Students can Download Class 10 Maths Chapter 5 Areas Related to Circles Ex 5.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.
Karnataka State Syllabus Class 10 Maths Chapter 5 Areas Related to Circles Ex 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°.
Answer:
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 \(=\frac{\theta}{360} \times \pi r^{2}\)
Question 2.
Find the area of a quadrant of a circle whose circumference is 22 cm.
Answer:
Circumference of a circle, 2πr = 22 cm.
Area of a quadrant of a circle =?
2πr = 22
r = \(\frac{7}{2}\)
Now, Area of a Quardrant of a circle
= \(9 \frac{5}{8}\) cm2 (or) 9.625cm2
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.
Answer:
Length of the minute hand of a clock, r = 14 cm.
Area swept by the minute hand in 5 minutes \(=360 \times \frac{5}{60}\) = 30°
Angle of segment of a circle, θ = 30°
∴ Area swept by the minute hand in 5 minute,
= 51\(\frac{1}{3}\) cm2
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.
The angle at the centre of the circle, θ = 90°
1) Area of minor segment =?
2) Area of Major segment =?
(i) Area of minor segment ∆OAB
= 78.5cm2 – 50cm2
= 28.5cm2
ii) Area of major segment = Area of circle – Area of minor sector
= πr2 – 78.5
= 3.14 x (10)2 – 78.5
= 314 – 78.5
= 235.5 cm2
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.
Answer:
Radius of the circle, r = 21 cm.
Angle at the centre of the circle, θ = 60°
1) the length of the arc?
2) Area of the sector formed by the arc?
3) Area of the segment.
1) The length of arc OAPB
= 22 cm
2) Area of the sector formed by the arc
= 231 sq. cm.
3) Area of the segment APB:
= Area of the sector OAPB – Area of the equilateral ∆OAB
= \(\frac{θ}{360}\) × πr2 – area of equilateral ∆le
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 \(\sqrt{3}=1.73\)).
Answer:
Radius of circle, r = 15 cm.
Angle at the centre, θ = 60°
Area of the minor segment – Area of the minor sector – Area of ∆OAB
Area of min or segment = 20.44 cm2
area of minor segment = area of circle – area of minor segment
= πr2 – 20.14
= 3.14 x (15)2 – 20.44
= 3.14 x 225 – 20.44
= 706.5 – 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 \(\sqrt{3}=1.73\)).
Answer:
Radius of the circle, r = 12 cm.
Angle at the centre, θ = 120°
Area of the minor segment – Area of the minor sector – Area of ∆OAB
= 12 ×7.37
= 88..44 cm2
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)
Answer:
Let the horse be tied at point O and the length of the rope is OH
(i) The area of the part of the field in which the horse can graze = Area of the Quadrant of a circle ( OAHB)
= \(\frac{\pi r^{2}}{4}\) = \(\frac{1}{4}\) ×3.14 ×(5)2
= \(\frac{78.5}{4}\) = 19.625m2
(ii) r = 10
Required area = \(\frac{\pi r^{2}}{4}\) = \(\frac{1}{4}\) ×3.14 ×(10)2
= \(\frac{314}{4}\) = 78.5 m2
Increase in grazing area = 78.5 – 19.625
= 58.875 m2
Question 9.
A brooch is made with silver wire in the form of a circle with diameter of 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.
Answer:
Diameter, d = 35mm θ = \(\frac{360^{\circ}}{10}\) = 36°
Radius, r = \(\frac{35}{2}\) mm
(i) Length of the silver wire required :
Length of the wire used in making 5 diameter + Circumference of Silver, wire required.
(ii) Area of each sector of the brooch = ?
Diameter of the brooch, d = 35mm
Radius, r = \(\frac{35}{2}\) mm
Angle at the centre, θ = \(\frac{360^{\circ}}{10}\) = 36°
The Area of each sector of the Brooch:
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.
Answer:
r = 45 cm,
θ = \(\frac{360}{8}\) = 45° [8 equally spaced ribs].
Area between two consecutive ribs
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.
Answer:
Radius of the wiper of the car, r = 25 cm.
Angle at the centre, θ = 115°
Area of the sector =?
Area of the blade when it is swept once
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)
Answer:
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
= \(\frac{θ}{360}\) × πr2
= \(\frac{80}{360}\) × 3.14 ×(16.5)2
= 189.97 km2.
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)
Answer:
r = 28 cm, θ = \(\frac{360}{6}\) = 60°
Area of design
= Area of sector – area of equilateral ∆le
Total Area of design = 77.47 x 6
= 464.82 cm2
Post of making design = 464.82 x 0.35
= 162.68 Rs.
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
a) \(\frac{p}{180}\) × 2πR
b) \(\frac{p}{180}\) × πR2
c) \(\frac{p}{360}\) × 2πR
d) \(\frac{p}{720}\) × 2πR2
Answer:
Area of a sector of angle p° of a circle with radius
Area of a sector = \(\frac{θ}{360}\) × πr2
Area = \(\frac{p}{360}\) πR2 = \(\frac{P}{2(360)}\) 2πR2
= \(\frac{p}{720}\) × 2πR2
∴ Correct answer is d) \(\frac{p}{720}\) × 2πR2