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)

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 = R² ⇒ R = 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.

SOLUTION:

Soln. : Diameter of the innermost region = 21 cm
Radius of the innermost (Gold Scoring) region
∴ Area of Gold region = π(10.5)² cm²


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]
= [(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
= [(31.5 + 21) (31.5 - 21)]cm²


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


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

= 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 =
So, circumference of the wheel

⇒ Distance covered by a wheel in one revolution

Distance travelled by the car in 1 hour
= 66 km = 66 × 1000 × 100 cm
∴ Distance travelled in 10 minutes

Now, number of revolutions


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°

∴ The Area of a sector

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


Here θ = 90°
∴ Area of the quadrant of the circle,

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°


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




Now, (i) Area of the minor segment
= [Area of the minor sector] -
[Area of right ∠AOB]

= 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

= (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


∴ Length of arc APB


(ii) Area of the sector with sector angle 60°

= 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

...(2)
From (1) and (2), we have
Area of segment

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 and)

SOLUTION:

Soln. : Here, radius (r) = 15 cm and
Sector angle (θ) = 60°
∴ Area of the sector


Since ∠O = 60° and OA = OB = 15 cm
∴ AOB is an equilateral triangle.
⇒ AB = 15 cm and ∠A = 60°
Draw OM ⊥ AB, in ∠AMO


Now, ar(∠AOB)


Now area of the minor segment
= (Area of minor sector) - (ar ∠AOB)
= (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

= 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 )

SOLUTION:

Soln. : Here θ = 120° and r = 12 cm
∴ Area of the sector


= 150.72 cm² ...(1)

Now, area of ∠AOB
[Q OM ⊥AB] ...(2)
In ∠OAB, ∠O = 120°
⇒ ∠A + ∠B = 180° - 120° = 60° Q OB = OA = 12 cm
⇒ ∠A = ∠B = 30°
So,
⇒



Now, from (2),
Area of ∠AOB

= 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)

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



(ii) When length of the rope is increased to 10 m, ∴ r = 10 m
⇒ Area of the new circular portion grazed


∴ 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) =

(i) Circumference = 2πr

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
⇒ Area of each sector

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.

SOLUTION:

Soln. : Here, radius (r) = 45 cm
Since circle is divided in 8 equal parts,
∴ Sector angle corresponding to each part

⇒ Area of a sector (part)


∴ The required area between the two ribs

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
[Q there are 2 blades]

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

Q 18.

A round table cover has six

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 )

SOLUTION:

Soln. : Here, r = 28 cm
Since, the circle is divided into six equal sectors.
∴ Sector angle
∴ Area of each sector


...(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 = BO ⇒ AB = 28 cm
Draw OM ⊥ AB
∴ In right ∠AOM, we have


∴ Area of ∠AOB

= 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) (B)
(C) (D)

SOLUTION:

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

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]

⇒ RQ² = 24² + 7² = 576 + 49 = 625

∴ radius of circle
∴ Area of ∠RPQ

= 12 × 7 cm² = 84 cm²
Now, area of semi-circle


∴ 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



Radius of the inner circle = 7 cm and θ = 40°
∴ Area of the sector BOD


Now, area of the shaded region
= Area of sector AOC - Area of sector BOD

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

⇒ Radius of each of the circles
∴ Area of the semi-circle APD

Area of the semi-circle BPC

∴ 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

Area of equilateral triangle, having side
a = 12 cm, is given by


Q Each angle of an equilateral triangle = 60°
∴ ∠AOB = 60°
∴ Area of sector COD


Now, area of the shaded region,
= [Area of the circle] + [Area of the equilateral triangle] - [Area of the sector COD]

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.

∴ Area of all the 4 quadrant squares

Diameter of the middle circle = 2 cm
⇒ Radius of the middle circle = 1 cm
∴ Area of the middle circle

Now, area of the shaded region
= [Area of the square ABCD] -
[(Area of the 4 quadrant circles)
+ (Area of the middle circle)]

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.

SOLUTION:

Soln. : Area of the circle = πr²

'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 OM ⊥AB, then


Also,



Now, area of ∠AOB,

Since area of ∠ABC = 3 ×[area of ∠AOB]

Now, area of the design = [Area of the circle] - [Area of the equilateral triangle]

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

Now, area of a sector of radius 7 cm and sector angle θ as 90°

Area of 4 sectors

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.

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

= BC + EH +




(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)

= 1060 m² + 1060 m²


[(40 + 30) × (40 - 30)] m²

= 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

⇒ AB = 2OA = 2 × 7 = 14 cm
OC = OA = 7 cm
Q AB and CD are perpendicular to each other
⇒ OC ⊥AB
∴ Area of ∠ABC

Again OD = OA = 7 cm
∴ Radius of the small circle

∴ Area of the small circle

Radius of the big circle
Area of the semi-circle OACB

= 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]

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 = 1.73205).

SOLUTION:

Soln. : Area of ∠ABC = 17320.5 cm²
∴ ∠ABC is an equilateral triangle and area of an equilateral

∴




⇒ (side)² = 40000
⇒ (side)² = (200)² ⇒ side = 200 cm
∴ Radius of each circle
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.


Area of 3 equal sectors

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.

Now, area of one circle

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,

(ii) shaded region.

SOLUTION:

Soln. : (i) Here, centre of the circle is O and radius = 3.5 cm.
∴ Area of the quadrant OACB

(ii)

∴ Area of the shaded region
= (Area of the quadrant OACB)
- (Area of ∠BOD)

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

= 400 + 400 = 800

⇒ Radius of the circle
Now, area of the quadrant

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



Again, radius of the smaller circle, r = 7 cm
Also, the sector angle is 30°
∴ Area of the sector OCD

∴ Area of the shaded region

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


= 22 × 7 cm² = 154 cm²
Area of right ∠ABC
⇒ 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
∴ Radius of the semi-circle BQC

∴ Area of the semi-circle BQC

= 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.

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


Similarly, area of the quadrant

Sum of the two quadrant


Now, area of design
= [Sum of the area of the two quadrant] -
[Area of the square ABCD]