RD Sharma 2022 Mcqs Solutions for Class 10 Maths Chapter 13 Areas Related To Circles are provided here with simple step-by-step explanations. These solutions for Areas Related To Circles are extremely popular among class 10 students for Maths Areas Related To Circles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2022 Mcqs Book of class 10 Maths Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2022 Mcqs Solutions. All RD Sharma 2022 Mcqs Solutions for class 10 Maths are prepared by experts and are 100% accurate.

Page No 191:

Question 1:

If the circumference and the area of a circle are numerically equal, then diameter of the circle is

(a) π2

(b) 2π

(c) 2

(d) 4

Answer:

We have given that circumference and area of a circle are numerically equal.

Let it be x.

Let r be the radius of the circle, therefore, circumference of the circle is and area of the circle will be.

Therefore, from the given condition we have,

………(1)

………(2)

Therefore, from equation (1) get . Now we will substitute this value in equation (2) we get,

Simplifying further we get,

Cancelling x we get,

Now we will cancel

………(3)

Now we will multiply both sides of the equation (3) by we get,

We can rewrite this equation as given below,

Comparing equation (4) with equation (1) we get r = 2.

Therefore, radius of the circle is 2. We know that diameter of the circle is twice the radius of the circle.

Therefore, diameter of the circle is.

Hence, option (d) is correct.

Page No 191:

Question 2:

If the difference between the circumference and radius of a circle is 37 cm, then using π = 227, the circumference (in cm) of the circle is

(a) 154

(b) 44

(c) 14

(d) 7

Answer:

We know that circumference; C of the circle with radius r is equal to.

We have given difference between circumference and radius of the circle that is 37 cm.

Substituting we get,

Dividing both sides of the equation by, we get,

Therefore, circumference of the circle will be 
2πr=2×227×7=44 cm2

Hence, the correct choice is (b).

Page No 191:

Question 3:

A wire can be bent in the form of a circle of radius 56 cm. If it is bent in the form fo a square, then its area will be

(a) 3520 cm2

(b) 6400 cm2

(c) 7744 cm2

(d) 8800 cm2

Answer:

We have given that a wire is bent in the form of circle of radius 56 cm. If we bent the same wire in the form of square of side a cm, the perimeter of the wire will not change.

perimeter of the circle = perimeter of the square

We know that r = 56 cm.

Now we will substitute the value of r in the equation,

………(1)

Dividing both sides of the equation by 4 we get,

Now we obtained side of the square. Now we can calculate the area of the square as given below.

Hence, the area of the square is .

Therefore, the correct answer is (c).

Page No 191:

Question 4:

If a wire is bent into the shape of a square, then the area of the square is 81 cm2 . When wire is bent into a semi-circular shape, then the  area of the semi-circle will be

(a) 22 cm2

(b) 44 cm2

(c) 77 cm2

(d) 154 cm2

Answer:

We have given that a wire is bent in the form of square of side a cm such that the area of the square is . If we bent the same wire in the form of a semicircle with radius r cm, the perimeter of the wire will not change.

perimeter of the square = perimeter of semi circle

………(1)

We know that area of the square = .

Now we will substitute the value of a in the equation (1),

Now we will substitute.

Multiplying both sides of the equation by 7 we get,

Now we will divide both sides of the equation by 36 we get, r = 7

Therefore, radius of the semi circle is 7cm.

Now we will find the area of the semicircle.

Area of the semicircle

Therefore, area of the semicircle is .

Hence the correct answer is option (c).

Page No 191:

Question 5:

A circular park has a path of uniform width around it. The difference between the outer and inner circumferences of the circular path is 132 m. Its width is

(a) 20 m

(b) 21 m

(c) 22 m

(d) 24 m

Answer:

Let OA = r be the radius of the inner circle and OB = r be the radius of the outer circle.

Therefore, circumference of the inner circle and circumference of the outer circle

Here we have to find the width of the circular park that is we have to find .

We have given the difference between the circumferences of outer circle and inner circle.

Substituting we get,

Now we will multiply both sides of the equation by .

Therefore, the width is.

Hence the correct answer is option (b).

Page No 191:

Question 6:

The radius of a wheel is 0.25 m. The number of revolutions it will make to travel a distance of 11 km will be

(a) 2800  

(b) 4000

(c) 5500

(d) 7000

Answer:

We have given the radius of the wheel that is 0.25 cm.

We know that distance covered by the wheel in one revolution.

Distance covered by the wheel in one revolution is equal to the circumference of the wheel.

………(1)

Distance moved is given as 11 km so we will first convert it to m.

11 km = 11000 m

Now we will substitute the values in equation (1),

Now we will substitute.

Simplifying equation (1) we get,

Therefore, it will make revolutions to travel a distance of 11 km.

Hence, the correct answer is option (d).

Page No 191:

Question 7:

The ratio of the outer and inner perimeters of a circular path is 23 : 22. If the path is 5 metres wide, the diameter of the inner circle is

(a) 55 m

(b) 110 m

(c) 220 m

(d) 230 m

Answer:

Let OA = r be the radius of the inner circle and OB = r be the radius of the outer circle.

Therefore, circumference of the inner circle and circumference of the outer circle

Here we have to find the diameter of the inner circle.

We have given the ratio of outer and inner perimeters of a circular path.

Simplifying the above equation we get,

……..(1)

We have also given that the path is 5 meters wide, that is we have given

……..(2)

We are asked to find the diameter of the inner circle hence, we will eliminate r using equations (1) and (2) for that we will multiply equation (2) by 22.

……..(3)

Subtracting equation (1) from equation (3) we get,

Therefore, radius of the inner circle is 110 meters.

Therefore, diameter of the inner circle meters

Therefore, diameter of the inner circle is .

Hence, the correct answer is option (c).

Page No 191:

Question 8:

The circumference of a circle is 100 m. The side of a square inscribed in the circle is

(a) 502

(b) 50π

(c) 502π

(d) 1002π

Answer:

We have given the circumference of the circle that is 100 cm. If d is the diameter of the circle, then its circumference will be.

We obtained diameter of the circle. Look at the figure, diameter of the circle is also the diagonal of the square ABCD.

We know that if we have diagonal of the circle we can calculate the side of the square, using the formula given below,

Substituting the value of diagonal we get,

Therefore, side of the inscribed square is.

Hence, the correct answer is option (d).

Page No 191:

Question 9:

The area of the incircle of an equilateral triangle of side 42 cm is

(a) 223 cm2

(b) 231 cm2

(c) 462 cm2

(d) 924 cm2

Answer:

Let ABC be the equilateral triangle such that AB = BC = CA = 42 cm. Also, let O be the centre and r be the radius of its incircle.

AB, BC and CA are tangents to the circle at M, N and P.

∴ OM = ON = OP = r

Area of ΔABC = Area (ΔOAB) + Area (ΔOBC) + Area (ΔOCA)

Area of the circle = 

Hence, the correct answer is option (c).



Page No 192:

Question 10:

The area of incircle of an equilateral triangle is 154 cm2 . The perimeter of the triangle is

(a) 71.5 cm

(b) 71.7 cm

(c) 72.3 cm

(d) 72.7 cm

Answer:

Area of incircle of equilateral triangle is 154

We have to find the perimeter of the triangle. So we will use area to get,

As triangle is equilateral so,

So,

So,

Therefore perimeter of the triangle is,

Therefore the answer is (d).

Page No 192:

Question 11:

If an arc of a circle forms 90° at the centre of the circle, then the ratio of its length to the circumference of the circle is
(a) 1 : 4
(b) 3 : 4
(c) 1 : 3
(d) 2 : 3

Answer:

Let r be the radius of the circle.

Length of the arc formed by 90° at the centre of the circle = 90°360°×2πr

Circumference of circle = 2πr

Ratio of its length to circumference = 90°360°×2πr2πr=14

Hence, the correct answer is option (a).

Page No 192:

Question 12:

The perimeter of a triangle is 30π cm and the circumference of its in circle is 88 cm. The area of the triangle is
(a) 70 cm2
(b) 140 cm2
(c) 660 cm2
(d) 420 cm2

Answer:

Perimeter of a triangle = 30π cm
Let the radius of the circle be r.



Circumference of in-circle = 88 cm
2πr=88r=44π

Area of ABC=Area of AOB+Area of AOC+Area of BOC=12×OF×AB+12×OE×AC+12×OD×BC=12×r×AB+BC+AC       OD=OE=OF=r=12×44π×30π=660 cm2

Hence, the correct answer is option (c).

Page No 192:

Question 13:

The area of a circle is 220 cm2. The area of ta square inscribed in it is

(a) 49 cm2

(b) 70 cm2

(c) 140 cm2

(d) 150 cm2

Answer:

Let BD be the diameter and diagonal of the circle and the square respectively.

We know that area of the circle

Area of the circle

Multiplying both sides of the equation by 7 we get,

Dividing both sides of the equation by 22 we get,

As we know that diagonal of the square is the diameter of the square.

………(1)

Substituting in equation (1) we get,

Area of the square

Therefore, area of the square is.

Hence, the correct answer is option (c).

Page No 192:

Question 14:

If the circumference of a circle increases from 4π to 8π, then its area is

(a) halved

(b) doubled

(c) tripled

(d) quadrupled

Answer:

Let the circumference

Therefore, area of the circle when radius of the circle is 2 can be calculated as below,

……..(1)

Now when circumference is, then the radius of the circle is calculated as below,

Therefore, area of the circle when radius of the circle is 2 can be calculated as below,

……..(2)

Therefore, from equation (1) and (2) we can say that its area is.

Hence, the correct answer is option (d).

Page No 192:

Question 15:

If the radius of a circle is diminished by 10%, then its area is diminished by

(a) 10%

(b) 19%

(c) 20%

(d) 36%

Answer:

Let x be the initial radius of the circle.

Therefore, its area is ……..(1)

It is given that the radius is diminished by 10%, therefore, its new radius is calculated as shown below,

Now we will find the percentage decreased in the area.

Therefore, its area is diminished by.

Hence, the correct answer is option (b).

Page No 192:

Question 16:

If the area of a square is same as the area of a circle, then the ratio of their perimeters, in terms of π, is

(a) π : 3

(b) 2 : π

(c) 3 : π

(d) π : 2

Answer:

We have given that area of a circle of radius r is equal to the area of a square of side a.

We have to find the ratio of the perimeters of circle and square.

………(1)

Now we will substitute in equation (1).

Therefore, ratio of their perimeters is.

Hence, the correct answer is (b).

Page No 192:

Question 17:

An arc of length 15.7 cm subtends a right angle at the centre of the circle. The radius of the circle is use π=227
(a) 20 cm
(b) 10 cm
(c) 15 cm
(d) 12 cm

Answer:

Let radius of the circle be r.

Length of arc=θ360°×2πr15.7=90°360°×2×227×rr=15.7×4×744r10 cm

Hence, the correct answer is option (b).

Page No 192:

Question 18:

The ratio of the areas of a circle and an equilateral triangle whose diameter and a side are respectively equal, is

(a) π:2

(b) π:3

(c) 3:π

(d) 2:π

Answer:

We are given that diameter and side of an equilateral triangle are equal.

Let d and a are the diameter and side of circle and equilateral triangle respectively.

We know that area of the circle

Area of the equilateral triangle

Now we will find the ratio of the areas of circle and equilateral triangle.

We know that radius is half of the diameter of the circle.

Now we will substitute in the above equation,

Therefore, ratio of the areas of circle and equilateral triangle is.

Hence, the correct answer is option (b).

Page No 192:

Question 19:

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  r12+r22
(a) >r2

(b) =r2

(c) <r2

(d) None of these

Answer:

We have given area of the circle of radius r1plus area of the circle of radius r2 is equal to the area of the circle of radius r.

Therefore, we have,

Cancelling, we get

Therefore, .

Hence, the correct answer is option (b).

Page No 192:

Question 20:

If the perimeter of a semi-circular protractor is 36 cm, then its diameter is

(a) 10 cm

(b) 12 cm

(c) 14 cm

(d) 16 cm

Answer:

We know that perimeter of a semi-circle of radius r ………(1)

We have given the perimeter of the semi-circle and we are asked to find the diameter of the semi-circle.

Therefore, substituting the perimeter of the semi-circle in equation (1) we get,

Multiplying both sides of the equation by 2 we get,

Substituting we get,

Now we will multiply both sides of the equation by 7.

Adding like terms we get,

Dividing both sides of the equation 72 we get,

Therefore, radius of the semi circle is 7cm.

Now we will find the diameter.

Hence, diameter of the semi-circle is.

Therefore, the correct answer is (c).

Page No 192:

Question 21:

The perimeter of the sector OAB shown in the following figure,  is

(a) 643 cm

(b) 26 cm

(c) 645 cm

(d) 19 cm

Answer:

We know that perimeter of a sector of radius r ………(1)

We have given sector angle and radius of the sector and we are asked to find perimeter of the sector OAB.

Therefore, substituting the corresponding values of the sector angle and radius in equation (1) we get,

………(2)

We will simplify equation (2) as shown below,

Substituting we get,

Now we will make the denominator same.

Therefore, perimeter of the sector is.

Hence, the correct answer is option (a).

Page No 192:

Question 22:

If the perimeter of a sector of a circle of radius 6.5 cm is 29 cm, then its area is

(a) 58 cm2

(b) 52 cm2

(c) 25 cm2

(d) 56 cm2

Answer:

We know that perimeter of a sector of radius r ………(1)

We have given perimeter of the sector and radius of the sector and we are asked to find the area of the sector. For that we have to find the sector angle.

Therefore, substituting the corresponding values of perimeter and radius in equation (1) we get,

………(2)

We will simplify equation (2) as shown below,

Dividing both sides of the equation by , we get,

Subtracting 1 from both sides of the equation we get,

………(3)

We know that area of the sector

From equation (3), we get

Area of the sector

Substituting we get,

Area of the sector

Therefore, area of the sector is .

Hence, the correct answer is option (b).

Page No 192:

Question 23:

If the area of a sector of a circle bounded by an arc of length 5π cm is equal to 20π cm2, then its radius is

(a) 12 cm

(b)16 cm

(c) 8 cm

(d) 10 cm

Answer:

We have given length of the arc and area of the sector bounded by that arc and we are asked to find the radius of the circle.

We know that area of the sector .

Now we will substitute the values.

……..(1)

……..(2)

Now we will divide equation (1) by equation (2),

Now we will cancel the like terms.

Therefore, radius of the circle is.

Hence, the correct answer is option (c).

Page No 192:

Question 24:

The area of the circle that can be inscribed in a square of side 10 cm is

(a) 40 π cm2

(b) 30 π cm2

(c) 100 π cm2

(d) 25 π cm2

Answer:

We know that ABCD is a square of length 10 cm. A circle is inscribed in the square therefore, all the sides of the square are become tangents of the circle.

By, the tangent property, we have

If we join PR then it will be the diameter of the circle of 10 cm.

Therefore, radius of the circle = 5cm

Therefore, area of the circle is.

Hence, the correct answer is option (d).

Page No 192:

Question 25:

If the difference between the circumference and radius of a circle is 37 cm, then its area is

(a) 154 cm2

(b) 160 cm2

(c) 200 cm2

(d) 150 cm2

Answer:

We have given the difference between circumference and radius of the circle.

Let C be the circumference, r be the radius and A be the area of the circle.

Therefore, from the given condition we have

Now we will substitute.

Now we will substitute the value of r in.

Now we will substitute.

Therefore, area of the circle is.

Hence, the correct answer is option (a).

Page No 192:

Question 26:

If the area of a circle is equal to the sum of the areas of two circles of diameters 10 cm and 24 cm, then diameter of the large circle (in cm) is

(a) 34

(b) 26

(c) 17

(d) 14

Answer:

Let the diameter of the larger circle be d
Now, Area of larger circle = Area of circle having diameter 10 cm + Area of circle having diameter 24 cm
πd22=π1022+π2422d22=52+122

d22=25+144d22=132d2=13d=26 cm
Hence, the correct answer is option (b).



Page No 193:

Question 27:

The area of the circle that can be inscribed in a square of side 6 cm is
(a) 36 π cm2
(b) 18 π cm2
(c) 12 π cm2
(d) 9 π cm2

Answer:

Side of square = 6 cm
∴ Diameter of a circle = Side of square = 6 cm
⇒ Radius of the circle = diameter2=62 = 3 cm

∴  Area of the circle = πr2 
                                 = π(3)2
                                 = 9π cm2

Hence, the correct answer is option (d).

Page No 193:

Question 28:

It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be
(a) 10 m
(b) 15 m
(c) 20 m
(d) 24 m

Answer:

Let D1 be the diameter of the first circular park = 16 m
∴ Radius of first circular park (r1) = 8 m

Let D2 be the diameter of the second circular park = 12 m
∴ Radius of second circular park (r2) = 6 m

Area of first circular park = πr12 = π(8)2 = 64π m2
Area of second circular park = πr22 = π(6)2 = 36π m2

Let R be the radius of the single circular park.

Given that,
Area of single circular park = Area of first circular park + Area of second circular park
πR2 = 64π + 36π      
πR2 = 100π
R2 = 100
R = 10
∴ Radius of the single circular park will be 10 m.

Hence, the correct answer is option (a).

Page No 193:

Question 29:

If diameter of a circle is increased by 40%, then its area increase by

(a) 96%

(b) 40%

(c) 80%

(d) 48%

Answer:

If d is the original diameter of the circle, then the original radius is.

If diameter of the circle increases by 40%, then new diameter of the circle is calculated as shown below,

That is new diameter

So, new area will be.

Now we will calculate the change in area.

Therefore, its area is increased by.

Hence, the correct answer is option (a).

Page No 193:

Question 30:

In the following figure, the shaded area is

(a) 50 (π−2) cm2

(b) 25 (π−2) cm2

(c) 25 (π+2) cm2

(d) 5 (π−2) cm2

Answer:

Area of the shaded region is-

So the answer is (b).

Page No 193:

Question 31:

In the following figure, the area of the segment PAQ is

(a) a24π+2

(b) a24π-2

(c) a24π-1

(d) a24π+1

Answer:

We have to find area of segment PAQ.

We know that.

Substituting the values we get,

Substituting and we get,

Now we will make the denominator same.

Therefore, area of the segment PAQ is.

Hence, the correct answer is option (b).

Page No 193:

Question 32:

In the following figure, the area of segment ACB is

(a) π3-32r2

(b) π3+32r2

(c) π3-23r2

(d) None of these

Figure

Answer:

We have to find area of segment ACB.

We know that.

Substituting the values we get,

Substituting and we get,

Therefore, area of the segment ACB is.

Hence, the correct answer is option (d).

Page No 193:

Question 33:

In the following figure, the ratio of the areas of two sectors S1 and S2 is

(a) 5 : 2

(b) 3 : 5

(c) 5 : 3

(d) 4 : 5

figure

Answer:

Area of the sector,

Area of the sector,

Now we will take the ratio,

Now we will simplify the ratio as below,

Substituting the values we get,

Therefore, ratio of the areas of the two sectors is .

Hence, the correct answer is option (d).

Page No 193:

Question 34:

If the area of a sector of a circle bounded by an arc of length 5π  cm is equal to 20π cm2, then the radius of the circle

(a) 12 cm

(b) 16 cm

(c) 8 cm

(d) 10 cm

Answer:

We have given length of the arc and area of the sector bounded by that arc and we are asked to find the radius of the circle.

If l is the length of the arc, A is the area of the arc and r is the radius of the circle, then we know the expression of the area of the sector in terms of the length of the arc and radius of the circle.

Now we will substitute the corresponding values of length of the arc and area of the sector.

Multiplying both sides of the equation by 2 we get,

Dividing both sides of the equation by we get,

Therefore, radius of the circle is .

Hence, the correct answer is option (c).

Page No 193:

Question 35:

If the area of a sector of a circle is 518 of the area of the circle, then the sector angle is equal to

(a) 60°

(b) 90°

(c) 100°

(d) 120°

Answer:

We have given that area of the sector is of the area of the circle.

Therefore, area of the sector area of the circle

Now we will simplify the equation as below,

Now we will multiply both sides of the equation by 360,

Therefore, sector angle is.

Hence, the correct answer is option (c).

Page No 193:

Question 36:

If he area of a sector of a circle is 720 of the area of the circle, then the sector angle is equal to

(a) 110°

(b) 130°

(c) 100°

(d) 126°

Answer:

We have given that area of the sector is of the area of the circle.

Therefore, area of the sector area of the circle

Now we will simplify the equation as below,

Now we will multiply both sides of the equation by 360,

Therefore, sector angle is.

Hence, the correct answer is option (d).

Page No 193:

Question 37:

In the following figure, If ABC is an equilateral triangle, then shaded area is equal to

(a) π3-34r2

(b) π3-32r2

(c) π3+34r2

(d) π3+3r2

Figure

Answer:

We have given that ABC is an equilateral triangle.

As we know that,.

Area of the shaded region = area of the segment BC.

Let

Substituting the values we get,

Substituting and we get,

Therefore, area of the shaded region is . Hence, the correct answer is option (a).



Page No 194:

Question 38:

In the following figure, the area of the shaded region is

(a) 3π cm2
 
(b) 6π cm2

(c) 9π cm2

(d) 7π cm2
 

Answer:

In the figure,

and,

Therefore, area of the shaded region is.

Hence, the correct answer is option (a).

Page No 194:

Question 39:

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

(a) 13 : 22

(b) 14 : 11

(c) 22 : 13

(d) 11 : 14

Answer:

We have given that perimeter of circle of radius r is equal to square of side a.

perimeter of the circle = perimeter of the square

Now we will substitute value of r in the following equation

Substituting we get,

Hence, ratio of the areas of the circle and square is.

Therefore, the correct answer is (b).

Page No 194:

Question 40:

The radius of a circle is 20 cm. It is divided into four parts of equal area by drawing three concentric circles inside it. Then, the radius of the largest of three concentric circles drawn is

(a) 105 cm

(b) 103 cm

(c) 105 cm

(d) 102 cm

Answer:

The circle can be divided into four parts of equal area by drawing three concentric circles inside it as,

It is given that OB = 20 cm. Let OA = x.

Since the circle is divided into four parts of equal area by the three concentric circles, we have,

Area of the fourth region = Area of the given circle

Therefore, the correct answer is (b).

 

Page No 194:

Question 41:

The area of a sector whose perimeter is four times its radius r units, is

(a) r24 sq. units

(b) 2r2  sq. units

(c) r2 sq.units

(d) r22 sq. units

Answer:

We know that perimeter of the sector .

We have given that perimeter of the sector is four times the radius.

Subtracting 2r from both sides of the equation,

Dividing both sides of the equation 2r we get,

……….(1)

Let us find the area of the sector.

Substituting we get,

Hence, area of the sector is.

Therefore, the correct answer is (c).

Page No 194:

Question 42:

If a chord of a circle of  radius 28 cm makes an angle of 90 ° at the centre, then the area of the major segment is

(a) 392 cm2

(b) 1456 cm2

(c) 1848 cm2

(d) 2240 cm2

Answer:

Area of major segment,

So the answer is (d)

Page No 194:

Question 43:

If area of a circle inscribed in an equilateral triangle is 48π square units, then perimeter of the triangle is

(a) 173 units

(b) 36 units

(c) 72 units

(d) 483 units

Answer:

Let the circle of radius r be inscribed in an equilateral triangle of side a.

Area of the circle is given as 48π.

⇒ πr2 = 48π

r2 = 48

Now, it is clear that ONBC. So, ON is the height of ΔOBC corresponding to BC.

Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB = 3 × Area of ΔOBC

Thus, perimeter of the equilateral triangle = 3 × 24 units = 72 units

So the answer is (c).

Page No 194:

Question 44:

The hour hand of a clock is 6 cm long. The area swept by it between 11.20 am and 11.55 am is

(a) 2.75 cm2

(b) 5.5 cm2

(c) 11 cm2

(d) 10 cm2

Answer:

Hour hand movesin one minute.

So, area,

So the answer is (b)

Page No 194:

Question 45:

ABCD is a square of side 4 cm. If E is a point in the interior of the square such that ΔCED is equilateral, then area of Δ ACE is

(a) 23-1 cm2

(b) 43-1 cm2

(c) 63-1 cm2

(d) 83-1 cm2

Answer:

We have the following diagram.

Since is equilateral,

Therefore,

Now,

Since AC is diagonal of sqr.ABCD

Therefore,

Therefore we get,

Now, in , draw a perpendicular EM to the base AC.

So in ,

Therefore,

Now in ,

So the answer is (b)

Page No 194:

Question 46:

The area of a circular path of uniform width h surrounding a circular region of radius r is
(a) π(2r + h)r
(b) π(2r + h)h
(c) π(h + r)r
(d) π(h + r)h

Answer:



Radius of inner circle is r.
Radius of outer circle is r + h.

Area of circular path=Area of outer circle-Area of inner circle=πr+h2-πr2=πr2+h2+2rh-r2=πh2+2rh=π2r+hh
Hence, the correct answer is option (b).

Page No 194:

Question 47:

If AB is a chord of length 53 cm of a circle with centre O and radius 5 cm, then area of sector OAB is

(a) 3π8cm2

(b) 8π3cm2

(c) 25πcm2

(d) 25π3cm2

Answer:

We have to find the area of the sector OAB.

We have,

So,

Hence,

Therefore area of the sector,

So answer is (d)

Page No 194:

Question 48:

The area of a circle whose area and circumference are numerically equal, is

(a) 2π sq. units

(b) 4 π sq. units

(c) 6π sq. units

(d) 8π sq. units

Answer:

We have given that circumference and area of a circle are numerically equal.

Let it be x.

Let r be the radius of the circle, therefore, circumference of the circle is and area of the circle will be.

Therefore, from the given condition we have,

………(1)

………(2)

Therefore, from equation (1) get . Now we will substitute this value in equation (2) we get,

Simplifying further we get,

Cancelling x we get,

Now we will cancel

………(3)

Now we will multiply both sides of the equation (3) by we get,

Therefore, area of the circle is.

Hence, option (b) is correct.

Page No 194:

Question 49:

If an arc of a circle of radius 14 cm subtends an angle of 45° at the centre of the circle, then it is
(a) a minor arc of length 5.5 cm
(b) a major arc of length 77 cm
(c) a major arc of length 38.5 cm
(d) a minor arc of length 11 cm

Answer:

Given that the radius of circle is 14 cm and an arc subtends an angle of 45° at the centre of the circle.

Length of minor arc=45°360°×2πr=18×2×227×14=11 cm

Length of Major Arc=2πr-11 cm=2×227×14-11=88-11=77 cm

Disclaimer: Both the options (b) and (d)  are correct.

Page No 194:

Question 50:

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) r = r1 + r2

(b) r12+r22=r2

(c) r1 + r2 < r

(d) r12+r22<r2

Answer:

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

πr12+πr22=πr2πr12+r22=πr2r12+r22=r2

Hence, the correct answer is option (b).



Page No 195:

Question 51:

Aarushi's mother is making a table cover for a round table. Aarushi painted a design on it. It has six equal designs as shown in the given figure. The radius of the cover is 28 cm. Take 3=1.73 answer the following questions:

(i) In the given figure, measure of ∠AOB is

(a) 60°
(b) 70°
(c) 30°
(d) 65°

(ii) The area of ∆AOB is
(a) 332 cm2
(b) 322 cm2
(c) 333 cm2
(d) 339.08 cm2

(iii) The area of sector OAPB is
(a) 410.66 cm2
(b) 407.66 cm2
(c) 244.20 cm2
(d) 246.40 cm2

(iv) The area of design is
(a) 411 cm2
(b) 414 cm2
(c) 429.48 cm2
(d) 455 cm2

(v) The cost of making the design at the rate of â‚¹0.50 per m2, is
(a) ₹214.74
(b) ₹210.74
(c) ₹215.74
(d) ₹274.14

Answer:


(i)It can be observed that these designs are segments of the circle.

Consider segment APB. Chord AB is a side of the hexagon. Each chord will substitute at the centre of the circle.
⇒∠AOB = 60°
Hence, the correct answer is option (a).


(ii) In ΔOAB,
OA = OB
∠OAB = ∠OBA             (∵ In a triangle angles corresponding to equal sides are also equal)

In ΔOAB,
∠OAB + ∠OBA + ∠AOB = 180° (∵ Angle sum property of triangle)
⇒∠OAB + ∠OAB + 60° = 180°
⇒2∠OAB = 120°
​⇒∠OAB = 60°

​⇒∠OBA = 60°
⇒ΔOAB is an equilateral triangle       (∵ ∠OAB = ∠OBA = ∠AOB = 60°)

∴ Area of ΔOAB =
=34×282=1.734×28×28=1.73×7×28=339.08 cm2

Hence, the correct answer is option (d).

(iii)

Area of sector OAPB =


= 410.66 cm2

Hence, the correct answer is option (a).

(iv)

Area of segment APB = Area of sector OAPB − Area of ΔOAB
= 410.66 - 339.08
= 71.58 cm2

Area of designs = 6 × 71.58 = 429.48 cm2


Hence, the correct answer is option (c).

(v) Disclaimer: The cost of making the design is at the rate of â‚¹0.50 per cm2.

Cost of making 1 cm2 designs = ₹0.50

Cost of making 429.48 cm2 designs = 0.50 × 429.48 = ₹214.74

Therefore, the cost of making such designs is ₹214.74.

Hence, the correct answer is option (a).

Page No 195:

Question 52:

In a class activity Sikha did a block painting on a square handkerchief as shown in the given figure. She made nine designer circles each of radius 7 cm. On the basis of above information answer each of the following questions:

(i) The perimeter of the nine circles is

(a) 44 cm
(b) 396 cm
(c) 198 cm
(d) 168 cm

(ii) The perimeter of the square is
(a) 168 cm
(b) 84 cm
(c) 198 cm
(d) 126 cm

(iii) The area of the square is
(a) 1674 cm2
(b) 1746 cm2
(c) 1764 cm2
(d) 1476 cm2

(iv) The area of the remaining portion of the square is
(a) 837 cm2
(b) 378 cm2
(c) 738 cm2
(d) 783 cm2

(v) If each block design costs ₹4 and the cost to paint the remaining area of the handkerchief is ₹0.10 per cm2, then the total cost is
(a) ₹73
(b) ₹36
(c) ₹37.8
(d) ₹73.80

Answer:



(i) Perimeter of 9 circles = 9 × 2πr
=9×2×227×7=396 cm

Hence, the correct answer is option (b).

(ii) Side of square = 42 cm
Perimeter of square = 4 × side
= 4 × 42
= 168 cm

Hence, the correct answer is option (a).

(iii) Area of square = (Side)2 = (42)2 = 1764 cm2

Hence, the correct answer is option (c).

(iv) Area of each circle = πr2

Area of 9 circles = 9 × 154 = 1386 cm2

Area of the remaining portion of the handkerchief = 1764 − 1386 = 378 cm2

Hence, the correct answer is option (b).

(v) Cost of 9 block designs = ₹9 × 4 = ₹36
Cost of the remaining portion of the handkerchief = ₹0.10 × 378 = ₹37.8
Total cost = ₹37.8 + ₹36 = ₹73.8

Hence, the correct answer is option (d).

Page No 195:

Question 53:

In the given figure, depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Based on the above information answer the following questions:

(i) The circumferences of the regions representing Gold, Red, Blue, Black and White scoring areas are in the ratio

(a) 1 : 1 : 1 : 1 : 1
(b) 1 : 2 : 3 : 4 : 5
(c) 2 : 3 : 4 : 5 : 6
(d) 1 : 3 : 5 : 7 : 9

(ii) If the area of the region representing Gold scoring area is A, then the area of the region representing Red scoring area is
(a) 2 A
(b) A
(c) 3 A
(d) 4 A

(iii) If the area of the region representing Gold scoring area is A, then the area of the region representing Blue score area is
(a) 2 A
(b) 3 A
(c) 4 A
(d) 5 A

(iv) If the area of the region representing Gold scoring area is A, then the area of the region representing Black score area is
(a) 8 A
(b) 7 A
(c) 6 A
(d) 5 A

(v) If the area of the region representing Red scoring area is A, then the area of the region representing Blue score area is
(a) 53 A

(b) 35 A

(c) 2 A

(d) 73 A

Answer:

Radius (r1) of gold region (i.e., 1st circle)

Given that each circle is 10.5 cm wider than the previous circle.

Radius (r2) of 2nd circle = 10.5 + 10.5 = 21 cm

Radius (r3) of 3rd circle = 21 + 10.5 = 31.5 cm

Radius (r4) of 4th circle = 31.5 + 10.5 = 42 cm

Radius (r5) of 5th circle = 42 + 10.5 = 52.5 cm

(i) Circumferences of the regions representing Gold, Red, Blue, Black and White scoring areas,
r1 : 2πr2 : 2πr3 : 2πr4 : 2πr5
= r1r2r3r4r5
= 10.5 : 21 : 31.5 : 42 : 52.5
= 1 : 2 : 3 : 4 : 5

Hence, the correct answer is option (b).

(ii)

Area of gold region (A) = Area of 1st circle

Area of red region = Area of 2nd circle − Area of 1st circle


= 3 A

Hence, the correct answer is option (c).

(iii)
Area of blue region = Area of 3rd circle − Area of 2nd circle


= 5 A

Hence, the correct answer is option (d).

(iv)

Area of black region = Area of 4th circle − Area of 3rd circle


= 7 A

Hence, the correct answer is option (b).

(v) Area of red region (A) = 1039.5 cm2
Area of blue region = 1732.5 cm2
= 53A

Hence, the correct answer is option (a).



Page No 196:

Question 54:

A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the given figure. 
Using the above information answer the following questions:

(i) The circumference of the brooch is

(a) 100 mm
(b) 120 mm
(c) 110 mm
(d) 95 mm

(ii) The area of the brooch is
(a) 962.5 mm2
(b) 926.5 mm2
(c) 960.5 mm2
(d) 956.5 mm2

(iii) The total length of silver wire used in making the brooch is
(a) 258 mm
(b) 285 mm
(c) 185 mm
(d) 385 mm

(iv) The area of each sector of the brooch is
(a) 69.25 mm2
(b) 92.65 mm2
(c) 86.25 mm2
(d) 96.25 mm2

(v) The sector angle of each sector of the brooch is
(a) 60°
(b) 18°
(c) 36°
(d) 30°

Answer:

Radius of circle (r) =

(i) Circumference of brooch = 2πr

= 110 mm

Hence, the correct answer is option (c).

(ii)
Area of brooch=πr2=227×3522=22×52×352=962.5 mm2

Hence, the correct answer is option (a).

(iii) Total length of wire required will be the length of 5 diameters and the circumference of the brooch.

Length of wire required = circumference of the brooch + 5 × diameters
= 110 + 5 × 35
= 110 + 175
= 285 mm

Hence, the correct answer is option (b).

(iv) Area of each sector =

= 96.25 mm2

Hence, the correct answer is option (d).

(v) Since all the sectors are equal.
​∴ Each of 10 sectors of the circle subtends equal angle at the centre of the circle of 360°10=36° as shown in the figure.

 
Hence, the correct answer is option (c).

Page No 196:

Question 55:

​​​​​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.

(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): If the areas of two circles are in the ratio 9 : 16, then their circumferences are in the ratio 3 : 4.
Statement-2 (Reason):  If the areas of two circles are in the ratio A1 : A2, then their circumferences are in the ratio A1: A2.
 

Answer:


Statement-2 (Reason):  If the areas of two circles are in the ratio A1 : A2, then their circumferences are in the ratio A1: A2.
Let r1 and r2 be the radii of the circles.
A1 and Aare the areas of two circles.
Area of circle 1Area of circle 2=A1A2πr12πr22=A1A2r1r2=A1A2               .....1

Circumference of circle 1Circumference of circle 2=2πr12πr2=r1r2=A1A2                  From 1
Thus, statement-2 is true.

Statement-1 (Assertion): If the areas of two circles are in the ratio 9 : 16, then their circumferences are in the ratio 3 : 4.
Let r1 and r2 be the radii of the circles and A1 and Aare the areas of two circles.

Area of circle 1Area of circle 2=916A1A2=916

Then according to the Statement-1, if the areas of two circles are in the ratio A1 : A2, then their circumferences are in the ratio A1: A2.​
Circumference of circle 1Circumference of circle 2=A1A2=916=34

Thus, statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Page No 196:

Question 56:

​​​​​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.

(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): If areas of two circles are in the ratio 9 : 25, then their radii are in the 3 : 5.
Statement-2 (Reason):  If A1, A2 are the areas of two circles, then their radii are in the ratio A1: A2.
 

Answer:

Statement-2 (Reason):  If A1A2 are the areas of two circles, then their radii are in the ratio A1: A2.
​Let r1r2 be the radii of the circles and A1A2 be the areas of two circles.

Area of circle 1Area of circle 2=A1A2πr12πr22=A1A2r1r2=A1A2

Thus, statement-2 is true and a correct explanation of statement-1.

Statement-1 (Assertion): If areas of two circles are in the ratio 9 : 25, then their radii are in the 3 : 5.
​Let r1r2 be the radii of the circles and A1A2 be the areas of two circles.

Area of circle 1Area of circle 2=916A1A2=925
Then according to the Statement-2, if A1A2 are the areas of two circles, then their radii are in the ratio A1: A2.
r1r2=A1A2r1r2=925r1r2=35

Thus, statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Page No 196:

Question 57:

​​​​​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.

(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): A track is in the form of a circular ring. If the inner and outer circumferences of the track are 352 m and 396 m respectively, then the width of the track 14 metres.
Statement-2 (Reason):  If the inner and outer circumferences of a circular annulus are C1 and C2 respectively, then the width of the annulus is C2-C12π.

Answer:

Statement-2 (Reason):  If the inner and outer circumferences of a circular annulus are C1 and C2 respectively, then the width of the annulus is C2-C12π.
Let r be the radius of the inner circumference of a circular annulus and h be the width.
Then, the radius of the outer circumference of a circular annulus is r + h.

C1=2πrC2=2πr+hC2-C1=2πr+2πh-2πrC2-C1=2πhh=C2-C12π

Thus, Statement-2 is true.

Statement-1 (Assertion): A track is in the form of a circular ring. If the inner and outer circumferences of the track are 352 m and 396 m respectively, then the width of the track 14 metres.

If the inner and outer circumferences of the track are 352 m and 396 m respectively, then the width is
h=C2-C12π        From Statement1=396-3522×227=4444×7=7 m

Thus, Statement-1 is false.
So, Statement-1 is false, Statement-2 is true.

Hence, the correct answer is option (d).

Page No 196:

Question 58:

​​​​​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): A circle circumscribe a square of side 'a', then the area of the region outside the square and enclosed by the circle is π2-1a2 sq. units
Statement-2 (Reason):  A circle is inscribed in a square of side 2a units. If the circle touches all the sides of the square, then the area of the region outside the circle and inside the square is (4 – π)a2 sq. units.

Answer:

Statement-2 (Reason):  A circle is inscribed in a square of side 2a units. If the circle touches all the sides of the square, then the area of the region outside the circle and inside the square is (4 – π)a2 sq. units.
​
Side of the square  cm

 Diameter of the circle  cm

So, radius of the circle = a cm

 Area of the region outside the circle and inside the square 

Thus, statement-2 is true.
 

Statement-1 (Assertion): A circle circumscribe a square of side 'a', then the area of the region outside the square and enclosed by the circle is π2-1a2 sq. units
Side of the square cm

Diameter of the circle cm

So, radius of the circle = a2 cm

Area of the region outside the square and enclosed by the circle

Thus, statement-1 is false.

Hence, the correct answer is option (d).

Page No 196:

Question 59:

​​​​​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.

(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): If the length of the minute hand of a clock is 7 cm, then the area swept by it in 5 minutes is 776 cm2.
Statement-2 (Reason):  The length of an arc of a sector of angle θ and radius r is given by l=θ360×2πr
 
 

Answer:

Statement-2 (Reason):  The length of an arc of a sector of angle θ and radius r is given by l=θ360×2πr
The length of an arc of a sector of angle θ and radius r is given by l=θ360×2πr.
Thus, Statement-2 is true.

Statement-1 (Assertion): If the length of the minute hand of a clock is 7 cm, then the area swept by it in 5 minutes is 776 cm2.
If the length of the minute hand of a clock is 7 cm.
r = 7 cm
Angle subtended by minute hand in 5 minutes(θ) = 360°60×5=30°

Area swept by minute hand in 5 min is given by θ360°×πr2.
=30360×227×7×7=776 cm2

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is option (b).



Page No 197:

Question 60:

​​​​​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.

(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): The area of the minor segment of a circle is always less than the area of the corresponding sector of the circle.
Statement-2 (Reason):  The area of the major segment of a circle is always less than the area of the corresponding sector of the circle.
 
 

Answer:

Statement-2 (Reason):  The area of the major segment of a circle is always less than the area of the corresponding sector of the circle.
​
The major segment of a circle is completely not a part of the correcsponding sector of the circle.
So, the area of the major segment of a circle is always more than the area of the corresponding sector of the circle.

Thus, statement-2 is false.

Statement-1 (Assertion): The area of the minor segment of a circle is always less than the area of the corresponding sector of the circle.

The minor segment of a circle is a part of the correcsponding sector of the circle.
So, the area of the minor segment of a circle is always less than the area of the corresponding sector of the circle.

Thus, statement-1 is true.
So, Statement-1 is true, Statement-2 is false.

Hence, the correct answer is option (c).

Page No 197:

Question 61:

​​​​​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.

(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): If three sectors whose sector angles are 35°, 65° and 80° are cut from a circle, then the sum of their sector areas is half of the area of the circle.
Statement-2 (Reason):  If 3 sectors of sector angles θ1°, θ2°, and θ3° are cut from a circle of radius, r, then the sum of the areas of these sectors is π360 θ1+ θ2 + θ3r2.
 

Answer:

Statement-2 (Reason):  If 3 sectors of sector angles θ1°, θ2°, and θ3° are cut from a circle of radius, r, then the sum of the areas of these sectors is π360 θ1+ θ2 + θ3r2.
If 3 sectors of sector angles θ1°, θ2°, and θ3° are cut from a circle of radius, r, then the sum of the areas of these sectors 

=θ1°360°×πr2+θ2°360°×πr2+θ3°360°×πr2=π360°×θ1°+θ2°+θ3°r2

Thus, Statement-2 is true.

Statement-1 (Assertion): If three sectors whose sector angles are 35°, 65° and 80° are cut from a circle, then the sum of their sector areas is half of the area of the circle.
Let θ1° = 35°, θ2° = 65°, and θ3° = 80°.
Then, sum of areas of their sectors is given by π360 θ1+ θ2 + θ3r2.
=π360°×35°+65°+80°×r2=π360°×180°×r2=π2×r2=12×πr2=12×Area of the circle

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Page No 197:

Question 62:

​​​​​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.

(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): If a motorcycle wheel covers 22 km distance in 5000 revolutions, then the radius of the wheel is 70 cm.
Statement-2 (Reason):  If a chord of a circle of radius r subtends a right angle at the centre of the circle, then area of the corresponding segment is  14π-2r2 sq. units.

Answer:

Statement-2 (Reason):  If a chord of a circle of radius subtends a right angle at the centre of the circle, then area of the corresponding segment is  14π-2r2 sq. units.
Consider a chord of a circle of radius r subtends an angle of 90° at the centre.



The area of the segment formed is
=Area of sector formed-Area of right triangle=90°360°×πr2-12×r2=14π-2r2

Thus, Statement-2 is true.

Statement-1 (Assertion): If a motorcycle wheel covers 22 km distance in 5000 revolutions, then the radius of the wheel is 70 cm.

The distance covered by a motorcycle wheel of radius r in 1 revolution is 2πr.
So, the distance covered by the motorcycle wheel of radius r in 5000 revolutions is 2πr × 5000.

2πr×5000=22 km2×227×r×5000=22×1000×100 cmr=22×1000×100×72×22×5000 cmr=70 cm

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is option (b).



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