CBSE Class 10

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.2. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-14-ex-14-2/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 14
Chapter Name	Statistics
Exercise	Ex 14.2
Number of Questions Solved	6
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.2

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year.

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:
For Mode:

For Mean:

We conclude that the maximum number of patients in the hospital are of the age 36.8 years. While on an average the age of patient admitted to the hospital is 35.37 years.

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetimes of the components.
Solution:

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution:

Question 4.
The following distribution gives the state-wise teacher- student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures

Solution:

Hence, we conclude that most states/U.T. have a student teacher ratio of 30.6 and on an average this ratio is 29.2.

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one – day international cricket matches.

Find the mode of the data.
Solution:

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:

Solution:

We hope the NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.2, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-12-ex-12-3/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 12
Chapter Name	Areas Related to Circles
Exercise	Ex 12.3
Number of Questions Solved	16
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3

Question 1.
Find the area of the shaded region in the given figure, if PQ = 24cm, PR = 7cm and O is the centre of the circle.
Solution:

Question 2.
Find the area of the shaded region in the given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 400.
Solution:
∠AOC = 40° (given)
Radius of the sector AOC = 14 cm

Question 3.
Find the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
Solution:
ABCD is a square

Given: side of the square = 14 cm
∴ Area of the square = (side)² = (14)² = 196 cm²
Radius of the semicircle APD = \(\frac { 1 }{ 2 }\)(side of square) = \(\frac { 1 }{ 2 }\) x 14 = 7 cm
Area of the semicircle APD = \(\frac { 1 }{ 2 }\) πr² = \(\frac { 1 }{ 2 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 = 11 × 7 = 77cm²
Similarly, area of the semicircle BPC = 77 cm²
Total area of both the semicircles = 77 + 77 = 154 cm²
Area of the shaded region = Area of square – area of both semicircles
= 196 – 154 = 42 cm²

Question 4.
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:
Area of the equilateral triangle OAB

Question 5.
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:
Given: side of the square ABCD = 4 cm
Area of the square ABCD = 4 x 4 = 16 cm²
Radius of the quadrant at corner = 1 cm

∴ Area of the remaining portion = Area of the square – Area to be cut from square
= 16 – (\(\frac { 44 }{ 7 }\)) = 16 – \(\frac { 44 }{ 7 }\)
= \(\frac { 112-44 }{ 7 }\) = \(\frac { 68 }{ 7 }\)cm²

Question 6.
In a circular table cover of the radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure. Find the area of the design (shaded region).

Solution:
Radius of the circle(r) = 32cm
Area of the circle = πr²
= \(\frac { 22 }{ 7 }\) × 32 × 32 = \(\frac { 22528 }{ 7 }\)cm²
∴ An equilateral triangle is formed in the circle as shown
Angle subtended by et ch side at centre

Question 7.
In the 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:
Side of the square ABCD = 14 cm
Area of the squat e = (side)² = 14 x 14 = 196²

Question 8.
The given figure depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments 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:

Area of the tracks at both semicircular ends = 2 x 1100 = 2200 cm²
Area of the 2 rectangular portions = 2 x l x h = 2 x 106 x 10 = 2120 cm²
Total area of the track = area of the track at semicircular ends + area of the rectangular portions
= 2200 + 2120 = 4320 cm²

Question 9.
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:
Given: OA = 7 cm
Radius of the semicircle ABC = OA = 7 cm
Area of the semicircle ABC = \(\frac { 1 }{ 2 }\)πr² = \(\frac { 1 }{ 2 }\) x \(\frac { 22 }{ 7 }\) x 7 x 7 = 11 x 7 = 77 cm²
Diameter AB = 2(OA) = 2 x 7 = 14 and OA = OC = 7 cm (radius)

Question 10.
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 \(\sqrt{3}\) = 1.73205).

Solution:
Given: area of an equilateral triangle ABC = 17320.5 cm²
Let side of the triangle AB’C be ‘a’
∴ Area of the ∆ABC = \(\frac { \sqrt { 3 } }{ 4 } { a }^{ 2 }\)
\(\frac { \sqrt { 3 } }{ 4 } { a }^{ 2 }\) = 17320.5

Now,
area of the all 3 sectors = \(\frac { 3×31400 }{ 6 }\) = 15700 cm²
∴ Area of the shaded portion = Area of the equilateral triangle
– Area of the three sectors formed at each vertex)
= 17320.5 – 15700 = 1620.5 cm²

Question 11.
On a square handkerchief, nine circular designs each of the radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.

Solution:
Radius of the one circular design = 7 cm
Area of the one circular design = πr² = \(\frac { 22 }{ 7 }\) x 7 x 7 = 154 cm²
Now, area of the 9 circular designs = 9 x 154 = 1386 cm²
Diameter of the circular design = 7 x 2 = 14 cm
Side of the square = 3(diameter of one circle) = 3 x 14 = 42 cm
Area of the square = 42 x 42 = 1764 cm²
Area of the remaining portion of handkerchief
= Area of the square – (Area of the 9 circular designs)
= 1764 – 1386
= 378 cm²

Question 12.
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:
(i) Radius of the quadrant OACB = 3.5 cm

(ii) OD = 2cm and OB = 3.5 cm

Question 13.
In the figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use re = 3.14)
Solution:
Given: side of the square OABC = OA = 20 cm
Area of the square = 20 x 20 = 400 cm²
(Diagonal of the square)² = (side of the square)² + (side of the square)² (By pythagoras theorem)
Diagonal of the square = \(\sqrt{2}\) x (side of the square)
= \(\sqrt{2}\) x (20) = 20\(\sqrt{2}\)cm
Radius of the quadrant of circle = Diagonal of square = 20\(\sqrt{2}\)

Question 14.
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:
Given: ∠AOB = 30°
Radius of the sector AOB = 21 cm

Question 15.
In the figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Solution:
Radius of the quadrant of the circle = 14 cm

Area of the shaded region = Area of the semicircular region – Area of the region I
= 154 cm² – 56 cm² = 98 cm²

Question 16.
Calculate the area of the designed region in the figure common between the two quadrants of the circles of the radius 8 cm each.
Solution:
Side of the square = 8 cm
Area of the square = 8 x 8 = 64 cm²
Radius of the quadrant (formed at vertex) = 8 cm

We hope the NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.2. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-12-ex-12-2/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 12
Chapter Name	Areas Related to Circles
Exercise	Ex 12.2
Number of Questions Solved	14
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.2

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 the sector (r) = 6 cm
Central angle of the sector = 60°

Question 2.
Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
Let radius of the circle = r
∴ Circumference of the circle = 2πr

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 of the clock = 14 cm

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 segment (Use ? = 3.14)
Solution:
Given: radius of the circle = 10 cm
Angle subtended by chord at centre = 90°
(i) Area of the minor segment

(ii) Area of the major segment = Area of the circle – Area of the minor segment
= πr² – 28.5 = 3.14 x 10 x 10-28.5
= 314-28.5 = 285.5 cm²

Question 5.
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) length of the arc.
(ii) area of the sector formed by the arc.
(iii) area of the segment formed by the corresponding chord.
Solution:
Radius of the circle = 21 cm
Angle at the centre = 60°

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 the circle = 15 cm
Angle subtended by chord at centre = 60°
Area of the sector = \(\frac{\pi r^{2} \theta}{360^{\circ}}\) = 3.14 x \(\frac{15 \times 15 \times 60^{\circ}}{360^{\circ}}\) = 117.75 cm²
Area of the triangle formed by radii and chord = \(\frac { 1 }{ 2 }\)r²θ
= \(\frac { 1 }{ 2 }\)(15)² sin 60° = \(\frac { 1 }{ 2 }\) x 15 x 15 x \(\frac{\sqrt{3}}{2}\) = 97.31 cm²
Area of the minor segment = Area of the sector – Area of the triangle formed by radii and chord
= 117.75 – 97.31 = 20.44 cm²
Area of the circle = πr² = 3.14 x 15 x 15 = 706.5 cm²
Area of the circle – Area of the minor segment
= 706.5 – 20.44 = 686.06 cm²

Question 7.
A chord of a circle of the 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:
Radius of the circle = 12 cm
Angle subtended by chord at centre = 120°

Area of the corresponding segment = Area of the sector – Area of A formed by radii and chord
= 150.72 – 62.28 cm² = 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 n = 3.14)
Solution:
(i) Length of the rope = Radius of the sector grazed by horse = 5 m
Here, angle of the sector = 90°

Length of the rope is increased from 5 m to 10 m
New radius of sector grazed by horse = 10 m

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 figure.

Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
Solution:
Length of one diameter = 35 mm
Total length of 5 diameters = 5 x 35 mm = 175 mm
Circumference of the circle = 2π = 2 x \(\frac { 22 }{ 7 }\) x \(\frac { 35 }{ 2 }\) = 110 mm
(i) Total length of the wire used = length of 5 diameters + circumference of brooch
= 175 + 110 = 285 mm

(ii) Total sectors are 10.

Question 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:
Radius of the circle = 45 cm
Number of ribs = 8

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:
Given: length of blade of wiper = radius of sector sweep by blade = 25 cm
Area cleaned by each sweep of the blade = area of sector sweep by blade
Angle of the sector formed by blade of wiper =115°

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:
Angle of the sector = 80°
Distance covered = 16.5 km
Radius of the sector formed = 16.5 km

Question 13.
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 ₹ 0.35 per cm². (Use \(\sqrt{3}\) = 1.7)

Solution:
Radius of the cover = 28 cm
∵ There are six equal designs

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) \(\frac{p}{180^{\circ}}\) × 2πR
(b) \(\frac{p}{180^{\circ}}\) × πR²
(c) \(\frac{p}{360^{\circ}}\) × 2πR
(d) \(\frac{p}{720^{\circ}}\) × 2πR²
Solution:
Sector angle isp in degrees
Radius of the circle = R
Area of the sector = \(\frac{\pi \mathrm{R}^{2} p}{361^{6}}\) = \(\frac{\left(\pi R^{2} p\right) 2}{720^{\circ}}\)
= \(\frac{p}{720^{\circ}}\) × 2πR²

We hope the NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.2, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.2, drop a comment below and we will get back to you at the earliest.