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Page No 712:

Question 1:

Find the circumference and the area of a circle of diameter 35 cm.

Answer:

Given:
Diameter, d = 35 cm
Thus, we have:
r=d2r=352 cm

Now,
Circumference of the circle = 2πr
                                             =2×227×352=110 cm

Area of the circle = πr2
                             =352×352×227=962.5 cm2

Page No 712:

Question 2:

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 712:

Question 3:

The area of a circle is 301.84 cm2. Find its circumference.

Answer:

Area of the circle = 301.84 cm2
We know:
Area of a circle=πr2

301.84=227r2301.84×722=r2r2=96.04r=9.8 cm

Now,
Circumference of the circle = 2πr
                                             =2×227×9.8=61.6 cm

Page No 712:

Question 4:

The circumference of a circle exceeds its diameter by 16.8 cm. Find the circumference of the circle.

Answer:

Circumference of a circle = 2πr
Diameter, d = 2r
Thus, we have:
d+16.8 = 2πr
2r +16 .8= 2πr2πr -2r =16.82rπ-1=16.8r=16.82π-1r=16.82227-1r=16.8222-77r=16.8×744-14r=117.630r=3.92 cm

Now,
Circumference of the circle = 2×227×3.92

                                             = 24.64 cm

Page No 712:

Question 5:

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

Answer:

Let the radius of the circle be r cm.
According to the question, we have:

2πr-r=37
r2π-1=37r=372×227-1=37×737=7 cm

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

Page No 712:

Question 6:

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 712:

Question 7:

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 712:

Question 8:

The length of a chain used as the boundary of a semicircular park is 90 m. Find the area of the park.

Answer:

Let the radius of the park be r m.
Thus, we have:
 πr+2r=90r(π+2)=90

 r=90π+2r=90227+2r=90×736r=17.5

Now,
Area of the park = 12πr2
                           =12×227×17.5×17.5=481.25 m2
​                   

Page No 712:

Question 9:

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 712:

Question 10:

The areas of two concentric circles are 962.5 cm2 and 1386 cm2. Find the width of the ring.

Answer:

Let the radii of the bigger and smaller circles be R cm and r cm, respectively.

Now,
Area of the bigger circle = 1386 cm2
Area = πR2
1386=227×R21386×722=R2R2=441R=21

Area of the smaller circle = 962.5 cm2
Area = πr2
962.5=227r2r2=962.5×722r2=306.25r=17.5

∴ Width of the ring = R -r
                                =21-17.5=3.5 cm

Page No 712:

Question 11:

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 712:

Question 12:

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

Answer:

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
                  



Page No 713:

Question 13:

The inner circumference of a circular track is 440 m, and the track is 14 m wide. Calculate the cost of levelling the track at 25 paise/m2. Also, find the cost of fencing the outer boundary of the track at Rs 5 per metre.

Answer:

Let the radius of the inner circle be r m.
Now,
Inner circumference = 440 m
2πr=440
2×227×r=440r=440×744r=70 

We know that the track is 14 m wide.
∴ Outer radius (R) = (70 + 14) = 84 m
                
Area of the track=π(R2- r2)= π 842-702=227×7056-4900= 6776m2

Cost of levelling at 25 paise per square metre =6676×25=169400 paise
Or,
Rs 169400100= Rs 1694
​
∴ Outer circumference = 2πR
=2×227×84=528 m

Rate of fencing = 5 per metre
∴ Total cost of fencing =528×5=Rs 2640
                                      

Page No 713:

Question 14:

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 713:

Question 15:

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 713:

Question 16:

In the given figure, AB is a diameter of a circle with centre O and OA = 7 cm. Find the area of the shaded region.

Answer:

AB is the diameter of the circle.
Here,
OA = 7 cm
OB = 7 cm
OA is the radius of the circle.
∴ OA = OB = OC =OD
Also,
OA is the diameter of the smaller circle.
∴ Radius of the smaller circle = OA2= 3.5 cm
Area of the smaller circle having diameter AO = πr2
                                                                           =227×3.5×3.5=38.5 cm2

Area of the triangle CBD = 12×b×h

                                            =12×14×7=49 cm2

Area of the semicircle having diameter CD = 12πr2
                                                                     =12×227×7×7=77 cm2

Now,
Area of the shaded region = Area of the semicircle - Area of the triangle CBD
                                           = 77 - 49
                                           =28 cm2

Also,
Area of the full-shaded region = 28 + 38.5
                                                 = 66.5 sq cm

Page No 713:

Question 17:

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 713:

Question 18:

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 713:

Question 19:

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 714:

Question 20:

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 714:

Question 21:

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 714:

Question 22:

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 714:

Question 23:

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 714:

Question 24:

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 714:

Question 25:

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 714:

Question 26:

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 714:

Question 27:

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 714:

Question 28:

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

Answer:

Diameter of the wheel = 140 cm
Radius = 70 cm
Circumference = 2πr
                         =2×227×70=440 cm
Speed of the wheel = 72.6 km per hour
Distance covered by the wheel in 1 minute = 72.6×1000×10060 = 121000 cm
​Number of revolutions made by the wheel in 1 minute =Total distance covered Circumference 
                                                                                            
                                                                                       =121000440=275

Hence, the wheel makes 275 revolutions per minute.

Page No 714:

Question 29:

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

Answer:

Circumference of the circle = 2πr
                                            22=2×227×rr=22×72×22r=3.5 cm
Area of the circle = πr2
                             =227×3.5×3.5=38.5 cm2

Area of the quadrant = 38.54=9.625 cm2

Page No 714:

Question 30:

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 714:

Question 31:

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 714:

Question 32:

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 715:

Question 33:

Find the area of the shaded region shown in the given figure. The four corners are circle quadrants, and at the centre, there is a circle.

Answer:

Area of the square = (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 715:

Question 34:

A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radius 3.5 m have been cut. Find the area of the remaining part.

Answer:

Area of the quadrant of one circle = 14×πr2
                                                        =14×227×3.5×3.5=9.625 m2

Area of the quadrants of four circles = 4×9.625=38.5 cm2
Area of the rectangle = 15×20=300 m2

∴ Area of the remaining part = Area of the rectangle - Area of the quadrants of four circles
                                                = 300 - 38.5
                                                = 261.5 m2

Page No 715:

Question 35:

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 715:

Question 36:

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

Answer:

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

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

Area of the shaded region = Area of the AOD - Area of the sector AOB

                                           =9.625-3.50 = 6.125 cm2

Page No 715:

Question 37:

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 715:

Question 38:

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 715:

Question 39:

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 716:

Question 40:

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 716:

Question 41:

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 716:

Question 42:

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 1.25 per m2.

Answer:


Area of the square = Side2
1600=Side2Side = 40 m

Each piece of land is a semicircle, area = 12×3.14×20×20=22×10×207=628 m2

Total area = 4×628.57 = 2512 m2

Cost of turfing at Rs 1.25 per sq. m = 1.25×2512=Rs 3140

Page No 716:

Question 43:

A lawn is rectangular in the middle, and it has semicircular portions along the shorter sides of the rectangle. The rectangular portion measures 50 m by 35 m. Find the area of the lawn.

Answer:


Area of the rectangle = 50×35=1750 m2
Radius of the circle = 17.5 m
Area of the semicircle = 12πr2
          
                                    =12×227×17.5×17.5=481.25 m2

∴ Total area of the lawn = Area of the circle + Area of the two semicircles
                                        =1750+481.25+481.25=2712.50 m2

Page No 716:

Question 44:

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 716:

Question 45:

In the given figure ∆ABC is right-angled at A, with AB = 6 cm and AC = 8 cm. A circle with centre O has been inscribed inside the triangle. Find the value of r, the radius of the inscribed circle.

Answer:


Join OC, OA and OB. This gives triangles OAC, OAB and OCB.

Consider CAB. Using Pythagoras' theorem, we have:

CA2+AB2=BC282+62=64+36=100BC =10 cm

Also,
ArABC=12×6×8=24 cm2

Further, we have:

ArABC=ArOAB+ArOCB+ArOCA24=12r×6+12r×8+12r×1024=12r6+8+1024=12×r×2424=12r 12r = 24  r=2 cm

Page No 716:

Question 46:

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 716:

Question 47:

Find the area of the region ABCDEFA shown in the given figure, given that ABDE is a square of side 10 cm, BCD is a semicircle with BD as diameter, EF = 8 cm, AF = 6 cm and ∠AFE = 90°.

Answer:


Join AE.
Now, AEDB is a square.
Area of the square = Side2 = 102=100 sq. cm

Area of semi-circle = 12πr2=12×3.14×5×5=39.25 cm2

Area of EFA = 12×B×H=12×8×6=24 cm2

Area of the region ABCDEFA = Area of the square + Area of the semicircle - Area of EFA
                                                = 100 + 39.25 - 24                                      
                                                = 115.25 sq. cm



Page No 717:

Question 48:

In the given figure, ABCD is a square of side 14 cm. Find the area of the shaded region in the given figure.

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 717:

Question 49:

Find the perimeter of the shaded region, where ADC, AEB and BFC are semicircles on diameters AC, AB and BC respectively.

Answer:

We know:
Perimeter (circumference of a circle) = 2πr
Perimeter of a semicircular arc = πr

Now,
For the arc ADC, radius is 2.1 cm.
∴ Perimeter of the arc ADC = πr=2.1π cm

For the arc AEB, radius is 1.4 cm.
∴ ​Perimeter of the arc AEB = πr=1.4π cm

For the arc BFC, radius is 0.7 cm.
∴ Perimeter of the arc BFC = πr=0.7π cm

Thus, we have:
Perimeter of the shaded region = 2.1π+1.4π+0.7π
                                                   =4.2π cm
                                                   =4.2×227=13.2 cm

Page No 717:

Question 50:

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 717:

Question 51:

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 717:

Question 52:

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 718:

Question 53:

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 718:

Question 54:

The area of an equilateral triangle is 493 cm2. Taking each angular point as centre, circle as 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.

Answer:


Let the side of the equilateral triangle be a cm.

Thus, we have:

34a2=493a2=196a=14

The radius of each circle is 7 cm. The angle at the vertex of each triangle is 60°.

Area of the sector with angle 60° and radius 7 cm:
60360×227×7×7=16×22×7=25.67 cm2

There are three such sectors.
Total area = 25.6666667×3=77 cm2

Area not included in the circles = Area of the triangle - Area of the three sectors
                                                   =34×14×14-77=7.77 cm2

Page No 718:

Question 55:

In the given figure, ∆ABC is a right-angled triangle with ∠B = 90°, AB = 48 cm and BC = 14 cm. With AC as diameter a semicircle is drawn and with BC as radius, A quadrant of a circle is drawn. Find the area of the shaded region.

Answer:

Consider the triangle ABC.

AC=AB2+BC2 =482+142=2304+196=2500=50 cm

Now,

Area of the shaded part =AreaABC+Areasemicircle having diameter AC-Areasector BCD=12×14×48+12×π×25×25-90360×π×14×14=336+π225×25-14×142=336+2214625-98=1164.14 cm2

Page No 718:

Question 56:

Calculate the area other than the area common between two quadrants of circles of radius 16 cm each, which is shown as the shaded region in the given figure.

Answer:

Area of the shaded region=Arsquare ABCD-Arquadrant ABED+Arsquare ABCD-Arquadrant CDFB=162-14πr2+162-14πr2=512-12π×16×16=512-402.3=109.7 cm2

Page No 718:

Question 57:

In a circular table cover of radius 70 cm, a design is formed leaving an equilateral ∆ABC in the middle, as shown in the figure. Find the total area of the design.

Answer:

ABC is equilateral.
Thus, we have:

A=B=C=60°

OB bisects B. Therefore, OBD=30°.

OBD is a right-angled triangle.

We have:

cos 30=BDBOBD=32BOBD=32×70BD=353

Now,
BC=2BD=353×2=703=AB=AC


Area of the ABC=34×703×703=4900×334=6357.75 cm2

Area of the circle = 227×70×70=15400 cm2

Area of the shaded part = Area of the circle - Area of the triangle
                                       =15400-6357.75=9042.25 cm2

Page No 718:

Question 58:

Find the area of the sector of a circle of radius 14 cm with central angle 45°.

Answer:

Area of the sector = θ360×πr2
                          
                              =45360×227×14×14=18×22×2×14=77 cm2



Page No 719:

Question 59:

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 719:

Question 60:

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 719:

Question 61:

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 719:

Question 62:

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 719:

Question 63:

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 719:

Question 64:

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 719:

Question 65:

The circumference of a circle is 88 cm. Find the area of the sector whose central angle is 72°.

Answer:

Given:
Circumference of the circle = 88 cm
θ=72
Area of the sector = ?

We know:c=2πr88=2×227×rr=88×744r=14 cm

Area of the sector = πr2θ360

                              =227×14×14×72360=123.2 cm2

Page No 719:

Question 66:

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 719:

Question 67:

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 719:

Question 68:

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 the sector with central angle 150°Area of the circle=π×62×150360π×62=150360=512So, the sector is 512 of the whole circle.Required ratio = Area of the sector with central angle 90°:Area of the sector with central angle 120°:Area of the sector with with central angle 150°

= 36π×90360:36π×120360:36π×150360=14:13:112=3:4:5

Page No 719:

Question 69:

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 719:

Question 70:

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 719:

Question 71:

The radius of a circle with centre O is 6 cm. Two radii OA and OB are drawn at right angles to each other. Find the areas of the minor and major segments.

Answer:


The triangle OAB is a right isosceles triangle.

Area of triangle OAB = 12×6×6=18 cm2

Now,
Area of the minor segment:
Area of the sector - Area of the triangle90360×π×36-18=10.26 cm2

Area of the major segment:
Area of the circle - Area of the minor segment  π×36-9π-18=27π+18=102.78 cm2

Page No 719:

Question 72:

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 719:

Question 73:

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 719:

Question 74:

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 719:

Question 75:

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 720:

Question 76:

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 726:

Question 1:

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 727:

Question 2:

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 727:

Question 3:

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 727:

Question 4:

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 727:

Question 5:

The circumferences of two circles are in the ratio 2 : 3. The ratio between their areas is
(a) 2 : 3
(b) 4 : 9
(c) 9 : 4
(d) none of these

Answer:

(b) 4:9
Let r1 cm and r2 cm be the radii of the two circles.
Thus, we have:
Perimeter of the first circle=2πr1
And,
Perimeter of the second circle=2πr2
Now,
2πr12πr2=23r1r2=23
Also,
Area of the first circle=πr12
And,
Area of the second circle=πr22
Thus, we have:
πr12πr22=r12r22
        
       =2232    r1r2=23=49=4:9

Hence, the ratio of the areas of the two circles is 4:9.

Page No 727:

Question 6:

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 727:

Question 7:

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 727:

Question 8:

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 727:

Question 9:

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

Answer:

(a) 2:3
Let r1 and r2 be the radii of the two circles.
Now,
Area of the first circle=πr12
And,
Area of the second circle=πr22
Thus, we have:
πr12πr22==49r12r22=49
Also,
Perimeter of the first circle=2πr1
And,
Perimeter of the second circle =2πr2
Thus, we have:
   2πr12πr2=r1r2r1r2=23   Since r12r22=49
         =2:3

Hence, the ratio of their circumferences is 2:3.

Page No 727:

Question 10:

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 727:

Question 11:

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 727:

Question 12:

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 727:

Question 13:

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 728:

Question 14:

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 728:

Question 15:

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 728:

Question 16:

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 728:

Question 17:

If the perimeter of a square is equal to the circumference of a circle, then the ratio of their areas when π=227 is.
(a) 14 : 11
(b) 11 : 14
(c) 22 : 7
(d) 7 : 22

Answer:

(b) 11:14
Let P be the perimeter of the square.
Now,
Each side of the square=P4
Let r be the radius of the circle.
We know:
Circumference of the circle=2πr
Now,
2πr=Pr=P4π
∴ Area of the square=P42
                                  =P216

Also,
Area of the circle =πr2
                             =π×P2π2=π×P24π2=P24π
∴ Required ratio=P216×4πP2
                           =π4=227×4=1114

                           =11:14

Page No 728:

Question 18:

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 728:

Question 19:

The area of the sector of a circle of radius R making a central angle of x° is
(a) x180×2πR
(b) x360×2πR
(c) x180×πR2
(d) x360×πR2

Answer:

(d) x360×πR2

Page No 728:

Question 20:

The length of an arc of the sector of a circle of radius R making a central angle of x° is
(a) 2πRx180
(b) 2πRx360
(c) πR2x180
(d) πR2x360

Answer:

(b) 2πRx360

Page No 728:

Question 21:

A chord of a circle of radius 28 cm subtends an angle of 45° at the centre of the circle. The area of the minor segment is
(a) 30.256 cm2
(b) 30.356 cm2
(c) 30.456 cm2
(d) 30.856 cm2

Answer:

(d) 30.856 cm2
Let r be the radius of the circle and θ be the angle.
∴ Area of the minor segment =πr2θ360-12rsin θ cm2

                                              =227×28×28×45360-12×28×28×sin 45° cm2=308-392×0.707 cm2=308-277.144 cm2=30.856 cm2

Page No 728:

Question 22:

A chord of a circle subtends an angle of 60° at the centre of the circle. If the length of the chord is 10 cm, then the area of the major segment is
(a) 305 cm2
(b) 295 cm2
(c) 310 cm2
(d) 335 cm2

Answer:

(a) 305 cm2
Let AB be the chord of a circle with centre O.
Now,
OA = OB = AB = 10 cm
Thus, we have:
r=10 cm and θ=60°
Area of the minor segment=πr2θ360-12r2sinθcm2
                                           =3.14×10×10×60360-12×10×10×sin 60° cm2=1573-50×32 cm2=1573-25×1.732 cm2=1573-43310 cm2=27130 cm2=9 cm2


Area of the major segment
=Area of the circle - Area of the minor segment

                                            =π×10×10-9 cm2=3.14×10×10-9 cm2=314-9 cm2=305 cm2

Page No 728:

Question 23:

The perimeter of a sector of a circle with central angle 90° is 25 cm. The area of the minor segment of the circle is
(a) 14 cm2
(b) 16 cm2
(c) 18 cm2
(d) 24 cm2

Answer:

(a) 14 cm2
Let r be the radius of the circle and θ be the angle.
Now,
Perimeter of the sector=2r+2πrθ360
                                     =2r+2×227×r×90360=2r+11r7=25r7
Also,

25r7=25r=25×725r=7 cm
Area of the minor segment=πr2θ360-12r2sinθcm2
                                           =227×7×7×90360-12×7×7×sin 90° cm2=772-492 cm2=282 cm2=14 cm2



Page No 729:

Question 24:

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 729:

Question 25:

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 729:

Question 26:

Match the following columns:

Column I Column II
(a) The circumference of a circle is
44 cm. The area of this circle is
.........cm2.
(p) 1936
(b) A wire is looped in the form of a
circle of radius 28 cm. It is bent
into a square. The area of the
square is .......cm2
(q) 10
(c) The radii of two circles are 9 cm
and 19 cm respectively. The radius
of the circle whose circumference
is equal to the sum of the
circumferences of the given
circles is .........cm2
(r) 154
(d) The radii of two circles are 8 cm
are 6 cm respectively. The radius
of the circle having its area of the
given circles is .........cm2
(s) 28

Answer:

(a) Let r be the radius of the circle.
Now,
Circumference of the circle=2πr
We have:

2πr=442×227×r=44r=44×744 cmr=7 cm

Also,
Area of the circle=πr2

                            =227×7×7 cm2=154 cm2

∴ ar 

(b) Let r be the radius of the circle.
Length of the circle = Circumference of the circle
    ​                            =2πr=2×227×28 cm=176 cm
Perimeter of the square = Length of the wire
∴ Side of the square =1764 cm
                                  = 44 cm
Area of the square=44×44 cm2
                              =1936 cm2
bp

(c) Let r be the radius of the circle whose circumference is equal to the sum of the circumferences of the given circles.
Thus, we have:
2πr=2π×9+2π×19 cm2πr=56π cm2r=56 cmr=562 cmr=28 cm
∴ cs

(d) Let r be the radius of the circle with area equal to the sum of the areas of the given circles.
Thus, we have:
πr2=π×82+π×62 cm2πr2=64π+36π cm2πr2=100π cm2r2=100 cm2r=10 cm
dq

Page No 729:

Question 27:

Match the following columns:

Column I Column II
(a) In a circle of radius 6 cm, the
angle of a sector is 60°. The area
of the sector is ..........cm2
(p) 25π6
(b) In a circle with centre O and
radius 5 cm, AB is a chord of
length 53 cm. The area of sector
OAB is .........cm2
(q) 44.8
(c) A chord of a circle of radius
14 cm subtends a right angle at
the centre. The area of the
sector is ........cm2
(r) 154
(d) The perimeter of the sector of
a circle of radius 5.6 cm is 27.2 cm.
The area of the sector is ..........cm2
(s) 1867

Answer:


(a) Area of the sector=πr2θ360
                                  =227×6×6×60360 cm2=1327 cm2=1867 cm2
∴ as

(b) Draw OD such that ODAB.
Now,
DB=12AB
    =532 cm
From the right ODB, we have:
    OD2=OB2-DB2OD2=53-5322 cm2OD2=25-754 cm2OD2=254 cm2OD=52 cm

∴ Area of AOB=12×AB×OD

                         =12×53×52 cm2=2534 cm2

Also,
Area of AOB=12×5×5×sin θ
                     =252 sin θ
We have:

252 sin θ=2534sin θ=2534×225sin θ=32θ=60°

Area of the sector OAPB =πr2θ360
                                        =π×5×5×60360 cm2=25π6 cm2
bp

(c) Area of the sector=πr2θ360
                                  =227×14×14×90360 cm2=154 cm2
∴ cr

(d) Let O be the centre of the circle of radius 5.6 cm and OACB be its sector with perimeter 27.2 cm.
Now,
OA+OB+arc AB=27.25.6+5.6+arc AB=27.2arc AB=16 cm

Area of the sector OACBO =12×Radius×Arc square units
                                           =125.6×16 cm2=44.8 cm2

∴ dq



Page No 730:

Question 28:

Match the following columns:

Column I Column II
(a) If the perimeter of a semi-
circular protractor is 66 cm,
then its radius is.........cm.
(p) 35
(b) Each wheel of a car makes
450 complete revolution in
covering 0.99 km. The radius
of each wheel is ........m.
(q) 32
(c) A bicycle wheel makes 5000
revolutions in covering 11 km.
The diameter of its wheel is
.........cm2
(r) 1256
(d) The given figure is a sector
of a circle of radius 10.5 cm.
Perimeter of the sector is
............cm2
figure
(s) 70

Answer:

(a)
Let the radius of the protractor be r cm.
Then, perimeter=πr+2r
                         =π+2r=227+2r cm=367r cm

Therefore,

   367r=66r=66×736 r=776 cmr=1256 
Hence, ar

(b)
Distance covered in 1 revolution=0.99×1000450 m
                                                    =990450 m=115 m
Let r be the radius of the wheel. Then,
Circumference of the wheel =2πr
                                           2×227×r=115r=115×744r=720 mr=720×100 cmr=35 cm
Hence, bp

(c)
Distance covered in 1 revolution=11×10005000 m
                                                    =115 m
Let the diameter of the wheel be d m.
Then,
    πd=115227×d=115d=115×722 d=710 
d=710 md=710×100 cmd=70 cm

Hence, cs

(d)
Let r be the radius of the arc. Then,
Arc length=2πrθ360
                =2×227×10.5×60360 cm=11 cm
Therefore, Perimeter = OA+OB+arc AB
                                  = (10.5 + 10.5 +arc AB) cm
                                  = 32 cm
Hence, dq



Page No 731:

Question 29:

Assertion (A)
The area of the quadrant of a circle having a circumference of 22 cm is 958cm2.

Reason (R)
The area of a sector of a circle of radius r with central angle x° is x×πr2360.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) Reason (R) true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Assertion (A):
Let r be the radius of the circle.
Now,
Circumference of the circle =2πr
We have:
2πr=222×227×r=22r=22×744 cmr=72 cm
Area of the quadrant=90°360°πr2
                                  =14×227×72×72 cm2=778 cm2=958 cm2
Hence, assertion (A) is true.

Reason (R):
The given statement is true.

Assertion (A) is true and reason (R) is the correct explanation of assertion (A).

Page No 731:

Question 30:

Assertion (A)
An arc of a circle of length 5π cm bounds a sector whose area is 20π cm2. Then, the radius of the circle is 4 cm.

Reason (R)
A chord of a circle of radius 12 cm subtends an angle of 60° at the centre of the circle. The area of the minor segment of the circle is 13.08 cm2.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) Reason (R) true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(d) Assertion (A) is false and reason (R) is true.
Assertion (A):
Let r be the radius of the circle.
We have:
Arc length=πrθ180
Now,

 πrθ180=5πrθ180=5rθ=5×180

Area of the sector=πr2θ360
Thus, we have:
πr2θ360=20πr2θ360=20

r2θ=20×360

Now,
r2θrθ=20×3605×180
r = 8 cm

Hence, assertion (A) is false.

Reason (R):
Let r be the radius of the circle.
Now,
Area of the minor segment=πr2θ360-12r2sin θ
=3.14×12×12×60360-12×12×12×sin 60° cm2=75.36-36×1.73 cm2=75.36-62.28 cm2=13.08 cm2

Hence, reason (R) is true.

Page No 731:

Question 31:

Assertion (A)
If the circumferences of two circles are in the ratio 2 : 3, then the ratio of their areas is 4 : 9.

Reason (R)
The circumference of a circle of radius r is 2πr.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) Reason (R) true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(b) Both assertion (A) and reason (R) are true, but reason (R) is not the correct explanation of assertion (A).

Assertion (A):
Let r1 and r2 be the radii of two circles.
Now,
Circumference of the first circle=2πr1
Circumference of the second circle=2πr2
Thus, we have:
2πr12πr2=23r1r2=23
Also,
Area of the first circle=πr12
Area of the second circle=πr22
Thus, we have:
πr12πr22=r12r22
        
       =2232    r1r2=23=49=4:9

Hence, the ratio of their areas is 4:9.
Hence, assertion (A) is true.

Reason (R):
The given statement is true.
Hence, both assertion (A) and reason (R) are true, but reason (R) is not the correct explanation of assertion (A).



Page No 738:

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 738:

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 738:

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 738:

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 738:

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 738:

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 738:

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 738:

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 738:

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 738:

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 739:

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 739:

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 739:

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 739:

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 739:

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 739:

Question 16:

The minute hand of a clock is 12 cm long. Find the area of the face of the clock described by the minute hand in 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 described by the minute hand in 35 minutes = Area of the sector where r=12 cm and θ=210° 
                                                                                                   =πr2θ360=227×12×12×210360 cm2=264 cm2

Page No 739:

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 739:

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 739:

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 740:

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



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