**Maths Class 10
Exercise 12.2**

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

**Solution: **Here, radius (r) = 6 cm and Î¸ = 60°

= 132/7 cm^{2}

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

**Solution: **It is given that, circumference (2Ï€r) = 22 cm

⇒
2 × 22/7 × r = 22

⇒
r = 7/2 cm

We
know that for a quadrant of the circle, Î¸ = 90°

^{2}

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

**Solution: **Here, r = 14 cm and Î¸ = 90°/3 = 30°

Therefore, the area swept by the minute hand =

^{2}

**4. 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 ****Ï€ = 3.14)**** **

**Solution: (i) **Here, r = 10 cm and Î¸ = 90°

= 78.5 cm^{2}

Area
of ∆OAB = ½ × Base × Height

= ½ × 10 × 10

= 50 cm^{2}

So, the area
of minor segment = area of minor sector – area of ∆OAB

= 78.5 – 50

= 28.5 cm^{2}

**(ii)** For major sector, radius = 10 cm
and Î¸ = 360° − 90° = 270°

=
235.5 cm^{2}

**5. 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: **Given, r = 21 cm and Î¸ = 60°

=
22 cm

=
231 cm^{2}

**(iii) **Area of segment formed by corresponding chord

= Area of the sector – Area
of ∆OAB

⇒
Area of segment =
231 – Area of ∆OAB ……….
(i)

In
right-angled ∆OMA and ∆OMB,

OA
= OB [Radii of the same
circle]

OM
= OM [Common]

∆OMA **≅** ∆OMB
[RHS congruence criterion]

AM
= BM [By CPCT]

Therefore, M
is the mid-point of AB and ∠AOM
= ∠BOM

⇒ ∠AOM
= ∠BOM

= ½
∠AOB = ½ × 60° = 30°

Therefore,
in right-angled ∆OMA,

cos
30° = OM/OA

⇒ √3/2 = OM/21

⇒ OM = 21√3/2 cm

Also,
sin 30° = AM/OA

⇒ ½ = AM/21

⇒ AM = 21/2 cm

Therefore, AB
= 2 AM

= 2 × 21/2 cm

= 21 cm

Area
of ∆OAB = ½ × AB × OM

=
½ × 21 × 21√3/2

=
441√3/4 cm^{2}

From
equation (i), we have

Area
of segment formed by corresponding chord = (231 – 441√3/4) cm^{2}

**6. A chord of a circle of radius 15 cm
subtends an angle of ****60°**** at the centre. Find the
area of the corresponding segment of the circle. (Use ****Ï€**** = 3.14 and √3 = 1.73)**

**Solution: **Here, r = 15 cm and Î¸ = 60°

=
117.75 cm^{2}

To
find the area of ∆AOB, draw OM **⊥ **AB.

In
right-angled ∆OMA and ∆OMB, we have

OA
= OB [Radii of the
same circle]

OM
= OM [Common]

Therefore, ∆OMA
**≅** ∆OMB
[RHS congruence criterion]

So, AM = BM
[By CPCT]

⇒ AM = BM = ½ AB and ∠AOM = ∠BOM
= ½ ∠AOB = ½ × 60° = 30°

In
right-angled ∆OMA, cos 30° = OM/OA

⇒
√3/2 = OM/15

⇒ OM = 15√3/2 cm

Also,
sin 30° = AM/OA

⇒
½ = AM/15

⇒ AM = 15/2 cm

⇒ 2 AM = 2 × 15/2 = 15 cm

⇒ AB = 15 cm

Since, area
of ∆AOB = ½ × AB × OM

= ½ × 15 ×
15√3/2 = 225√3/4

= (225 ×
1.73)/4

= 97.3125
cm^{2}

Therefore, the area of minor segment = Area of minor
sector – Area of ∆AOB

= 117.75 – 97.3125 = 20.4375 cm^{2}

And,
area of major segment = Ï€r^{2} – Area
of minor segment

= 3.14 × 15 ×15 – 20.4375

= 706.5 – 20.4375

= 686.0625 cm^{2}

**7. 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 ****Ï€ = 3.14 and √3 = 1.73)**

**Solution: **Here, r = 12 cm and Î¸ = 120°

= 150.72
cm^{2}

To
find the area of ∆AOB, draw OM **⊥ **AB.

In
right-angled ∆OMA and ∆OMB, we have

OA
= OB [Radii of the same circle]

OM
= OM [Common]

∆OMA
**≅** ∆OMB
[RHS congruence criterion]

Therefore, AM
= BM [By CPCT]

⇒ AM = BM = ½ AB and ∠AOM = ∠BOM
= ½ ∠AOB = ½ × 120° = 60°

In
right-angled ∆OMA, cos 60° = OM/OA

⇒
½ = OM/12

⇒ OM = 6 cm

Also,
sin 60° = AM/OA

⇒ √3/2 = AM/12

⇒ AM = 6√3 cm

⇒ 2 AM = 2 × 6√3 = 12√3 cm

⇒ AB = 12√3 cm

Now,
the area of ∆AOB = ½ × AB × OM

=
½ × 12√3 × 6

= 36√3

= 36
× 1.73

=
62.28 cm^{2}

Therefore,
the area of
corresponding segment = Area of corresponding sector – Area of ∆AOB

=
150.72 – 62.28 = 88.44 cm^{2}

**8. 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 ****Ï€ = 3.14****)**

**Solution: (i) **The
horse can graze the area of a quadrant with radius 5 m.** **

Area of the quadrant with 5 m radius =

= 19.625 m^{2}

**(ii)** Area of quadrant with radius 10 m =

= 78.5 m^{2}^{}

Then, the
increase in grazing area = 78.5 – 19.625

= 58.875 m^{2}

**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 figure.
Find:**

**(i) the total length of the silver wire required.**

**(ii) the area of each sector of the brooch.**

**Solution: (i) **Diameter of the circular brooch = 35
mm

⇒ Radius of the circular brooch = 35/2 mm

Circumference
of the circular brooch = 2Ï€r = 2 × 22/7
× 35/2

= 110 mm

The
length of 5 diameters = 35 × 5 = 175
mm

The
total length of the silver wire required = Circumference + Length of 5
diameters

= 110 + 175 = 285 mm

**(ii) **To find the area of
each sector, r = 35/2 mm and
Î¸ = 360°/10 = 36°

The area of each sector of the brooch =

=
385/4 mm^{2}

**10. 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:**Here, r = 45 cm and Î¸ = 360°/8 = 45°

Area
between two consecutive ribs of the umbrella = Area of a sector of the circle

**11. 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: **The area cleaned by
the blade is in the shape of a sector.

Here, r =
25 cm and Î¸ = 115°

The
total area cleaned at each sweep of the blades

**12.
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 ****Ï€ = 3.14****)**

**Solution: **Here, r = 16.5 km and Î¸ = 80°

The area of sea over which the ships are warned =

^{2}

**13. A round table cover has six equal designs
as shown in 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 cm ^{2}. **

**(Use ****√3 = 1.7)**

**Solution: **Here,** **r = 28 cm and Î¸ = 360°/6 = 60°

=
1232/3

= 410.67
cm^{2}

To
find the area of ∆AOB, draw OM **⊥ **AB.

In
right-angled ∆OMA and ∆OMB, we have

OA
= OB [Radii of the same circle]

OM
= OM [Common]

∆OMA
**≅** ∆OMB [RHS congruence criterion]

Therefore, AM
= BM [By CPCT]

⇒ AM = BM = ½ AB and ∠AOM = ∠BOM
= ½ ∠AOB = ½ × 60° = 30°

In
right-angled ∆OMA, cos 30° = OM/OA

⇒
√3/2 = OM/28

⇒ OM = 14√3 cm

Also,
sin 30° = AM/OA

⇒ ½ = AM/28

⇒ AM = 14 cm

⇒ 2 AM = 2 × 14 = 28 cm

⇒ AB = 28 cm

Now,
the area of ∆AOB = ½ × AB × OM

= ½ ×
28 × 14√3

= 196√3

= 196
× 1.7

= 333.2
cm^{2}

Therefore,
the area of minor
segment = Area of minor sector – Area of ∆AOB

= 410.67 – 333.2 = 77.47 cm^{2}

So,
the area of one design = 77.47 cm^{2}

Area
of six designs = 77.47 × 6 = 464.82 cm^{2}

Cost
of making the designs = 464.82 × 0.35 = Rs. 162.68

**14. Tick the correct answer in the following:**

**Area of a sector of angle p (in
degrees) of a circle with radius R is:**

**Solution: (D)** Given, r = R and Î¸ = *p*