KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2

KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.2

{Unless stated otherwise, use \(\pi=\frac{22}{7}\)}

Question 1.
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Solution:
Radius of a sector of a circle, r = 6 cm.
Angle of a sector of a circle, θ = 60°
Area of a sector of a circle = ?
Area of a sector of a circle \(=\frac{\theta}{360} \times \pi r^{2}\)
ex 5.2 class 10 Maths kseeb Solutions Chapter 5 Areas Related to Circles
= 18.8 sq.cm.

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 2.
Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
Circumference of a circle, 2πr = 22 cm.
Area of a quadrant of a circle = ?
2πr = 22
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 2
∴ Area of a quadrant of a circle
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 3
\(=\frac{77}{8}\)
= 9.62 sq. cm.

Question 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:
Length of minute hand = radius of the circle (r) = 14 cm
∵ Angle swept by the minute hand in 60 minutes = 360°
∴ Angle swept by the minute hand in 360°
5 minutes = \(\frac{360^{\circ}}{60}\) × 5 = 30°
Now, area of the sector with r = 14 cm and θ = 30°
= \(\frac{\theta}{360^{\circ}}\) × πr2 = \(\frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7}\) × 14 × 14 cm2
= \(\frac{11 \times 14}{3} \mathrm{cm}^{2}=\frac{154}{3} \mathrm{cm}^{2}\)
Thus, the required area swept by the minute hand in 5 minutes = \(\frac{154}{3}\) cm2

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 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 π = 3.14)
Solution:
Radius of the circle, r = 10 cm.
Angle at the centre of circle, θ = 90°
1) Area of minor segment = ?
2) Area of Major segment = ?
(i) Area of minor segment OAPB
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 5

ii) Area of Minor Segment, APB :
= Area of OAPB – Area of AOAB
\(=78.5 \mathrm{cm}^{2} \cdot-\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\)
\(=78.5-\frac{1}{2} \times 10 \times 10\)
= 78.5 – 50
= 28.5 sq. cm.

(iii) Area of the Major Segment (AQBO):
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 6

Question 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:
Here, radius(r) = 21 cm and θ = 60°
(i) Circumference of the circle = 2πr
= 2 × \(\frac{22}{7}\) × 21 cm = 2 × 22 × 3 cm = 132 cm
MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 4
∴ Length of the arc APB
MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 5

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 6.
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 \(\sqrt{3}=1.73\)).
Solution:
Radius of circle, r = 15 cm.
Angle at the centre, θ = 60°
1) Area of the minor segment =?
2) Area of the Major segment =?
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 11
(i) Area of Minor segment, APB :
= Area of sector OAPB – Area of ∆OAB
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 12
= 117.75 sq. cm. – 97.31
= 20.44 sq. cm.

ii) Area of the Major segment:
= Area of Circle – Area of a segment of circle OAPB
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 13
= 3.14 × 225 – 20.44
= 707.14 – 20.44
= 686.06 cm2.

Question 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 π = 3.14 and \(\sqrt{3}=\) = 1.73 )
Solution:
Here θ = 120° and r = 12 cm
MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 8
Draw OM ⊥ AB
In ∆AOB, ∠O = 120°
By angle sum property,
∠A + ∠B + ∠O = 180°
⇒ ∠A + ∠B = 180° – 120° = 60°
∵ OB = OA = 12 cm
⇒ ∠A = ∠B = 30°
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 Q7
= 36 × 1.73 cm2 = 62.28 cm2
∴ Area of the minor segment = [Area of sector] – [Area of ∆AOB]
= [150.72 cm2] – [62.28 cm2] = 88.44 cm2

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 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 π = 3.14)
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 17
Solution:
Length of the rope, r = 5 m, θ = 90°
Area of segment =?
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 18

(ii) Length of the rope, r = 10 cm, θ = 90°
Area of segment =?
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 19
∴ Area of the field which is increased in the grazing
= Equation (2) – Equation (1)
= 78.5 – 19.63
= 58.87 sq. cm.

Question 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.
MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 12
Solution:
Diameter of the circle = 35 mm
∴ Radius (r) = \(\frac{35}{2}\) mm
(i) Circumference = 2πr
= 2 × \(\frac{22}{7} \times \frac{35}{2}\) mm = 22 × 5 mm = 110 mm
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 mm = 175 mm
∴ Total length of the silver wire = (110 + 175) mm = 285 mm

(ii) Since the circle is divided into 10 equal sectors,
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 Q9

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 10.
An umbrella has 8 ribs which are equally spaced (see the figure). Assuming the umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 23
Solution:
Radius of circle, r = 45 cm.
Angle at the centre, θ = \(\frac{360^{\circ}}{8}\) = 45°
Area between the two consecutive ribs of the umbrella =?
Area of each segment of the circle,
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 24

Question 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:
Here, radius (r) = 25 cm
Sector angle (θ) = 115°
∴ Total area cleaned at each sweep of the blades
MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 16

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 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 π = 3.14)
Solution:
Radius, r = 16.5 km. = \(\frac{33}{2}\) km.
Angle, θ = 80°
Area of the Sector =?
Area of the sea over which the ships are warned
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 27

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 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 cm2.
(Use \(\sqrt{3}=\) = 1.7)
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 28
Solution:
Each angle at the centre,
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 29
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 30
Area of segment APB design 1 :
= Area of OAPB – Area of ∆AOB
= 410.6 – 333.2
= 77.46 sq.cm.
∴ Total are of SIX designs = 6 × 77.46
= 464.8 sq.cm.
Cost of designing for 1 sq.cm, is Rs. 0.35
∴ Cost of designing for 464.8 sq.cm ………….
= 464.8 × 0.35
= Rs. 162.68

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 14.
Tick the correct answer in the following:
Area of a sector of angle p (in degrees) of a circle with radius R is
MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 21
Solution:
(D) Here, radius = R
Angle of a sector (θ) = p
∴ Area of the sector
MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 22

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