champstreet
www.champstreet.com
Areas Related to Circles - NCERT Questions
Q 1.

The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

SOLUTION:

Soln. : We have, r1 = 19 cm and r2 = 9 cm
∴ Circumference of circle-I = 2πr1 = 2π(19) cm
and circumference of circle-II = 2πr2 = 2π (9) cm
Sum of the circumference of circle-I
and circle-II = 2π(19) + 2π(9) = 2π(19 + 9) cm = 2π(28) cm
Let R be the radius of the circle-III.
∴ Circumference of circle-III = 2πR
According to the condition, 2πR = 2π(28)
ncert
Thus, the radius of the new circle = 28 cm.

Q 2.

The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

SOLUTION:

Soln. : We have,
Radius of circle-I, r1 = 8 cm
Radius of circle-II, r2 = 6 cm
∴ Area of circle-I = πr12 = π(8)2 cm2
Area of circle-II = πr22 = π(6)2 cm2
Let the radius of the circle-III be R
∴ Area of circle-III = πR2
Now, according to the condition,
πr12 + πr22 = πR2
⇒ π(8)2 + π(6)2 = πR2
⇒ π(82 + 62) = πR2
⇒ 82 + 62 = R2
⇒ 64 + 36 = R2
⇒ 100 = R2
⇒ 102 = R2R = 10
Thus, the radius of the new circle = 10 cm.

Q 3.

The given figure depicts an archery target marked with its five scoring regions 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. Find the area of each of the five scoring regions.
ncert

SOLUTION:

Soln. : Diameter of the innermost region = 21 cm
Radius of the innermost (Gold Scoring) region ncert
∴ Area of Gold region = π(10.5)2 cm2
ncert
ncert
Area of the Red region = π(10.5 + 10.5)2 - π(10.5)2 = π(21)2 - π(10.5)2 = π[(21)2 - (10.5)2]
= ncert [(21 + 10.5) (21 - 10.5)]cm2


Since each band is 10.5 cm wide
∴ Radius of Gold and Red region
= (10.5 + 10.5) = 21 cm.
Area of Blue region
= π[(21 + 10.5)2 -(21)2]cm2
[(31.5)2 - (21)2] cm2
= ncert [(31.5 + 21) (31.5 - 21)]cm2


Similarly,
Area of Black region
= π[(31.5 + 10.5)2 - (31.5)2] cm2
= ncert [(42)2 - (31.5)2] cm2
=ncert [(42 - 31.5) (42 + 31.5)] cm2
ncert
ncert
Area of White region
= π[(42 + 10.5)2 - (42)2] cm2
= π[(52.5)2 - (42)2] cm2
= π[(52.5 + 42)(52.5 - 42)] cm2
ncert
= 3118.5 cm2

Q 4.

The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

SOLUTION:

Soln. : Diameter of a wheel = 80 cm
∴ Radius of the wheel = ncert
So, circumference of the wheel
ncert
⇒ Distance covered by a wheel in one revolution
ncert
Distance travelled by the car in 1 hour
= 66 km = 66 × 1000 × 100 cm
∴ Distance travelled in 10 minutes
ncert
Now, number of revolutions
ncert
ncert
Thus, the required number of revolutions
= 4375.

Q 5.

Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units
(B) π units
(C) 4 units
(D) 7 units

SOLUTION:

Soln. : (A) We have
[Numerical area of the circle]
= [Numerical circumference of the circle]
⇒ πr2 = 2πr
⇒ πr2 - 2πr = 0
r2 - 2r = 0
r(r - 2) = 0
r = 0 or r = 2
But r cannot be zero
r = 2 units.
Thus, the radius of circle is 2 units.

Q 6.

Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

SOLUTION:

Soln. : Here r = 6 cm and θ = 60°
ncert
∴ The Area of a sector
ncert
ncert

Q 7.

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

SOLUTION:

Soln. : Let radius of the circle = r
∴ 2πr = 22
ncert
ncert
Here θ = 90°
∴ Area of the ncertquadrant of the circle,
ncert
ncert

Q 8.

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

SOLUTION:

Soln. : Length of minute hand = radius of the circle ⇒ r = 14 cm
Q Angle swept by the minute hand in
60 minutes = 360°
∴ Angle swept by the minute hand in 5 minutes
Now, area of the sector with r = 14 cm
and θ = 30°
ncert
ncert
Thus, the required area swept by the minute hand in 5 minutes

Q 9.

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding (i) minor segment (ii) major sector. (Use π = 3.14)

SOLUTION:

Soln. : Length of the radius (r) = 10 cm Sector angle θ = 90°
Area of the sector ncert
ncert
ncert
ncert
ncert
Now, (i) Area of the minor segment
= [Area of the minor sector] -
[Area of right ∠AOB]
ncert
= 78.5 cm2 - 50 cm2 = 28.5 cm2.
(ii) Area of the major segment
= [Area of the circle] -
[Area of the minor segment]
= πr2 - 78.5 cm2
ncert
= (314 - 78.5) cm2 = 235.5 cm2.

Q 10.

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) the length of the arc
(ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord

SOLUTION:

Soln. : Here, radius = 21 cm and θ = 60°
(i) Circumference of the circle = 2πr
ncert
ncert
∴ Length of arc APB
ncert
ncert
(ii) Area of the sector with sector angle 60°
ncert
= 11 × 21 cm2 = 231 cm2
(iii) Area of the segment APB = [Area of the sector AOB] - [Area of ∠AOB] ...(1)
In ∠AOB, OA = OB = 21 cm
∴ ∠A = ∠B = 60° [Q ∠O = 60°]
AOB is an equilateral ∠.
AB = 21 cm
ncert
ncert ...(2)
From (1) and (2), we have
Area of segment ncert
ncert

Q 11.

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 andncert)

SOLUTION:

Soln. : Here, radius (r) = 15 cm and
Sector angle (θ) = 60°
∴ Area of the sector
ncert
ncert
Since ∠O = 60° and OA = OB = 15 cm
AOB is an equilateral triangle.
AB = 15 cm and ∠A = 60°
Draw OMAB, in ∠AMO
ncert
ncert
ncert Now, ar(∠AOB) ncert
ncert
ncert
Now area of the minor segment
= (Area of minor sector) - (arAOB)
= (117.75 97.3125) cm2 = 20.4375 cm2
Area of the major segment
= [Area of the circle] -
[Area of the minor segment]
= πr2 - 20.4375 cm2
ncert
= 706.5 - 20.4375 cm2 = 686.0625 cm2.

Q 12.

A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.
(Use π = 3.14 and ncert)

SOLUTION:

Soln. : Here θ = 120° and r = 12 cm
∴ Area of the sector ncert
ncert
ncert
= 150.72 cm2 ...(1)
ncert
Now, area of ∠AOB ncert
[Q OMAB] ...(2)
In ∠OAB, ∠O = 120°
⇒ ∠A + ∠B = 180° - 120° = 60° Q OB = OA = 12 cm
⇒ ∠A = ∠B = 30°
So, ncert
ncert
ncert
ncert
ncert
Now, from (2),
Area of ∠AOBncert
ncert
= 36 × 1.73 cm2 = 62.28 cm2 ...(3)
From (1) and (3)
Area of the minor segment
= [Area of sector] - [Area of ∠AOB]
= [150.72 cm2] - [62.28 cm2] = 88.44 cm2.

Q 13.

A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see figure). Find :
(i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m.
(Use π = 3.14)
ncert

SOLUTION:

Soln. : Here, Length of the rope = 5 m
∴ Radius of the circular region grazed by the horse = 5 m
(i) Area of the circular portion grazed
ncert
ncert
ncert
(ii) When length of the rope is increased to 10 m, ∴ r = 10 m
⇒ Area of the new circular portion grazed
ncert
ncert
∴ Increase in the grazing area
= (78.5 - 19.625) m2 = 58.875 m2.

Q 14.

9. 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 figure. Find:
(i) the total length of the silver wire required. (ii) the area of each sector of the brooch.

SOLUTION:

Soln. : Diameter of the circle = 35 mm
∴ Radius (r) =
ncert
(i) Circumference = 2πr
ncert
Length of 1 piece of wire used to make diameter to divide the circle into 10 equal sectors = 35 mm
∴ Length of 5 pieces = 5 × 35 = 175 mm
∴ Total length of the silver wire
= 110 + 175 mm = 285 mm
(ii) Since the circle is divided into 10 equal sectors,
∴ Sector angle θ ncert
⇒ Area of each sector
ncert
ncert

Q 15.

An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
ncert

SOLUTION:

Soln. : Here, radius (r) = 45 cm
Since circle is divided in 8 equal parts,
∴ Sector angle corresponding to each part
ncert
⇒ Area of a sector (part)
ncert
ncert
∴ The required area between the two ribs
ncert

Q 16.

A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

SOLUTION:

Soln. : Here, radius (r) = 25 cm.
Sector angle (θ) = 115°
∴ Area cleaned by each sweep of the blades
ncert [Q there are 2 blades]
ncert
ncert

Q 17.

To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned.
(Use π = 3.14)

SOLUTION:

Soln. : Here, Radius (r) = 16.5 km and Sector angle (θ) = 80°
∴ Area of the sea surface over which the ships are warned
ncert
ncert

Q 18.

A round table cover has six
ncert
equal designs as shown in the figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs 0.35 per cm2. (Use ncert)

SOLUTION:

Soln. : Here, r = 28 cm
Since, the circle is divided into six equal sectors.
∴ Sector angle
∴ Area of each sector
ncert
ncert
ncert ...(1)
Now, area of 1 design
= Area of segment APB
= Area of sector ABO - Area of ∠AOB ...(2)
In ∠AOB, ∠AOB = 60°, OA = OB = 28 cm
∴ ∠OAB = 60° and ∠OBA = 60°
⇒ ∠AOB is an equilateral triangle.
AB = AO = BOAB = 28 cm
Draw OMAB
∴ In right ∠AOM, we have
ncert
ncert
∴ Area of ∠AOB ncert
ncert
= 14 × 14 × 1.7 cm2 = 333.3 cm2 ...(3)
Now, from (1), (2) and (3), we have:
Area of segment APB = 410.67 cm2 - 333.2 cm2
= 77.47 cm2
⇒ Area of 1 design = 77.47 cm2
∴ Area of the 6 equal designs
= 6 × (77.47) cm2 = 464.82 cm2
Cost of making the design at the rate of
₹0.35 per cm2
= ₹ 0.35 × 464.82 = ₹ 162.68.

Q 19.

Tick the correct answer in the following :
Area of a sector of angle π (in degrees) of a circle with radius R is
(A) ncert (B) ncert
(C) ncert (D) ncert

SOLUTION:

Soln. : (D) Here, radius (r) = R
Angle of sector (θ) = p°
∴ Area of the sector
ncert
ncert

Q 20.

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

SOLUTION:

Soln. : Since O is the centre of the circle,
QOR is a diameter.
⇒ ∠RPQ = 90° [Angle in a semi-circle]
Now, in right ∠RPQ, RQ2 = PQ2 + PR2
[Pythagoras theorem]
ncert
RQ2 = 242 + 72 = 576 + 49 = 625
ncert
∴ radius of circle ncert
∴ Area of ∠RPQ
ncert
= 12 × 7 cm2 = 84 cm2
Now, area of semi-circle
ncert
ncert
∴ Area of the shaded portion
= 245.54 cm2 - 84 cm2 = 161.54 cm2.

Q 21.

Find the area of the shaded region in figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and
AOC = 40°.

SOLUTION:

Soln. : Radius of the outer circle = 14 cm
and θ = 40°
∴ Area of the sector AOC
ncert
ncert
ncert
Radius of the inner circle = 7 cm and θ = 40°
∴ Area of the sector BOD
ncert
ncert
Now, area of the shaded region
= Area of sector AOC - Area of sector BOD
ncert
ncert

Q 22.

Find the area of the shaded region in the figure, if ABCD is a square of side 14 cm and APD and BPC are semi-circles.

SOLUTION:

Soln. : Side of the square = 14 cm ∴ Area of the square ABCD = 14 × 14 cm2
= 196 cm2
Now, diameter of the circle
= (Side of the square) = 14 cm
ncert
⇒ Radius of each of the circles ncert
∴ Area of the semi-circle APD
ncert
Area of the semi-circle BPC
ncert
∴ Area of the shaded region
= Area of the square - [Area of semi-circle APD
+ Area of semi-circle BPC]
= 196 - [77 + 77] cm2 = (196 - 154) cm2
= 42 cm2.

Q 23.

Find the area of the shaded region in the figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

SOLUTION:

Soln. : Area of the circle with radius 6 cm
ncert
Area of equilateral triangle, having side
a = 12 cm, is given by
ncert
ncert
Q Each angle of an equilateral triangle = 60°
∴ ∠AOB = 60°
∴ Area of sector COD
ncert
ncert
Now, area of the shaded region,
= [Area of the circle] + [Area of the equilateral triangle] - [Area of the sector COD]
ncert

Q 24.

From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the figure. Find the area of the remaining portion of the square.

SOLUTION:

Soln. : Side of the square = 4 cm ∴ Area of the square ABCD = 4 × 4 cm2
= 16 cm2
Q Each corner has a quadrant circle of radius 1 cm.
ncert
∴ Area of all the 4 quadrant squares
ncert
Diameter of the middle circle = 2 cm
⇒ Radius of the middle circle = 1 cm
∴ Area of the middle circle
ncert
Now, area of the shaded region
= [Area of the square ABCD] -
[(Area of the 4 quadrant circles)
+ (Area of the middle circle)]
ncert
ncert

Q 25.

In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in figure. Find the area of the design.
ncert

SOLUTION:

Soln. : Area of the circle = πr2
ncert
'O' is the centre of the circle,
AO = OB = OC = 32 cm
⇒ ∠AOB = ∠BOC = ∠AOC = 120°
Now, in ∠AOB, ∠1 + ∠2 = 60°
and OA = OB ⇒ ∠1 = ∠2
∴ ∠1 = 30°
If OMAB, then
ncert
ncert
Also, ncert
ncert
ncert
ncert
Now, area of ∠AOB,
ncert
Since area of ∠ABC = 3 ×[area of ∠AOB]
ncert
Now, area of the design = [Area of the circle] - [Area of the equilateral triangle]
ncert

Q 26.

In figure, ABCD is a square of side 14 cm. With centres A, B, C and D four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

SOLUTION:

Soln. : Side of the square ABCD = 14 cm
∴ Area of the square ABCD = 14 × 14 cm2
= 196 cm2.
Q Circles touch each other
Radius of the circle ncert
ncert
Now, area of a sector of radius 7 cm and sector angle θ as 90°
ncert
Area of 4 sectors
ncert
Area of the shaded region =
[Area of the square ABCD]
- [Area of the 4 sectors]
= 196 cm2 - 154 cm2 = 42 cm2.

Q 27.

The figure depicts a racing track whose left and right ends are semicircular.
ncert
The distance between the two inner parallel line segment is 60 m and they are each 106 m long.
If the track is 10 m wide, find:
(i) the distance around the track along its inner edge
(ii) the area of the track.

SOLUTION:

Soln. : (i) Distance around the track along its inner edge
ncert
= BC + EH + ncert
ncert
ncert
ncert
ncert
(ii) Now, area of the track = Area of the shaded region = (Area of rectangle ABCD) + (Area of rectangle EFGH) + 2 [(Area of semi-circle of radius 40 m) (Area of semi-circle of radius 30 cm)] [Q The track is 10 m wide]
⇒ Area of the track
= (106 × 10 m2) + (106 × 10 m2)
ncert
= 1060 m2 + 1060 m2
ncert
ncert
[(40 + 30) × (40 - 30)] m2
ncert
= 2120 m2 + 2200 m2 = 4320 m2.

Q 28.

In the figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

SOLUTION:

Soln. : O is the centre of the circle, OA = 7 cm
ncert
AB = 2OA = 2 × 7 = 14 cm
OC = OA = 7 cm
Q AB and CD are perpendicular to each other
OCAB
∴ Area of ∠ABC
ncert
Again OD = OA = 7 cm
∴ Radius of the small circle
ncert
∴ Area of the small circle
ncert
Radius of the big circle ncert
Area of the semi-circle OACB
ncert
= 11 × 7 cm2 = 77 cm2
Now, Area of the shaded region
= [Area of the small circle] + [Area of the big semi-circle OABC] - [Area of ∠ABC]
ncert
ncert

Q 29.

The area of an equilateral triangle ABC is 17320.5 cm2. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see figure). Find the area of the shaded region.
(Use π = 3.14 and ncert = 1.73205).

SOLUTION:

Soln. : Area of ∠ABC = 17320.5 cm2
∴ ∠ABC is an equilateral triangle and area of an equilateral
ncert
ncert
ncert
ncert
ncert
ncert
⇒ (side)2 = 40000
⇒ (side)2 = (200)2 ⇒ side = 200 cm
∴ Radius of each circle ncert
Since each angle of an equilateral triangle is 60°,
∴ ∠A = ∠B = ∠C = 60°
∴ Area of a sector having angle of sector as 60° and radius 100 cm.
ncert
ncert
Area of 3 equal sectors
ncert
Now, area of the shaded region
= [Area of the equilateral triangle ABC]
- [Area of 3 equal sectors]
= 17320.5 cm2 - 15700 cm2 = 1620.5 cm2.

Q 30.

On a square handkerchief, nine circular designs each of radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.

SOLUTION:

Soln. : Q The circles touch each other. ∴ The side of the square ABCD
= 3 × diameter of a circle
= 3 × (2 × radius of a circle) = 3 × (2 × 7cm)
= 42 cm
⇒ Area of the square ABCD = 42 × 42 cm2
= 1764 cm2.
ncert
Now, area of one circle
ncert
Q There are 9 squares
∴ Total area of 9 circles = 154 × 9 = 1386 cm2
∴ Area of the remaining portion of the handkerchief = (1764 - 1386) cm2 = 378 cm2.

Q 31.

In the figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the
(i) quadrant OACB,
ncert
(ii) shaded region.

SOLUTION:

Soln. : (i) Here, centre of the circle is O and radius = 3.5 cm.
∴ Area of the quadrant OACB
ncert
(ii) ncert
ncert
∴ Area of the shaded region
= (Area of the quadrant OACB)
- (Area of ∠BOD)
ncert

Q 32.

In the figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region (Use π = 3.14)

SOLUTION:

Soln. : OABC is a square such that its side
OA = 20 cm
OB2 = OA2 + AB2
OB2 = 202 + 202
ncert
= 400 + 400 = 800
ncert
⇒ Radius of the circle ncert
Now, area of the quadrant
ncert
Area of the square OABC = 20 × 20 cm2
= 400 cm2
∴ Area of the shaded region
= 628 cm2 - 400 cm2 = 228 cm2.

Q 33.

AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If ∠AOB = 30°, find the area of the shaded region.

SOLUTION:

Soln. : Q Radius of bigger circle R = 21 cm and sector angle θ = 30°
∴ Area of the sector OAB
ncert
ncert
ncert
Again, radius of the smaller circle, r = 7 cm
Also, the sector angle is 30°
∴ Area of the sector OCD
ncert
∴ Area of the shaded region
ncert
ncert

Q 34.

In the figure, ABC is a quadrant of a circle of radius 14 cm and a semi-circle is drawn with BC as diameter. Find the area of the shaded region.

SOLUTION:

Soln. : Radius of the quadrant = 14 cm
Therefore, area of the quadrant ABPC
ncert
ncert
= 22 × 7 cm2 = 154 cm2
Area of right ∠ABC ncert
⇒ Area of segment BPC = 154 cm2 - 98 cm2
= 56 cm2
Now, in right ∠ABC,
AC2 + AB2 = BC2
⇒ 142 + 142 = BC2
⇒ 196 + 196 = BC2
BC2 = 392 ncert
∴ Radius of the semi-circle BQC
ncert
∴ Area of the semi-circle BQC
ncert
= 11 × 7 × 2 cm2 = 154 cm2
Now, area of the shaded region
= [Area of semi-circle BQC]
- [Area of segment BPC]
= 154 cm2 - 56 cm2 = 98 cm2.

Q 35.

Calculate the area of the designed region in the figure, common between the two quadrants of circles of radius 8 cm each.
ncert

SOLUTION:

Soln. : Q Side of the square = 8 cm
∴ Area of the square (ABCD) = 8 × 8 cm2
= 64 cm2
Now, radius of the quadrant ADQB = 8 cm
∴ Area of the quadrant ADQB
ncert
ncert
Similarly, area of the quadrant
ncert
Sum of the two quadrant
ncert
ncert
Now, area of design
= [Sum of the area of the two quadrant] -
[Area of the square ABCD]
ncert