3 cm2=2744.7 cm2

Page No 740:

Question 19:

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

Answer:


Let r be the radius of the circle.
Thus, we have:
r=502 m
 = 25 m
Area left ungrazed = (Area of the square) - 4(Area of the sector where r=25 m and θ=90°)

                               =50×50-43.14×25×25×90360 m2=2500-4×1962.54 m2=2500-1962.5 m2=537.5 m2



Page No 741:

Question 20:

A square tank has an area of 1600 cm2. There are four semicircular plots around it. Find the cost of turfing the plots at Rs 12.50 per m2.

Answer:

Let a m be the side of the square.
Area of the square=a2
Thus, we have:
a2=1600a=40 
Area of the plots = 4(Area of the semicircle of radius 20 m)
                           =412πr2 m2=412×3.14×20×20 m2=2512 m2

∴ Cost of turfing the plots at 12.50 per m2=Rs 2512×12.50
                                                               = Rs 31400



View NCERT Solutions for all chapters of Class 10