ncert class 10 math solutions

NCERT Solutions for Class 10 Maths are solved by experts of LearnCBSE.in in order to help students to obtain excellent marks in their board examination. All the questions and answers that are present in the CBSE NCERT Books have been included on this page. We have provided all the Class 10 Maths NCERT Solutions with a detailed explanation i.e., we have solved all the questions with step-by-step solutions in understandable language. So students having great knowledge of NCERT Solutions Class 10 Maths can easily make a grade in their board exams. Read on to find out more about NCERT Solutions for Class 10 Mathematics. You can also practice Extra Questions for Class 10 Maths.

NCERT Solutions For Class 10 Maths Chapter 15 Probability Ex 15.1

Get Free NCERT Solutions for Class 10 Maths Chapter 15 Ex 15.1 PDF. Probability Class 10 Maths NCERT Solutions are extremely helpful while doing your homework. Exercise 15.1 Class 10 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 15 Maths Class 10 Probability Exercise 15.1 provided in NCERT TextBook.

Topics and Sub Topics in Class 10 Maths Chapter 15 Probability:

Section NameTopic Name
15Probability
15.1Introduction
15.2A Theoretical Approach
15.3Summary

NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1.

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 15
Chapter NameProbability
ExerciseEx 15.1
Number of Questions Solved20
CategoryNCERT Solutions

Ex 15.1 Class 10 Maths Question 1.
Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = ………
(ii) The probability of an event that cannot happen is ……… Such an event is called ………
(iii) The probability of an event that is certain to happen is ………. Such an event is called ………
(iv) The sum of the probabilities of all the elementary events of an experiment is ………..
(v) The probability of an event is greater than or equal to …………. and less than or equal to ………..
Solution:
Probability Class 10 Maths NCERT Solutions Ex 15.1 pdf download Q1

You can also download the free PDF of  Ex 15.1 Class 10 Probability NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

Download NCERT Solutions For Class 10 Maths Chapter 15 Probability PDF

Ex 15.1 Class 10 Maths Question 2.
Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Solution:
Probability Class 10 Maths NCERT Solutions Ex 15.1 pdf download Q2

Ex 15.1 Class 10 Maths Question 3.
Why is tossing a coin considered to be a fair way of deciding which team should get the bail at the beginning of a football game?
Solution:
Probability Class 10 Maths NCERT Solutions Ex 15.1 pdf download Q3

Ex 15.1 Class 10 Maths Question 4.
Which of the following cannot be the probability of an event?
(A) 23
(B) -1.5
(C) 15%
(D) 0.7
Solution:

Probability Class 10 Maths NCERT Solutions Ex 15.1 pdf download Q4

Ex 15.1 Class 10 Maths Question 5.
If P (E) = 0.05, what is the probability of ‘not E’?
Solution:
Probability Class 10 Maths NCERT Solutions Ex 15.1 pdf download Q5

Ex 15.1 Class 10 Maths Question 6.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Solution:
Probability Class 10 Maths NCERT Solutions Ex 15.1 pdf download Q6

Ex 15.1 Class 10 Maths Question 7.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Solution:
Probability Class 10 Maths NCERT Solutions Ex 15.1 pdf download Q7

Ex 15.1 Class 10 Maths Question 8.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is
(i) red?
(ii) not red?
Solution:
Probability Class 10 Maths NCERT Solutions Ex 15.1 pdf download Q8

Ex 15.1 Class 10 Maths Question 9.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red?
(ii) white?
(iii) not green?
Solution:
Ex 15.1 Class 10 Maths NCERT Solutions pdf download Q9

Ex 15.1 Class 10 Maths Question 10.
A piggy bank contains hundred 50 p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin
(i) will be a 50 p coin?
(ii) will not be a ₹ 5 coin?
Solution:
Ex 15.1 Class 10 Maths NCERT Solutions pdf download Q10

Ex 15.1 Class 10 Maths Question 11.
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see figure). What is the probability that the fish taken out is a male fish?
NCERT Solutions For Class 10 Maths Chapter 15 Probability Ex 15.1 Q11
Solution:
Ex 15.1 Class 10 Maths NCERT Solutions pdf download Q11

Ex 15.1 Class 10 Maths Question 12.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see figure.), and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?
NCERT Solutions For Class 10 Maths Chapter 15 Probability Ex 15.1 Q12
Solution:
Ex 15.1 Class 10 Maths NCERT Solutions pdf download Q12

Ex 15.1 Class 10 Maths Question 13.
A die is thrown once. Find the probability of getting
(i) a prime number
(ii) a number lying between 2 and 6
(ill) an odd number
Solution:
Ex 15.1 Class 10 Maths NCERT Solutions pdf download Q13

Ex 15.1 Class 10 Maths Question 14.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds
Solution:
Ex 15.1 Class 10 Maths NCERT Solutions pdf download Q14

Ex 15.1 Class 10 Maths Question 15.
Five cards – the ten, jack, queen, king and ace of diamonds, are well shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is
(a) an ace?
(b) a queen?
Solution:
Ex 15.1 Class 10 Maths NCERT Solutions pdf download Q15

Ex 15.1 Class 10 Maths Question 16.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
Ex 15.1 Class 10 Maths NCERT Solutions pdf download Q16

Ex 15.1 Class 10 Maths Question 17.
(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:
Ch 15 Maths Class 10 NCERT Solutions ex 15.1 pdf download Q17

Ex 15.1 Class 10 Maths Question 18.
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two digit number.
(ii) a perfect square number.
(iii) a number divisible by 5.
Solution:
Ch 15 Maths Class 10 NCERT Solutions ex 15.1 pdf download Q18

Ex 15.1 Class 10 Maths Question 19.
A child has a die whose six faces show the letters as given below:
NCERT Solutions For Class 10 Maths Chapter 15 Probability Ex 15.1 Q19
The die is thrown once. What is the probability of getting
(i) A?
(ii) D?
Solution:
Ch 15 Maths Class 10 NCERT Solutions ex 15.1 pdf download Q19

Ex 15.1 Class 10 Maths Question 20.
Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1 m?
NCERT Solutions For Class 10 Maths Chapter 15 Probability Ex 15.1 Q20
Solution:
Ch 15 Maths Class 10 NCERT Solutions ex 15.1 pdf download Q20

Ex 15.1 Class 10 Maths Question 21.
A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) she will buy it?
(ii) she will not buy it?
Solution:
Ch 15 Maths Class 10 NCERT Solutions ex 15.1 pdf download Q21

Ex 15.1 Class 10 Maths Question 22.
Two dice, one blue and one grey, are thrown at the same time. Now
(i) Complete the following table:
NCERT Solutions For Class 10 Maths Chapter 15 Probability Ex 15.1 Q22
(ii) A student argues that-there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 111. Do you agree with this argument? Justify your answer.
Solution:
Ch 15 Maths Class 10 NCERT Solutions ex 15.1 pdf download Q22

Ex 15.1 Class 10 Maths Question 23.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result, i.e. three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
Ch 15 Maths Class 10 NCERT Solutions ex 15.1 pdf download Q23

Ex 15.1 Class 10 Maths Question 24.
A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment.]
Solution:
Ch 15 Maths Class 10 NCERT Solutions ex 15.1 pdf download Q24

Ex 15.1 Class 10 Maths Question 25.
Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes- two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 13.
(ii) If a die is thrown, there are two possible outcomes- an odd number or an even number. Therefore, the probability of getting an odd number is 12.
Solution:
Ch 15 Maths Class 10 NCERT Solutions ex 15.1 pdf download Q25

Class 10 Maths Probability Mind Maps

Probability – An Experimental (Empirical) Approach

Let n be the total number of trails. The empirical probability of an event E happening, is given by
NCERT Solutions For Class 10 Maths Chapter 15 Probability Ex 15.1 Mind Map 1

(i) Experiment : An operation which can produce some well defined outcomes is known as experiment.
(ii) Trail : Performing of an experiment is called trial.
(iii) Equally likely outcomes : Outcomes of trial are equally likely if there is no reason to accept one in preference to the others.
(iv) Sample space : The set of all possible outcomes of an experiment is called sample space.
(v) Elementary event : An event having only one outcome
Note that the sum of probabilities of all the elementary events of an experiment is 1.

Probability – A Theoretical Approach(Classical Probability)

If an event ‘A’ can happen is ‘m’ ways and does not happen in ‘n’ ways, then the probability of occurrence of event ‘A’ denoted by P(A) is given by
Number of favourable outcomes m
NCERT Solutions For Class 10 Maths Chapter 15 Probability Ex 15.1 Mind Map 2

Probability of Impossible and Sure Events
The probability of an event which is impossible to occur is 0 and such an event is called impossible event, i.e; for impossible event T, P(I) = 0
The probability of an event which is sure or certain to occur is 1 and such an event is called sure event or certain event.
i.e; for sure event or certain event ‘s’, P(s) = 1

Range of the Probability of an Event

From the definition of the probability P(E), we see that the numerator (number of outcomes favourable to the event E) is always equal or greater than 0 but less than or equal to the denominator (the number of all possible outcomes). Therefore,
0 ≤ P(E) ≤ 1

Complementary Events

The event representing (‘not E) is called the complement of event ‘E’ and we say that the events E and E¯¯¯¯ are complementary events,
NCERT Solutions For Class 10 Maths Chapter 15 Probability Ex 15.1 Mind Map 3

NCERT Solutions for Class 10 Maths Chapter 15 Probability (Hindi Medium) Ex 15.1

NCERT Solutions for Class 10 Maths Chapter 15
NCERT Solutions for Class 10 Maths Chapter 15 Exercise 15.1 Probability
10 maths 15.1 optional ex.
NCERT Solutions for Class 10 Maths Chapter 15 Exercise 15.1 Probability in pdf form
NCERT Solutions for Class 10 Maths Chapter 15 Exercise 15.1 Probability in english medium
NCERT Solutions for Class 10 Maths Chapter 15 Exercise 15.1
10 Maths Exercise 15.1
NCERT Solutions for Class 10 Maths Chapter 15 Exercise 15.1 in free pdf form
10 Maths Exercise 15.1 solutions in pdf
NCERT Solutions for Class 10 Maths Chapter 15 Exercise 15.1 in english medium
NCERT Solutions for Class 10 Maths Chapter 15 Exercise 15.1 free download
10 Maths Exercise 15.1 answers guide free
NCERT Solutions for Class 10 Maths Chapter 15 Exercise 15.1 in Hindi medium for cbse and up board.
NCERT Solutions for Class 10 Maths Chapter 15 Exercise 15.1 in Hindi medium
NCERT Solutions for Class 10 Maths Chapter 15 Exercise 15.1 updated for 2018-19 up board
NCERT Solutions for Class 10 Maths Chapter 15 Exercise 15.1 updated as per new syllabus 2018-19.
10 Maths Chapter 15 Exercise 15.1 Probability solutions
10 Maths Chapter 15 Exercise 15.1 Probability solutions in pdf form
10 Maths Chapter 15 Exercise 15.1 Probability solutions for up board
10 Maths Chapter 15 Exercise 15.1 Probability solutions in english medium
10 Maths Chapter 15 Exercise 15.1 Probability solutions free guide
Class 10 Maths Chapter 15 Exercise 15.1 in Hindi medium
10 Maths Chapter 15 Exercise 15.1 Probability solutions all question answers
10 Maths Chapter 15 Exercise 15.1 Probability solutions free to download
10 Maths Chapter 15 Exercise 15.1 Probability solutions in hindi medium
10 Maths Chapter 15 Exercise 15.1 Probability solutions hindi me
10 Maths Chapter 15 Exercise 15.1 Probability solutions for mp board
10 maths ex. 15.1pdf

NCERT Solutions for Class 10 Maths

NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.1

Get Free NCERT Solutions for Class 10 Maths Chapter 14 Ex 14.1 PDF. Statistics Class 10 Maths NCERT Solutions are extremely helpful while doing your homework. Exercise 14.1 Class 10 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 14 Maths Class 10 Statistics Exercise 14.1 provided in NCERT TextBook.
Topics and Sub Topics in Class 10 Maths Chapter 14 Statistics:

Section NameTopic Name
14Statistics
14.1Introduction
14.2Mean Of Grouped Data
14.3Mode Of Grouped Data
14.4Median Of Grouped Data
14.5Graphical Representation Of Cumulative Frequency Distribution
14.6Summary

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1.

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 14
Chapter NameStatistics
ExerciseEx 14.1
Number of Questions Solved9
CategoryNCERT Solutions

Ex 14.1 Class 10 Maths Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.1 Q1
Which method did you use for finding the mean, and why?
Solution:
Statistics Class 10 Maths NCERT Solutions Ex 14.1 PDF Download Q1

You can also download the free PDF of  Ex 14.1 Class 10 Statistics NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

Download NCERT Solutions For Class 10 Maths Chapter 14 Statistics PDF

Ex 14.1 Class 10 Maths Question 2.
Consider the following distribution of daily wages of 50 workers of a factory.
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.1 Q2
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
Statistics Class 10 Maths NCERT Solutions Ex 14.1 PDF Download Q2

Ex 14.1 Class 10 Maths Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f.
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.1 Q3
Solution:
Statistics Class 10 Maths NCERT Solutions Ex 14.1 PDF Download Q3

Ex 14.1 Class 10 Maths Question 4.
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.1 Q4
Solution:
Ex 14.1 Class 10 Maths NCERT Solutions PDF Download Q4

Ex 14.1 Class 10 Maths Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.1 Q5
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution:
Ex 14.1 Class 10 Maths NCERT Solutions PDF Download Q5

Ex 14.1 Class 10 Maths Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.1 Q6
Find the mean daily expenditure on food by a suitable method.
Solution:
Exercise 14.1 Class 10 Maths NCERT Solutions PDF Download Q6

Ex 14.1 Class 10 Maths Question 7.
To find out the concentration of SO2 in the air (in parts per million, i.e. ppm), the data was collected for 30 localities in a certain city and is presented below:
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.1 Q7
Find the mean concentration of SO2 in the air.Solution:
Exercise 14.1 Class 10 Maths NCERT Solutions PDF Download Q7

Ex 14.1 Class 10 Maths Question 8.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.1 Q8
Solution:
Chapter 14 Maths Class 10 NCERT Solutions Ex 14.1 PDF Download Q8

Ex 14.1 Class 10 Maths Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.1 Q9
Solution:
Chapter 14 Maths Class 10 NCERT Solutions Ex 14.1 PDF Download Q9

Class 10 Maths Statistics Mind Maps

Basic Terms

Class limits : Suppose mark obtained by all of the students are divided into class intervals 25 – 35, 35 – 45 and so interval on.
In class interval 25 – 35, 25 is called lower class limit and 35 is called upper class limit.
Class size : The difference between upper and lower class limit.
Class mark : It is given by
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.1 Mind Map 1

Ungrouped and Grouped Data

The data obtained in original form are called raw data or ungrouped data.
To put the data in a more condensed form, we make groups of suitable size, and mention the frequency of each group. Such a table is called grouped data.

Mean

For Ungrouped:
Consider ‘n’ observations in ungrouped data as : x1, x2, x3, …. xn. The mean of these observations is:
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.1 Mind Map 2
(i) Direct method:
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.1 Mind Map 3
where ‘fi‘ is the frequency corresponding to the class mark ‘xi
(ii) Assumed mean method:
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.1 Mind Map 4
Where A is assumed mean and di = xi – A
(iii) step deviation method:
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.1 Mind Map 5

Mode

(i) For Ungrouped Data:
The mode is that observation which occurs most frequently, i.e., an observation with maximum frequency.
(ii) For Grouped Data:
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.1 Mind Map 6
l =lower limit of the modal class,
f1 = frequency of the modal class
f 2 = frequency of the class suceeding the modal class,
h = size of the class interval,
f0 = frequency of the class preceeding the modal class.
The class with the maximum frequency is called modal class.

Median

Median is a measure of central tendency which gives the value of the middle-most observation in the data.
(i) For Ungrouped Data:
Arrange the data in ascending order.
If number of data n is odd, then the median is (n+12)th observation.
If number of data n is even, then the median is the average of (n2)th and (n2+1)th observations.

(ii) For Grouped Data:
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.1 Mind Map 7
l = lower limit of the median class,
c.f = cumulative frequency of the class proceeding the median class,
f = frequency of the median class,
h = class size,
n = number of observations.
The class whose cumulative frequency is greater than (and nearest to) n2 is called median class.

Relationship Between Mean, Mode and Median
3 Median = Mode + 2 Mean

Cumulative Frequency Curve (Ogive)

(i) The smooth free hand curve is formed by joining the points (xi, fi) where x; is the upper limit of a class and f is the corresponding c.f. The curve so obtained is called a cumulative frequency curve, or an ogive of the less than type.
(ii) The smooth free hand curve is formed by joining the points (xi, fi) where Xi is the lower limit of a class and f is the corresponding c.f. The curve so obtained is called a cumulative frequency curve, or an ogive of the more than type.

Median by Graph

(i) Draw the ogive of the less than type and ogive of the more than type on the same axis. The two ogives will intersect each other at a point. From this point, if we draw a perpendicular on the x-axis, the point at which it cuts the x-axis gives us the median.
(ii) Draw the ogive of the less than type, then locate n2 on the y-axis (n = number of observations).

From this point on y-axis, draw a line parallel to x-axis cutting the less than ogive at a point. From this point draw a perpendicular on the x-axis, the point at which the perpendicular cuts the x-axis gives us the median.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics (Hindi Medium) Ex 14.1

NCERT Solutions for class 10 Maths Chapter 14
NCERT Solutions for class 10 Maths Chapter 14 Statistics Exercise 14.1 in pdf form
NCERT Solutions for class 10 Maths Chapter 14 Exercise 14.1
NCERT Solutions for class 10 Maths Chapter 14 Exercise 14.1 in free pdf
NCERT Solutions for class 10 Maths Chapter 14 Exercise 14.1 in english medium
NCERT Solutions for class 10 Maths Chapter 14 Exercise 14.1 for mp, up board
NCERT Solutions for class 10 Maths Chapter 14 Exercise 14.1
NCERT Solutions for class 10 Maths Chapter 14 Exercise 14.1 in hindi medium free
NCERT Solutions for class 10 Maths Chapter 14 Exercise 14.1 in pdf form free
NCERT Solutions for class 10 Maths Chapter 14 Exercise 14.1 for up board
NCERT Solutions for class 10 Maths Chapter 14 Exercise 14.1 guide in hindi
NCERT Solutions for class 10 Maths Chapter 14 Exercise 14.1 for mp board
Statistics for class 10 - TA 1

NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

Get Free NCERT Solutions for Class 10 Maths Chapter 13 Ex 13.1 PDF. Surface Areas and Volumes Class 10 Maths NCERT Solutions are extremely helpful while doing your homework. Exercise 13.1 Class 10 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 13 Maths Class 10 Surface Areas and Volumes Exercise 13.1 provided in NCERT TextBook.

Topics and Sub Topics in Class 10 Maths Chapter 13 Surface Areas and Volumes:

Section NameTopic Name
13Surface Areas And Volumes
13.1Introduction
13.2Surface Area Of A Combination Of Solids
13.3Volume Of A Combination Of Solids
13.4Conversion Of Solid From One Shape To Another
13.5Frustum Of A Cone
13.6Summary

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1.

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 13
Chapter NameSurface Areas and Volumes
ExerciseEx 13.1
Number of Questions Solved9
CategoryNCERT Solutions

Ex 13.1 Class 10 Maths Question 1.
2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
NCERT Solutions for Class 10 Maths Chapter 13 Pdf Q1

Ex 13.1 Class 10 Maths Question 2.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:
NCERT Solutions for Class 10 Maths Chapter 13 Pdf Q2

You can also download the free PDF of  Ex 13.1 Class 10 Surface Areas and Volumes NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

Download NCERT Solutions For Class 10 Maths Chapter 13 Surface Areas and Volumes PDF

Ex 13.1 Class 10 Maths Question 3.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
Ex 13.1 Class 10 Maths NCERT Solutions Q3

Ex 13.1 Class 10 Maths Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Solution:
Exercise 13.1 Class 10 Maths NCERT Solutions PDF Q4

Ex 13.1 Class 10 Maths Question 5.A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter d of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:
Ch 13 Maths Class 10 Maths NCERT Solutions PDF Q5

Ex 13.1 Class 10 Maths Question 6.A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Solution:
Chapter 13 Maths Class 10 NCERT Solutions PDF Q6

Ex 13.1 Class 10 Maths Question 7.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2. (Note that the base of the tent will not be covered with canvas.)
Solution:
Ex 13.1 Class 10 Maths NCERT Solutions PDF Q7

Ex 13.1 Class 10 Maths Question 8.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Solution:
Ex 13.1 Class 10 Maths NCERT Solutions PDF Q8

Ex 13.1 Class 10 Maths Question 9.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Q9
Solution:
Exercise 13.1 Class 10 Maths NCERT Solutions PDF Q9

Class 10 Maths Surface Areas and Volumes Mind Maps

Surface Areas and Volumes of Solids

(i) Cuboid:

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Mind Map 1
Volume = l × b × h
Total surface area = 2 [lb + bh + hl]
Lateral surface area = 2 [bh + hl]
Diagonal of the coboid = ℓ2+b2+h2−−−−−−−−−−√

(ii) Cube:

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Mind Map 2
Volume = a3
Total surface area = 6a2
Lateral surface area = 4a2
Diagonal of a cube = √3a

(iii) Cylinder:
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Mind Map 3
(a) Right circular cylinder:

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Mind Map 4
Volume = πr2h
Curved Surface Area = 2nrh
Total Surface Area= 2πrh + 2πr2 = 2πr(r + h)

(b) Right circular hollow cylinder:

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Mind Map 5
Let r and R be internal & external radii.
Volume = n(R2 – r2)h
Curved Surface Area = 2π(R + r)h
Total Surface Area = 2π(R + r)h + 2π(R2 – r2)
= 2π(R + r)(h + R — r)

(iv) Right circular cone:

Slant height, l = r2+h2−−−−−−√
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Mind Map 6
Volume = 13πr2h
Curved Surface Area = πrl
Total Surface Area = πrl + πr2

Surface Areas and Volumes of Sphere and Hemisphere

(i) Sphere:

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Mind Map 7
Volume = 43≠r3
Surface area =4πr2

(ii) Hemisphere:

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Mind Map 8
Volume = 23≠r3
C.S.A = 2πr2
T.S.A = 3πr2

(iii) Hemispherical shell:

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Mind Map 9
Volume = 23π(R3−r3)
Curved Surface Area = 2π(R2 + r2)
Total Surface Area = 2π(r2 + R2) + n (R2 – r2)
= π (r2 + 3R2)

Surface Areas of a Combination of Solids

The surface area of a solid which is a combination of two or more solids is calculated by adding the surface areas of the individual solids which are visible in the new solid formed.
For Example:
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Mind Map 10
If we consider the surface of the newly formed object as given in the figure above, we would be able to see only the curved surfaces of the two hemispheres and the curved surface of the cylinder.
So, the total surface area of the new solid is the sum of the curved surface areas of each of the individual parts. This gives, TSA of new solid = CSA of one hemisphere + CSA of cylinder + CSA of other hemisphere

Volume of a Combination of Solids

Whenever solid is formed by combining two or more solids, then the amount of matter present in the new solid is equal to the sum of amounts of matter in the constituting solids. Volume of new solid = sum of the volumes of the individual solids

Conversion of Solid form One Shape to Another

(i) When a solid is converted from one shape to other, then its volume remains same only its shape and size changes.
(ii) If a solid is converted into a number of small identical solids, then Number of small items
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Mind Map 11

Frustum of a Cone [Very Important for Board Exam]

When we slice (or cut) through a cone with a plane parallel to its base (see below figure ) and remove the cone that is formed on one side of that plane, the part that is now left over on the other side of the plane is called a frustum of the cone.
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Mind Map 12
(i) Volume of the frustum of cone 13πh(r21+r22+r1r2)
(ii) C.S.A. of the frustum of cone = π(r1 + r2)l,
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Mind Map 13
where l = h2+(r1−r2)2−−−−−−−−−−−−√
(iii) T.S.A. of the frustum of cone
l = h2+(r1−r2)2−−−−−−−−−−−−√

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes (Hindi Medium) Ex 13.1

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.1 Surface Areas and Volumes
10 maths ex. 13.1
10 maths exercise 13.1
NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.1 Surface Areas and Volumes in pdf
class 10 maths ex. 13.1
13.1 Surface Areas and Volumes in pdf
NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.1 in PDF form.
NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.1 surface areas and volumes in English medium free for 2018-19.
NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.1 for CBSE and UP Board 2018-19.
Class 10 Maths Exercise 13.1 SolutionsClass 10 Maths Exercise 13.1 Solutions
Class 10 Maths Exercise 13.1 sols in Hindi medium.
Class 10 Maths Exercise 13.1 solutions for CBSE and UP Board 2018-19 updated.
Class 10 Maths Chapter 13 Exercise 13.1 in PDF form.
Class 10 Maths Chapter 13 Exercise 13.1 for CBSE and UP Board 2018-19.

NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1

Get Free NCERT Solutions for Class 10 Maths Chapter 12 Ex 12.1 PDF. Areas Related to Circles Class 10 Maths NCERT Solutions are extremely helpful while doing your homework. Exercise 12.1 Class 10 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 12 Maths Class 10 Areas Related to Circles Exercise 12.1 provided in NCERT TextBook.

Topics and Sub Topics in Class 10 Maths Chapter 12 Areas Related to Circles:

Section NameTopic Name
12Areas Related to Circles
12.1Introduction
12.2Perimeter And Area Of A Circle – A Review
12.3Areas Of Sector And Segment Of A Circle
12.4Areas Of Combinations Of Plane Figures
12.5Summary

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1 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 12.1

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 12
Chapter NameAreas Related to Circles
ExerciseEx 12.1
Number of Questions Solved5
CategoryNCERT Solutions

Ex 12.1 Class 10 Maths Question 1.
The radii of the two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.
Solution:
Ex 12.1 Class 10 Maths NCERT Solutions PDF Q1

Ex 12.1 Class 10 Maths Question 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:
Exercise 12.1 Class 10 Maths NCERT Solutions PDF Q2

You can also download the free PDF of  Ex 12.1 Class 10 Areas Related to Circles NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

Download NCERT Solutions For Class 10 Maths Chapter 12 Areas Related to Circles PDF

Ex 12.1 Class 10 Maths Question 3.
The given figure depicts an archery target marked with its five scoring regions from the center outwards as Gold, Red, Blue, Black, and White.
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1 Q3
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:
Exercise 12.1 Class 10 Maths NCERT Solutions PDF Q3
Exercise 12.1 Class 10 Maths NCERT Solutions PDF Q3.1

Formulae Handbook for Class 10 Maths and Science

Ex 12.1 Class 10 Maths Question 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:
Ch 12 Maths Class 10 NCERT Solutions Ex 12.1 PDF Q4

Ex 12.1 Class 10 Maths Question 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) n units
(c) 4 units
(d) 7 units
Solution:
Ex 12.1 Class 10 Maths NCERT Solutions PDF Q5

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles (Hindi Medium) Ex 12.1

NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1 in English medium
NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1 Areas related to circles for 2018-19.
NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1 in Hindi medium for cbse and up board.
NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1 updated as per new syllabus 2018-19.
Class 10 Maths Chapter 12 Exercise 12.1 Areas related to circles
Exercise 12.1 updated as per new syllabus 2018-19.
Class 10 Maths Chapter 12 Exercise 12.1
NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1

Class 10 Maths Areas Related To Circles Mind Maps

Terms Related To Circle

(i) Chord: A line segment joining any two points on a circle.
(ii) Arc: A piece of a circle between two points on the circle is called an arc.
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1 Mind Map 1
The arc less than the semicircular arc is called minor arc and the one greater than the semi-circular arc is called major arc.
(iii) Sector: The portion of a circular region enclosed by two radii and the corresponding arc is called a sector of the circle.
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1 Mind Map 2
Sector smaller than the semi-circle is called minor sector and the sector larger than the semi-circle is called major sector.
(iv) Segment: The portion of a circular region enclosed between a chord and the corresponding arc is called a segment of the circle.
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1 Mind Map 3
The segment bounded by the chord and the minor arc intercepted by the chord is called minor segment and the segment bounded by the chord and the major arc intercepted by the chord is called major segment.

Circle

The set of all points in a plane which are at a fixed distance from a fixed point in the plane is called circle. The fixed point is called centre and the fixed distance is called radius of the circle

Circumference and Area of a Circle
(i) The circumference of a circle is defined as distance covered by travelling once around a circle and is given by C = 2πr = πd
where r = radius of the circle and d = diameter of the circle.
(ii) The Area of a circle of radius r is given by,
A = πr2=π4d2 where, d = diameter of the circle.
(iii) Area of a circular ring:
The area of the circular path or ring is given by the difference of the area of outer circle and the area of inner circle.
Area of circular ring = n(R2 – r2)
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1 Mind Map 4

Length of an Arc and Area of Sector

(i) The length of an arc of a sector of an angle 9 is given by,
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1 Mind Map 5
(ii) The area of the sector AOB of angle 0 is given by,
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1 Mind Map 6
This is the area of minor sector.
∴ area of major sector AOB
= πr2 – Area of a minor sector AOB

Area of Segment

(i) Area of segment APB = Area (sector OAPB) – Area(∆OAB)
This is the area of minor segment.
∴ area of major segment AQB = πr2 – Area of minor segment APB

(ii) If θ is the central angle, then the area of segment APB
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1 Mind Map 7

More Resources

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

Get Free NCERT Solutions for Class 10 Maths Chapter 11 Ex 11.1 PDF. Constructions Class 10 Maths NCERT Solutions are extremely helpful while doing your homework. Exercise 11.1 Class 10 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 11 Maths Class 10 Constructions Exercise 11.1 provided in NCERT TextBook.

Topics and Sub Topics in Class 10 Maths Chapter 11 Constructions:

Section NameTopic Name
11Constructions
11.1Introduction
11.2Division Of A Line Segment
11.3Construction Of Tangents To A Circle
11.4Summary

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 11
Chapter NameConstructions
ExerciseEx 11.1
Number of Questions Solved5
CategoryNCERT Solutions

Ex 11.1 Class 10 Maths Question 1.
Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.
Solution:
NCERT Solutions For Class 10 Maths Chapter 11 Pdf Constructions Ex 11.1 Q1

Ex 11.1 Class 10 Maths Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 23 of the corresponding sides of the first triangle.
Solution:
NCERT Solutions For Class 10 Maths Chapter 11 Pdf Ex 11.1 Q2
NCERT Solutions For Class 10 Maths Chapter 11 Pdf Ex 11.1 Q2.1

You can also download the free PDF of  Ex 11.1 Class 10 Constructions NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

Download NCERT Solutions For Class 10 Maths Chapter 11 Constructions PDF

Ex 11.1 Class 10 Maths Question 3.
Construct a triangle with sides 5 cm, 6 cm, and 7 cm and then another triangle whose sides are 75 of the corresponding sides of the first triangle.
Solution:
Ex 11.1 Class 10 NCERT Solutions PDF Q3
Ex 11.1 Class 10 NCERT Solutions PDF Q3.1

Ex 11.1 Class 10 Maths Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 112 times the corresponding sides of the isosceles triangle.
Solution:
Exercise 11.1 Class 10 NCERT Solutions PDF Q4
Exercise 11.1 Class 10 NCERT Solutions PDF Q4.1

Ex 11.1 Class 10 Maths Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 34 of the corresponding sides of the triangle ABC.
Solution:
Constructions Class 10 NCERT Solutions Ex 11.1 PDF Q5 Constructions Class 10 NCERT Solutions Ex 11.1 PDF Q5.1

Ex 11.1 Class 10 Maths Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 43 times the corresponding sides of ∆ABC.
Solution:
Chapter 11 Maths Class 10 NCERT Solutions Exercise 11.1 PDF Q6
Chapter 11 Maths Class 10 NCERT Solutions Exercise 11.1 PDF Q6.1

Ex 11.1 Class 10 Maths Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 53F times the corresponding sides of the given triangle.
Solution:
Steps of Construction:
1. Construct a ∆ABC, such that BC = 4 cm, CA = 3 cm and ∠BCA = 90°
2. Draw a ray BX making an acute angle with BC.
3. Mark five points B1, B2, B3, B4 and B5 on BX, such that BB1 = B1B2 = B2B3 = B3B4 = B4B5.
4. Join B3C.
5. Through B5, draw B5C’ parallel to B3C intersecting BC produced at C’.
6. Through C’, draw C’A’ parallel to CA intersecting AB produced at A’.
Thus, ∆A’BC’ is the required right triangle.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 Q7
Justification:
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 Q7.1

Class 10 Maths Constructions Mind Maps

Construction

Construction implies drawing geometrical figures accurately such that triangles, quadrilateral and circles with the help of ruler and compass.

Division of a Line Segment

A line segment can be divided in a given ratio (both internally and externally)
Example:
Divide a line segment of length 12 cm internally in the ratio 3:2.
Solution :
Steps of construction :
(i) Draw a line segment AB = 12 cm. by using a ruler.
(ii) Draw a ray making a suitable acute angle ∠BAX with AB.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 Mind Map 1
(iii) Along AX, draw 5 ( = 3 + 2) arcs intersecting the ray AX at A1? A2, A3, A4 and A5 such that
AA1 = A1A2 = A2A3 = A3A4 = A4A5
(iv) Join BA5.
(v) Through A3 draw a line A3P parallel to A5B making ∠AA3P = ∠AA5B, intersecting AB at point P.
The point P so obtained is the required point, which divides AB internally in the ratio 3 : 2.

Similar Triangles

(i) This Construction involves two different situation.
(a) Construction of a similar triangle smaller than the given triangle.
(b) Construction of a similar triangle greater than the given triangle.
(ii) The ratio of sides of the triangle to be constructed with the corresponding sides of the given triangle is called scale factor.
Example:
Draw a triangle ABC with side BC = 7 cm. ∠B = 45°, ∠A = 105°. Construct a triangle whose sides are (4/3) times the corresponding side of ∆ABC.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 Mind Map 2
Solution :
Steps of construction :
(i) Draw BC = 7 cm.
(ii) Draw a ray BX and CY such that ∠CBX= 45° and
∠BCY = 180° – (45° + 105°) = 30°
Suppose BX and CY intersect each other at A.
∆ABC so obtained is the given triangle.
(iii) Draw a ray BZ making a suitable acute angle with BC on opposite side of vertex A with respect to BC.
(iv) Draw four (greater of 4 and 3 in 4/3) arcs intersecting the ray BZ at B1, B2, B3, B4 such that BB1 = B1B2 = B2B3 = B3B4.
(v) Join B3 to C and draw a line through B4 parallel to B3C, intersecting the extended line segment BC at C’.
(vi) Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’. Triangle A’BC’ so obtained is the required triangle.

Tangents to a Circle

Two tangents can be drawn to a given circle from a point outside it.
Example:
Draw a circle of radius 4 cm. Take a point P outside the circle. Without using the centre of the circle, draw two tangents to the circle from point P.
Solution :
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 Mind Map 3
Steps of construction :
(i) Draw a circle of radius 4 cm.
(ii) Take a point P outside the circle and draw a secant PAB, intersecting the circle at A and B.
(iii) Produce AP to C such that AP = CP.
(iv) Draw a semi-circle with CB as diameter.
(v) Draw PD ⊥ CB, intersecting the semi-circle at D.
(vi) With P as centre and PD as radius draw arcs to intersect the given circle at T and T’
(vii) Join PT and PT’. Then, PT and PT’ are the required tangents.
Note:
If centre of a circle is not given, then it can be located by finding point of intersection of perpendicular bisector, of any two nonparallel chords of a circle.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions (Hindi Medium) Ex 11.1

NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1 constructions in pdf
NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1 rachnaen
NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1 download in english medium
NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1 for up board
NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1 for cbse board
NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1 for mp board
NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1 free guide
NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1 all question answers
NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1 in hindi medium
NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1 in hindi pdf
NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1 all questions
NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1 ki kunji

NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1

Get Free NCERT Solutions for Class 10 Maths Chapter 10 Ex 10.1 PDF. Circles Class 10 Maths NCERT Solutions are extremely helpful while doing your homework. Exercise 10.1 Class 10 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 10 Maths Class 10 Circles Exercise 10.1 provided in NCERT TextBook.

Topics and Sub Topics in Class 10 Maths Chapter 10 Circles:

Section NameTopic Name
10Circles
10.1Introduction
10.2Tangent To A Circle
10.3Number Of Tangents From A Point On A Circle
10.4Summary

NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 10
Chapter NameCircles
ExerciseEx 10.1
Number of Questions Solved4
CategoryNCERT Solutions

Ex 10.1 Class 10 Maths Question 1.
How many tangents can a circle have?
Solution:
There can be infinitely many tangents to a circle.

Ex 10.1 Class 10 Maths Question 2.
Fill in the blanks:
(i) A tangent to a circle intersects it in ………… point(s).
(ii) A line intersecting a circle in two points is called a ………… .
(iii) A circle can have ………………. parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called ……….. .
Solution:
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 PDF Q2

You can also download the free PDF of  Ex 10.1 Class 10 Circles NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

Download NCERT Solutions For Class 10 Maths Chapter 10 Circles PDF

Ex 10.1 Class 10 Maths Question 3.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is
(a) 12 cm
(b) 13 cm
(c) 8.5 cm
(d) 119−−−√ cm
Solution:
Ex 10.1 Class 10 Maths Solutions NCERT PDF Q3

Note: PQ = √119

Ex 10.1 Class 10 Maths Question 4.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Solution:
Exercise 10.1 Class 10 Maths Solutions NCERT PDF Q4

Class 10 Maths Circles Mind Map

Introduction

A circle is a set of all points in a plane at a fixed distance from a fixed point in a plane. The fixed point is called the centre of the circle. The fixed distance is called the radius of the circle.

Line and a Circle

NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 Mind Map 1
In Fig. (i), the line PQ and the circle have no common point. In this case, PQ is called a non-intersecting line with respect to the circle. In Fig. (ii), there are two common points A and B that the line PQ and the circle have. In this case, we call the line PQ a secant of the circle. In Fig. (iii), there is only one point A which is common to the line PQ and the circle. In this case, the line is called a tangent to the circle.

Tangent

A tangent to a circle is a straight line which touches the circle at only one point. The point where the tangent touches the circle is called point of contact of the tangent to the circle.
A tangent to a circle is a special case of a secant, when the two ends points of its corresponding chord coincides.
Theorem: Tangent at any point on a circle is perpendicular to the radius through the point of contact.
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 Mind Map 2
CB is the tangent to the given circle touching at A and OA is the radius.
∴ ∠OAB = 90°
(i) At any point on the circle there can be one and only one tangent.
(ii) The line containing the radius through the point of contact is called the normal to the circle at the point.

Number of Tangents from a Point to Circle

(i) No tangent can be drawn from the point lying inside the circle, as shown in fig. (i)
(ii) One and only one tangent can be drawn from a point lying on the circle, as shown in fig. (ii)
(iii) Only two tangents can be drawn from an exterior point to a circle, as shown in fig. (iii)
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 Mind Map 3

Length of a Tangent

The length of the segment of a tangent from an external point to the point of contact with the circle is called the length of the tangent

NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 Mind Map 4
In the given figure, T1 and T2 are the points of contact of the tangents PT1 and PT2 respectively from the external point P.

Theorem Related to Length of Tangents From the External Points

The lengths of tangents drawn from an external point to a circle are equal.
i.e.,
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 Mind Map 5
Here, PQ and PR are the two tangents drawn from P to the circle

NCERT Solutions for Class 10 Maths Chapter 10 Circles (Hindi Medium) Ex 10.1

10 Maths Chapter 10 Exercise 10.1
10 Maths Chapter 10 Exercise 10.1 in Hindi medium
10 maths ex. 10.1 in hindi
More Resources for CBSE Class 10

NCERT Solutions For Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1

Get Free NCERT Solutions for Class 10 Maths Chapter 9 Ex 9.1 PDF. Some Applications of Trigonometry Class 10 Maths NCERT Solutions are extremely helpful while doing your homework. Exercise 9.1 Class 10 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 9 Maths Class 10 Some Applications of Trigonometry Exercise 9.1 provided in NCERT TextBook.

Topics and Sub Topics in Class 10 Maths Chapter 9 Some Applications of Trigonometry:

Section NameTopic Name
9Some Applications of Trigonometry
9.1Introduction
9.2Heights And Distances
9.3Summary

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex Ex 9.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Exercise 9.1.

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 9
Chapter NameSome Applications of Trigonometry
ExerciseEx 9.1
Number of Questions Solved16
CategoryNCERT Solutions

Hi all, We have also solved 68 questions of Chapter 12 – Some Applications of Trigonometry from RD Sharma Class 10 Maths textbook. You can download these solutions in PDF from the above link.

Ex 9.1 Class 10 Maths Question 1.
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.
Solution:
Ex 9.1 Class 10 Maths NCERT Solutions PDF Q1

Ex 9.1 Class 10 Maths Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Solution:
Ex 9.1 Class 10 Maths NCERT Solutions PDF Q2

You can also download the free PDF of  Ex 9.1 Class 10 Some Applications of Trigonometry NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

Download NCERT Solutions For Class 10 Maths Chapter 9 Some Applications of Trigonometry PDF

Ex 9.1 Class 10 Maths Question 3.
A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Solution:
Ex 9.1 Class 10 Maths NCERT Solutions PDF Q3

Ex 9.1 Class 10 Maths Question 4.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30°. Find the height of the tower.
Solution:
Ex 9.1 Class 10 Maths NCERT Solutions PDF Q4

Ex 9.1 Class 10 Maths Question 5.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Solution:
Exercise 9.1 Class 10 Maths NCERT Solutions pdf Q5

Ex 9.1 Class 10 Maths Question 6.
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution:
Exercise 9.1 Class 10 Maths NCERT Solutions pdf Q6

Ex 9.1 Class 10 Maths Question 7.
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Solution:
Exercise 9.1 Class 10 Maths NCERT Solutions pdf Q7

Ex 9.1 Class 10 Maths Question 8.
A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Solution:
Exercise 9.1 Class 10 Maths NCERT Solutions pdf Q8

Ex 9.1 Class 10 Maths Question 9.
The angle of elevation of the top of a building from the foot of a tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution:
Class 10 Maths Chapter 9 NCERT Solutions Ex 9.1 pdf Q9

Ex 9.1 Class 10 Maths Question 10.
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distance of the point from the poles.
Solution:
Class 10 Maths Chapter 9 NCERT Solutions Ex 9.1 pdf Q10

Ex 9.1 Class 10 Maths Question 11.
A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see the given figure). Find the height of the tower and the width of the CD and 20 m from pole AB.
NCERT Solutions For Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 Q11
Solution:
Chapter 9 Maths Class 10 NCERT Solutions Ex 9.1 pdf Q11

Ex 9.1 Class 10 Maths Question 12.
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Solution:
Chapter 9 Maths Class 10 NCERT Solutions Ex 9.1 pdf Q12

Ex 9.1 Class 10 Maths Question 13.
As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Solution:
Ch 9 Maths Class 10 NCERT Solutions Ex 9.1 pdf Q13

Ex 9.1 Class 10 Maths Question 14.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After sometime, the angle of elevation reduces to 30° (see figure). Find the distance travelled by the balloon during the interval.
NCERT Solutions For Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 Q14
Solution:
Ch 9 Maths Class 10 NCERT Solutions Ex 9.1 pdf Q14

Ex 9.1 Class 10 Maths Question 15.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Solution:
Some Applications Of Trigonometry Class 10 NCERT Solutions Pdf Ex 9.1 Q15

Ex 9.1 Class 10 Maths Question 16.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Solution:
Some Applications Of Trigonometry Class 10 NCERT Solutions Pdf Ex 9.1 Q16

Class 10 Maths Some Application Of Trigonometry Mind Maps

SOME APPLICATION OF TRIGONOMETRY

Introduction

The height or length of an object or the distance between two distinct objects can be determined with the help of trigonometric ratios.

Line of Sight and Angle of Elevation

NCERT Solutions For Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 Mind Map 1
In the above figure, the line AC drawn from the eye of an observer at A to the top of the pole ‘C’ is called the line of sight. The observer is looking at the top of the pole. The angle BAC, so formed by the line of sight with the horizontal, is called the angle of elevation of the top of the pole from the eye of an observer.

Angle of Depression

NCERT Solutions For Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 Mind Map 2
In the above figure, the line AC, is the line of sight as the observer is looking downwards from the top of the building at A towards the object at C. Here angle DAC, so formed by the line of sight with the horizontal, when the observer is lowering his/her head is called Angle of depression.

NCERT Solutions For Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 Mind Map 3
From the above figure, if we want to find the height CD of the pole without actually measuring it, we need the following information:
(i) Distance ED of the observer from the pole.
(ii) the angle of elevation ∠BAC, of the top of the pole.
(iii) the height AE of the observer if it is considerable.
Assuming that the above three conditions are known we can determine the height of the pole in the following way.
In the figure, CD = CB + BD. Here, BD = AE, which is the height of the observer.

To find BC, we will use trigonometric ratios of ∠BAC or ∠A.
In ∆ABC, the side BC is the opposite side to the known ∠A. Now we use either tan A or cot A, as these trigonometric ratios involve AB and BC to find BC.
Therefore, tan A = BCAB or cot A = ABBC, which on solving would give us BC. By adding AE to BC, you will get the height of the pole.

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry (Hindi Medium) Ex 9.1

NCERT Solutions for class 10 Maths Chapter 9
NCERT Solutions for class 10 Maths Chapter 9 Exercise 9.1NCERT Solutions for class 10 Maths Chapter 9 Exercise 9.1 in english medium
NCERT Solutions for class 10 Maths Chapter 9 Exercise 9.1 in pdf form
NCERT Solutions for class 10 Maths Chapter 9 Exercise 9.1 all questions guide
NCERT Solutions for class 10 Maths Chapter 9 Exercise 9.1 for up, mp and cbse board
NCERT Solutions for class 10 Maths Chapter 9 Exercise 9.1 download
10 Maths Exercise 9.1
10 Maths Exercise 9.1 in hindi medium
10 Maths Exercise 9.1 all question answers
10 Maths Exercise 9.1 pdf sols
10 Maths Exercise 9.1 download in pdf
10 Maths Exercise 9.1 guide full solve
10 Maths Exercise 9.1 hindi me
10 Maths Exercise 9.1 questions 12, 13, 14, 15, 16
applications of trigonometry class 10
application of trigonometry class 10
some applications of trigonometry class 10
some applications of trigonometry
ch 9 maths class 10
ncert solutions for class 10 maths chapter 9
applications of trigonometry class 10 ncert solutions
applications of trigonometry
trigonometry class 10 ncert solutions
ex 9.1 class 10
some applications of trigonometry class 10 ncert solutions pdf
class 10 maths chapter 9
extra questions for class 10 maths some application of trigonometry
ch 9 class 10 maths
application of trigonometry class 10 ncert solutions
class 10 ch 9 maths application of trigonometry
exercise 9.1 class 10

NCERT Solutions for Class 10 Maths

NCERT Solutions For Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1

Get Free NCERT Solutions for Class 10 Maths Chapter 8 Ex 8.1 Introduction to Trigonometry Class 10 Maths NCERT Solutions are extremely helpful while doing homework. Exercise 8.1 Class 10 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 8 Maths Class 10 Introduction to Trigonometry Exercise 8.1 Provided in NCERT Textbook.

Topics and Sub Topics in Class 10 Maths Chapter 8 Introduction to Trigonometry:

Section NameTopic Name
8Introduction to Trigonometry
8.1Introduction
8.2Trigonometric Ratios
8.3Trigonometric Ratios Of Some Specific Angles
8.4Trigonometric Ratios Of Complementary Angles
8.5Trigonometric Identities
8.6Summary

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1.

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 8
Chapter NameIntroduction to Trigonometry
ExerciseEx 8.1
Number of Questions Solved11
CategoryNCERT Solutions

Ex 8.1 Class 10 Maths Question 1.
In ∆ABC right angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A
(ii) sin C, cos C
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Trigonometry Exercise 8.1 Q1

Ex 8.1 Class 10 Maths Question 2.
In given figure, find tan P – cot R.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Trigonometry Exercise 8.1 PDF Download Q2

You can also download the free PDF of Chapter 8 Ex 8.1 Introduction to Trigonometry NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

Download NCERT Solutions For Class 10 Maths Chapter 8 Introduction to Trigonometry PDF

Ex 8.1 Class 10 Maths Question 3.
If sin A = 34 , calculate cos A and tan A.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Trigonometry Exercise 8.1 PDF Q3

Ex 8.1 Class 10 Maths Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Trigonometry Exercise 8.1 Free PDF Download Q4

Ex 8.1 Class 10 Maths Question 5.
Given sec θ = 1312 , calculate all other trigonometric ratios.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Trigonometry Exercise 8.1 Q5

Ex 8.1 Class 10 Maths Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:
Exercise 8.1 Class 10 NCERT Solutions Chapter 8 Trigonometry Q6

Ex 8.1 Class 10 Maths Question 7.
If cot θ = 78, evaluate:
(i) (1+sinθ)(1−sinθ)(1+cosθ)(1−cosθ)
(ii) cot²θ
Solution:
Exercise 8.1 Class 10 NCERT Solutions Chapter 8 Trigonometry Free PDF Download Q7

Ex 8.1 Class 10 Maths Question 8.
If 3 cot A = 4, check whether 1−tan2A1+tan2A = cos² A – sin² A or not.
Solution:
Exercise 8.1 Class 10 NCERT Solutions Chapter 8 Trigonometry Free PDF Download Q7

Ex 8.1 Class 10 Maths Question 9.
In triangle ABC, right angled at B, if tan A = 1√3, find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution:
Trigonometry Class 10 Chapter 8 Exercise 8.1 NCERT Solutions Free PDF Download Q9
Trigonometry Class 10 Chapter 8 Exercise 8.1 NCERT Solutions PDF Q9.1

Ex 8.1 Class 10 Maths Question 10.
In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:
Trigonometry Class 10 Chapter 8 Exercise 8.1 NCERT Solutions Q10

Ex 8.1 Class 10 Maths Question 11.
State whether the following statements are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = 125 for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = 43 for some angle.
Solution:
Trigonometry Class 10 Chapter 8 Exercise 8.1 NCERT Solutions PDF Download Q11

Class 10 Maths Introduction To Trigonometry

Trigonometry

Trigonometry is the study of relationships between the sides and angles of a right-angled triangle.

Trigonometric Ratios

Trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides.
Let ∆ABC be a triangle right angled at B. Then the trigonometric ratios of the angle A in right ∆ABC are defined as follows:
NCERT Solutions For Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 Q11

NCERT Solutions For Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 Q11.1

Note:
The values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains same.
NCERT Solutions For Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 Q11.2

Trigonometric Ratios for Complementary Angles

sin (90° – A) = cos A
cos (90° – A) = sin A
tan (90° – A) = cot A
cot (90° – A) = tan A
sec (90° – A) = cosec A
cosec (90° – A) = sec A
Note:
Here (90° – A) is the complementary angle of A.

Trigonometric Identities

An equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle(s) involved.
(i) sin2θ + cos2θ = 1 [for 0° ≤ θ ≤ 90°]
(ii) sec2θ – tan2θ = 1 [for 0° ≤ θ ≤ 90°]
(iii) cosec2θ – cot2θ = 1 [for 0° < θ ≤ 90°]

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry (Hindi Medium) Ex 8.1

NCERT Solutions for class 10 Maths Chapter 8 Exercise 8.1 Introduction to Trigonometry in ENGLISH MEDIUM
10 maths ex. 8.1
8.1 class 10 maths
trigo 8.1 class 10
class 10 maths trigo ex. 8.1 sols
class 10 maths exercise 8.1 in english
10 maths exercise 8.1 of trigonometry
class 10 chapter 8 ex. 8.1

NCERT Solutions for Class 10 Maths

NCERT Solutions For Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1

Get Free NCERT Solutions for Class 10 Maths Chapter 7 Ex 7.1 Coordinate Geometry Class 10 Maths NCERT Solutions are extremely helpful while doing homework. Exercise 7.1 Class 10 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 7 Maths Class 10 Coordinate Geometry Exercise 7.1 Provided in NCERT Textbook

Topics and Sub Topics in Class 10 Maths Chapter 7 Coordinate Geometry:

Section NameTopic Name
7Coordinate Geometry
7.1Introduction
7.2Distance Formula
7.3Section Formula
7.4Area of a Triangle
7.5Summary

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1.

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 7
Chapter NameCoordinate Geometry
ExerciseEx 7.1
Number of Questions Solved10
CategoryNCERT Solutions

Ex 7.1 Class 10 Maths Question 1.
Find the distance between the following pairs of points:
(i) (2, 3), (4, 1)
(ii) (-5, 7), (-1, 3)
(iii) (a, b), (-a, -b)
Solution:
Coordinate Geometry Class 10 Maths NCERT Solutions Ex 7.1 PDF Download
Coordinate Geometry Class 10 Maths NCERT Solutions Ex 7.1 PDF Download Q1

Ex 7.1 Class 10 Maths Question 2.
Find the distance between the points (0, 0) and (36, 15).
Solution:
Coordinate Geometry Class 10 Maths NCERT Solutions Ex 7.1 PDF Q2

You can also download the free PDF of Chapter 7 Ex 7.1 Coordinate Geometry NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

Download NCERT Solutions For Class 10 Maths Chapter 7 Coordinate Geometry PDF

Ex 7.1 Class 10 Maths Question 3.
Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Solution:
Coordinate Geometry Class 10 Maths NCERT Solutions Ex 7.1 Q3

Ex 7.1 Class 10 Maths Question 4.
Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Solution:
Ex 7.1 Class 10 Maths NCERT Solutions Coordinate Geometry PDF Download Q4

Ex 7.1 Class 10 Maths Question 5.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in the given figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
NCERT Solutions For Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 Q5
Solution:
Ex 7.1 Class 10 Maths NCERT Solutions Ch 7 Coordinate Geometry PDF Download Q5

Ex 7.1 Class 10 Maths Question 6.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.
(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)
(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution:
Ex 7.1 Class 10 Maths NCERT Solutions Ch 7 Coordinate Geometry PDF Q6Ex 7.1 Class 10 Maths NCERT Solutions Ch 7 Coordinate Geometry Q6.1

Ex 7.1 Class 10 Maths Question 7.
Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Solution:
Ch 7 maths Class 10 NCERT Solutions Ex 7.1 PDF Download Q7

Ex 7.1 Class 10 Maths Question 8.
Find the values of y for which the distance between the points P (2, -3) and Q (10, y) is 10 units.
Solution:
Ch 7 maths Class 10 NCERT Solutions Ex 7.1 PDF Q8

Ex 7.1 Class 10 Maths Question 9.
If Q (0, 1) is equidistant from P (5, -3), and R (x, 6), find the values of x. Also, find the distances QR and PR.
Solution:
Ch 7 maths Class 10 NCERT Solutions Ex 7.1 Q9

Ex 7.1 Class 10 Maths Question 10.
Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).
Solution:
Ch 7 maths Class 10 NCERT Solutions Coordinate Geometry Ex 7.1 Q10

Class 10 Maths Coordinate Geometry Mind Maths

Coordinate of a Point in XY – Plane

The Perpendicular distance of x a point from the y-axis is called its x-coordinate or abscissa. The perpendicular distance y of a point from the x-axis is called its y-coordinate or ordinate. The x and y taken together in order is called coordinte of point denoted by (x, y).
The coordinate of the points on x-axis are of the form (x, 0) and the points on the y-axis are of the form(0, y). Coordinate of origin is (0, 0).

Sign-conventions in the XY-Plane

The x and y-axis divide the plane into four parts known as quadrants denoted by I, II, III and IV. The sign of x and y-coordinates in each of the quadrant is shown below:
NCERT Solutions For Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 Mind Map 1

Distance Formula

The distance between any two points
P(x1, y1) and Q(x2, y2) in the plane is given by,
NCERT Solutions For Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 Mind Map 2
Also the distance of the point P(x1, y1) from the origin is
NCERT Solutions For Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 Mind Map 3

Section Formula

NCERT Solutions For Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 Mind Map 4
The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2) internally in the ratio m1 : m2 i.e., PAPB=m1m2
NCERT Solutions For Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 Mind Map 5
The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2) externally in the ratio, m1 : m2 i.e., PAPB=m1m2 are
NCERT Solutions For Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 Mind Map 6
(iii) If the ratio in which P divides AB is K : 1, then the coordinates of the point P will be
NCERT Solutions For Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 Mind Map 7

Mid-Point Formula

The coordinates of the mid point P of the line segment joining the points A(x1, y1) and B(x2, y2) is
NCERT Solutions For Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 Mind Map 8

Area of a Triangle

The area of ∆ABC formed by the vertices A(x1, y1), B(x2, y2) is given by
NCERT Solutions For Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 Mind Map 9
Note:
(i) Area of triangle = 12 × base × Altitude
(ii) Area of polygon can be calculated by dividing it into the triangular region.
(iii) If three points are collinear then area of the triangle formed by them is zero.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Hindi Medium) Ex 7.1

NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1 Coordinate Geometry
NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1 Coordinate Geometry in English medium PDF
NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1 Coordinate Geometry for Gujrat, UP and CBSE Board
NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1 question 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
10 Maths Chapter 7 Exercise 7.1 solutions in Hindi medium PDF
10 Maths Chapter 7 Exercise 7.1 sols for CBSE, Gujrat, UP Board session 2018-19.
10 Maths Chapter 7 Exercise 7.1 solutions in Hindi PDF download free
10 Maths Chapter 7 Exercise 7.1 solutions for 2018-2019 exams.
Class 10 Maths Chapter 7 Exercise 7.1 Coordinate Geometry solutions in Hindi medium pdf
Class 10 Maths Chapter 7 Exercise 7.1 Coordinate Geometry sols for CBSE, Gujrat, UP Board, Bihar and Uttarakhand
Class 10 Maths Chapter 7 Exercise 7.1 Coordinate Geometry solutions for 2018-19
Chapter 7 Exercise 7.1 Coordinate Geometry solutions for 2018-19
Exercise 7.1 Coordinate Geometry solutions in Hindi medium pdf

NCERT Solutions for Class 10 Maths

NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1

Get Free NCERT Solutions for Class 10 Maths Chapter 6 Ex 6.1 PDF. Triangles Class 10 Maths NCERT Solutions are extremely helpful while doing your homework. Exercise 6.1 Class 10 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 6 Maths Class 10 Triangles Exercise 6.1 provided in NCERT TextBook.

You can also download Maths Class 10 to help you to revise complete syllabus and score more marks in your examinations.

Free download NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.1 Triangles PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, AP SSC, TS SSC and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex Ex 6.1 are part of Class 10 Maths NCERT Solutions. Here we have given NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.1

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 6
Chapter NameTriangles
ExerciseEx 6.1
Number of Questions Solved3
CategoryNCERT Solutions

Ex 6.1 Class 10 Maths Question 1.
Fill in the blanks by using the correct word given in brackets.
(i) All circles are ……………. . (congruent/similar)
(ii) All squares are …………… . (similar/congruent)
(iii) All …………….. triangles are similar. (isosceles/equilateral)
(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are …………… and
(b) their corresponding sides are …………… (equal/proportional)
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.1 Triangles in English medium PDF

Ex 6.1 Class 10 Maths Question 2.
Give two different examples of pairs of
(i) similar figures.
(ii) non-similar figures.
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.1 updated for 2019-2020

Ex 6.1 Class 10 Maths Question 3.
State whether the following quadrilaterals are similar or not.
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q1
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.1 trinagles in hindi

Triangles Class 10 Ex 6.2

Ex 6.2 Class 10 Maths Question 1.
In the given figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q2
Solution:
Triangles Class 10 Exercise 6.2

Ex 6.2 Class 10 Maths Question 2.
E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR:
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Solution:
Triangles Class 10 Exercise 6.2

Ex 6.2 Class 10 Maths Question 3.
In the given figure, if LM || CB and LN || CD.
Prove that AMAB=ANAD∙
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q3
Solution:
exercise 6.2 class 10

Ex 6.2 Class 10 Maths Question 4.
In the given figure, DE || AC and DF || AE.
Prove that BFFE=BEEC∙
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q4
Solution:
class 10 maths triangles

Ex 6.2 Class 10 Maths Question 5.
In the given figure, DE || OQ and DF || OR. Show that EF || QR.
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q5
Solution:
Class 10 Triangles Ex 6.2

Ex 6.2 Class 10 Maths Question 6.
In the given figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q6
Solution:
triangles class 10

Ex 6.2 Class 10 Maths Question 7.
Using B.P.T., prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that your have proved it in class IX)
Solution:
ncert solutions for class 10 maths chapter 6

Ex 6.2 Class 10 Maths Question 8.
Using converse of B.P.T., prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.  (Recall that your have done it in class IX)
Solution:
ch 6 maths class 10

Ex 6.2 Class 10 Maths Question 9.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AOBO=CODO∙
Solution:
triangles class 10 ncert solutions

Ex 6.2 Class 10 Maths Question 10.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that AOBO=CODO∙ Show that ABCD is a trapezium.
Solution:
similar triangles class 10

Triangles Class 10 Ex 6.3

Ex 6.3 Class 10 Maths Question 1.
State which pairs of triangles in the given figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q7

NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q8
Solution:
ncert solutions for class 10 maths chapter 6 triangles
ncert solutions for class 10 maths chapter 6 pdf

Ex 6.3 Class 10 Maths Question 2.
In the given figure, ∆ODC ~ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q9
Solution:
ncert solutions for class 10 maths triangles

NCVT MIS

Ex 6.3 Class 10 Maths Question 3.
Diagonals AC and BD of a trape∠ium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OAOC=OBOD∙
Solution:
triangles class 10 solutions

Ex 6.3 Class 10 Maths Question 4.
In the given figure, QRQS=QTPR and ∠1 = ∠2. show that ∆PQR ~ ∆TQR.
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q10
Solution:
similarity of triangles class 10

Ex 6.3 Class 10 Maths Question 5.
S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3 Triangles in English medium PDF

Ex 6.3 Class 10 Maths Question 6.
In the given figure, if ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC.
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q11
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3 Tribhuj in Hindi medium PDF

Ex 6.3 Class 10 Maths Question 7.
In the given figure, altitudes AD and CE of ∆ABC intersect each other at the point P. Show that:
(i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q12
Solution:
Triangles Class 10 Ex 6.3

Ex 6.3 Class 10 Maths Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB.
Solution:
Triangles Class 10 Exercise 6.3

Ex 6.3 Class 10 Maths Question 9.
In the given figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q13
Solution:
exercise 6.3 class 10

Ex 6.3 Class 10 Maths Question 10.
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ~ ∆FEG, show that
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q14
Solution:
class 10 maths triangles ex 6.3
exercise 6.3 class 10 ncert solutions

Ex 6.3 Class 10 Maths Question 11.
In the given figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF.
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q15
Solution:
Class 10 Triangles Exercise 6.3

Ex 6.3 Class 10 Maths Question 12.
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR (see in given figure). Show that ∆ABC ~ ∆bPQR.
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q16
Solution:
Chapter 6 Maths Class 10 Ex 6.3 NCERT Solutions PDF

Ex 6.3 Class 10 Maths Question 13.
D is a point on the side BC of a triangle ABC, such that ∠ADC = ∠BAC. Show that CA² = CB.CD.
Solution:
similar triangles class 10 ex 6.3

Ex 6.3 Class 10 Maths Question 14.
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆PQR.
Solution:
chapter 6 maths class 10 ex 6.3

Ex 6.3 Class 10 Maths Question 15.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution:
class 10 maths chapter 6 ex 6.3

Ex 6.3 Class 10 Maths Question 16.
If AD and PM are medians of triangles ABC and PQR respectively, where
∆ABC ~ ∆PQR. Prove that ABPQ=ADPM∙
Solution:
ncert solutions for class 10 maths chapter 6 triangles ex 6.3

Triangles Class 10 Ex 6.4

Ex 6.4 Class 10 Maths Question 1.
Let ∆ABC ~ ∆DEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.
Solution:
ncert solutions for class 10 maths chapter 6 ex 6.4 pdf

Ex 6.4 Class 10 Maths Question 2.
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.
Solution:
cbse class 10 maths triangles ex 6.4 ncert solutions

Ex 6.4 Class 10 Maths Question 3.
In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that: ar(ABC)ar(DBC)=AODO
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q17
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.4 updated for 2019-2020

Ex 6.4 Class 10 Maths Question 4.
If the areas of two similar triangles are equal, prove that they are congruent.
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.4 Tribhuj in Hindi medium PDF

Ex 6.4 Class 10 Maths Question 5.
D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ABC. Find the ratio of the areas of ∆DEF and ∆ABC.
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.4 trinagles in hindi

Ex 6.4 Class 10 Maths Question 6.
Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution:
Triangles Class 10 Ex 6.4

Ex 6.4 Class 10 Maths Question 7.
Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Solution:
Triangles Class 10 Exercise 6.4

Ex 6.4 Class 10 Maths Question 8.
Tick the correct answer and justify
(i) ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is
(a) 2 :1
(b) 1:2
(c) 4 :1
(d) 1:4
ex 6.4 class 10
(ii) Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(a) 2 : 3
(b) 4 : 9
(c) 81 : 16
(d) 16 : 81
exercise 6.4 class 10

Triangles Class 10 Ex 6.5

Ex 6.5 Class 10 Maths Question 1.
Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
Solution:
solution of class 10 maths ex 6.5

Ex 6.5 Class 10 Maths Question 2.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM X MR.
Solution:
ncert solutions for class 10 maths chapter 6 triangles ex 6.5

Ex 6.5 Class 10 Maths Question 3.
In the given figure, ABD is a triangle right angled at A and AC i. BD. Show that
(i) AB2 = BC.BD
(ii) AC2 = BC.DC
(iii) AD2 = BD.CD
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q18
Solution:
cbse class 10 maths triangles ex 6.5 ncert solutions

Ex 6.5 Class 10 Maths Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.
Solution:
ncert solutions for class 10 maths triangles ex 6.5

Ex 6.5 Class 10 Maths Question 5.
ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2 , Prove that ABC is a right triangle.
Solution:
Triangles Class 10 Ex 6.5

Ex 6.5 Class 10 Maths Question 6.
ABC is an equilateral triangle of side la. Find each of its altitudes.
Solution:
Triangles Class 10 Exercise 6.5

Ex 6.5 Class 10 Maths Question 7.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Solution:
ex 6.5 class 10

Ex 6.5 Class 10 Maths Question 8.
In the given figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q19
Solution:
exercise 6.5 class 10 ncert solutions

Ex 6.5 Class 10 Maths Question 9.
A ladder 10 m long reaches a window 8 m above the ground. ind the distance of the foot of the ladder from base of the wall.
Solution:
Class 10 Triangles Ex 6.5

Ex 6.5 Class 10 Maths Question 10.
A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution:
Class 10 Triangles Exercise 6.5

Ex 6.5 Class 10 Maths Question 11.
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 112 hours?
Solution:
similar triangles class 10 ex 6.5

Ex 6.5 Class 10 Maths Question 12.
Two poles of heights 6 m and 11m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Solution:
chapter 6 maths class 10 ex 6.5

Ex 6.5 Class 10 Maths Question 13.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.
Solution:
class 10 maths chapter 6 ex 6.5

Ex 6.5 Class 10 Maths Question 14.
The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see the figure). Prove that 2AB2 = 2AC2 + BC2.
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q20
Solution:
Ch 6 Maths Class 10 Ex 6.5 NCERT Solutions PDF Q14

Ex 6.5 Class 10 Maths Question 15.
In an equilateral triangle ABC, D is a point on side BC, such that BD = 13BC. Prove that 9AD2 = 7AB2.
Solution:
Ch 6 Maths Class 10 Ex 6.5 NCERT Solutions PDF Q15

Ex 6.5 Class 10 Maths Question 16.
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Solution:
Class 10 Maths Chapter 6 Ex 6.5 NCERT Solutions PDF Q16

Ex 6.5 Class 10 Maths Question 17.
Tick the correct answer and justify : In ∆ABC, AB = 63–√cm, AC = 12 cm and BC = 6 cm. The angle B is:
(a) 120°
(b) 60°
(c) 90°
(d) 45
Solution:
Class 10 Maths Chapter 6 Ex 6.5 NCERT Solutions PDF Q17

Triangles Class 10 Ex 6.6

Ex 6.6 Class 10 Maths Question 1.
In the given figure, PS is the bisector of ∠QPR of ∆PQR. Prove that QSSR=PQPR
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q21
Solution:
ncert solutions for class 10 maths chapter 6 ex 6.6 pdf

Ex 6.6 Class 10 Maths Question 2.
In the given figure, D is a point on hypotenuse AC of ∆ABC, DM ⊥ BC and DN ⊥ AB. Prove that:
(i) DM2 = DN X MC
(ii) DN2 = DM X AN
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q22
Solution:
ncert solutions for class 10 maths triangles ex 6.6
triangles ex 6.6 class 10 solutions

Ex 6.6 Class 10 Maths Question 3.
In the given figure, ABc is triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2  + 2BC X BD
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q23
Solution:
similarity of triangles ex 6.6 class 10

Ex 6.6 Class 10 Maths Question 4.
In the given figure, ABC is atriangle in which ∠ABC 90° and AD ⊥ CB. Prove that AC2 = AB2 + BC2 – 2BC X BD
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.6 Q24
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 updated for 2019-2020

Ex 6.6 Class 10 Maths Question 5.
In the given figure, Ad is a median of a triangle ABC and AM ⊥ BC. Prove that
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.6 Q5
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 trinagles in hindi
Triangles Class 10 Ex 6.6

Ex 6.6 Class 10 Maths Question 6.
Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Solution:
Triangles Class 10 Exercise 6.6

Ex 6.6 Class 10 Maths Question 7.
In the given figure, two chords AB and CD intersect each other at the point P. Prove that:
(i) ∆APC ~∆DPB
(ii) AP X PB = CP X DP
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.6 Q7
Solution:
exercise 6.6 class 10

Ex 6.6 Class 10 Maths Question 8.
In the given figure, two chords Ab and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:
(i) ∆PAC ~ ∆PDB
(ii)PA X PB = PC X PD
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.6 Q8
Solution:
class 10 maths triangles

Ex 6.6 Class 10 Maths Question 9.
In the given figure, D is a point on side BC of ∆ABC, such that BDCD=ABAC∙ Prove that AD is the bisector of ∆BAC.
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.6 Q9
Solution:
Class 10 Triangles Ex 6.6

Ex 6.6 Class 10 Maths Question 10.
Nazima is fly fishing in a stream. The trip of her fishing rod is 1.8m above the surface of the water and the fly at the end of the string rests on the water 3.6m away and 2.4 m from a point directly under the trip of the rod. Assuming that her string (from the trip of the rod to the fly) is that, how much string does she have out (see the figure)? If she pills in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.6 Q10
Solution:
Chapter 6 Maths Class 10 Ex 6.6 NCERT Solutions PDF Q10

NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 in Hindi Medium

प्र. 1. कोष्ठकों में दिए शब्दों में से सही शब्दों का प्रयोग करते हुए, रिक्त स्थानों को भरिए :
(i) सभी वृत्त …….. होते है| (सर्वांगसम, समरूप)
(ii) सभी वर्ग…… होते हैं| (समरूप, सर्वांगसम)
(iii) सभी …….. त्रिभुज समरूप होते है | (समद्विबाहु, समबाहु)
(iv) भुजाओं की समान संख्या वाले दो बहुभुज समरूप होते हैं, यदि
(i) उनके संगत कोण ……..हो तथा
(ii) उनकी संगत ……भुजाएँ हों| (बराबर, समानुपाती|
हलः
(i) सभी वृत्त समरूप होते हैं।
(ii) सभी वर्ग समरूप होते हैं।
(iii) सभी समबाहु त्रिभुज समरूप होते हैं।
(iv) भुजाओं की समान संख्या वाले दो बहुभुजे समरूप होते हैं, यदि
(i) उनके संगत कोण बराबर हों तथा
(ii) उनकी संगत समानुपाती भुजाएँ हों।

प्र० 2. निम्नलिखित युग्मों के दो भिन्न-भिन्न उदाहरण दीजिएः
(i) समरूप आकृतियाँ
(ii) ऐसी आकृतियाँ जो समरूप नहीं हैं।
हलः
(i) (a) दो वृत्त परस्पर समरूप होते हैं।
(b) दो वर्ग परस्पर समरूप होते हैं।
(ii) (a) एक वृत्त और एक त्रिभुज समरूप नहीं होते हैं।
(b) एक समद्विबाहु त्रिभुज और एक विषमबाहु। त्रिभुज समरूप आकृतियाँ नहीं होती हैं।

प्र० 3. बताइए कि निम्न चतुर्भुज समरूप हैं या नहीं:

NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 in Hindi Medium

हलः 

संगत भुजाएँ समानुपाती हैं, परन्तु इनके संगत कोण समान नहीं हैं। ये आकृतियाँ समरूप नहीं हैं।

Class 10 Maths Triangles Mind Map

Similar Figures

Two figures having the same shape but not necessarily the same size are called similar figures
Two figures having the same shape as well as same size are called congruent figures
Note that all congruent figures are similar but the similar figures need not be congruent.

Similarity of Polygons

Two polygons of the same number of sides are similar if
(i) their corresponding angles are equal and
(ii) their corresponding sides are in the same ratio (or proportion)

Similarity of Triangles

Two triangles are similar if
(i) their corresponding angles are equal and
(ii) their corresponding sides are in the same ratio (or proportion)
Note : If the corresponding angles of two triangles are equal, then they are known as equiangular triangles.
The ratio of any two corresponding sides in two equiangular triangles is always the same.

Basic Proportionality Theorem (BPT) and its Converse

Basic Proportionality Theorem
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then other two sides are divided in the same ratio. Thus in ∆ABC, if DE || BC, then
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Mind Map 1

NCERT Solutions For Class 10 Maths Chapter 6 Triangles Mind Map 2
Converse of BPT
If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side,

Criteria For Similarity of Triangles

(i) AAA Similarity Criterion : If in two triangles, corresponding angles are equal then their corresponding sides are in the same ratio and hence the two triangles are similar.
(ii) AA Similarity Criterion : If in two triangles, two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are similar.
(iii) SSS Similarity Criterion : If in two triangles, corresponding sides are in the same ratio then their corresponding angles are equal and hence the triangles are similar.
(iv) SAS Similarity Criterion : If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio (proportion), then the two triangles are similar.

Areas of Similar Triangles

The ratio of the area of two similar triangles is equal to the ratio of the squares for their corresponding sides thus if ∆ABC – ∆PQR, then
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Mind Map 3

Pythagoras Theorem and its Converse

(i) If perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangles on both sides of the perpendicular are similar to the whole triangle and also to each other.
(ii) Pythagoras Theorem : In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Thus in right ∆ABC right angled at B
AC2 = AB2 + BC2
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Mind Map 4
(iii) Converse of Pythagoras Theorem : If in a triangle, square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

You can also download the free PDF of Class 10 Triangles NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

Topics and Sub Topics in Class 10 Maths Chapter 6 Triangles:

Section NameTopic Name
6Triangles
6.1Introduction
6.2Similar Figures
6.3Similarity Of Triangles
6.4Criteria For Similarity Of Triangles
6.5Areas Of Similar Triangles
6.6Pythagoras Theorem
6.7Summary

More Resources for CBSE Class 10

Formulae Handbook for Class 10 Maths and Science

NCERT Solutions for Class 10 Maths