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Page No 120:

Question 1:

Choose the correct answer from the given four options:
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) R12+R22=R2

(C) R1 + R2 < R

(D) R12+R22<R2

Answer:

We are given that
Area of circle = Area of first circle + Area of second circle
∴ πR2 = πR21 + πR22
⇒ R2 = R21 + R22
Hence, the correct answer is option B.

Page No 120:

Question 2:

Choose the correct answer from the given four options:
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) Nothing definite can be said about the relation among R1, R2 and R.

Answer:

We are given that,
Circumference of circle with radius R = Circumference of first circle with radius R1 + Circumference of second circle with radius R2
 2πR = 2πR1 + 2πR2
R = R1+ R2
Hence, the correct answer is option (A).



Page No 121:

Question 3:

Choose the correct answer from the given four options:
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) Nothing definite can be said about the relation between the areas of the circle and square.

Answer:

Let r be the radius of the circle and a be the side of square.
It is given that,
Circumference of a circle = Perimeter of square
 2πr = 4a
227r=2a
⇒ 11r = 7a
r=711a                            .....(1)
Now, area of circle, A1 = πr2 and area of square, A2 = a2
From (1), we have
A1 = π × 711a2

227×49121 a2
1411a2

∴     A11411 A2                    [ A2 = a2]
⇒ A1 > A2
   Area of the circle > Area of the square.
Hence, the correct answer is option B.

Page No 121:

Question 4:

Choose the correct answer from the given four options:
Area of the largest triangle that can be inscribed in a semi-circle of radius r units is
(A) r2 sq. units

(B) 12 r2 sq. units

(C) 2 r2 sq. units

(D) 2 r2 sq. units

Answer:

A largest triangle that can be inscribed in a semi-circle of radius r units is the triangle having its base as the diameter of the semi-circle and the two other sides are taken by considering a point C on the circumference of the semi-circle and joining it by the end points of diameter A and B.

∴ ∠ C = 90° (by the properties of circle)
So, ΔABC is right angled triangle with base as diameter AB of the circle and height be CD.
Height of the triangle = r
∴ Area of largest ΔABC = 12× Base × Height = 12 × AB × CD
= 12 × 2r × r = r2 sq. units
Hence, the correct answer is option (A).
 

Page No 121:

Question 5:

Choose the correct answer from the given four options:
If the perimeter of a circle is equal to that of a square, then the ratio of their areas is
(A) 22 : 7
(B) 14 : 11
(C) 7 : 22
(D) 11: 14

Answer:

Let r be the radius of the circle and a be the side of the square.
We are given that Perimeter of a circle = Perimeter of a square
⇒ 2πr = 4a
aπr2
Area of the circle = r2 and Area of the square = a2
Now, Ratio of their areas = Area of CircleArea of Square
=πr2a2
=πr2πr22
=πr2π2 r24
1411
Hence, the correct answer is option B.

Page No 121:

Question 6:

Choose the correct answer from the given four options:
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 = 8 m

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

Area of first circular park = πr2 = π(8)2 = 64 π m2
Area of second circular park =  πr2 = π(6)2 = 36 π m2
Now, we are given that,
Area of single circular park = Area of first circular park + Area of second circular park
∴ πR2 = 64π + 36π = 100π          (where R is the radius of the single circular park)
⇒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 121:

Question 7:

Choose the correct answer from the given four options:
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 = 3 cm

∴  Area of the circle = πr2 = π(3)2 = 9π cm2
Hence, the correct answer is option D.

Page No 121:

Question 8:

Choose the correct answer from the given four options:
The area of the square that can be inscribed in a circle of radius 8 cm is
(A) 256 cm2
(B) 128 cm2
(C) 642 cm2
(D) 64 cm2

Answer:


Let, the radius of circle, r = OC = 8 cm.
∴ Diameter of the circle = AC = 2 × OC = 2 × 8 = 16 cm
Let a be the side of the square.
Now, according to the given condition,
Diagonal of square = Diameter of the circle.
Now in right angled triangle ACB,
(AC)2 = (AB)2 + (BC)2                            (By Pythagoras theorem)
⇒ (16)2 = a2 + a2
⇒ 256 = 2a2
a2 = 128
∴ Area of the square = a2 = 128 cm2
Hence, the correct answer is option (B).

Page No 121:

Question 9:

Choose the correct answer from the given four options:
The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36 cm and 20 cm is
(A) 56 cm
(B) 42 cm
(C) 28 cm
(D) 16 cm

Answer:

Diameter of first circle, d1 = 36 cm
Diameter of second circle, d2 = 20 cm
∴ Circumference of first circle = πd1 = 36π cm
Circumference of second circle = πd2 = 20π cm
Now, we are given that,
Circumference of circle = Circumference of first circle + Circumference of second circle
πD = πd1 + πd2
⇒ πD = 36π + 20π
⇒ πD = 56π
D = 56
⇒ Radius = 562 = 28 cm
Hence, the correct answer is option (C).

Page No 121:

Question 10:

Choose the correct answer from the given four options:
The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is
(A) 31 cm
(B) 25 cm
(C) 62 cm
(D) 50 cm

Answer:

Area of first circle = πr2 = π(24)2 = 576π cm2
Area of second circle = πr2 = π(7)2 = 49π cm2
Now, we are given that,
Area of the circle = Area of first circle + Area of second circle
∴ πR2 = 576π +49π                (where, R is the radius of the new circle)
⇒ πR2 = 625π
R2 = 625
R = 25
∴ Radius of the circle = 25 cm
Thus, diameter of the circle = 2R = 50 cm.
Hence, the correct answer is option (D).



Page No 122:

Question 1:

Is the area of the circle inscribed in a square of side a cm, πa2 cm2? Give reasons for your answer.

Answer:

False

Let a be the side of square.
We are given that the circle is inscribed in the square.
∴ Diameter of circle = Side of square = a
 Radius of the circle = a2
Area of the circle = πa22=πa24

Hence, area of the circle is πa24 cm2.

Page No 122:

Question 2:

Will it be true to say that the perimeter of a square circumscribing a circle of radius a cm is 8a cm? Give reasons for your answer.

Answer:

True

Let radius of circle r = a cm
∴ Diameter of the circle = d = 2 × Radius = 2a cm
As the circle is inscribed in the square, therefore,
Side of a square = Diameter of circle = 2a cm
Hence, Perimeter of a square = 4 × (side) = 4 × 2a = 8a cm

Page No 122:

Question 3:

In Fig 11.3, a square is inscribed in a circle of diameter d and another square is circumscribing the circle. Is the area of the outer square four times the area of the inner square? Give reasons for your answer.

Answer:

False

Diameter of the circle = d
Therefore,
Diagonal of inner square EFGH = Side of the outer square ABCD = Diameter of circle = d
Let side of inner square EFGH be a
Now in right angled triangle EFG,
(EG)2 = (EF)2 + (FG)2                             (By Pythagoras theorem)
d2 = a2 +a2
d2 = 2a2
a2d22
∴ Area of inner square = a2d22
Also, Area of outer square = d2

∴ the area of the outer square is only two times the area of the inner square.
Thus, area of outer square is not equal to four times the area of the inner square.



Page No 123:

Question 4:

Is it true to say that area of a segment of a circle is less than the area of its corresponding sector? Why?

Answer:

False

It is not true because in case of major segment, area is always greater than the area of its corresponding sector. It is true only in the case of minor segment.

Page No 123:

Question 5:

Is it true that the distance travelled by a circular wheel of diameter d cm in one revolution is 2π d cm? Why?

Answer:

False

The distance travelled by a circular wheel of radius r in one revolution is equal to the circumference of the circle.
Circumference of the circle = πd; where d is the diameter of the circle.

Page No 123:

Question 6:

In covering a distance s metres, a circular wheel of radius r metres makes s2πr revolutions. Is this statement true? Why?

Answer:

True

The distance travelled by a circular wheel of radius r m in one revolution is equal to the circumference of the circle i.e. 2πr
∴ No. of revolutions completed in 2πr m distance = 1
No. of revolutions completed in 1 m distance = 12πr

No. of revolutions completed in s m distance = 12πr × s .

Page No 123:

Question 7:

The numerical value of the area of a circle is greater than the numerical value of its circumference. Is this statement true? Why?

Answer:

False
Let r be the radius of the circle.
Area of the circle = πr2
Circumference of the circle = 2πr
Case 1 :
Both are equal only when r = 2
Case 2 :
Numerical value of circumference is greater than numerical value of area of circle when 0 < r < 2 .
Case 3 :
Numerical value of area of circle is greater than the numerical value of the circumference of the circle when  r > 2 .

Page No 123:

Question 8:

If the length of an arc of a circle of radius r is equal to that of an arc of a circle of radius 2r, then the angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle. Is this statement false? Why?

Answer:

True.

Let P and Q be two circles with radius r and 2r respectively. Let C1 and C2 be the centers of the circles P and Q respectively.
Let AB be the arc length of P and CD be the arc length of Q
Let θ1 and θ2 be the angle subtended by the arc AB and CD respectively on the center.
Given that AB = CD = l (say)                                             .....(1)
Now, arc length = θ1360×2π (radius)
  arc(AB)=θ1360×2πr
and   arc(CD)=θ2360×4πr

θ1360×2πr = θ2360×4πr                                         [ Using (1) ]
θ1=2θ2

∴ Angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle.

Page No 123:

Question 9:

The areas of two sectors of two different circles with equal corresponding arc lengths are equal. Is this statement true? Why?

Answer:

False

It is true for arcs of the same circle. But in different circles, it is not possible.

Page No 123:

Question 10:

The areas of two sectors of two different circles are equal. Is it necessary that their corresponding arc lengths are equal? Why?

Answer:

False

It is true for arcs of the same circle. But in different circle, it is not possible.

Page No 123:

Question 11:

Is the area of the largest circle that can be drawn inside a rectangle of length a cm and breadth b cm (a > b) is π b2 cm2? Why?

Answer:

False

The largest circle that can be drawn inside a rectangle is possible when,
Diameter of the circle = Breadth of the rectangle = b
∴ Radius of the circle = b2
Hence, area of the circle = π r2πb22.

Page No 123:

Question 12:

Circumferences of two circles are equal. Is it necessary that their areas be equal? Why?

Answer:

True

We are given that
Circumference of circle with radius R1 = Circumference of circle with radius R2
⇒ 2πR1 = 2πR2
R1 = R2
⇒ π(R1)2 = π(R2)2
Hence the areas are also equal.

Page No 123:

Question 13:

Areas of two circles are equal. Is it necessary that their circumferences are equal? Why?

Answer:

True

We are given that,
Area of circle with radius R1 = Area of circle with radius R2
⇒ π(R1)2 = π(R2)2
R1= R2
⇒ 2πR1= 2πR2
Hence the circumferences are also equal.

Page No 123:

Question 14:

Is it true to say that area of a square inscribed in a circle of diameter p cm is p2 cm2? Why?

Answer:

False

When the square is inscribed in the circle, the diameter of a circle is equal to the diagonal of a square but not the side of the square.
Let side of square = a
p2 = a2a2
p2 = 2a2
a2=p22
Hence, area of square = p22.



Page No 125:

Question 1:

Find the radius of a circle whose circumference is equal to the sum of the circumferences of two circles of radii 15 cm and 18 cm.

Answer:

Radius of first circle = r1 = 15 cm
Radius of second circle = r2 = 18 cm
∴ Circumference of first circle = 2πr1 = 30π cm
Circumference of second circle = 2πr2 = 36π cm
Let R be the radius of the circle.
Now, we are given that,
Circumference of circle = Circumference of first circle + Circumference of second circle
R = 2πr1+ 2πr2
⇒ 2πR = 30π + 36π
⇒ 2R = 66
R = 33 cm
Hence, required radius of a circle is 33 cm.

Page No 125:

Question 2:

In Fig. 11.5, a square of diagonal 8 cm is inscribed in a circle. Find the area of the shaded region.

Answer:

Let a be the side of square.
∴ Diameter of a circle = Diagonal of the square = 8 cm
Now in right angled triangle ABC,
(AC)2 = (AB)2 + (BC)2                                    (By Pythagoras theorem)
∴ (8)2 = a2 +a2
⇒ 64= 2a2
a2= 32
Hence area of square = a2 = 32 cm2
∴ Radius of the circle = Diameter2 = 4 cm
∴ Area of the circle = πr2 = π(4)2 = 16 cm2
So, the area of the shaded region = Area of circle – Area of square
∴ the area of the shaded region = 16π – 32
= 18.286 cm2.



Page No 126:

Question 3:

Find the area of a sector of a circle of radius 28 cm and central angle 45°.

Answer:

Area of a sector of a circle = θ360×π×r2 ,
where r is the radius and, θ the angle subtended by the arc at the center of the circle
Here, radius of circle = 28 cm,
Angle subtended at the center = 45°
∴ Area of a sector of a circle = 45360×π×282
= 308 cm2
Hence, the required area of a sector of a circle is 308 cm2.

Page No 126:

Question 4:

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

Answer:

Radius of wheel = r = 35 cm
1 revolution of the wheel = Circumference of the wheel = 2πr
2×227×35
= 220 cm

But speed of the wheel = 66 km/h
66×1000×10060 cm/min
= 110000 cm/min

∴ Number of revolutions in 1 min = 110000220 = 500
Hence, required number of revolutions per minute is 500.

 

Page No 126:

Question 5:

A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions 20 m × 16 m. Find the area of the field in which the cow can graze.

Answer:


Let ABCD be a rectangular field.
Length of field = 20 m
Breadth of the field = 16 m
Suppose a cow is tied at a point A.
Let length of rope be AE = 14 m = l (say).
Angle subtended at the center of the sector = 90°
∴ Area of a sector of a circle = 90360×π×142
= 154 m2
Hence, the required area of a sector of a circle is 154 m2.

Page No 126:

Question 6:

Find the area of the flower bed (with semi-circular ends) shown in Fig. 11.6.

Answer:

Length and breadth of the rectangular portion AFDC of the flower bed are 38 cm and 10 cm respectively.
Area of the flower bed = Area of the rectangular portion + Area of the two semi-circles.


∴ Area of rectangle AFDC = Length × Breadth
= 38 × 10 = 380 cm2
Both ends of flower bed are semi-circle in shape.
∴ Diameter of the semi-circle = Breadth of the rectangle AFDC = 10 cm

∴ Radius of the semi circle = 102 = 5 cm

Area of the semi-circle = πr2225π2 cm2.

Since there are two semi-circles in the flower bed,
∴ Area of two semi-circles = 2 × πr22 = 25π cm2
Total area of flower bed = (380 + 25π) cm2.

Page No 126:

Question 7:

In Fig. 11.7, AB is a diameter of the circle, AC = 6 cm and BC = 8 cm. Find the area of the shaded region (Use π= 3.14).

Answer:

Given, AC = 6cm and BC = 8 cm
A triangle in a semi-circle with hypotenuse as diameter is right angled triangle.
∴ In right angled triangle ACB,
(AB)2 = (AC)2 + (CB)2                             (By Pythagoras theorem)
⇒  (AB)2 = (6)2 + (8)2
⇒ (AB)2 = 36 + 64
⇒ (AB)2 = 100
⇒ (AB)= 10
∴ Diameter of the circle = 10 cm
Thus, Radius of the circle = 5 cm
Area of circle = πr2
= π(5)2
= 25π cm2
= 25 × 3.14 cm2
= 78.5 cm2
Also, Area of the right angled triangle = 12 × Base × Height
12 × AC × CB

12 × 6 × 8 = 24 cm2
Now, Area of the shaded region = Area of the circle  –  Area of the triangle
= (78.5 - 24) cm2
= 54.5 cm2

Page No 126:

Question 8:

Find the area of the shaded field shown in Fig. 11.8.

Answer:

We can clearly see that the figure comprises of a rectangle and a semi-circle.

Now, area of the figure = Area of the semi-circle + Area of the rectangle.
Here, from the figure, Radius of the semi-circle = r = 6 - 4 = 2 m

∴ Area of the semi-circle = πr224π2 = 2π

Also, area of the rectangle = Length × Breadth
= AB × BC
= 4 × 8 = 32 m2
∴Area of shaded region = Area of rectangle ABCD + Area of semi-circle DEF
= (32 + 2π) m2
 



Page No 127:

Question 9:

Find the area of the shaded region in Fig. 11.9.

Answer:


Area of the shaded region = Area of the rectangle ABCD – (Area of the rect. EFGH + (Area of the semi-circle EFJ +Area of the semi-circle GHI)
Length and breadth of outer rect. ABCD are 26 m and 12 m respectively.
∴ Area of the rect. ABCD = Length × Breadth
= AB × BC = 26 × 12 = 312 m2
From the figure, length and breadth of inner rect. EFGH are (26 -- 5) m and (12 -- 4) m, i.e. 16 m and 4 m respectively.
∴ Area of the rect. EFGH = Length × Breadth
= EF × FG
= 16 × 4
= 64 m2

Breadth of the inner rectangle = Diameter of the semi-circle EJF = d = 4 m

∴ Radius of semi-circle EJF = r = 2 m

Area of the semi-circle EFJ = Area of the semi-circle GHI
πr22
4π2 = 2π m2

∴ Area of shaded region = 312 – (64 + 2π + 2π) m2
= ( 248 – 4π ) m2.

Page No 127:

Question 10:

Find the area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 60°.

Answer:

Let r be the radius of the circle = 14 cm
Angle subtended at the center of the sector = θ = 60°
In triangle AOB, ∠AOB = 60°, ∠OAB = ∠OBA = θ

∴ θ + θ + 60 = 180                                                 [Since, sum of all interior angles of a triangle is 180°]
⇒ 2θ = 120
⇒ θ = 60
∴ Each angle is of 60° and hence the triangle AOB is an equilateral triangle.
Now, Area of the minor segment = Area of the sector AOBC – Area of triangle AOB
Angle subtended at the center of the sector = 60°

Angle subtended at the center (in radians) = θ = 60π180π3

∴ Area of a sector of a circle = r2θ2
12×142×π3
3083 cm2

Area of the equilateral triangle = 34side2
34142
493 cm2

∴ Area of minor segment = 3083-493 cm2.

Page No 127:

Question 11:

Find the area of the shaded region in Fig. 11.10, where arcs drawn with centres A, B, C and D intersect in pairs at mid-points P, Q, R and S of the sides AB, BC, CD and DA, respectively of a square ABCD (Use π = 3.14).

Answer:

Since P, Q, R and S are mid-points of  AB, BC, CD and DA respectively.
∴ AP = PB = BQ = QC = CR = RD = DS = SA = 6 cm.
Given, side of a square BC = 12 cm
Area of the square = 12 × 12 = 144 cm2
Area of the shaded region = Area of the square - (Area of the four quadrants)
Area of one quadrant = πr24=3.14×624=113.044 cm2

Area of four quadrants = 113.04 cm2
Area of the shaded region = 144 - 113.04 = 30.96 cm2.

Page No 127:

Question 12:

In Fig. 11.11, arcs are drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm. to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of the shaded region (Use π = 3.14).

Answer:

Since D, E, F bisects BC, CA, AB respectively.
∴ AE = EC = CD = DB = BF = FA = 5 cm
Now area of the shaded region = (Area of the three sectors)
Since the triangle is an equilateral triangle, therefore each angle is of 60°
∴ Angle subtended at the center of each sector = 60°
Angle subtended at the center (in radians) = θ = 60×π180 = π3

Radius of each sector = 5 cm

∴ Area of a sector of a circle = 12×r2×θ
12×52×π3
25×3.146 cm2

∴ Area of three sectors of a circle = 3×78.56 cm2
= 39.25 cm2


∴ Area of shaded region = 39.25 cm2.
 



Page No 128:

Question 13:

In Fig. 11.12, arcs have been drawn with radii 14 cm each and with centres P, Q and R. Find the area of the shaded region.

Answer:

Let r be the radius of each sector = 14 cm
Area of the shaded region = Area of the three sectors
Let angles subtended at P, Q, R be x°, y°, z° respectively.

Angle subtended at P, Q, R (in radians) be xπ180, yπ180, zπ180 respectively.


∴ Area of a sector with central angle at P = 12×r2×θ
12×142×xπ180 = 196πx360 cm2

Similarly, 
Area of a sector with central angle at Q = 196πy360 cm2

Area of a sector with central angle at R = 196πz360 cm2

∴ Area of three sectors = 196πx360+196πy360+196πz360 cm2 .

Since, sum of all interior angles in any triangle is 180°
x + y + z = 180°  

Thus, Area of three sectors = 196π360×x+y+z = 196π360×180
= 308 cm2

Hence, the required area of the shaded region is 308 cm2.

Page No 128:

Question 14:

A circular park is surrounded by a road 21 m wide. If the radius of the park is 105 m, find the area of the road.
 

Answer:

Let r be the radius of the park = 105 m
Given that the circular park is surrounded by a road of width 21 m.
So, Radius of the outer circle (R)= (105+21) m = 126 m
Area of the road = Area of the outer circle – Area of the circular park
= πr2 – πR2

π1262 - 1052
227126 + 105×126 - 105
227×231×21
66×231
= 15246 m2
Hence, the required area of the road is 15246 m2.

Page No 128:

Question 15:

In Fig. 11.13, arcs have been drawn of radius 21 cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region.

Answer:

Let r be the radius of each sector = 21 cm
Area of the shaded region = Area of the four sectors
Let angles subtended at A, B, C and D be x°, y°, z° and w° respectively.

Angle subtended at A, B, C, D (in radians) be xπ180, yπ180, zπ180, 180

∴ Area of a sector with central angle at A = 12×r2×θ
12×212×xπ180 = 441πx360 cm2

Similarly, 
Area of a sector with central angle at B = 441πy360 cm2

Area of a sector with central angle at C = 441πz360 cm2

Area of a sector with central angle at D = 441πw360 cm2

∴ Area of four sectors = 441πx360+441πy360+441πz360+441πw360 cm2 .

Since, sum of all interior angles in any quadrilateral is 360°
∴ x + y + z +w = 360°
 

Thus, Area of four sectors = 441π360×x+y+z+w = 441π360×360
441π cm2

Hence, the required area of the shaded region is 1386 cm2.

 

Page No 128:

Question 16:

A piece of wire 20 cm long is bent into the form of an arc of a circle subtending an angle of 60° at its centre. Find the radius of the circle.

Answer:

Length of arc of circle = 20 cm
Here, central angle θ = 60°
 Length of arc = θ360×2πr
20 = 60360×2πrr=60π cm

Hence, the radius of the circle is 60π cm .



Page No 132:

Question 1:

The area of a circular playground is 22176 m2. Find the cost of fencing this ground at the rate of Rs 50 per metre.

Answer:

Area of the circular playground = 22176 m2                                              (Given)
Let r be the radius of the circle.
∴ πr2 = 22176
227 r2 = 22176
r2 = 22176 × 227
r2 = 7056
r = 84
∴ Radius of the circular playground = 84 m
Now, circumference of the circle = 2πr
= 2×227×84
= 528 m
Cost of fencing 1 meter of ground = Rs 50
∴ Cost of fencing the total ground = Rs 528 × 50 = Rs 26400

Page No 132:

Question 2:

The diameters of front and rear wheels of a tractor are 80 cm and 2 m respectively. Find the number of revolutions that rear wheel will make in covering a distance in which the front wheel makes 1400 revolutions.

Answer:

Diameter of front wheels = d1 = 80 cm
Diameter of rear wheels = d2 = 2 m = 200 cm


Let r1 be the radius of the front wheels = 802 = 40 cm

Let r2 be the radius of the rear wheels = 2002 = 100 cm

Now, Circumference of the front wheels = 2πr1
2×227×40
17607 cm

∴ Distance covered by the front wheel = 1400 × 17607 = 352000 cm
Circumference of the rear wheels = 2πr = 2 × 227 × 100 = 44007 cm

Number of revolutions made by rear wheel in covering a distance in which the front wheel makes 1400 revolutions
Distance covered by front wheelCircumference of rear wheel

35200044007
=2464044=560

Hence, the rear wheel makes 560 revolutions.

Page No 132:

Question 3:

Sides of a triangular field are 15 m, 16 m and 17 m. With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length 7 m each to graze in the field. Find the area of the field which cannot be grazed by the three animals.

Answer:

Sides of the triangle are 15 m, 16 m, and 17 m.
Now, perimeter of the triangle = (15+16+17) m = 48 m
∴ Semi-perimeter of the triangle = s 482 = 24 m

 Area of the triangle = 24×24-15×24-16×24-17                                [ By Heron's Formula]
= 109.982 m2



Let B, C and H be the corners of the triangle on which buffalo, cow and horse are tied respectively with ropes of 7 m each.
So, the area grazed by each animal will be in the form of a sector.
∴ Radius of each sector = r = 7 m
Let x, y, z be the angles at corners B, C, H respectively.

∴ Area of sector with central angle xx360°×π×r2 = x360°×π×72

∴ Area of sector with central angle y = y360°×π×r2 = y360°×π×72

∴ Area of sector with central angle z = z360°×π×r2 = z360°×π×72

Area of field not grazed by the animals = Area of triangle – (area of the three sectors)

= (109.982) - x360°×π×72 +y360°×π×72 + z360°×π×72 

= (109.982) - π×72360°× x+y+z 

= (109.982) - π×72360°× 180° 

= 109.892 – 77 = 32.982 cm2.

Hence, the area of the field which cannot be grazed by the three animals is 32.982 cm2.



Page No 133:

Question 4:

Find the area of the segment of a circle of radius 12 cm whose corresponding sector has a central angle of 60° (Use π = 3.14).

Answer:

Figure

Radius of the circle = r = 12 cm
∴ OA = OB = 12 cm
∠ AOB = 60°                                                                                       (given)
Since, triangle OAB is an isosceles triangle, ∴ ∠ OAB = ∠ OBA = θ (say)

∴ θ + θ + 60° = 180°                                                 [Sum of interior angles of a triangle is 180°]
⇒2θ = 120° ⇒ θ = 60°
Thus, the triangle AOB is an equilateral triangle.
∴ AB = OA = OB = 12 cm
Area of the triangle AOB = 34×a2
34×122
363 cm2
= 62.354 cm2

Now, Central angle of the sector OBCA  = 60° 
Area of sector OBCA = πr2360×θ
3.14×12×12360°×60°
= 75.36 cm2
 Area of the segment ABCA = Area of the sector OBCA – Area of the triangle AOB
= (75.36 – 62.354) cm2 = 13.006 cm2.
 

Page No 133:

Question 5:

A circular pond is 17.5 m is of diameter. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of Rs 25 per m2

Answer:

Figure
Diameter of the circular pond = 17.5 m
Let r be the radius of the park = 17.52 m = 8.75 m
Given that the circular pond is surrounded by a path of width 2 m.
So, Radius of the outer circle = R = (8.75 + 2) m = 10.75 m
Area of the road = Area of the outer circular path – Area of the circular pond
= πR2 – πr2
= 3.14 × (10.75)2 – 3.14 × (8.75)2
= 3.14 × ((10.75)2 – (8.75)2)
= 3.14 × ((10.75 + 8.75) × (10.75 – 8.75))
= 3.14 × 19.5 × 2 = 122.46 m2
Hence, the area of the path is 122.46 m2.
Now, Cost of constructing the path per m2 = Rs. 25
∴ Cost of constructing 122.46 m2 of the path = Rs. 25 × 122.46 = Rs. 3061.50

Page No 133:

Question 6:

In Fig. 11.17, ABCD is a trapezium with AB || DC, AB = 18 cm, DC = 32 cm and distance between AB and DC = 14 cm. If arcs of equal radii 7 cm with centres A, B, C and D have been drawn, then find the area of the shaded region of the figure.

Answer:

AB = 18 cm, DC = 32 cm
Distance between AB and DC  = 14 cm
Now, Area of the trapezium = 12 × (18+32) × 14 = 350cm2                                      [ 12× (Sum of parallel sides) × Height]


As AB ∥ DC,
∴ ∠ A +∠ D = 180°
and ∠ B +∠ C = 180°
Also, radius of each arc = 7 cm
Therefore,
Area of the sector with central angle A = 12×A180×π×r2

Area of the sector with central angle D = 12×D180×π×r2


Area of the sector with central angle B = 12×B180×π×r2


Area of the sector with central angle C = 12×C180×π×r2

Total area of the sectors = A360×π×r2 +D360×π×r2  + C360×π×r2  + B360×π×r2 
A+D360×π×r2  + C +B 360×π×r2  

180360×π×r2  + 180 360×π×r2  

= 77 + 77 = 154
∴ Area of shaded region = Area of trapezium – (Total area of sectors)
= 350 – 154 = 196 cm2
Hence, the required area of shaded region is 196 cm2.

Page No 133:

Question 7:

Three circles each of radius 3.5 cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these circles.

Answer:

Figure

The three circles are drawn in such a way that each of them touches the other two.
So, by joining the centers of the three circles, we get,
AB = BC = CA = 2(Radius) = 7 cm
Therefore, triangle ABC is an equilateral triangle with each side 7 cm.
∴ Area of the triangle = 34×a2
34×72
= 21.2176 cm2

Now, Central angle of each sector = 60° 
Area of each sector = 60360×π×3.52
= 6.4167 cm2
Total area of three sectors = 3 × 6.4167 = 19.25 cm2
∴ Area enclosed between three circles = Area of triangle ABC – Area of the three sectors
= 21.2176 – 19.25
= 1.9676 cm2
Hence, the required area enclosed between these circles is 1.967 cm2 (approx).

Page No 133:

Question 8:

Find the area of the sector of a circle of radius 5 cm, if the corresponding arc length is 3.5 cm.

Answer:

Radius of the circle (r) = 5 cm
Arc length of the sector (l) = 3.5 cm
Let the central angle be θ.
θ360×2πr = 3.5                     (Arc = θ360×2πr )
θ= 3.5×36010π      

Area of circle with angle θ = θ360×πr2 
3.5×36010π×π×52360

Hence, required area of the sector of a circle is 8.75 cm2.

Page No 133:

Question 9:

Four circular cardboard pieces of radii 7 cm are placed on a paper in such a way that each piece touches other two pieces. Find the area of the portion enclosed between these pieces.

Answer:

The four circles are placed in such a way that each piece touches the other two pieces.
Now, on joining the centers of the circles by a line segment, we get a square ABDC with sides as,
AB = BD = DC = CA = 2(Radius) = 2(7) cm = 14 cm
Now, Area of the square = (Side)2 = (14)2 = 196 cm2
∴ ∠ A = ∠ B = ∠ D = ∠ C = 90°                                                                   [ABDC is a square]
Also, Radius of each sector = 7 cm

Area of each sector = 90360×π72                                                                      [Area of sector = θ360×πr2]
772 cm2

∴ Area of the shaded portion = Area of square – Area of the four sectors
196 - 4×772
= 196 – 154
= 42 cm2
Hence, required area of the portion enclosed between these pieces is 42 cm2.

Page No 133:

Question 10:

On a square cardboard sheet of area 784 cm2, four congruent circular plates of maximum size are placed such that each circular plate touches the other two plates and each side of the square sheet is tangent to two circular plates. Find the area of the square sheet not covered by the circular plates.

Answer:

Since, Area of the square = 784 cm2
∴ Side of the square = √Area = √784 = 28 cm
Since the four circular plates are congruent, therefore diameter of each circular plate = 282 = 14 cm
∴ Radius of each circular plate = 7 cm
Area of the sheet not covered by plates = Area of the square – Area of the four circular plates
∴ Are of one circular plate = πr2 = 154 cm2
So, Area of four plates = 4×154 = 616 cm2            [Since all four circular plates are congruent, therefore area of all four plates will be equal.]
Area of the sheet not covered by plates = 784 – 616 = 168 cm2.

Page No 133:

Question 11:

Floor of a room is of dimensions 5 m × 4 m and it is covered with circular tiles of diameters 50 cm each as shown in Fig. 11.18. Find the area of floor that remains uncovered with tiles. (Use π = 3.14)

Answer:

Length of the floor (l) = 5 m
Breadth of the floor (b) = 4 m
∴ Area of the floor = l × b = 5 × 4 = 20 m2
Now, Diameter of each circular tile = 50 cm
∴ Radius of each circular tile (r) = 25 cm = 0.25 m
Area of one circular tile = πr2
= 3.14 × (0.25)2
= 0.19625 m2

Area of such 80 tiles = 80 × 0.19625
= 15.7 m2

Area of the floor that remains uncovered with tiles = Area of the floor – Area of all 80 circular tiles
= 20 – 15.7 = 4.3 m2
Hence, the required area of floor that remains uncovered with tiles is 4.3 m2.



Page No 134:

Question 12:

All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if area of the circle is 1256 cm2. (Use π = 3.14).

Answer:

Figure

Area of the circle = 1256 cm2
Let r be the radius of the circle.
∴ Area = πr2
⇒ 1256 = 3.14 × r2
r2 = 12563.14 = 400
r = 20 cm
∴ Diameter of the circle (d) = 2×20 = 40 cm
Since all the vertices of a rhombus lie on a circle, therefore the diagonals of the rhombus pass through the center of the circle and thus diagonals of the rhombus are equal to the diameter of the circle.
Let d1 and d2 be the diagonal of the rhombus.
Since, diagonals of a rhombus are equal, therefore d1 = d2 = d = 40 cm
Now, Area of the rhombus = 12  × d1 × d2
12 × 40 × 40 = 800 cm2
Hence, the required area of rhombus is 800 cm2.

Page No 134:

Question 13:

An archery target has three regions formed by three concentric circles as shown in Fig. 11.19. If the diameters of the concentric circles are in the ratio 1 : 2 : 3, then find the ratio of the areas of three regions.

Answer:

Diameters are in the ratio 1 : 2 : 3
So, let the diameters of the concentric circles be 2r, 4r and 6r.
∴ Radius of the circles be r, 2r, 3r respectively.
Now, Area of the outermost circle = π (Radius)2 = π (3r)2 = 9πr2
Area of the middle circle = π (Radius)2 = π (2r)2 = 4πr2
Area of the innermost circle = π (Radius)2 = π (r)2 = πr2
Now, Area of the middle region = Area of middle circle – Area of the innermost circle
= 4πr- πr2 = 3πr2
Now, Area of the outer region = Area of outermost circle – Area of the middle circle
= 9πr2 -r2 = 5πr2
Required ratio = Area of inner circle : Area of the middle region : Area of the outer region
= πr2 : 3πr2 : 5πr2
⇒ Required Ratio is 1 : 3 : 5

Page No 134:

Question 14:

The length of the minute hand of a clock is 5 cm. Find the area swept by the minute hand during the time period 6 : 05 am and 6 : 40 am.

Answer:

Length of the minute hand = 5 cm = Radius of the clock
Minutes between the time period 6:05 am to 6:40 am = 35 minutes
In 60 minutes, the minute hand completes one revolution, i.e. 360°.
∴ Angle made by minute hand in 1 minute = 360°60° = 60°
Thus angle made by minute hand in 35 minutes = 60° × 35 = 210°
∴ Area swept by minute hand in 35 minutes = 210360×π×r2
210360×π×52

2756
= 45.833 cm2
Hence, the required area swept by the minute land is 45.833 cm2.

Page No 134:

Question 15:

Area of a sector of central angle 200° of a circle is 770 cm2. Find the length of the corresponding arc of this sector.

Answer:

Let r be the radius of the circle.
Given, Central angle = 200°
Area of the sector = 770 cm2
770 = 200360×π×r2
r = 21 cm
Thus, radius of the sector = 21 cm
Now, Length of the corresponding arc = θ360×2πr
200360×2π×21
= 73.33 cm
Hence, the required length of the corresponding arc is 73.33 cm.



Page No 135:

Question 16:

The central angles of two sectors of circles of radii 7 cm and 21 cm are respectively 120° and 40°. Find the areas of the two sectors as well as the lengths of the corresponding arcs. What do you observe?

Answer:

Radius of one sector (r1) = 7 cm
Radius of second sector (r2) = 21 cm
Central angle of one sector = 120°
Central angle of second sector = 40°
Area of first sector = θ360×π×r12
120360×π×72

1543
= 51.33 cm2

Area of second sector = θ360×π×r22
40360×π×212
= 154 cm2

Now, arc length of first sector = θ360×2πr1
120360×2π× 7
443 cm

Arc length of second sector = θ360×2πr2
40360×2π×21
443 cm

Hence, we observe that arc lengths of two sectors of two different circles may be equal but their area need not be equal.
 

Page No 135:

Question 17:

Find the area of the shaded region given in Fig. 11.20.

Answer:


Figure
Area of square ABCD = (side)2
= 142
= 196 cm2
As,
14 = 3 + r + 2r + r + 3
⇒ 14 = 6 + 4r
⇒ 14 – 6 = 4r
⇒ 8 = 4r
r = 2 cm

Area of internal portion = Area of 4 semi-circles + Area of square JKLM                             .....(1)
Area of semi-circle = πr22
π(2)22
447 cm2
Area of 4 semicircles = 4×447 
1767 cm2
Side of square = 2r = 2(2)= 4 cm
Area of square = 42
= 16 cm2
From (1), Area of the internal portion = 1767+16
Area of shaded region = Area of square ABCD - Area of the internal portion
196-1767+16
10847
= 154.85 cm2
 

Page No 135:

Question 18:

Find the number of revolutions made by a circular wheel of area 1.54 m2 in rolling a distance of 176 m.

Answer:

Let r be the radius of the circular wheel.
Area = 1.54 m2
∴πr2 = 1.54
r2=0.49
⇒ r = 0.7 m
Thus, radius of the circular wheel = 0.7 m
Circumference of the wheel = 2πr = 2 × (3.14) × 0.7 = 4.4 m
Distance travelled by wheel in one revolution = Circumference of circular wheel = 4.4 m
Since, distance travelled by a circular wheel = 176 m
Total distance covered by the wheel = No. of revolutions made by wheel × (Distance covered in one revolution)
∴ No. of revolutions made by wheel = Total Distance Distance in one revolution
1764.4

= 40
Hence, the required number of revolutions made by a circular wheel is 40.

Page No 135:

Question 19:

Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm subtending an angle of 90° at the centre.

Answer:

Length of the chord = 5 cm (Given)
Let r be the radius of the circle.
Then, OA = OB = r cm
Now, angle subtended at the center of the sector OABO = 90°
∴ Triangle AOB is a right-angled triangle.
So, by Pythagoras theorem, (AB)2 = (OA)2 + (OB)2
⇒ 25 = 2r2
r=52 cm
Also, AOB is an isosceles triangle.
Since, line segment OD is perpendicular on AB, therefore it divides AB into two equal parts. Thus, AD = DB = 52 = 2.5 cm
Let AD = h cm

So, in right angled triangle AOD, 
(AO)2 = (AD)2 + (OD)2                                 [by Pythagoras theorem]

522=522+h2
h 52 = 2.5 cm

∴ Area of the isosceles triangle AOB = 12 × Base × Height
12 × 5 × 52254 cm2

Now, Area of the minor sector = θ360×π×r2
90360×π×522
25π8 cm2

Area of the minor segment = Area of the minor sector – Area of the isosceles triangle
25π8-254 cm2
Area of the major segment = Area of the circle – Area of the minor segment
πr2 -25π8-254
π522 -25π8-254
25π2 -25π8-254
75π8+254 cm2
∴ Difference of the areas of two segments of a circle = |Area of major segment – Area of minor segment|
75π8+254-25π8-254
=25π4+252 cm2
Hence, the required difference of the areas of two segments is 25π4+252 cm2 .


Page No 135:

Question 20:

Find the difference of the areas of a sector of angle 120° and its corresponding major sector of a circle of radius 21 cm.

Answer:

Radius of the circle (r) = 21 cm
Area of the circle = πr2 = π(21)2=1386 cm2
Central angle of the sector AOBA = 120°
Area of the minor sector AOBA = 120360×π×r2 
120360×π×212
= = 462 cm2
Area of the major sector ABOA = Area of the circle – Area of the sector AOBA
= 1386 – 462 = 924 cm2

Now, Difference of the areas of a sector AOBA and its corresponding major sector ABOA
= |Area of major sector ABOA – Area of minor sector AOBA|
= |924 - 462| = 462 cm2
Hence, the required difference of two sectors is 462 cm2.
 



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