# CBSE Class 10

## Arithmetic Progressions Class 10 Extra Questions Maths Chapter 5 with Solutions Answers

Here you will find Arithmetic Progressions Class 10 Extra Questions Maths Chapter 5 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction.

## Extra Questions for Class 10 Maths Arithmetic Progressions with Answers Solutions

Extra Questions for Class 10 Maths Chapter 5 Arithmetic Progressions with Solutions Answers

### Arithmetic Progressions Class 10 Extra Questions Objective Type

Question 1.
Common difference for arithmetic progression (A.P) 3, 1, -1, -3 ….. will be:
(a) 1
(b) 2
(c) -2
(d) 3
(c) -2
Solution.
Common difference = a2 – a1
= 1 – 3 = -2
Hence, choice (iii) is correct

Question 2.
Common difference of 2, 0.5, -1, -2.5, -4,….. is:
(a) -1.5
(b) -1.3
(c) 2.4
(d) 2.6
(a) -1.5

Question 3.
100th term of $$\frac {1}{2}$$, 1, $$\frac {3}{2}$$, 2, ….is:
(a) 50
(b) 60
(c) 70
(d) 40
(a) 50

Question 4.
rth term of a + 2b, a – b, a – 46, … is:
(a) a + (5 – 3r)b
(b) a + (4 – 3r)b
(c) a + (6 – r)b
(d) a + (2 – r)b
(a) a + (5 – 3r)b

Question 5.
The sum of first 16 terms of the AP 10, 6, 2, …. is:
(a) 320
(b) -320
(c) -352
(d) -400
(b) -320

Question 6.
The sum of first 20 terms of the AP 1, 3, 5, 7, 9, … is
(a) 264
(b) 400
(c) 472
(d) 563
(b) 400

Question 7.
(5 + 13 + 21 + … + 181) =?
(a) 2476
(b) 2337
(c) 2219
(d) 2139
(d) 2139

Question 8.
The sum of n terms of the AP √2, √8, √18, √32, … is
(a) 1
(b) 2n(n + 1)
(c) $$\frac {1}{2}$$n(n + 1)
(d) $$\frac{1}{\sqrt{2}}$$n(n + 1).
(d) $$\frac{1}{\sqrt{2}}$$n(n + 1)

Question 9.
How many terms of the AP 3, 7, 11, 15, ….will make the sum 406?
(a) 10
(b) 12
(c) 14
(d) 20
(c) 14

Question 10.
The sum of first 100 natural numbers is
(a) 4950
(b) 5050
(c) 5000
(d) 5150
(b) 5050

Question 11.
The sum of all odd numbers between 100 and 200 is
(a) 3750
(b) 6200
(c) 6500
(d) 7500
(d) 7500

Question 12.
The sum of all even natural numbers less than 100 is
(a) 2272
(b) 2352
(c) 2450
(d) 2468
(c) 2450

Question 13.
The sum of first fifteen multiples of 8 is
(a) 840
(b) 960
(c) 872
(d) 1080
(b) 960

Question 14.
In an AP, the first term is 22, nth term is 11 and the sum of first n terms is 66. The value of n is
(a) 10
(b) 12
(c) 14
(d) 16
(b) 12

Question 15.
In an AP, the first term is 8, nth term is 33 and the sum of first n terms is 123. Then, d =?
(a) 5
(b) – 5
(c) 7
(d) 3
(a) 5

Question 16.
The sum of n terms of an AP is given by Sn = (2n2 + 3n). What is the common difference of the AP?
(a) 3
(b) 4
(c) 5
(d) 9
(b) 4

### Arithmetic Progressions Class 10 Extra Questions Very Short answer Type

Question 1.
For the following APs, write the first term and the common difference.
(i) 3, 1, -1, -3, …
(ii) – 5, -1, 3, 7, …..,
(iii) $$\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}$$ ,…..
(iv) 0.6, 1.7, 2.8, 3.9, …
Solution.
(i) First term a = t1 = 3, Common difference
d = 2nd term – 1st term
= 1 – 3 = -2

(ii) First term a = t1 = -5, Common difference
d = 2nd term – 1st term
= -1 -(-5) = 1 + 5 = 4

(iii) First term a = t1 = $$\frac {1}{3}$$ Common difference
d = 2nd term – 1st term = $$\frac{5}{3}-\frac{1}{3}=\frac{4}{3}$$

(iv) First term a = t1 = 0.6 Common difference
d = 2nd term – 1st term
= 1.7 – 0.6 = 1.1

Question 2.
For what value of k will k + 9, 2k – 1, and 2k + 7 are the consecutive terms of an arithmetic progression (A.P)
Solution.
Given, k + 9, 2k – 1 and 2k + 7 are the consecutive term
∴ d = a2 – a1 = a3 – a2
⇒ (2k – 1) – (k + 9) = (2k + 7) – (2k – 1)
⇒ 2k – 1 – k – 9 = 2k + 7 – 2k + 1
k – 10 = 8
∴ k = 8 + 10 = 18

Question 3.
For what value of k, will the points (k, -1), (2, 1) and (4, 5) lie on a line.
Solution.
If point (k, -1), (2, 1) and (4, 5) lie on a line, i.e., the points are collinear.
x1 = k, y1 = -1, x2 = 2, y2 = 1, x3 = 4, y3 = 5
The area of A formed by these points are zero.
∴ 0 = $$\frac {1}{2}$$ [x1 (y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
⇒ k(1 -5) + 2(5 + 1) + 4 (-1 -1) = 0
⇒ -4k + 12 – 8 = 0
⇒ -4k = -4
⇒ k = 1

Question 4.
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution.
let a be the first term and d the common difference of an AP. Now, the nth term of an AP is
an = a + (n – 1)d
a11 = a + 10d = 38
[∵ a11 = 38 (given)] …(i)
and a16 = a + 15d = 73
[∵ a16 = 73 (given)] ..(ii)
On subtracting Eq. (i) from EQuestion (ii) we get
5d = 35 = d = $$\frac {35}{5}$$ = 7
From Eq. (i)
a + 10 × 7 = 38
⇒ a = 38 – 70 = – 32
∴ The 31st term of an AP
a31 = a + 30d
= -32 + 30 × 7
= – 32 + 210 = 178

Question 5.
Find the sum of the following AP’s
(i) 2, 7, 12, …, to 10 terms
(ii) -37, -33, -29, …, to 12 terms
(iii) 0.6, 1.7, 2.8, … to 100 terms
(iv) $$\frac{1}{15}, \frac{1}{12}, \frac{1}{10}$$ ……., to 11 terms.
Solution.
(i) Let a be the first term and d be the common difference of the given AP.
Then, we have a = 2 and d = 7 – 2 = 5
∵ Sum of n terms of AP,
Sn = $$\frac {n}{2}$$[2a + (n – 1)d]
Putting a = 2, d = 5, n = 10, we get
S10 = $$\frac {10}{2}$$[2 × 2 + (10 – 1)5]
= 5 (4 + 9 × 5)
= 5 (4 + 45) = 5 × 49 = 245
(ii) let a be the first and d be the common difference of the given AP,
Then, we have a = -37,
d = -33 -(-37) = – 33 + 37
= 4
∵ Sum of n terms of an AP,
Sn = $$\frac {n}{2}$$[2a + (n – 1)d], we get
S12 = [2 × (-37) + (12 – 1)4]
= 6(-74 + 11 × 4)
= 6 – 74 +44)
= 6 × (-30) = – 180.
(iii) Let a be the first term and d be the common difference of the given AP.
Then, we have a = 0.6, d = 1.7 -0.6 = 1.1
∵ Sum of n terms of an AP,
Sn = $$\frac {n}{2}$$[2a + (n – 1)d], we get
S100 = $$\frac {100}{2}$$ [2 × 0.6 + (100 – 1) 1.1]
= 50 (1.2 + 99 × 1.1)
= 50 (1.2 + 108.9)
= 50 × 110.1
= 5505
(iv) Let a be the first term and d be the common difference of the given AP.

Question 6.
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution.
Let a be the first term and d be the common difference of an AP.
Give, first term a = 5, last term l = 45 and sum of n terms Sn = 400

∴ The number of terms is 16 and the common difference is $$\frac {8}{3}$$.

Question 7.
The first and the last terms of an AP are 17 and 350, respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution.
Let a be the first term and d be the common difference.
Given, first term a = 17, last term l = an = 350, common difference d = 9
∵ l = an = 350
⇒ a + (n – 1)d = 350
[∵ l = an = a + (n – 1)d]
⇒ 17 + (n – 1)9 = 350 =
⇒ 9(n – 1) = 350 – 17 = 333
⇒ n – 1 = $$\frac {333}{9}$$ = 37
⇒ n = 37 + 1 = 38
Putting a = 17,1 = 350, n = 38
∵ Sum of n terms
Sn = $$\frac {n}{2}$$(a + l), we get
S39 = $$\frac {38}{2}$$ (17 + 350)
= 19 (367) = 6973
So, there are 38 terms in the AP having their sum as 6973.

Question 8.
Which term of the AP 3, 8, 13, 18, …. is 78?
Solution.
Let nth term, be 78. Given, 3, 8, 13, 18 …. are in AP.
First term a, = 3, common difference, d = 8 – 3 = 5
∵ nth term an = 78
∴ a + (n – 1) )d = 78
⇒ 3 + (n – 1) 5 = 78
⇒ (n – 1) 5 = 78 – 3 5
⇒ (n – 1) = 75 n – 1 = 15
⇒ n = 15 + 1
⇒ n = 16
Hence, 16th term be 78.

Question 9.
Find the number of terms in each of the following APs
(i) 7, 13, 19, …., 205
(ii) 18, 15 $$\frac {1}{2}$$, 13, …., -47
Solution.
(i) Suppose, there are n terms in the given AP. Then nth term an = 205, first term a = 7, common difference d = 13 – 7 = 6.
∵ a + (n – 1) d = an
∴ a + (n – 1)d = 205
⇒ 7 + (n – 1)6 = 205
⇒ 6(n – 1) = 205 – 7
⇒ 6(n – 1) = 198
⇒ n – 1 = $$\frac {198}{6}$$ = 33
n = 33 + 1 = 34
⇒ Hence, the given AP contains 34 terms.

(ii) Suppose, there are n terms in the given AP
Given, first term a = 18, common difference

Then nth term an = -47

= -13 × -2 = 26
⇒ n = 26 + 1 = 27
Hence, the given AP contains 27 terms.

### Arithmetic Progressions Class 10 Extra Questions Short answer Type

Question 1.
Find the sum of all odd integers between z and 100 which are divisible by 3.
Solution.
The odd numbers which are divisible by 3, between 2 and 100 are 3, 9, 15, 21, 27, 33,…..99 which clearly form on A.P. here a = 3, d = 6 and l = 99
∴ 99 = a + (n – 1)d
⇒ 99 = 3 + (n – 1)d
⇒ (n – 1)6 = 99 – 3 = 96
⇒ n = $$\frac {96}{6}$$ + 1 = 17
Now, to find the sum of 17 terms

= 17 × 51
= 867.

Question 2.
Which term of arithmetic progression 21, 18, 15…. is – 81? Is zero any term of this arithmetic progress?
Solution.
Given AP be 21, 18, 15…… Let nth term of this progression is – 81
a = 21, d = -3
∴ -81 = 21 + (n – 1) × -3
⇒ -81 – 21 = -3(n – 1)
⇒ n – 1 = $$\frac {102}{3}$$ = 34
∴ n = 34 + 1
= 35
Hence 35th term of this progression is 81
Let Tn = 0 = 21 + (n – 1)(-3)
⇒ 21 – 3n + 3 = 0
∴ 3n = 24
n = 8
then a 8th term of the progression is 0.

Question 3.
How many terms should taken from the series 9, 17, 25, 33… to get a total of 636?
Solution.
The given series an A.P. where a = 9, d 25 – 17 = 8 and Sn = 636.
⇒ Sn $$\frac {n}{2}$$[2a + (n – 1)d]
⇒ 636 = $$\frac {n}{2}$$[2 × 9 + (n – 1) 8]
⇒ 636 = n [9 + 4n – 4]
⇒ 4n2 + 5n – 636 = 0
⇒ 4n2 + (53 – 48) n – 636 = 0
⇒ 4n2 + 53n – 48n – 636 = 0
⇒ n(4n + 53)- 12 (4n + 53) = 0
⇒ (4n + 53) (n – 12) = 0
∴ n = 12 or – $$\frac {53}{4}$$
as number of terms are not infraction and – ve
∴ n = 12
Hence sum of 12 term of the series = 636.

Question 4.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Solution.
Since, the penalty for each succeeding day is 50 more than the preceding day, therefore the penalties for the first day, the second day, the third day, etc. will form an AP.
Let us denote the penalty for the nth day by an, then
a1 = 200, a2 = ₹ 250, a3 = ₹ 300
Here, a = ₹ 200, d = ₹ 250 – ₹ 200 = ₹ 50 and n = 30.
∴ The money the contractor has to pay penalty, if he delayed the work by 30 days.

= 15 (400 + 29 × 50)
= 15 (400 + 1450) = 15 × 1850.
= 27750
So, a delay of 30 days costs is ₹ 27750

Question 5.
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Solution.
Suppose, the respective prizes are a + 60, a + 40, a + 20, a, a – 20, a – 40, a – 60
According to question,
a + 60 + a + 40 + a + 20 + a + a – 20 + a – 40 + a – 60 = 700
⇒ 7a = 700
⇒ a = $$\frac {700}{7}$$ = 100
Hence, the seven prizes are 100 + 60, 100 + 40, 100 + 20, 100, 100 – 20, 100 – 40, 100 – 60 .’i.e., 160, 140, 120, 100, 80, 60, 40 .

Question 6.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution.
Let the Two APs a1, a2, a3…….an and b1, b2, b3, …. bn
Also, let d be the same common difference of two APs. then
the nth term of first AP
On = a1 + (n-1)d and
the nth term of second AP
bn = b1 + (n – 1)d
Now, an – bn = [a1 + (n – 1)d] – [b1 + (n – 1)d]
⇒ an – bn = a1 – by for all n ∈ N
⇒ a100 – b100 = a1 – b1 = 100 (Given)
∴ a1000 – b1000 = a1 – b1
⇒ a1000 – b1000 = 100 [∵ a1 – b1 = 100]
Hence, the difference between their 1000th terms is also 100 for all n ∈ N.

Question 7.
Have many three digit numbers are divisible by 7?
Solution.
We know that, 105 is the first and 994 is the last 3 digit number divisible by 7. Thus, we have to determine the number of terms in the list 105, 112, 119, …, 994.
Clearly, the successive difference of terms is constant with common difference
d = 112 – 105 = 7
So, it forms an AP.
Let there be n terms, in the AP, then nth term = 994
∵ an = a + (n – 1)d
⇒ 105 + (n – 1)7 = 994
⇒ 7(n – 1) = 994 – 105
⇒ 7(n – 1) = 889
⇒ n – 1 = $$\frac {889}{7}$$
= 127
⇒ n = 127 + 1 = 128
So, there are 128 numbers of three digit which are divisible by 7.

Question 8.
How many multiple of 4 lie between 10 and 250?
Solution.
We see that 12 is the first integer between 10 and 250, which is a multiple of 4. Also, when we divide 250 by 4, the remainder is 2. Therefore, 250 – 2 = 248 is the greatest integer divisible by 4 and lying between 10 and 250. Thus, we have to find the number of terms in an AP whose first term = 12, last term = 248 and common difference = 4.
Let the n term of the AP, is
an = 248
= 12 + (n – 1) 4 = 248
[∵ an= a + (n – 1)d]
⇒ 4(n – 1) = 248 – 12
⇒ 4(n – 1) = 236
⇒ n – 1 = $$\frac {236}{4}$$ = 59
⇒ n = 59 + 1 = 60
Hence, there are 60 multiples of 4 lie between 10 and 250.

### Arithmetic Progressions Class 10 Extra Questions Long answer Type

Question 1.
A spiral is made up of successive semicircles with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, …. as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semi-circles?
[Take, π = $$\frac {22}{7}$$ ]

Solution.
Length of spiral made up of thirteen consecutive semi-circles
= (π × 0.5 + π × 1.0 + π × 1.5 + π × 2.0 + … + π × 6.5)
= π × 0.5 (1 + 2 + 3 + … + 13)
which form an AP with first term, a = 1,
common difference, d = 2 – 1 = 1
and number of term, n = 13

= 143 cm.

Question 2.
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see figure). In how many rows are the 200 logs placed and how many logs are in the top row?

Solution.
Since, logs are stacked in each row form a series 20 + 19 + 18 + 17 + …. Clearly, it is an AP with first term, a = 20 and common difference, d = 19 – 20 = -1
Suppose, Sn = 200
∵ Sn = $$\frac {n}{2}$$[2a + (n – 1) d]
⇒ 200 = $$\frac {n}{2}$$[2 × 20 + (n – 1) (-1)]
⇒ 400 = n(40 – n + 1)
⇒ n2 – 41n + 400 = 0
⇒ n2 – 25n – 16n + 400 = 0
(By factorization method)
⇒ n(n-25) – 16 (n – 25) = 0
⇒ (n – 25) (n – 16) = 0
⇒ n = 16
or n = 25
Hence, the number of rows is either 25 or 16.
When n = 16, tn = a + (n – 1)
= 20 + (16 – 1) (-1)
= 20 – 15
= 5
When n = 25,
tn = a+ (n – l) d
= 20 + (25 – 1) (-1)
20 – 24 = -4 (Not possible)
Hence, the number of row is 16 and number of logs in the top row = 5.

Question 3.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in lines (see figure) A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

Solution.
According to question, a competitor pick up the Ist potato, second potato, third potato, fourth potato ….
The distances sum by competitor are 2 × 5, 2 × (5 + 3), 2 × (5 + 3 + 3), 2 × (5 + 3 + 3 + 3) i.e., 10, 16, 22, 28, …..
Clearly, it is an AP with first term, a = 10
and common difference, d = 16 – 10 = 6
∵ The sum of n terms,
Sn = $$\frac {n}{2}$$[2a + (n – 1) d]
∴ The sum of 10 terms,
S10 = $$\frac {10}{2}$$[2 × 10 + (10 – 1) × 6]
[∵ n = 10 (given)]
= 5 (20 + 54)
= 5 × 74
= 370
Hence, the total distance the competitor has to run = 370 m.

## Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Here you will find Triangles Class 10 Extra Questions Maths Chapter 6 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction.

## Extra Questions for Class 10 Maths Triangles with Answers Solutions

Extra Questions for Class 10 Maths Chapter 6 Triangles with Solutions Answers

### Triangles Class 10 Extra Questions Objective Type

Question 1.
In the figure, a line segment PQ is drawn || to the base BC of ∆ABC If PQ : BC = 1: 3, then the ratio of AP and PB will be :
(a) 1 : 2
(b) 1 : 3
(c) 1 : 4
(d) 2 : 3.
(c) 1 : 4
Solution.
Given PQ || BC and PQ : BC = 1 : 3

∴ AP : PB = 1 : 2
Hence Choice (c) is correct

Question 2.
In the figure AB = 3 cm, AC = 6 cm, BD = 2 cm and CD = 4 cm. Find the ratio ∠BAD and ∠CAD is
(a) 2 : 4
(b) 1 : 1
(c) 3 : 6
(d) 6 : 3
(b) 1 : 1
Solution.
In ∆ADB, sin ∠BAD = $$\frac{B D}{B A}$$ = $$\frac{2}{3}$$

In ∆ADC, sin ∠CAD =$$\frac{4}{6}$$ = $$\frac{2}{3}$$
∴ ratio is 1 : 1 choice (b) is correct

Question 3.
In a ∆ABC, if DE is drawn parallel to BC, cutting AB and ACat Dand E respectively such that AB = 7.2 cm, AC = 6.4 cm and AD = 4.5 cm. Then AE = ?

(a) 5.4 cm
(b) 4 cm
(c) 3.6 cm
(d) 3.2 cm
(b) 4 cm

Question 4.
In ∆ABC, DE || BC so that AD = (7x – 4) cm, AE = (5x – 2) cm, DB = (3x + 4) cm and EC = 3x cm. Then, we have :

(a) x = 3
(b) x = 5
(c) x = 4
(d) x = 2.5
(c) x = 4

Question 5.
In ∆ABC, DE || BC such that $$\frac{AD}{DB}$$ = $$\frac{3}{5}$$. If AC = 5.6 cm, then AE = ?

(a) 4.2 cm
(b) 3.1 cm
(c) 2.8 cm
(d) 2.1 cm
(d) 2.1 cm

Question 6.
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of ∆ABC and ∆BDE is :
(a) 2 : 1
(b) 1 : 2
(c) 4 : 1
(d) 1 : 4.
(c) 4 : 1

Question 7.
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.
(d) 16 : 81.

Question 8.
Hari goes 18 m due east and then 24 m due north. The distance from the starting point is :
(a) 40 m
(b) 30 m
(c) 26 m
(d) 42 m.
(b) 30 m

Question 9.
The length of the second diagonal of a rhombus, whose side is 5 cm and one of the diagonals is 6 cm is :
(a) 7 cm
(b) 8 cm
(c) 9 cm
(d) 12 cm.
(b) 8 cm

Question 10.
In ∆ABC, AB = 6√3, AC = 12 cm and BC = 6 cm. The angle B is :
(a) 120°
(b) 60°
(c) 90°
(d) 45°
(c) 90°

### Triangles Class 10 Extra Questions Very Short Answer Type

Question 1.
In the figure, DE || BC. If AD = x, BD = x – 2, AE = x + 2 and EC = x – 1, find the value of x.
Solution.
In ∆ABC, DE || BC

⇒ (x – 2) (x + 2) = x (x – 1)
⇒ x2 – 4 = x2 – x
∴ x = 4

Question 2.
State whether the following quadrilaterals are similar or not 3 cm C

Solution.
The two quadrilaterals, in figure are not similar because their corresponding angles are not equal. It is clear from the figure that, ∠A is 90° but ∠P is not 90°.

Question 3.
E and Fare points on the sides PQ and PR respectively of a 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.
(i) In figure,

⇒ EF is not parallel QR because converse of basic proportionality theorem is not satisfied.

(ii) In figure,

⇒ EF || QR because converse of basic proportionality of theorem is satisfied

(iii) In figure,

⇒ EF || QR because converse of basic proportionality theorem is satisfied

Question 4.
In the figure, if $$\frac{A O}{O C}=\frac{B O}{D O}=\frac{1}{2}$$ and AB = 5 cm find the value of DC.

Solution.
In ∆OAB and ∆OCD
∠AOB = ∠COD (vertically opposite angles)

∴ DC = 5 × 2 = 10 cm.

Question 5.
Perimeters of two similar triangles are 40 cm and 60 cm respectively. Find ratio among their areas.
Solution.
Given that the triangles are similar
∴ ratio of the perimeters of triangle = 40 : 60
= 2 : 3
So the ratio of the area = (2)2 : (3)2
= 4 : 9

Question 6.
Using theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (recall that you done it in Class IX).
Solution.
In ∆ABC, D and E are midpoints of side AB and AC, respectively.

(By converse of basic proportionality theorem)

Question 7.
In figure, E is a point on side CB produced of an isosceles ∆ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF
Solution.
In figure, we are given that ∆ABC is isosceles

and AB = AC =
⇒ ∠B = ∠C …..(i)
For ∆ABD and ∆ECF,
∠ABD = ∠ECF [From Eq. (i)]
and ∠ADB = ∠EFC [Each = 90°]
⇒ ABD ~ AECF
(AAA similarity criterion)

Question 8.
D is point on the side BC of a ∆ABC such that ∠ADC = ∠BAC. Show that CA2 = CB · CD.
Solution.
Draw a ∆ABC such that D is a point on BC and join AD.

For ∆ABC and ∆DAC, we have
and ∠ACB = ∠DCA (Common ∠C)
(AAA similarity criterion)
⇒ $$\frac{A C}{C B}$$ $$\frac{C D}{C A}$$
⇒ $$\frac{C A}{C D}$$ $$\frac{C B}{C A}$$
⇒ CA × CA = CB × CD
⇒ CA2 = CB × CD

Question 9.
Let ∆ABC ~ ∆DEF and their areas be, 64 cm2 and 121 cm2, respectively. If EF = 15.4 cm, find BC.
Solution.
∆ABC ~ ∆DEF (Given)

(Using property of area of similar triangles).

Question 10.
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD. Find the ratio of the area of ∆AOB and ∆COD.
Solution.

Question 11.
If the areas of two similar triangles are equal, prove that they are congruent.
Solution.
Let ∆ABC ~ ∆PQR and ar
(∆ABC) = ar (∆PQR) (Given)

(Using property of area of similar triangles)
⇒ AB = PQ, BC = QR
and CA = PR
(SSS proportionality criterion)
⇒ ∆ABC ≅ ∆PQR.

Question 12.
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.
Draw ABCD is a square having sides of length = a

Then, the diagonal, BD = a√2
We construct equilateral ∆PAB and ∆QBD.
∆PAB ~ ∆QBD
(Equilateral triangles are similar)

Question 13.
ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.
Solution.
Draw ABC is an isosceles triangle right angled at C.

and AC = BC …..(i)
By Pythagoras theorem, we have
AB2 = AC2+ BC2 = AC2 + AC2 = 2AC2
[∵ BC = AC by Eq. (i)]

Question 14.
A ladder 10 m long reaches a window 8 cm above the ground. Find the distance of the foot of the ladder from base of the wall.
Solution.
Let B be the position of the window and CB be the length of the ladder.

Then, AB = 8 cm (Height of window)
BC = 10 cm (Length of ladder)
Let AC = x m be the distance of the foot of the ladder from the base of the wall.
Using Pythagoras theorem in ∆ABC, we get
AC2 + AB2 = BC2
∴ x2 + (8)2 = (10)2
⇒ x2 = 100 – 64 = 36
⇒ x = 6, i.e., AC = 6 cm

### Triangles Class 10 Extra Questions Short Answer Type

Question 1.
In figures, (i) and (ii), DE || BC. Find EC in figure (i) and AD in figure (ii).

Solution.

Question 2.
In ∆ABC, if the side AD is perpendicular to side BC and AD2 = BD × CD. Prove that ∆ABC is a right angle ∆.
Solution.
In ∆ABD and ∆ADC, ∠D = 90°
∴ AB2 = AD2 + BD2 …(i)
and AC2 = AD2 + CD2 … (ii)

AB2 + AC2 = 2AD2 +BD2+CD2
= 2(BD × CD) + BD2+CD2
= BD2 + 2BD × CD + CD2
= 2(BD + CD)2
(Given AD2 = BD × CD).
= BC2
∴ ∠BAC = 90°
Hence ∆BAC is a right angle ∆.

Question 3.

Prove that AB2 + CD2 = BD2 + AC2
Solution :
∴ AB2 = AD2 + BD2 …(i)
again in ∆ADC, ∠D = 90°
AC2 = AD2 + CD2 …(ii)
Subtract (ii) from (i)
AB2 – AC2 = BD2 – CD2
⇒ AB2 + CD2 = AC2 + BD2

Question 4.
Using theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. :
Solution.
In ∆ABC, D is the mid-point of AB.

i.e., $$\frac{A D}{D B}$$ = 1 ……(i)
As straight line l || BC.
Line l is drawn through D and it meets AC at E.
By basic proportionality theorem,

⇒ E is the mid-point of AC.

Question 5.
The diagonals of a quadrilateral ABCD intersect each other at the point O. such $$\frac{A O}{B O}=\frac{C O}{D O}$$ Show that ABCD is a trapezium.
Solution.

⇒ OE || CD(By converse of basic proportionality theorem)
Now, we have BA || OE and OE || CD = AB || CD
⇒ Quadrilateral ABCD is a trapezium.

Question 6.
S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.
Solution.
Draw a ∆RPQ such that S and T are points on PR and QR and joining them.

In figure, we have ∆RPQ and ∆RTS in which
∠RPQ = ∠RTS (Given)
∠PRQ = ∠SRT (Each = ∠R)
Then, by AAA similarity criterion, we have
∆RPQ ~ ∆RTS
Note: If any two corresponding angles of the triangles are equal, then their third corresponding angles are also equal by AAA.

Question 7.
In figure, altitudes AD and CE of ∆ABC intersect each other at the point P. Show that

(i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iv) ∆PDC ~ ∆BEC
Solution.
(i) In figure, ∠AEP = ∠CDP (Each = 90°)
and ∠APE = ∠CPD
(Vertically opposite angles)
⇒ ∠AEP ~ ∆CDP
(By AAA similarity criterion)

(ii) In figure,
∠ADB = ∠CEB (Each = 90°)
and ∠ABD = ∠CBE (Each = ∠B)
⇒ ∆АВD ~ ∆СВЕ
(By AAA similarity criterion)

(iii) In figure,
∠AEP = ∠ADB (Each = 90°)
and ∠PAE = ∠DAB (Common angle)
(By AAA similarity criterion)

(iv) In figure,
∠PDC = ∠BEC (Each = 90°)
and ∠PCD = ∠BCE (Common angle)
⇒ ΔPDC ~ ΔBEC
(By AAA similarity criterion)

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.
Draw a parallelogram ABCD and produce a line AD to AE and joining BE. In parallelogram ABCD,
∠A = ∠C …(i)
Now, for ∆ABE and ∆CFB,
∠EAB = ∠BCF [From Eq. (i)],

∠ABE = ∠BFC
(Alternate angles as AB || FC)
∆ABE ~ ∆CFB (AAA similarity)

Question 9.
In figure, ABC and AMP are two right triangles, right angled at B and M, respectively. Prove that

(i) ∆ABC ~ ∆AMP
(ii) $$\frac{C A}{P A}=\frac{B C}{M P}$$
Solution.
(i) In figure, we have ∠ABC = ∠AMP (Each = 90°)
Because the ∆ABC and ∆AMP are right angled at B and M, respectively.
Also, ∠BAC = ∠PAM
(Common angle ∠A)
⇒ ABC – ∆AMP
(By AAA similarity criterion)

(ii) As ∆ABC ~ ∆AMP,
$$\frac{A C}{A P}=\frac{B C}{M P}$$
(Ratio of the corresponding sides of similar triangles)
⇒ $$\frac{C A}{P A}=\frac{B C}{M P}$$

Question 10.
In a triangle ABC, AD is the median of BC and E is mid-point of AD. If BE produced, it meets AC in figure. Prove that AF = $$\frac {1}{3}$$ AC.
Solution.
Given: In ∆ABC, AD is the median and E is the mid-point of AD.BE is joined and produced intersect AC at F.
T.P.T. – AF = $$\frac {1}{3}$$ AC

Construction-Draw a line parallel to BF through B, which intersect AC at G. EF || DG
Proof – In ∆ADC, EF || DG
∴ AF = FG (Mid-point theorem) …(i)
In ABFC, EF || DG
∴ FG = GC (Mid-point theorem) ….(ii)
From Eq. (i) & (ii)
AF = FG = GC Now
AC = AF + FG + GC
= AF + AF + AF
= 3AF
AF = $$\frac {1}{3}$$AC.

Question 11.
BL and CM are the medians of a right triangle ABC right angle at A. Prove that 4(BL2 + CM)2 = 5 BC2.
Solution.
Given,
In ∆BAC, ∠A = 90° and BL and CM are the medians.

T.P.T. – 4(BL2+cm2) = 5BC2
Proof – In ∆BAC, ∠A = 90°
AB2 + AC2 = BC2 ….(i)
In ∆BAL, ∠A = 90°
BL2 = AB2 + AL2

In ∆MAC,  ∠A = 90°
CM2 = AM2 = AC2

⇒ 4(BL2 + cm2) = 5(AB2 + AC2)
= 5BC2 From Eq. (i)
(AB2 + AC2 = BC2)

Question 12.
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.
Soultion.
In figure (i), AB is a pole behind it a Sun is risen which casts a shadow of length BC = 4 cm and makes an angle θ to the horizontal and in figure it, PM is a height of the tower and behind a Sun risen which casts a shadow of length, NM = 28 cm.

Question 13.
In figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O,
show that

Solution.
Draw AL ⊥ BC and DM ⊥ BC (See figure)
∠ALO = ∠DMO = 90°
and ∠AOL = ∠DOM
(Vertically opposite angle)
∴ ΔOLA ~ ΔΟΜD
(AAA similarity criterion)

Question 14.
D, E and Fare respectively the midpoint of sides AB, BC and CA of ∆ABC. Find the ratio of the areas of ∆DEF and ∆ABC.
Solution.
Draw a ∆ABC taking mid-points D, E and F on AB, BC and AC. Join them.
Here, DF = $$\frac {1}{2}$$BC, DE = $$\frac {1}{2}$$CA
and EF = $$\frac {1}{2}$$AB …….(i)
(∵ D, E and Fare mid-points of sides AB, BC and CA respectively)

[∵ ar (∆CAB)= ar (∆ABC)]
Hence, the required ratio is 1 : 4.

Question 15.
Equilateral triangles are drawn on the sides of a right angled triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.
Solution.
Given. A right angled triangle ABC, with right angle at B. Equilateral triangles PAB, QBC and RAC are described on sides AB, BC and CA respectively.

To prove. Area (∆PAB) + area (∆QBC) = area (∆RAC)
Proof. Since, triangles PAB, QBC and RAC are equilateral. Therefore, they are equiangular and hence similar.

Question 16.
In adjoining Fig. ABC and BCD are two triangles on the same base BC. If AD intersects BC at O, show that

Solution.
Given. ∆ABC and ABCD are on the same base and AD intersects BC at O.

Construction. Draw AM ⊥ BC and DN
Proof. In ∆AMO and ∆DNO,
M = ∠N
[Each 90°, by construction]
∠AOM – ∠DON [same angles]
∆AMO ~ ∆DNO [AA corollary]
⇒ $$\frac{A O}{D O}=\frac{A M}{D N}$$
[Corresponding sides are proportional in similar triangles]

Question 17.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM × MR.
Solution.
In ∆PQR and ∆MPQ,

∠1 + ∠2 = ∠2 + ∠4 (Each = 90°)
⇒ ∠1 = ∠4
Similarly, ∠2 = ∠3
and ∠PMR = ∠PMQ (each 90°)
∆QPM ~ ∆PRM (AAA criterion)

⇒ PM2 = QM × RM
or PM2 = QM × MR

Question 18.
ABC is an isosceles triangle with AC = BC. If AB2 = 2AC?, prove that ABC is a right triangle.
Solution.
Draw an isosceles ∆ABC with AC =BC.

In ∆ABC, we are given that
AC = BC …..(i)
and AB2 = 2AC2 …(ii)
Now, AC2 + BC2-AC2 + AC2 [By Eq. (i)]
= 2AC2 = AB2 (By Eq. (ii)]
i.e., AC2 + BC2 = AB2
Hence, by the converse of the Pythagoras theorem, we have ∆ABC is right angled at C.

Question 19.
ABC is an equilateral triangle of side 2a. Find each of its altitude.
Solution.
Draw equilateral ∆ABC, each side is 2a.

BD = CD = $$\frac {1}{2}$$BC = a
(∵ In an equilateral triangle altitude AD is the perpendicular bisector of BC).
Now, from ∆ABD by Pythagoras theorem, we get
⇒ (2a)2 = AD2 + a2

Question 20.
Prove that the sum of the square of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Solution.
Draw ABCD is a rhombus in which AB = BC = CD = DA = a (Say)

Its diagonal AC and BD are right angled bisector of each other at O.
In ∆OAB, ∠AOB = 90°,
OA = $$\frac {1}{2}$$AC and OB = $$\frac {1}{2}$$BD
In ∆AOB, use Pythagoras theorem, we have
OA2 + OB2 = AB2

⇒ AC2 + BD2 = 4AB2
or 4AB2 = AC2 + BD2
⇒ AB2 + BC2 + CD2 + DA2 = AC2 + BD2
(∵ AB = BC = CD = DA)
Hence proved.

Question 21.
In figure, O is a point in the interior of a ∆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
Solution.
In ∆ABC, from point O join lines OB, OC and OA.
(i) In right angled ∆OFA,
OA2 = OF2 + AF2
(By Pythagoras therorem)
⇒ OA2 – OF2 = AF2 …..(i)
Similarly, in ∆OBD,
OB2 – OD2 = BD2 …(ii)
and in ∆OCE,
OC2 – OE2 = CE2 …(iii)
On adding Eqs. (i), (ii) and (iii), we get
OA2 + OB2 + OC2 – OD2 – OE2 – OF2
= AF2 + BD2 + CE2

(ii) From part Eqs. (i), we get
OA2 + OB2 + OC2 – OD2 – OE2 – OF2
= AF2 + BD2 + CE2 …(iv)
Similarly,
OA2 + OB2 + OC2 – OD2 – OE2 – OF2
= BF2 + CD2 + AE2 …(v)
From Eqs. (iv) and (v), we have
AF2 + BD2 + CE2 = AE2 + CD2 + BF2

Question 22.
A guy wire attached to a vertical pole of height 18m is 24m 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.
Let AB be the vertical pole of height 18 m, A guy wire is of length BC = 24 m.

Let AC = x m be the distance of the stake from the base of the pole.
Using Pythagoras theorem in ∆ABC, we get
i.e., AC2 + AB2 = BC2
∴ x2 + (18)2 = (24)2
⇒ x2 = (24)2 – (18)2
= 576 – 324 = 252
⇒ x = √252 m
(∵ We take positive sign because cannot be negative)
⇒ x = 6 √7 m

Question 23.
An aeroplane leaves an airport and flies due North at a speed of 1000 kmh-1. At the same time, another aeroplane leaves the same airport and flies due West at a speed of 1200 kmh-1. How far apart will be two planes after 1$$\frac {1}{2}$$h?
Solution.
The first plane travels distance BC in the direction of North in 1$$\frac {1}{2}$$h at speed of 1000 km/h.

Question 24.
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the f00t of the poles is 12 m, find the distance between their tops.
Solution.
Let BC and AD be the two poles of heights 11 m and 6 m.
Then, CE = BC – AD
= 11 – 6
= 5 cm

Let distance between tops of two poles DC m = x cm
Using Pythagoras theorem in ∆DEC, we get
i.e., DC2 = DE2 + CE2
⇒ x2 = (12)2 + (5)2 = 169
⇒ x = 13
Hence, distance between their tops = 13 m.

Question 25.
D and E are points on the sides CA and CB, respectively of a ∆ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.
Solution.
Draw a right ∆ABC at C. Take D and E points on the sides CA and BC and join ED, BD and EA.

In right angled ∆ACE,
AE2 = CA2 + CE2 ……(i)
(By pythagoras theorem)
and in right angled ABCD,
BD2 = BC2 + CD2 …(ii)
On adding Eqs. (i) and (ii), we get
AE2 + BD2 = (Ca2 + CE2) + (BC2 + CD2)
= (BC2 + Ca2) + (CD2 + CE2)
(∵ In ∆ABC, Ba2 = BC2 + CA2 and in ∆ECD, DE2 = CD2 + CE2)
= BA2 + DE2
(By Pythagoras theorem),
∴ AE2 + BD2 = AB2 + DE2
Hence proved.

Question 26.
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.
Draw ∆ABC is an equilateral triangle of side a.(Say)

Now, BD = CD = $$\frac {1}{2}$$BC = $$\frac {1}{2}$$a
(In an equilateral triangle altitude AD is a perpendicular bisector of BC)
In right angled ∆ABD,
⇒ a2 = x2 + ($$\frac {1}{2}$$a)2
⇒ a2 = x2 + $$\frac {1}{4}$$a2
⇒ 4a2 = ax2 + a2
⇒ 3a2 = 4x2
Hence proved.

### Triangles Class 10 Extra Questions Long Answer Type

Question 1.
In figure, if LM || CB and LN || CD,
prove that $$\frac{A M}{A B}=\frac{A N}{A D}$$

Solution.

Question 2.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that $$\frac{A O}{B O}=\frac{C O}{D O}$$.
Solution.
We draw,
EOF ||AB (Also || CD)
In ∆ACD, OE || CD

Question 3.
Prove that If the corresponding sides of two triangles are proportional then their corresponding angles are equal, and hence the two triangles are similar.
Solution.
Given. ∆ABC and ∆DEF in which
$$\frac{A B}{D E}$$ = $$\frac{B C}{B F}$$ = $$\frac{A C}{D F}$$
To Prove. ∆ABC ~ ∆DEF.
Construction. Let us take ∆ABC and ∆DEF such that
$$\frac{A B}{D E}$$ = $$\frac{B C}{B F}$$ = $$\frac{A C}{D F}$$(<1).
Cut DP = AB and DQ = AC. Join PQ.

⇒ BC = PQ
Thus, AB = DP, AC = DQ and DC = PQ
∴ ∆ABC ≅ ∆DPQ
[By SSS-congruence]
∴ ∠A = ∠D, ∠B = P = ∠E
and C = ∠Q = ∠F
⇒ ∠A = ∠D, ∠B = ∠E and
∠C = ∠F.
Thus, the given triangles are equiangular and hence similar.

Question 4.
Prove that If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional then the two triangles are similar.
Given. ∆ABC and ∆DEF in which
∠A = ∠D and $$\frac{A B}{D E}$$ = $$\frac{A C}{D F}$$
To prove. ∆ABC ~ ∆DEF.
Construction. Let us take ∆ABC and ∆DEF such that
$$\frac{A B}{D E}$$ = $$\frac{A C}{D F}$$ (<1) and ∠A = ∠D.
Cut DP = AB and DQ = AC. Join PQ.

Proof. In ∆ABC and ADPQ, we have :
AB = DP [By construction]
∠A = ∠D (Given)
AC = DQ [By construction]
∆ABC ≅ ∆DPQ [By SAS-congruence]
∴ ∠A = ∠D, ∠B = ∠P
and ∠C = ∠Q.

[∵ AB = DP and AC = DQ)
⇒ PQ || EF
[By the converse of Thales’ theorem)
⇒ ∠P = ∠E and ∠Q = ∠F
[Corresponding ∠s]
∴ ∠A = ∠D, ∠B = P = ∠E
and C = 2Q = ∠F.
Thus, ∠A = ∠D, ∠B = ∠E
and C = ∠F.
So, the given triangles are equiangular and hence similar.

Question 5.
Prove that If two triangles are equiangular, prove that the ratio of their corresponding sides is the same as the ratio of the corresponding medians.
Solution.
Given. ∆ABC and ADEF in which ∠A = ∠D, ∠B = ∠E and ∠C = ∠F, and AL and DM are the medians.

Question 6.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other the point O. Using a similarity criterion for two triangles, show that
$$\frac{O A}{O C}=\frac{O B}{O D}$$
Solution.
Draw ABCD is a trapezium and AC and BD are diagonals intersect at O.

In figure, AB || DC (Given)
⇒ ∠1 = ∠3, ∠2 = 24
(Alternate interior angles)
Also, ∠DOC = ∠BOA
(Vertically opposite angles)
⇒ ∆OCD ~ ∆QAB (Similar triangle)
⇒ $$\frac{O C}{O A}=\frac{O D}{O B}$$
(Ratios of the corresponding sides of the similar triangles)
⇒ $$\frac{O A}{O C}=\frac{O B}{O D}$$ (Taking reciprocals)
Hence proved.

Question 7.
Sides AB and AC and median AD of a ∆ABC are respectively proportional to sides PQ and PR and median PM of another ∆PQR. Show that ∆ABC ~ ∆PQR.
Solution.
Given, in ∆ABC and ∆PQR,
AD and PM are their medians, respectively.

To Prove. ∆ABC ~ ∆PQR
Construction. Produce AD to E such that AD = DE and produce PM to N such that PM = MN. Join BE, CE, QN and RN.

Proof. Quadrilaterals ABEC and PQNR are parallelograms because their diagonals bisect each other at D and M, respectively.

From Eqs. (ii) and (iii), we have
$$\frac{A B}{P Q}=\frac{B E}{Q N}=\frac{A E}{P N}$$
⇒ ΔΑΒΕ ~ ΔΡΩΝ
⇒ ∠1 = ∠2 ….(iv)
Similarly, we can prove that
∆ACE ~ ∆PRN
∠3 = ∠4 …..(v)
On adding Eqs. (iv) and (v), we have
∠1 + ∠3 = ∠2 + ∠4
⇒ ∠A = ∠P
⇒ ∆BC ~ ∆PQR
(SAS similarity criterion)

Question 8.
In figure, if ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC.

Solution.
In figure,
∆ABE ≅ ∆ACD (Given)
⇒ AB = AC and AE = AD (CPCT)

and also,
∠DAE = ∠BAC (Each = ∠A)
∆ADE ~ ∆ABC (By SAS similarity criterion)
Hence proved.

Question 9.
If AD and PM are medians of ∆ABC and ∆PQR, respectively, where ∆ABC ~ ∆PQR,
prove that $$\frac{A B}{P Q}=\frac{A D}{P M}$$
Solution.
Draw two ∆ABC and APQR taking D and M points on BC and QR such that AD and PM are the medians of the ∆ABC and ∆PQR.

∆ABC – ∆PQR (Given)
⇒ $$\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}$$ ; ∠A = ∠P,
∠B = ∠Q, ∠C = ∠R …(i)
Now, BD = CD = $$\frac {1}{2}$$BC
andQM = RM = $$\frac {1}{2}$$QR …..(ii)
(∵ D is mid-point of BC and M is mid point of QR)
From Eq. (i),

Question 10.
Using the properties of similar triangles, prove that sum of square of two sides is equal to square of its hypotenuse in a right triangle.
Solution.
Given : In right angled ∆ABC, ∠B= 90°

To Prove: AC2 = AB2 + BC2
Construction : Draw BD ⊥ AC.
Pr00f : ∆ADB ~ ∆ABC(by A similarly criterion)
∴ $$\frac{A D}{A B}$$ = $$\frac{A B}{A C}$$
⇒ AB2 = AD × AC
∆BDC ~ ∆ABC(by A similarty criterion)
∴ $$\frac{B C}{D C}$$ + $$\frac{A C}{B C}$$
⇒ BC2 = AC × DC …(ii)
AB2 + BC2 = AD × AC + AC × DC
= AC × AC
= AC2

Question 11.
Prove that the ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution.

In figure, AD is a median of ∆ABC and PM is a median of ∆PQR. Here, D is mid-point of BC and M is mid-point of QR. Now, we have,
∆ABC ~ APQR
⇒ ∠B = ∠Q
(Corresponding angles are equal) …(i)
Also, $$\frac{A B}{P Q}$$ + $$\frac{B C}{Q R}$$
(Ratio of corresponding sides are equal)

Question 12.
The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3 CD (see in figure). Prove that 2AB2 = 2AC2 + BC2.

Solution.
Given, DB = 3CD

= CD = $$\frac{1}{4}$$BC …(i)
and DB = $$\frac{3}{4}$$ BC
In ∆ABD, AB2 = DB2 + AD2 …(ii)
In ∆ACD, AC2 = CD2 + AD2 …(iii)
(By Pythagoras as theorem)
On subtracting Eq. (iii) from Eq. (ii), we get
AB2 – AC2 = DB2 – CD2

⇒ 2AB2 – 2AC2 = BC2
⇒ 2AB2 = 2AC2 + BC2
Hence proved.

Question 13.
In an equilateral ∆ABC, D is a point on side BC such that BD = $$\frac{1}{3}$$BC. Prove that 9AD2 = 7AB2
Solution.
Draw ABC is an equilateral triangle, D is a point on side BC such that BD = $$\frac{1}{3}$$BC. Draw a line AE is perpendicular to BC.

AB = BC = CA = a (Say)
(By property of equilateral triangle)
BD = $$\frac{1}{3}$$BC = $$\frac{1}{3}$$a
⇒ CD = $$\frac{2}{3}$$BC = $$\frac{2}{3}$$a
∵ AE ⊥ BC
⇒ BE = EC = $$\frac{1}{2}$$a
(∵ In an equilateral triangle altitude AE is perpendicular bisector of BC.)
DE = BE – BD = $$\frac{1}{2}$$a – $$\frac{1}{3}$$a – $$\frac{1}{6}$$a
= AB2 – BE2 + DE2
(∵ Right ∆ABE, AE2 = AB2 – BE2)

⇒ 9AD2 = 7AB2 Hence proved.

## Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Solutions Answers

Here you will find Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction.

## Extra Questions for Class 10 Maths Coordinate Geometry with Answers Solutions

Extra Questions for Class 10 Maths Chapter 7 Coordinate Geometry with Solutions Answers

### Coordinate Geometry Class 10 Extra Questions Objective Type

Question 1.
The point (-13, -14) lies in :
(d) None of these.

Question 2.
Point (8, -8) is in which quadrant?
(a) First
(b) Second
(c) Third
(d) Fourth.
(d) Fourth.

Question 3.
The position of the point with abscissa = -5 and ordinate = +4 will be :
(d) None of these.

Question 4.
Distance of point (-6, 8) from origin is :
(a) 8
(b) 2√7
(c) 10
(d) 6
(c) 10
Solution.
Distance of point (-6, 8) from origin (0, 0) is

∴ Choice (c) is correct.

Question 5.
Coordinates of a point on X-axis which is equal distance from the points A(2, -5) and B(-2, 9) will be :
(a) (-7, 0)
(b)(0, 7)
(c) 0, -7)
(d) (7, 0).
(a) (-7, 0)
Solution.
Any point on X-axis is P(x, 0)
∴ AP = BP ⇒ AP2 = BP2
= (x – 2)2 + (-5 – 0)2 = (x + 2)2 + (9 – 0)2
= x2 – 4x + 4 + 25 = x2 + 4x + 4 + 81
x2 – 4x + 29 = x2 + 4x + 85
– 8x = 85 – 29 = 56
x = $$\frac{56}{-8}$$ = -7
∴ point P on x – axis is (-7, 0)
Hence choice (a) is correct.

Question 6.
The coordinate of a point on Y-axis which is equidistant from the point A(6, 5) and B(-4, 3) will be :
(a) (0, 9)
(b) (0, -9)
(c) (0, 5)
(d) (0, 3)
(a) (0, 9
Solution.
Let point P is on Y-axis whose coordinate is (0, y)
PA = PB
⇒ PA2 = PB2
(6 – 0)2 + (5 – y)2 = (-4 -0)2 + (3 – y)2
⇒ 36 + 25 + y2 – 10y = 16 + 9 + y2 – 6y
⇒ 61 – 10y = 25 – 6y
⇒ 61 – 25 = – 6y + 10y
⇒ 4y = 36
⇒ y = 9
Required point P(0, 9)
Hence choice (a) is correct. (a)

Question 7.
A point lies in third quadrant. Co-ordinates of this point may be :
(a) (5, 5)
(b) (5, -5)
(c) (-5, -5)
(d) (-5, 5).
(c) (-5, -5)

Question 8.
The distance between the points (5, – 3) and (8, 1) is:
(a) 5 units
(b) 6 units
(c) 25 units
(d) 10 units.
(a) 5 units

Question 9.
The distance between the points (2, 0) and (-1, 4) is :
(a) 5 units
(b) 25 units
(c)- 25 units
(d) 10 units.
(a) 5 units

Question 10.
The distance between the points (-6, 5) and (-1, 7) is :
(a) 169 units
(b) √119 units
(c) √13 units
(d) None of these.
(d) None of these.

Question 11.
The co-ordinates of a point Mare (3, 4). Its distance from the origin is :
(a) 7 units
(b) 11 units
(c) 5 units
(d) 12 units.
(c) 5 units

Question 12.
The distance between points (7, 3) and (-5, -2) is :
(a) 10 units
(b) 13 units
(c) 16 units
(d) 18 units.
(b) 13 units

Question 13.
The distance between origin and (15, 8) is :
(a) 18 units
(b) 15 units
(c) 17 units
(d) 28 units.
(c) 17 units

Question 14.
The distance between (-1, -3) and (3, 0) is :
(a) 4 units
(b) 5 units
(c) 13 units
(d) 16 units.
(b) 5 units

Question 15.
The distance between the points (-3, 4) and (0, 0) will be :
(a) 5
(b) 4
(c) 13
(d) 1
(a) 5

Question 16.
The third vertex of an equilateral triangle whose other two vertices are (1, 1) and (1, -1) respectively is :
(a) (√3, – √3)
(b) (- √3, √3)
(c) both (a) and (b)
(d) none of these.
(c) both (a) and (b)

Question 17.
The coordinates of two points are (6, 0) and (0, 8). The coordinates of the mid-point are :
(a) (3, 4)
(b) (6, 8)
(c) (0, 0)
(d) None of these.
(a) (3, 4)

Question 18.
The co-ordinates of two points are (9, 4) and (3, 8). The co-ordinates of the mid-point are :
(a) (6, 0)
(b) (0, 6)
(c) (6, 6)
(d) (-6, -6).
(c) (6, 6)

Question 19.
The co-ordinates of two points are (-8, 0) and (0, -8). The co-ordinates of the midpoint of the line segment joining them will be :
(a) (-8, 4)
(b) (4, -8)
(c) (-4, -4)
(d) (-4, 4)
(c) (-4, -4)

Question 20.
The abscissa of the point which divides the line joining points (0, 4) and (0, 8) internally in the ratio of 3 : 1 will be :
(a) 0
(b) 4
(c) 6
(d) 5
(a) 0

### Coordinate Geometry Class 10 Extra Questions Very Short Answer Type

Question 1.
Find the coordinate of a point A. Where AB is the diameter of a circle whose centre is (2, -3) and coordinate of B is (1, 4).
Solution.
As centre is the mid point of AB.

Let coordinate of A is (x, y)
∴ $$\frac{x+1}{2}$$ = 2
⇒ 2 x + 1 = 4
⇒ x = 4 – 1 = 3
and $$\frac{y+4}{2}$$ = -3 = y + 4 = -6
y = – 10
Hence coordinate of A is (3, -10)

Question 2.
What are the ordinate of the point whose abscissa is 10 and whose distance from (2, -3) is 10 units?
Solution.
Let y be the ordinate, then the point is (10, y). The distance between (10, y) and (2, -3)

⇒ y2 + 6y + 73 = 100
⇒ y2 + 6y – 27 = 0
⇒ (y +9)(y – 3) = 0
⇒ y = -9, y = 3.

Question 3.
Find the distance between the points A(a sin θ, a cos θ) and B(a cos θ, – a sin θ).
Solution.
Here x1 = a sin θ, x2 = a cos θ y1 = a cos θ, y2 = – a sin AB
Distance AB

Question 4.
Prove that the point (4, 1) is the centre of the circle on whose circumference lie the points (-2, 9), (10, -7) and (12, -5).
Solution.
Let A(-2, 9), B(10, -7), C(12, – 5) and O(4, 1). Then,

∴ OA = OB = 0C.
Hence O is the centre and A, B, C are the points on the circumference of the circle.

Question 5.
Find that ratio in which the point $$P\left(\frac{3}{4}, \frac{5}{12}\right)$$ divides the line segment joining the points $$A\left(\frac{1}{2}, \frac{3}{2}\right)$$ and B(2, -5).
Solution.
Let P divides AB in the ratio λ : 1

⇒ 3λ + 3 = 8λ + 2
⇒ 5λ = 1
λ = $$\frac {1}{5}$$
Hence P divides AB in ratio 1 : 5

Question 6.
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B.
Solution.
Let M (0, 0) and N (36, 15) be the given points.
Here, x1 = 0, y1 = 0 and x2 = 36, y2 = 15

Since, the position of towns A and B are given (0, 0) and (36, 15), respectively and so, the distance between them is 39 km.

Question 7.
Find the coordinates of that point which divides the line segment joining two given points (3, 5) and (8, 10) internally in the ratio of 2 : 3.
Solution.
Here, x1 = 3, x2 = 8, y1 = 5, y2 = 10, m = 2, n = 3.

Hence, the co-ordinates of the required point are (5, 7).

Question 8.
If the area of ∆ABC formed by A(x, y), B(1, 2) and C(2, 1) is 6 square units, then prove x + y = 15.
Solution.
Area of ∆ = $$\frac{1}{2}$$ [x1 (y2 – y3) + x2(y3 – y1)+ x3 y1 – y2)]
⇒ 6 = $$\frac{1}{2}$$[x(2 – 1) + 1(1 – y) + 2 (y – 2)]
⇒ 12 = x – y + 1 + 2y – 4
⇒ 12 = x + y = 3
∴ x + y = 15

Question 9.
Find the coordinates of point which divide line segment AB joining points A(-2, 2) and B( 2, 8) into four equal parts.
Solution.
Clearly point a is the midpoint of AB

Question 10.
The coordinates of one endpoint of a diameter of a circle are (3, 5). If the coordinates of the centre be (6, 6), find the coordinates of the other endpoint of the diameter.
Solution.
Let AB be the given diameter and O the centre of the circle. Then, the coordinates of A and O and (3, 5) and (6, 6) respectively. Let (a, b) be the coordinates of B.
Then $$\frac{3+a}{2}$$ = 6 and $$\frac{5+b}{2}$$ = 6
∴ a = 9 and 6 = 7.
Hence, the required point is (9, 7).

Question 11.
If the points (6, 9),(0, x)and (-6, -7) are collinear then find the value of x.
Solution.
x1 = 6, y1 = 9
x2 = 0, y2 = x
x3 = -6, y3 = -7
The three points are collinear if the area of the triangle formed by them is zero.
Now,
area = $$\frac{1}{2}$$[x1(y2 – y3) + x2(y3 – y1)
+ x3(y1 – y2)] = 0
⇒ $$\frac{1}{2}$$[6(x + 7) + 0(-7 – 9) + (-6)(9 – x)) = 0
⇒ 6x + 42 – 54 + 6x = 0
⇒ 12x = 12
⇒ x = 1.

Question 12.
Find the value of the points (a, 1), (1, -1) and (11, 4) are collinear.
Solution.
Here, A(a, 1) = (x1, y1) B(1, -1) = (x2, y2,) andC(11, 4) = (x3, y3)
Condition for collinear points
⇒ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)
⇒ a(-1 -4) + 1(4 – 1) + 11(1 + 1) = 0
⇒ -5a + 3 + 22 = 0
⇒ -5a = -25
⇒ a = 5
∴ The given three points are collinear.

### Coordinate Geometry Class 10 Extra Questions Short Answer Type

Question 1.
Show that the points (2, 1),(5, 2), (6, 4) and (3, 3) form a parallelogram.
Solution.
Let A(2, 1), B(5, 2), C(6, 4) and D(3, 3).

Thus, AB = CD, BC = DA.
Since, opposite sides are equal, hence ABCD is a parallelogram.

Question 2.
Prove that the points A(1, -2), B(3, 0) and C(1, 2) are the vertices of an isosceles right-angled triangle.
Solution.
Let A = (1, -2), B = (3, 0) and C = (1, 2).

CA = 4
Now, AB2 + BC2 = (2√2)2 + (2√2)2
= 8 + 8 = 16
AB2 + BC2 = (AC)2 …(4)
Hence, from result (1), (2) and (4) ∆ABC is an isosceles right-angled triangle.

Question 3.
Find the centre of the circle which passes through the points (-3, -2), (-2, 3) and (3, 2).
Solution.
Let O(x, y) be the centre of the circle and the points
A(-3, – 2), B(-2, 3) and C(3, 2)
lie on the circle, Then, we have
OA2 = (x + 3)2 + (y + 2)2
OB2 = (x + 2)2 + (y – 3)2
and OC2 = (x – 3)2 + (y – 2)2
∵ OA = OB
⇒ OA2 = OB2
∴(x + 3)2 + (y + 2)2 = (x + 2)2 + (y – 3)2
⇒ x2 + y2 + 6x + 4y + 13
= x2 + y2 + 4x – 6y + 13
⇒ 6x + 4y = 4x – 6y
⇒ 2x + 10y = 0
⇒ x + 5y = 0 ……(i)
Again, OB = OC
⇒ OB2 = OC2
∴ (x + 2)2 + (y – 3)2 = (x – 3)2 + (y – 2)2
⇒ x2 + y2 + 4x – 6y + 13
= x2 + y2 – 6x – 4y + 13
⇒ 4x – 6y = – 6x – 4y
⇒ 10x – 2y = 0
⇒ 5x – y = 0 …..(ii)
On solving (i) and (ii), we get
x = 0 and y = 0.
Hence, the centre of the circle is (0, 0).

Question 4.
Prove that the triangle whose vertices are respectively (a, a), (-a, -a) and (-a√3, a√3) is an equilateral triangle.
Solution.
Let points P(a, a), Q(-a, -a) and R(-a√3, a√3) are given, then
and

= a√8 = 2a√2
Thus, PQ = QR = RP
∴ ∆PQR is an equilateral triangle.

Question 5.
Find the point on the X-axis which is equidistant from (2, -5) and (-2, 9).
Solution.
Since, the point on the X-axis.
∴ it’s ordinate = 0
So, A(x, 0) is any point on the X-axis.
Since, A(x, 0) is equidistant from B(2, -5) and C(-2, 9).
AB = AC
⇒ AB2 = AC2
⇒ (x – 2)2 + (0 + 5)2 = (x + 2)2 + (0 – 9)2

⇒ x2 – 4x + 4 + 25 = x2 + 4x + 4 + 81
⇒ -4x – 4x = 81 – 25
⇒ -8x = 56
⇒ x = –$$\frac{56}{8}$$ = -7
So, the point equidistant from given points on the x-asis is (-7, 0).

Question 6.
Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Solution.
According to question,
PQ = 10

Squaring both sides, we get
⇒ y2 + 6y + 73 = 100
⇒ y2 + 6y – 27 = 0
⇒ y2 + 9y – 3y – 27 = 0
⇒ y(y + 9) – 3 (y + 9) = 0
⇒ (y + 9) (y – 3) = 0
⇒ y = -9 or y = 3
⇒ y = -9 or 3

Question 7.
Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).
Solution.
Let the point A(x, y) be equidistant from the points B(3, 6) and (-3, 4).
AB = AC
AB2 = AC2
⇒ (x – 3)2 + (y – 6)2 = (x + 3)2 + (y – 4)2

⇒ x2 – 6x + 9 + y2 – 12y + 36
⇒ x2 + 6x + 9 + y2 – 8y + 16
⇒ -6x – 6x – 12y + 8y + 36 – 16 = 0
⇒ -12x – 4y + 20 = 0
⇒ -4(3x + y – 5) = 0
⇒ 3x + y – 5 = 0
(∵ -4 ≠ 0)

Question 8.
If the vertices of a ∆ABC, are (sin2 θ, cos2 θ), (cos2 θ, sin2 θ) and (0, 0) then find the coordinates of the centroid of the triangle.
Solution.
Here A = (sin2 θ, cos2 θ)
= (x1, y1)
B = (cos2 θ, sin2 θ)
= (x2, y2)
C = (0, 0)
= (x3, y3)
The co-ordinates of the centroid be

Question 9.
Find the ratio in which the point (11, 15) divides the line segment joining the points (15, 5) and (9, 20).
Solution.
A(x1, y1) = (15, 5)
B(x2, y2) = (9, 20)
P(x, y) = (11, 15)
m1 : m2 = ?

⇒ 11(m1 + m2) = 9m1 + 15m2
⇒ 15(m1 + m2) = 20m1 + 5m2
⇒ 2m1 = 4m2
⇒ 5m1 = 10m2
⇒ m1 = 2m2
⇒ m1 = 2m2
⇒ $$\frac{m_{1}}{m_{2}}$$ = $$\frac{2}{1}$$
= i.e., m1 = m2 = 2 : 1.

Question 10.
Find the coordinates of the point which divides the line segment joining the points (a, b) and (b, a) internally in the ratio of (a – b) : (a + b).
Solution.
Formula,

Question 11.
The mid-points of the sides of a triangle are (1, 2), (0, -1) and (2, -1). Find the coordinates of the vertices of the triangle.
Solution.
Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of ∆ABC. Let (1, 2), (0, -1) and (2, -1) be the mid-points of AB, BC and CA respectively.
Then

x1 + x2 = 2 …..(i)
x2 + x3 = 0 …..(ii)
x3 + x1 = 4 …..(iii)
y1 + y2 = 4 ……(iv)
y2 + y3 = -2 …..(v)
y3 + y1 = -2 ……(vi)
Adding (i), (ii) and (iii) we get
x1 + x2 + x3 = 3.
∴ x3 = 1, x1 = 3 and x2 = -1.
Adding (iv), (v) and (vi) we get
y1 + y2 + y3 = 0.
∴ y3= – 4, y1 = 2 and y2 = 2.
Hence the vertices of the triangle are A(3, 2), B(-1, 2) and C(1, -4) respectively.

Question 12.
The three vertices of a parallelogram taken in order are (0, -1), (1, 3) and (2, 2) respectively. Find the coordinates of the forth vertex.
Solution.
Let A(0, -1), B(1, 3), C(2, 2) and D(x, y) be the vertices of a parallelogram. Then the mid-points of the diagonals AC and BD have the same co-ordinates.

Hence, the forth vertex of the parallelogram is (1, -2).

Question 13.
Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.
Solution.
Let the coordinates of a point are (x, y).
we have, x1 = -1, y1 = 7;
x2 = 4, y2 = -3
and m1 = 2, m2 = 3
∴ By using section formula,

Hence, coordinates of the point are (1, 3)

Question 14.
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Solution.
Let A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram.

Since, ABCD is a parallelogram.
∴ AC and BD will bisect each other.
Hence, mid-point of AC and mid-point of BD are same point.

⇒ 1 + x = 7 and 8 = 5 + y
∴ x = 6 and y = 3.

Question 15.
Prove that the three points (1, 2), (3, 3) and (-1, 1) are collinear.
Solution.
Three points are collinear if the area of the triangle formed by them is zero.
Here
x1 = 1, y1 = 2
x2 = 3, Y2 = 3
x3 = -1, y3 = 1
Now, area

∴ The given three points are collinear.

Question 16.
If the points A(x, y), B(2, 3) and C(3, 4) are collinear, then prove that x + 5y = 17.
Solution.
Here x1 = x, x2 = 2, x3 = -3, y1 = y, y2 = 3, y3 = 4

The points A, B, C are collinear, therefore ∆ = 0.
∴ $$\frac{1}{2}$$ (- x – 5y + 17) = 0
⇒ – x – 5y + 17 = 0
⇒ x + 5y = 17.

Question 17.
For what value of a, the points (1, 4) (a, -2) and (-3, 16) are collinear?
Solution.
The given points are collinear if the area of triangle formed by these three points is zero.
Here, x1 = 1, x2 = a, x3 = -3
y1 = 4, y2 = -2, y3 = 16

Question 18.
Find the area of the triangle whose vertices are
(i) (2, 3), (-1, 0), (2, -4)
(ii) (-5, -1), (3, -5), (5, 2)

Question 19.
Find the equation of the locus of points equidistant from (-2, 2) and (-4, -2).
Solution.
Let A(-2, 2) and B(-4, -2) be the given points and let P(x, y) be the variable point, Then.
PA = PB PA2 = PB2
⇒ (x + 2)2 + (y – 2)2 = (x + 4)2 + (y + 2)2
⇒ x2 + 2x + 4 + y2 – 2y + 4 = x2 + 8x + 16 + y2 + 4y + 4
⇒ 2x – 2y + 8 = 8x + 4y + 20
⇒ – 6x – 6y – 12 = 0
⇒ x + y + 2 = 0
which is the required equation of the locus.

Question 20.
Find the equation of the locus of all points equidistant from (3, 5) and the x-axis.
Solution.
Let P(x, y) be the variable point, then its distance from x-axis is y and its distance from (3, 5) is

By the given condition we have

Squaring on both sides we get
y2 = x2 – 6x + 9 + y2 – 10y + 25
0 = x2 – 6x – 10y + 25
∴ x2 – 6x – 10y + 25 = 0
which is the required equation of the locus.

Question 21.
Prove that the equation of the locus of a point which moves in such a way that its distance from a point (-g, -f) is always equal to a, is
x2 + y2 + 2gx + 2fy + c = 0,
where x = (g2 + f2 – a2).
Solution.
Let P(x, y) be the variable point on the locus. Then,

⇒ (x + g)2 + (y + f)2 = a2
⇒ x2 + y2 + 2gx + 2fy + (g2 + f2 – a2) = 0
⇒ x2 + y2 + 2gx + 2fy + c = 0,
where c = (g2 + f2 – a2).

### Coordinate Geometry Class 10 Extra Questions Long Answer Type

Question 1.
In a classroom, 4 friends are seated at the points, A, B, C and Das shown in figure. Champ 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.

Solution.
From the given figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6, 1).

Since, the four sides and diagonals are equal. Hence, ABCD is a square.
So, Champa is correct.

Question 2.
If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x. Also, find the distance QR and PR.
Solution.
Since, the point Q (0, 1) is equidistant from P(5, 3) and R(x, 6)
∴ QP = QR
∴ QP2 = QR2
= (5 – 0)2 + (-3 – 1)2 = (x – 0)2 + (6 – 1)2

⇒ 52 + 42 = x2 + 52
⇒ 25 + 16 = x2 + 25
⇒ x2 = 16
⇒ x = ±4
Thus, R is (4, 6) or (-4, 6).
Now, QR = Distance between Q(0, 1) and R(4, 6)

Question 3.
Find the third vertex of the triangle ABC, if its two vertices are (3,5), (-4, -6) and the coordinates of the centroid are (4, 3).

Solution.
Let the co-ordinates of the third vertex be C(x, y).
Here x1 = 3, x2 = – 4, x3 = x
y1 = 5, y2 = -6, y3 = y

⇒ -1 + x = 12; – 1 + y = 9
⇒ x = 13; y = 10
∴ Co-ordinates of Care (13, 10).

Question 4.
The endpoints of a line segment AB are A(2, 5) and B(1, 9). In what ratio the line segment is drawn through A and B is divided by x-axis and y-axis?
Solution.
For x-axis :
The coordinates of any point on the x-axis are (x, 0).
Let the required ratio be m1 : m2

Hence, the required ratio is 5 : 9. The negative sign represents external division.
For y-axis :
The coordinates of any point on the y-axis are (0, y).
Let the required ratio be m1 : m2

Hence, the required ratio is 2 : 1. The negative sign represents external division.

Question 5.
A point (2, 3) divides the distance between the points (x, y) and (-5, 4) internally in the ratio 1 : 2. Find the values of x and y.
Solution.

Question 6.
Use analytical geometry to prove that the mid-point of the hypotenuse of a right-angled triangle is equidistant from its vertices.
Solution.
Let AOB be the given right-angled triangle with base OA taken along the x-axis and the perpendicular OB taken along the y-axis.
Let OA = a and OB = b.
Let D be the mid-point of hypotenuse AB.
Then, the coordinates of these points are

⇒ DA = DB = DO.

Question 7.
If the centroid of any triangle ABC be G, then using analytical geometry, prove that AB2 + BC2 + CA2 = 3(GA2 + GB2 + GC2).
Solution.
Let B be the origin, BC be X-axis and through B perpendicular on BC be Y-axis. Then coordinates of B are (0, 0).

Let BC = a, then co-ordinates of Care (a, 0). Let the coordinates of A be (x1, y1), then co-ordinates of the centroid G of ∆ABC are :

Question 8.
Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Solution.
Let A(4, -1) and B(-2, -3) be the line segments and points of trisection of the line segment be P and Question Then,
AP = PQ = BQ = k (Say)
∴ PB = PQ + QB = 2k
and AQ = AP + PQ = 2k

⇒ AP : PB = k : 2k = 1 : 2
and AQ : QB = 2k : k = 2 : 1
Since, P divides AB internally in the ratio 1 : 2. So, the coordinate of P are

and Q divides AB internally in the ratio 2 : 1. So, the coordinates of Q are

So, the two points of trisection are

Question 9.
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along with AD, as shown in figure. Niharika runs $$\frac{1}{4}$$th the distance AD on the 2nd line and posts a green flag. Preet runs $$\frac{1}{5}$$th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Solution.
From the above figure, the position of green flag posted by Niharika is M

Let P be the position of the blue flag posted by Rashmi in the halfway of line segment MN.

Hence, the blue flag is on the fifth line at a distance of 22.5 m above it.

Question 10.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by ( 1, 6).
Solution.
Let the point A(-1, 6) divide the line joining B(-3, 10) and C(6, -8) in the ratio k : 1. Then, the coordinates of A are

⇒ 6k – 3 = – k – 1 and – 8k + 10 = 6k + 6
⇒ 6k + k = – 1 + 3 and – 8k – 6k = 6 – 10
⇒ 7k = 2 and – 14k = -4 ⇒ k = $$\frac{2}{7}$$
So, the point A divides BC in the ratio 2 : 7.

Question 11.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and Bis (1, 4).
Solution.
Suppose, AB be a diameter of the circle having its centre at C(2, -3) and coordinates of endpoint B are (1, 4).
Let the coordinates of A be (x, y).

Since, AB is diameter.
∴ C is the mid-point of AB.

⇒ y + 4 = -6
⇒ y = -10

Question 12.
If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that
AP = $$\frac{3}{7}$$AB and P lies on the line segment AB.
Solution.
According to the question,

Suppose, P(x, y) be the point which divides the line segment joining the points A (-2, -2) and B (2, -4) in the ratio 3 : 4.

Question 13.
Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
Solution.
Let P, Q and R be the points on line segment AB such that
AP = PQ = QR = RB
Let AP= PQ = QR = RB = k .

Q is the mid-point of AB.

Question 14.
The coordinates of three vertices of a ∆ABC are A(3, 4), B(2, -1) and C(4, -6) and the middle points of the three sides BC, CA and AB are respectively, P, Q and R. Prove that:
Area of ∆ABC = 4 (area of ∆PQR).
Solution.
The middle point of BC is

From (i) and (ii), we get
Area of ∆ABC = 4 x area of ∆PQR.

Question 15.
Find the area of a triangle, the coordinates of the mid-points of whose sides are (-2, -1), (1, 6) and (5, 3) respectively.
Solution.
Let ABC be the given triangle and D(-2, -1), E(1, 6) and F(5, 3) be the midpoints of AB, BC and AC respectively.

The co-ordinates of mid-points D, E and F are
Here
x1 = -2, x2 = 1, x3 = 5
y1 = -1, y2 = 6, y3 = 3
Area of ∆DEF

= $$\frac{37}{2}$$ square unit.
Area of ∆ABC = 4 × ar (∆DEF)
= 4 × $$\frac{37}{2}$$
= 74 square units.

Question 16.
Median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A (4, -6), B (3, – 2) and C (5, 2).
Solution.
According to the question, AD is the median of ∆ABC, therefore D is the midpoint of BC.

(∵ Area of triangle is positive) .
∵ Area of ∆ADC = Area of ∆ABD
Hence, the median of the triangle divides it into two triangles of equal areas.

Question 17.
Find the locus of the point P(x, y) when :
(i) x = 3
(ii) y = 5
(iii) x = y
Solution.
(i) The given condition is x = 3.
As y is not restricted, it can take any real value between – ∞ to ∞ i.e., – ∞ < y < ∞.

Hence, in the figure the line AB is the required locus.

(ii) The condition given in the question is y = 5. As x is not restricted, it can take any real value between -∞ and ∞ i.e., -∞ < x < ∞.

Hence, in the figure the line PQ is the required locus.

(iii) The given condition is x = y. If x takes the values – 2, -1, 0, 1, 2, then y will also take the values -2, -1, 0, 1, 2 respectively. Therefore, the points (-2, -2), (-1, -1), (0, 0), (1, 1) and (2, 2) will satisfy the above condition. Hence, AB will be the required locus.

## Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers

Here you will find Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction.

## Extra Questions for Class 10 Maths Introduction to Trigonometry with Answers Solutions

Extra Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry with Solutions Answers

### Introduction to Trigonometry Class 10 Extra Questions Objective Type

Question 1.
The value of $$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}$$ will be:
(i) sin 60°
(ii) cos 60°
(iii) tan 60°
(iv) sin 30°
(ii) cos 60°
Solution.

= 2 sin 30o. cos 30°
= sin 60°
Hence, Choice (ii) is correct

Question 2.
If sin θ = 1, the value of sin 20 will be :
(i) -1
(ii) 0
(iii) 1
(iv) 2
(ii) 0

Question 3.
The value of cot (- 1470°) will be :
(i) –$$\frac{1}{\sqrt{3}}$$
(ii) – √3
(iii) $$\frac{1}{\sqrt{3}}$$
(iv) √3
(ii) – √3

Question 4.
In triangle ABC, the value of sin (B + C) will be :
(i) sin B + sin C
(ii) sin A
(iii) 0
(iv) cos A
(ii) sin A

Question 5.
The value of cos 15° is :
(i)
(ii)
(iii)
(iv)
(i)

Question 6.
If sin θ = cosec θ and ≤ θ ≤ π then value of 0 will be :
(i) π
(ii) $$\frac{\pi}{2}$$
(iii) $$\frac{\pi}{4}$$
(iv) 0
(ii) $$\frac{\pi}{2}$$

Question 7.
The value of cosec 810 will be :
(i) -1
(ii) 0
(iii) 1
(iv) ∞
(iii) 1

Question 8.
Find the value of $$\frac{\sin ^{2} 15^{\circ}-\cos ^{2} 15^{\circ}}{\sin ^{2} 15^{\circ}+\cos ^{2} 15^{\circ}}$$.
(i) 1
(ii) -1
(iii) $$\frac{2}{\sqrt{3}}$$
(iv) $$\frac{-\sqrt{3}}{2}$$
(iv) $$\frac{-\sqrt{3}}{2}$$

Question 9.
The value of will be :
(i) $$\frac{1}{\sqrt{2}}$$
(ii) √2 – 1
(iii) √2
(iv) $$\frac{\sqrt{2}+1}{\sqrt{2}-1}$$
(iii) √2

Question 10.
The value of sin 570° will be :
(i) $$\frac{-1}{2}$$
(ii) $$\frac{-\sqrt{3}}{2}$$
(iii) $$\frac{1}{2}$$
(iv) $$\frac{\sqrt{3}}{2}$$
(i) $$\frac{-1}{2}$$

Question 11.
If tan A = $$\frac{1}{\sqrt{3}}$$ and tan B = √3 then the value of tan (A + B) will be :
(i) 0
(ii) $$\frac{1}{\sqrt{3}}$$
(iii) 1
(iv) ∞
(iv) ∞

Question 12.
If sin α = cos α then the value of α will be:
(i) 30°
(ii) 45°
(iii) 60°
(iv) 90°
(ii) 45°

Question 13.
The value of sin2 θ +$$\frac{1}{1+\tan ^{2} \theta}$$ is :
(i) cos2 θ
(ii) sin2 θ
(iii) 1
(iv) sec2 θ
(iii) 1

Question 14.
If sec θ = 2 then the value of θ will be :
(i) $$\frac{\pi}{2}$$
(ii) $$\frac{\pi}{3}$$
(iii) $$\frac{\pi}{4}$$
(iv) $$\frac{\pi}{6}$$
(ii) $$\frac{\pi}{3}$$

Question 15.
The value of cos (-1920°) will be :
(i) – 1/2
(ii) 0
(iii) 1/2
(iv) 1
(i) – 1/2

Question 16.
If 2 cos 30 = 1. Then the value of a will be :
(i) 30°
(ii) 45°
(iii) 60°
(iv) 20°
(iv) 20°

Question 17.
The value of $$\frac{\sin 20^{\circ}}{\cos 70^{\circ}}$$ will be :
(i) more than 1
(ii) 1
(iii) 0
(iv) less than 1
(ii) 1

Question 18.
The value of sin (- 570°) is :
(i) –$$\frac{\sqrt{3}}{2}$$
(ii) $$\frac{\sqrt{3}}{2}$$
(iii) $$\frac{1}{2}$$
(iv) –$$\frac{1}{2}$$
(iii) $$\frac{1}{2}$$

Question 19.
The value of sin 840° will be:
(i) $$\frac{1}{2}$$
(ii) –$$\frac{1}{2}$$
(iii) $$\frac{\sqrt{3}}{2}$$
(iv) –$$\frac{\sqrt{3}}{2}$$
(iii) $$\frac{\sqrt{3}}{2}$$

Question 20.
The value of sin 3270°
(i) $$\frac{1}{\sqrt{2}}$$
(ii) $$\frac{1}{3}$$
(iii) $$\frac{1}{2}$$
(iv) $$\frac{\sqrt{3}}{2}$$
(iii) $$\frac{1}{2}$$

Question 21.
The value of cos 2940° is :
(i) $$\frac{1}{\sqrt{3}}$$
(ii) $$\frac{1}{2}$$
(iii) $$\frac{1}{3}$$
(iv) $$\frac{\sqrt{3}}{2}$$
(ii) $$\frac{1}{2}$$

Question 22.
The value of cos2 61° + cos2 29° will
(i) 0
(ii) 1
(iii) -1
(iv) 2
(ii) 1

Question 23.
If sin θ = $$\frac{\sqrt{3}}{2}$$ and 0° < θ < 90°, then the value of tan 2θ will be :
(i) -√3
(ii) –$$\frac{1}{\sqrt{3}}$$
(iii) $$\frac{1}{\sqrt{3}}$$
(iv) √3
(i) -√3

Question 24.
The value of cos 2A is :
(i) cos2 A – sin2A
(ii) 1 – 2 cos2A
(iii) cos2A + sin2A
(iv) 2 sin2 A – 1.
(i) cos2 A – sin2A

Question 25.
The palue of sin 2A is :
(i) $$\frac{1+\tan ^{2} \mathrm{A}}{1-\tan ^{2} \mathrm{A}}$$
(ii) $$\frac{1-\tan ^{2} \mathrm{A}}{1+\tan ^{2} \mathrm{A}}$$
(iii) $$\frac{2 \tan \mathrm{A}}{1-\tan ^{2} \mathrm{A}}$$
(iv) $$\frac{2 \tan \mathrm{A}}{1+\tan ^{2} \mathrm{A}}$$
(iv) $$\frac{2 \tan \mathrm{A}}{1+\tan ^{2} \mathrm{A}}$$

Question 26.
The value of cos (-405°) is :
(i) –$$\frac{1}{\sqrt{2}}$$
(ii) $$\frac{1}{\sqrt{2}}$$
(iii) $$\frac{1}{2}$$
(iv) –$$\frac{1}{2}$$
(ii) $$\frac{1}{\sqrt{2}}$$

Question 27.
If sin θ = $$\frac{\sqrt{3}}{2}$$ and 0° < θ < 90°, the value of cot 2θ will be :
(i) –$$\frac{1}{\sqrt{3}}$$
(ii) -√3
(iii) $$\frac{1}{\sqrt{3}}$$
(iv) √3
(i) –$$\frac{1}{\sqrt{3}}$$

Question 28.
The value will be :
(i) -1
(ii) 0
(iii) 1
(iv) 3
(i) -1

Question 29.
The value will be :
(i) -∞
(ii) -1
(iii) +1
(iv) ∞
(ii) -1

Question 30.
The value of sec 70° sin 20° – cos 20° . cosec 70° will be :
(i) -1
(ii) 0
(iii) 1
(iv) infinity.
(ii) 0

Question 31.
The value of sin 12° cos 78° + sin 78° cos 12° will be:
(i) 2
(ii) 1
(iii) 0
(iv) -1.
(ii) 1

Question 32.
The maximum value of sin x for the value of x is :
(i) x = $$\frac{\pi}{4}$$
(ii) x = $$\frac{\pi}{2}$$
(iii) x = π
(iv) x = $$\frac{3 \pi}{2}$$
(ii) x = $$\frac{\pi}{2}$$

Question 33.
Value of $$\frac{\sin 31^{\circ}}{\cos 59^{\circ}}$$ will be :
(i) -1
(ii) 0
(iii) 1
(iv) 2.
(iii) 1

Question 34.
If cosθ = $$\frac{1}{2}$$, the value of cosec2θ is :
(i) $$\frac{1}{2}$$
(ii) $$\frac{\sqrt{3}}{2}$$
(iii) $$\frac{3}{4}$$
(iv) $$\frac{4}{3}$$
(iv) $$\frac{4}{3}$$

Question 35.
The value of sec θ cosec θ tan θ is :
(i) sec2θ
(ii) cosec2θ
(iii) cos2θ
(iv) cosθ.
(i) sec2θ

Question 36.
The value of cos 240° is :
(i) –$$\frac{\sqrt{3}}{2}$$
(ii) –$$\frac{1}{2}$$
(iii) $$\frac{1}{2}$$
(iv) $$\frac{\sqrt{3}}{2}$$
(ii) –$$\frac{1}{2}$$

Question 37.
The value of $$\frac{2 \tan 15^{\circ}}{1+\tan ^{2} 15^{\circ}}$$ is :
(i) $$\frac{\sqrt{3}}{2}$$
(ii) $$\frac{1}{\sqrt{2}}$$
(iii) $$\frac{1}{2}$$
(iv) $$\frac{1}{\sqrt{3}}$$
(iii) $$\frac{1}{2}$$

Question 38.
The value of  is :
(i) $$\frac{1}{2}$$
(ii) $$\frac{\sqrt{3}}{2}$$
(iii) $$\frac{2}{\sqrt{3}}$$
(iv) $$\frac{\sqrt{2}}{1}$$
(iii) $$\frac{2}{\sqrt{3}}$$

Question 39.
The value of sin (-300°) will be :
(i) $$\frac{\sqrt{3}}{2}$$
(ii) –$$\frac{\sqrt{3}}{2}$$
(iii) $$\frac{1}{2}$$
(iv) –$$\frac{1}{2}$$
(i) $$\frac{\sqrt{3}}{2}$$

Question 40.
The value of sin (-1080°) will be:
(i) -1
(ii) 0
(iii) 1
(iv) ∞
(ii) 0

Question 41.
The measure of the angle subtended at the centre of a circle of radius 10 min radian by the arc of measure 5π m will be :
(i) $$\frac{\pi}{2}$$
(ii) $$\frac{\pi}{4}$$
(iii) $$\frac{\pi}{5}$$
(iv) $$\frac{\pi}{10}$$
(i) $$\frac{\pi}{2}$$

Question 42.
If tan α = sin α, then the value of α will be:
(i) 90°
(ii) 60°
(iii) 45°
(iv) 0°
(iv) 0°

Question 43.
The value of 9 sec2 θ – 9 tan2 θ is :
(i) 1
(ii) 8
(iii) 9
(iv) 10
(iii) 9
Solution.
9 sec2 θ – 9 tan2 θ = 9 ( sec2 θ – tan2 θ ) = 9 × 1 = 9
Hence, choice (iii) is correct

Question 44.
It tan θ = $$\frac{a}{b}$$, then the value of $$\frac{b \sin \theta-a \cos \theta}{b \sin \theta+a \cos \theta}$$ will be :
(i) 1
(ii) $$\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$$
(iii) $$\frac{b^{2}-a^{2}}{b^{2}+a^{2}}$$
(iv) 0
(iv) 0

Question 45.
The value of $$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30}$$ will be
(i) cos 60°
(ii) sin 60°
(iii) tan 60°
(iv) cot 60°
(iii) tan 60°
Solution.

= √3 = tan 60°
Hence choice (iii) is correct. (iii)

### Introduction to Trigonometry Class 10 Extra Questions Very Short Answer Type

Question 1.
Find the value of tan $$\left(\frac{11 \pi}{6}\right)$$ :
Solution :
tan $$\left(\frac{11 \times 180}{6}\right)$$
= tan (11 × 30)
= tan 330°
= tan (360° – 30°)
= tan 30°
= – $$\frac{1}{\sqrt{3}}$$

Question 2.
If 3x1 = cosec θ and $$\frac{3}{x_{2}}$$ = cot θ find the value of
Solution.
Given, 3x1 = cosec θ
∴ x1 = $$\frac{1}{3}$$ cosec θ

Question 3.
Prove that

Solution.

Question 4.
Find the value of $$\frac{\sin 27^{\circ}}{\cos 63^{\circ}}$$.
Solution.

Question 5.
If cos A = $$\frac{\sqrt{3}}{2}$$ , then find the value of sin 2A.
Solution.
Given, cos A = $$\frac{\sqrt{3}}{2}$$ = cos 30°
∴ A = 30°
sin 2A = sin 2 × 30° = sin 60° = $$\frac{\sqrt{3}}{2}$$

Question 6.
If tan 2A = cot (A – 18°), where 2A is an acute angle find the value of A.
Solution.
tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°) [as cot (90 – θ) = tan θ]
⇒ 90° – 2A = A – 18°
⇒ 3A = 90° + 18° = 108°
∴ A = $$\frac{108}{3}$$ = 36°

Question 7.
Solve the equation

Solution.

∴ θ = 60°

Question 8.
Prove that sin4 θ – cos4 θ = 2 sin2 θ – 1.
Solution.
L.H.S. = sin4 θ – cos4 θ
= (sin2 θ)2 – (cos2 θ)2
= (sin2 θ + cos2 θ) (sin2 – cos2 θ)
= 1[sin2 – (1 – sin a )]
= 2 sin2 θ – 1 = R.H.S.

Question 9.
If cosec A = 2, Find the value of
Solution.
Given, cosec A = 2 = cosec 30°
∴ A = 30° sin A

= √3 + 2 – √3
= 2

Question 10.
In ∆ABC, Prove :

Solution.
In ∆ABC
A + B + C = 180°(angle sum property of ∆)
= A + B = 180° – C

[∴ cos (90° – θ) = sin θ]

Question 11.
In a right ∆ABC, right angled at point C. If tan A = 1 prove that 2 sin A cos A = 1.

Solution.
tan A = 1 = $$\frac{BC}{CA}$$
∴ CA = BC = x (let)
By pythagoras theorem
BA2 = BC2 + CA2
= x2 + x2 = 2x2
∴ BA = x √2

∴ L.H.S. = 2. sin A cos A
= 2.$$\frac{1}{\sqrt{2}}$$ × $$\frac{1}{\sqrt{2}}$$
= 1 = R.H.S

Question 12.
Prove

Solution.
(i) R.H.S. = (sec θ – tan θ)2

(ii) R.H.S. (cosec θ + cot θ)2

Question 13.
Prove (1 – sin θ) (1 + sin θ) (1 + tan2 θ) = 1
Solution.
L.H.S. = (1 – sin θ) (1 + sin θ) (1 + tan2 θ)
= (1 – sin2 θ) (sec2 θ)
[∵ 1 – sin2 = cos2 θ, 1 + tan2 θ = sec2 θ]
= cos2 θ × sec2 θ
= cos2θ × $$\frac{1}{\cos ^{2} \theta}$$
= 1 = R.H.S.

Question 14.
Prove : (1 + cot θ + tan θ) (sin θ – cos θ)
=
Solution.
L.H.S. = (1 + cot θ +, tan θ) (sin θ – cos θ)

Question 15.
Prove that:
cot A – cot 2A = cosec 2A.
Solution :
L.H.S. = cot A – cot 2A

= cosec 2A = R.H.S.

Question 16.
Find the value of :
(i) Find the value of top $$\frac{13 \pi}{6}$$
(ii) Find the value of sec $$\frac{23 \pi}{4}$$
(iii) Find the value of cot $$\frac{13 \pi}{3}$$
(iv) Find the value of (cos 75° + cos 15°).
Solution.

Question 17.
Prove that :

Solution :
L. H. S.

= R.H.S.

Question 18.
If cos (-840°) = –$$\frac{1}{2}$$ then find the value of sin (-840°).
Solution.
cos (- 840°) = cos (840°)
= cos (120°) = –$$\frac{1}{2}$$
Hence
sin (-840°) = – sin 120° = $$\frac{-\sqrt{3}}{2}$$

Question 19.
If sin θ = $$\frac{4}{5}$$, then find the value of cos 2θ.
Solution.
sin θ = $$\frac{4}{5}$$
cos 2θ = 1 – 2 sin2θ
= 1 – 2($$\frac{4}{5}$$)2 = –$$\frac{7}{25}$$

Question 20.
Find the value of tan 35° tan 40° tan 45° tan 50° tan 55°
Solution.
tan 35o tan 40° tan 45o tan 50° tan 55°
= tan (90° – 55°) tan (90° – 50°) .1.tan 50° tan 55°
= cot 55° cot 50°.1.tan 50° tan 55°
= 1.

Question 21.
Prove that :

Solution :
R. H. S.

Question 22.
Prove.

Solution.
L. H. S.

Question 23.
Prove that : (cos A + cos B)2 + (sin A + sin B)2

Solution :
L.H.S.
cos2 A + cos2 B + 2 cos A cos B
+ sin2 A + sin2 B + 2 sinA sinB
⇒ 1 + 1 + 2 (cos A cos B + sin A sin B)
⇒ 2 + 2 cos (A – B)
⇒ 2 [1 + cos (A – B)]

Question 24.
Prove the following:
(i)
(ii)
(iii)
(iv) Find the value of

(v) Find the value of

(vi) Find the value of sin 3A in terms of sin A. Hence find the value of sin 135°. If A = 45°.
(vii) If sin θ = $$\frac{3}{25}$$, find the value of sin 2θ
Solution.
(i)

(ii)

(iii)

(iv)

(v)

(vi) sin 3A = sin(2A + A)
= sin 2A. cos A + cos 2A.sin A
= 2sin A.cos A. cos A + (1 – 2 sin2A).sin A
= 2 sin A(1 – sin2A) + sin A – 2 sin3A
= 2 sin A – 2 sin3A + sin A – 2 sin3A
= 3 sin A – 4 sin3A.
∴ sin 135° = sin(3 × 45°)
= 3 sin 45° – 4 sin3 45°

(vii) As sin 2 θ = 2 sin θ . cos θ

Question 25.
Prove that : sec θ (1 – sin θ) (sec θ + tan θ) = 1
Solution :
L.H.S.

Question 26.
Prove that :
cos 33°cos 27 – cos 57°cos 63° = $$\frac{1}{2}$$
Solution :
L.H.S.
cos 33° cos 27° – cos (90°-33°) cos (90° – 27°)
∴ cos 33° cos 27° – sin 33° sin27°
∴ cos (33° + 27°) = cos (60°)
= $$\frac{1}{2}$$

Question 27.
Find the value of :
cos 80o.cos 70° – cos 10° cos 20°.
Solution.
cos 80° cos 70° – cos 10° cos 20°
= cos(90° – 10°) cos(90° – 20°) – cos 10° cos 20°
= sin 10° sin 20° – cos 10° cos 20°
= – (cos 10° cos 20° – sin 10° sin 20°)
= -cos(10° + 20°)
= -cos 30° = –$$\frac{-\sqrt{3}}{2}$$.

Question 28.
Prove :

Solution.

L.H.S. sin 5A – 2 sin 3A + sin A cos 5A – 2 cos 3A + cos A
sin 5A +sin A – 2 sin 3A , cos 5A + cos A – 2 cos 3A
2sin 3A cos 2A – 2sin 3A 2 cos 3A cos 2A – 2 cos 3A 2sin 3A(cos 2A – 1)
2cos 3A (cos 2A – 1) = tan 3A = R.H.S.

Question 29.
Prove cos2 π/8 + cos23π + cos2 5π/8 + cos2 7π/8 = 2.
Solution :

1 + 1 = 2 = R.H.S.

Question 30.
Prove :
cos 4A = 1 – 8 sin2A + 8 sin4A
Solution :
L. H. S.
cos 4A = cos 2(2A)
= 1 – 2 sinθ 2A
= 1 – 2 (2 sin A cosA)2
= 1 – 8 sin2 A cos2A
= 1 – 8 sin – A (1 – sin2 A)
= 1 – 8 sin2 A + 8 sin4A
= R.H.S.

### Introduction to Trigonometry Class 10 Extra Questions Short Answer Type

Question 1.
Prove that :

Solution :
L.H.S.

= 1 + 1 = 2 = R.H.S.

Question 2.
Prove that:

Solution :
L. H. S.

= 2 × cosec A = R.H.S.

Question 3.
Prove that :
$$\frac{\sin 8 A}{\sin A}$$ = 8 cos A cos 2A cos 4A
Solution :

= 8 cos A cos 2A cos 4A = R.H.S

Question 4.
Prove that : (1 – sin θ)(1 + sin θ)( 1+ tan2θ)= 1
Solution :
L.H.S.
= (1- sin A)(1 + sin )( 1+ tan2θ)
= (1 – sin2 θ) × sec2θ
= cos2θ × $$\frac{1}{\cos ^{2} \theta}$$ = 1
= R.H.S.

Question 5.
Prove that : sin A (1 + tan A) + cos A (1 + cot A) = sec A + cosec A
Solution.

= sec A + cosec A R.H.S.

Question 6.
Find the value of

Solution :
(cos 15° + sin 15°)

Question 7.
Prove that : 16 cos 20° cos 40° cos 60° cos 80° = 1
Solution:
L.H.S.= 16 cos 20° cos 40° cos 60° cos 80°
= 8 × $$\frac{1}{2}$$ (2 cos 80° cos 40°) cos 20°
= 4 (cos 120° cos 40°) cos 200 = 2. 2. – cos 20° + cos 40ocos 2007
= 2 [-cos 20° + cos 60° + cos 20°]
= 2 × $$\frac{1}{2}$$ = 1 = R.H.S.

Question 8.
Prove that:

Solution :
L.H.S.
= sec2A – sec2A – tan2A × tan2A
= sec4 A – tan4A
= (sec2A -tan2A)(sec2A + tan2A)
= 1 (1 + tan2A + tan2A)
= 1 + 2 tan2A = R.H.S.

Question 9.
Prove that :

Solution :
L.H.S. =

= 2 cot θ = R.H.S.

Question 10.
Prove that :
(i) If 2 tan P= 3 tan Q, then prove that:
tan(P – Q) = $$\frac{\sin 2 Q}{5-\cos 2 Q}$$
(ii) If cos A + sin A = √2 cos A, prove that cos A – sin A = √2 sin A . sin 63° + cos 630

Solution.

(ii) Given cos A + sin A = √2cos A
∴ sin A = (√2 – 1) cos A Now, multiply both sides by (√2 +1)
∴ (√2 + 1)sin A = (√2 + 1) (√2 – 1) cosA
= √2 sin A + sin A = cos A
Hence, √2sin A = cos A – sin A.

(iii)

(iv) L.H.S.

Question 11.
If cosec A = $$\frac{17}{15}$$, then find the value of sec A.
Solution.
cosec A = $$\frac{17}{15}$$
sin A = $$\frac{15}{17}$$
cos A = $$\frac{8}{17}$$
∵ sec A = $$\frac{17}{8}$$

Question 12.
Prove that

(i) θ, (ii) cos2 θ + cos2 θ.cot2 θ = cot2 θ
Solution.

(ii) L.H.S. = cos2 θ + cos2 x cot2 θ
= cos2 θ[1 + cot2 θ]
= cos2 θ × cosec2θ
= cos θ × $$\frac{1}{\sin ^{2} \theta}$$ = cot2 θ = R.H.S.

Question 13.
Solve the equation :

Solution.

Question 14.
If (A + B) = 45°, then prove that (1 + tan A) (1 + tan B) = 2
Solution.
A + B = 45°
⇒ tan(A + B) = tan 45o
⇒ $$\frac{\tan A+\tan B}{1-\tan A \tan B}$$ = 1
⇒ tan A + tan B = 1 – tan A tan B
⇒ tan A + tan B + tan A tan B = 1
⇒ 1 + tan A + tan B + tan A tan B = 1 + 1
(1 + tan A) + tan B(1 + tan A) = 2
(1 + tan A)(1+tan B) = 2

Question 15.
Prove that tan 75° = 2 + √3.
Solution.
tan 75°
= tan (45° + 30°)

Question 16.
Prove that

Solution.

Question 17.
Find the value of $$\frac{\sin 65^{\circ}}{\sin 115^{\circ}}$$
Solution.
We have,

Question 18.
Prove that

Solution.

Question 19.
Prove that sec 70° sin 20° + cos 20° cosec 70° = 2
Solution.
L. H. S.
sec 70° sin 20° + cos 20° cosec 70°

Question 20.
Prove that :
(i) cos 2A = $$\frac{1-\tan ^{2} A}{1+\tan ^{2} A}$$
(ii)
Solution.
(i) R.H.S

(ii)

Question 21.
If sec θ + tan θ = p, prove that
$$\frac{p^{2}-1}{p^{2}+1}$$ = sin θ.
Solution.

Question 22.
If A = 30° and B = 60°, then prove that sin(A + B) = sin A cos B + cos A sin B.
Solution.
L.H.S. = sin (A + B)
= sin (30° + 60°)
= sin 90° = 1
R.H.S. = sin A cos B + cos A sin B
= sin 30°. cos 60° + cos 30°.sin 60°

L.H.S. = R.H.S.

Question 23.
Find the value of sin 65o cos 35° – cos 65° sin 35°.
Solution.
sin 65° cos 35° – cos 65° sin 35°
= sin(65° – 35°)
= sin 30° = $$\frac{1}{2}$$

Question 24.
If sinθ = $$\frac{9}{41}$$, then find the value of tan θ – cosec θ.
Solution.

Question 25.
Find the value of sin A cos(90° – A) + cos A sin(90° – A).
Solution.
sin A cos (90° – A) + cos A sin (90° – A)
= sin A sin A + cos A cos A
= sin2A + cos2A = 1

Question 26.
Prove that :
$$\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}$$ = cosec θ – cot θ
Solution.

cosec θ – cot θ = R.H.S.

Question 27.
(a) Prove that :

Or
sin 4θ – cos 4θ = 2 tan2θ – 1.
(b) Prove that :
cos4θ – sin2θ = cos 2θ.
(c) Find the value of cos 375°.
Solution.
(a) Do yourself.
(b) Do yourself.
(c) cos 375° = cos (360° + 15°)
= cos 15o = cos (45° – 30°)
= cos 45°.cos 30° + sin 45o.sin 30°

Question 28.
Prove that: sin 3A cos3A + cos 3A. sinA = $$\frac{3}{4}$$ sin 4A.
Solution.
Do yourself.

Question 29.
Prove that :

Solution.
Do yourself.

Question 30.
(a) Prove that:

Solution.

Question 31.
Prove :
$$\frac{\sec A-\tan A}{\sec A+\tan A}$$
= 1 – 2 sec A tan A + 2 tan2A
Solution.
L.H.S. – $$\frac{\sec A-\tan A}{\sec A+\tan A}$$

= 1 + tan2A + tan2 – A – 2sec A tan A
= 1 – 2 sec A. tan A + 2 tan2A
= R.H.S.

Question 32.
Prove :
2cos A + cos 3A + cos 5A
= 4 cos A cos2 2A.
Solution.
L.H.S. = 2 cos A + cos 3A + cos 5A
= 2 cos A + 2 cos 4A cos (-A)
= 2 cos A + 2 cos 4A cos A
= 2 cos A(1 + cos 4A)
= 2 cos A(1 + 2 cos22A – 1)
= 4 cos A cos22A = R.H.S.

Question 33.
Prove : sin2A – sin2B = sin(A + B) sin(A – B).
Solution.
L.H.S. = sin2A – sin2B
= (sin A – sin B) (sin A + sin B)

= sin(A + B).sin(A – B)
= R.H.S.

Question 34.
Prove :

Solution.
L.H.S.

L.H.S. = R.H.S.

Question 35.
Prove :
Proved.
sec A – tan A = $$\frac{1}{\sec A+\tan A}$$
Solution.
L.H.S. = sec A – tan A

So, L.H.S. = R.H.S.

Question 36.
Prove that :

Solution.
L.H.S.

### Introduction to Trigonometry Class 10 Extra Questions Long Answer Type

Question 1.
Prove that:
(cos A + cos B)2 + (sin A – sin B)2
= 4 cos2 $$\left(\frac{A+B}{2}\right)$$
Solution.
L.H.S.
cos2A + cos2B + 2cos A cos B + sin2A
+ sin2B – 2 sin A sinB
⇒ (cosA + sinoA) + (cos’B + sino B)
+ 2 (cosA cos B – sin A sin B)
⇒ 1 + 1 + 2 cos (A – B)
⇒ 2[1 + cos (A – B)]

= 4cos2 $$\frac{A-B}{2}$$
= R.H.S.

Question 2.
Prove that:
$$\sqrt{\frac{1-\sin A}{1+\sin A}}$$= sec A – tan A
Solution.

= sec A – tan A
= R.H.S.

Question 3.
Prove that :

Solution :
L.H.S.

= sec A- sec A + tan A
= tan A ……(ii)
From (i) and (ii)
L.H.S. = R.H.S.

Question 4.
Prove that : cos3A cos3A + sin3A sin3A = cos32A.
Solution :

= cos3 2A = R.H.S.

Question 5.
If $$\frac{\cos \alpha}{\cos \beta}$$ = n, $$\frac{\sin \alpha}{\sin \beta}$$ = m prove that
(m2 – n2) sin2β = 1 – n2.
Solution.
cos α = n cos β and sin α = m sin β
cos2α + sin2 α = n2 cos2β + m2 sin2 β.
⇒ 1 = n2 (1 – sin2β) + m2 sin2 β
⇒ 1 = n2 – n2 sin2 β + m2 sino β
⇒ 1 – n2 = sin2β (m2 – n2)

Question 6.
Prove that :
sin 20° sin 40° sin 80° = $$\sqrt{\frac{3}{8}}$$
Solution :

Question 7.
Prove :

Solution :

Question 8.
Prove :

Solution.
L.H.S.

= (cosec θ + cot θ)2 = R.H.S. Proved

Question 9.
Prove:
tan2 θ + cot2 θ + 2 = sec2 θ.cosec2 θ
Solution.
tan2θ + cot2 θ + 2
= (1 + tan2θ) + (1 + cot2θ)
= sec2 θ + cosec2 θ

= sec2 θ cosec2 θ = R.H.S.

Question 10.
If A + B + C = 180°, Prove that :
(i) sin A + sin B + sin C = 4 cos$$\frac{A}{2}$$ . cos$$\frac{B}{2}$$ . cos$$\frac{C}{2}$$
(ii) cot A . cot B + cot B . cot C + cot C . cot A = 1
(iii) sin 2A + sin 2B + sin 2C = 4 sin A sin B . sin C
(iv) cos A + cos B + cos C = 1 + 4 sin$$\frac{A}{2}$$ . sin$$\frac{B}{2}$$ . sin$$\frac{C}{2}$$
(v) sin A + sin B – sin C = 4 sin$$\frac{A}{2}$$ . sin$$\frac{B}{2}$$ . cos$$\frac{C}{2}$$
(vi) Prove :

Solution.
(i) L.H.S.
= sin A + sin B + sin C

(ii) We have,
A + B + C = 180°
⇒ A + B = 180°- C
∴ cot(A + B) = cot(180° – C)
⇒ $$\frac{\cot A \cdot \cot B-1}{\cot B+\cot A}$$ = -cot C
⇒ cot A . cot B – 1 = -cot B . cot C – cot A . cot C
⇒ cot A . cot B + cot B . cot C + cot C . cot A = 1
= R.H.S.

(iii) L.H.S. = sin 2A + sin 2B + sin 2C

= 2 sin(A + B) . cos(A – B) + 2 sin C.cos C…(i)
∵ A+B+C = 180°
∴ A + B = 180° – C
⇒ sin(A + B) = sin(180° – C)
= sin C
and cos (A + B) = cos (180° – C)
= -cos C
Putting value in (i),
= 2 sin C.cos(A – B) – 2 sin C. cos(A + B)
= 2 sin C[cos (A – B) – cos(A + B)]

= 2 sin C(2 sin A.sin B]
= 4 sin A . sin B . sin C
= R.H.S.

(iv) L.H.S. = cos A + cos B + cos C

= R.H.S.

(v) Do yourself.

(vi) L.H.S.

Question 11.
Prove that:

= cos 2θ – tan 3θ . sin 2θ.
(ii) sin 10° . sin 30° . sin 50° . sin 70° = $$\frac{1}{16}$$
(iii) cos 20° . cos 40° cos 60°. cos 80° = $$\frac{1}{16}$$
(iv)
(v) sin 20°.sin 40° sin 60°.sin 80° = $$\frac{3}{16}$$
(vi) tan 70° = tan 20° + 2 tan 50°.
(vii) tan 3A – tan 2A – tan A = tan 3A . tan 2A tan A.
(viii) If A + B + C = 90°, prove that
tan A tan B + tan B tan C + tan C tan A = 1
(ix) Prove that :

(x) Prove :

+ (sin A + sin B)2
Solution.
(i) L.H.S.

= cos 2θ – tan 3θ.sin 2θ
= R.H.S.

(ii) L.H.S. = sin 10° sin 50° sin 70° sin 30°

(iii) L.H.S. = cos 20° cos 40°.cos 80°.cos 60°

(iv)

(v) L.H.S.
= sin 20° sin 40° sin 60° sin 80°
= sin 60° sin 20° sin 40° sin 80°

L.H.S. = R.H.S.

(vi) tan 70° = tan(50° + 20°) .

⇒ tan 70° – tan 70° tan 50° tan 20°
= tan 50° + tan 20°
⇒ tan 70° – tan(90° – 20°) tan 50°
tan 20° = tan 50° + tan 20°
⇒ tan 70° – cot 20° tan 50° tan 20°
= tan 50° + tan 20°
⇒ tan 70° – $$\frac{1}{\tan 20^{\circ}}$$ tan 50° tan 20°
= tan 50° + tan 20°
⇒ tan 70° – tan 50° = tan 50° + tan 20°
⇒ tan 70° = tan 50° + tan 20°+ tan 50°
⇒ tan 70°= tan 20° + 2tan 50°

(vii) We have
tan 3A = tan(A + 2A)
tan A + tan 2A
⇒ tan 3A = $$\frac{\tan A+\tan 2 A}{1-\tan A \tan 2 A}$$
⇒ tan 3A – tan 3A tan A tan 2A = tan A + tan 2A
⇒ tan 3A – tan 2A – tan A
= tan 3A tan 2A tan A.

(viii) Given
A + B + C = 90°
⇒ A + B = 90° – C
⇒ tan(A + B) = tan(90° – C)
⇒ $$\frac{\tan A+\tan B}{1-\tan A \tan B}$$ = cot C = $$\frac{1}{\tan C}$$
⇒ tan =tan A . tan C + tan B . tan C
= 1 – tan A tan B
⇒ tan A tan B + tan B . tan C + tan C . tan A = 1

(ix)

= sec A + tan A
= R.H.S.

(x) R.H.S. = (cos A-cos B)2 + (sin A + sin B)2
= cos2A + cos2B – 2cos A cos B
+ sin2A + sin2B + 2 sin A sin B
= (cos2 A +sin2 A) + (cos2B + sin2B)
– 2{cos A cos B – sin A sin B]
= 1 + 1 – 2[cos A.cos B – sin A.sin B]
= 2[1 – cos(A + B)]

= L.H.S.

Question 12.
Solve :

Solution.

⇒ cos θ = cos 60°
∴ θ = 60°

Question 13.
Prove : (cos A + sec A)2 + (sin A + cosec A)2 7 + tan2 A + cot2 A.
Solution.
L.H.S.
= (cos A + sec A)2 + (sin A + cosec A)2
= cos2A + sec2A + 2.cos A sec A + sin2 A + cosec2A + 2sin A cosec A
= 1 + 2 + 2 + sec2A + cosec2 A
= 5 + 1 + tan2A + 1 + cot2 A
= 7 + tan2A + cot2A
= R.H.S.

Question 14.
Prove :

Solution :

Question 15.
Prove that:
tan (A + B) tan (A – B)

Solution.
tan (A + B)

Question 16.

Solution.
L. H. S.

= tan 5 A = R.H.S.

Question 17.
Prove :
cos A cos (60° + A). Cos (60° – A) = $$\frac{1}{4}$$ cos 3A
Solution.
L.H.S.
= cos A. cos (60° + A). cos (60° – A)
= cos A (cos 60°. cos A-sin 60° sin A)
(cos 60°. cos A + sin 60°sin A)

Question 18.
Prove :

Solution.
L.H.S.

= tan (A – B)
= R.H.S.

Question 19.
Prove :
sin A sin 2 A + sin 3A sin 6A = sin 4A sin 5A.
Solution.
L.H.S.
= sin A sin 2A + sin3 A sin 6A
= $$\frac{1}{2}$$ [2 sin A sin 2A + 2 sin3A sin 6A]
= $$\frac{1}{2}$$ [cos A – cos 3A + cos 3 A – cos 9A]
= $$\frac{1}{2}$$ [cos A – cos 9A]
= $$\frac{1}{2}$$ × 2 sin $$\frac{9 \mathrm{A}+\mathrm{A}}{2}$$ sin $$\frac{9 \mathrm{A}-\mathrm{A}}{2}$$
= sin 5A sin 4A
= sin 4A sin 5A = R.H.S.

Question 20.
Prove :

Solution.
L.H.S.

= sec θ + tan θ = R.H.S.

## NCERT Solutions for Class 12 Political Science Chapter 5 Contemporary South Asia

Detailed, Step-by-Step NCERT Solutions for Class 12 Political Science Chapter 5 Contemporary South Asia Questions and Answers were solved by Expert Teachers as per NCERT (CBSE) Book guidelines covering each topic in chapter to ensure complete preparation.

## Contemporary South Asia NCERT Solutions for Class 12 Political Science Chapter 5

### Contemporary South Asia Questions and Answers Class 12 Political Science Chapter 5

Question 1.
Identify the country :
(a) The struggle among pro-monarchy, pro-democracy groups and extremists created an atmosphere of political instability.
(b) A landlocked country with multi-party competition.
(c) The first country to liberalise its economy in the South Asian region.
(d) In the conflict between the military and pro-democracy groups, the military as prevailed over democracy.
(e) Centrally located and shares borders with most of the South Asian countries.
(f) Earlier the island had the Sultan as the head of state. Now, it is a republic.
(g) Small savings and Credit cooperatives in the rural areas have helped in reducing poverty.
(h) A landlocked country with a monarchy.
(a) Nepal
(b) Nepal
(c) Sri Lanka
(d) Pakistan
(e) India
(f) Maldives
(h) Bhutan.

Question 2.
Which among the following statements about South Asia is wrong ?
(a) All the countries in South Asia are democratic.
(b) Bangladesh and India have signed an agreement on river water sharing.
(c) SAFTA was signed at the 12th SAARC Summit in Islamabad.
(d) The US and China played an influential role in South Asian Politics.
(a) Wrong
(b) True
(c) True
(d) True.

Question 3.
What are the some of the commonalities and differences between Bangladesh and Pakistan in their democratic experiences ?
Both Pakistan and Bangladesh have experienced both civilian and military rulers.
The Military and Democracy in Pakistan : After the framing of Pakistan constitution General Ayub Khan took over the administration and later on got himself elected as President of Pakistan. But masses were not happy with his rule.

Ultimately, military rule was established and General Yahya Khan faced Bangladesh Crisis. Bangladesh emerged independent nation in 1971. After this democratic government was established in Pakistan and Zulfikar Ali Bhutto became the Prime Minister from 1971 to 1977.

But in 1977 Zulfikar Bhutto was removed by General Zia-ul-Haq. It was again in 1988 that democratic government was established under the leadership of Benazir Bhutto. From 1988 to 1999 democracy remained in Pakistan. I

n 1999 Prime Minister Nawaz Sharif was removed and General Musharraf became the ruler of the Pakistan. In 2001 General Musharraf got himself elected as the President of Pakistan. Elections were held in Sept. 2008 and democratic government was established.

Many factors are responsible for the failure of democracy and in establishing stable and strong democracy. The social dominance of the military, clergy and landowing aristocracy are responsible for overthrowing the democratic government.

Wars with India have made military rulers and promilitary groups very powerful. Inspite of the fact that democracy has not succeeded in Pakistan, there has been a strong pro-democracy sentiments in country. Moreover, America and other western countries have encouraged military rulers for their own interests.

Military Rule and Democracy in Bangladesh.
Bangladesh was a part of Pakistan and was known as East Pakistan (1947 to 1971). East Bengal was not given fair treatment by the rulers of Pakistan and it was made virtually a colony. The people of this region resented the domination of Western Pakistan and the imposition of the Urdu language. In an election held early 1971 in Pakistan, Sheikh Mujib’s Awami League got majority in Pakistan Parliament.

But Sheikh Mujib was not called to form a government and he was arrested. East Bengal declared independence and the liberation war started. War took place between India and Pakistan in December, 1971 and Pakistan was defeated in the war. India was the first country to grant recognition to the People’s Republic Bangladesh. Bangladesh drafted its Constitution and declared full faith in Democracy,

Secularism and Socialism. Sheikh Mujib was the first President of Bangladesh. In 1975 Constitution of Bangladesh was amended and presidential form of govt, was adopted in place of parliamentary government. Sheikh Mujib abolished all parties except his own party i.e., Awami League.

He was assassinated in a military uprising in August 1975. Military ruler Zia Rehman formed his own party and won elections in 1979. He was assassinated and Lt. General H.M. Ershad became the ruler of Bangladesh. He was later elected the President of Bangladesh. President Ershad resigned in 1990. Elections were held in 1991. Since then democracy is working in Bangladesh.

Question 4.
List three challenges to democracy in Nepal.

• Writing of constitution for Nepal.
• Maoist believe in armed insurrection against monarch and ruling nobility.
• Restoration of Parliament and free and fair elections.

Question 5.
Name the principal players in the ethnic conflict in Sri Lanka. How do you assess the prospects of the resolution of this conflict ? (Imp.)
Despite cordial relations there has been occasional tension between India and Sri Lanka and the cause of tension was the problem of nearly one million people of origin in Sri Lanka. Sri Lanka was not prepared to grant full citizenship rights to all Indian immigrants in Sri Lanka. The government of Sri Lanka passed the India and Pakistan Residents (Citizenship) Act in 1949.

About 8 lakhs people of Indian origin applied for citizenship but only one lakh 34 thousands were able to secure citizenship (upto Oct., 1964). The rest were asked to go back to India. But India’s stand was that those who were living in Sri Lanka for generations or had been born there, are the citizens of Sri Lanka and not of India. Ultimately, Lai Bahadur Shastri and Smt. Bandarnaike of Sri Lanka reached an agreement on the question of citizenship of Indian people there.

In June, 1985 Prime Minister Mr. Rajiv Gandhi and the Sri Lanka President J.R. Jayawardene held discussions to find a political solution to the ethnic problem in the island nation. On July 21, 2000 Sri Lanka President Mrs. Chandrika Kumaratunga agreed to give Sri Lankan citizenship to those Indian origin

Tamils who were forced to take Indian citizenship in 1964 as a result of the Indian-Sri Lanka pact on the stateless people of Indian origin in Sri Lanka. In 2002 Norway and Iceland have been trying to bring the warring groups back to negotiations. In 2009, LTTE leader Prabhakaran was killed. There for the long lasting conflict, seemed to be end.

Question 6.
Mention some of the recent agreements between India and Pakistan. Can we be sure that the two countries are well on their way to a friendly relationship ?
Indian Prime Minister A.B. Vajpayee visited Islamabad in January 2004. Both the countries decided to improve their relations. The bus services, train services and air services have been resumed between Indo-Pak. On 7th April, 2005 bus service between Srinagar and Muzaffarabad started.

Both countries agree to start a bus service between Amritsar and Lahore and also to religious places such as Nankana Sahib. In February, 2007 India and Pakistan signed an agreement on reducing the risk from accident relating to nuclear weapons. Both the countries have agreed not to attack each other’s nuclear facilities.

Question 7.
Mention two areas each of co-operation and disagreement between India and Bangladesh.
Two Areas of Co-operation :

•  India and Bangladesh entered a new era of bilateral relationship with the launch of bus service linking Kolkata with Dhaka on June 19, 1999.
• On December 12, 1996 India and Bangladesh signed the Ganga Water sharing treaty leaving behind a long period of mutual distrust and suspicion.

Two Areas of disagreement :

• Dispute started between India and Bangladesh over sharing of Ganga and Brahmaputra river waters.
• A major irritant in Indo-Bangladesh relations was Tin Bigha Corridor. The two countries have not succeeded in resolving their boundary dispute.

Question 8.
How are the external powers influencing bilateral relation in South Asia ? Take any one example to illustrate your points.
No state or region exists in a vacuum. It is influenced by outside powers and events. South Asia is also influenced by big powers such as China and U.S.A. India’s relations with China are improving rapidly but China’s friendly relations with Pakistan is a major irritant. After Cold War American involvement in South Asia has increased. America is having friendly relations with both India and Pakistan.

Many times America has played the role of moderator between India and Pakistan. Due to liberalisation American involvement in economic activities of both the countries i.e., India and Pakistan has increased. The South Asian market is very big and American participation in the South Asian market is increasing very fast. Thus U.S.A. is concerned with the regional security and peace.

Question 9.
Write a short note on the role and limitations of SAARC as a forum for facilitating economic co-operation among the South Asian countries. (Imp.)
The South Asian Association of Regional co-operation was formally inaugurated in December, 1985. SAARC has played a very significant role in the sphere of economic development of South Asian countries. SAPTA and SAFTA are the steps taken in directions of economic co-operation among SAARC nations. The co-operation in economic area among SAARC nations has a great importance.

• The economic co-operation will raise the standard of living of the people of South Asia.
• It has accelerated the speed of economic development of the region.
• The economic co-operation will promote collective self-reliance among the countries of South Asia.
• Economic co-operation will bring SAARC nations close and strengthen the mutual trust.
• The areas of mistrust and conflict will be minimised.
• The economic co-operation among SAARC nations will minimise the involvement of outside forces in region.
• The economic co-operation will bring honourable life to the region.
• Economic co-operation will lead to co-operation in other fields also such as social, cultural, educational, etc.

Thus, co-operation in core economic area among SAARC nations is a definite way to the development of region. The SAARC nations have recognised this fact and has signed SAPTA and SAFTA to boost their co-operations.

But SAARC has not much success due to persisting political differences. Moreover some of our neighbours fear that SAFTA is a way for India to capture their markets and to influence their societies and politics through economic activities and trade. However, India think that all countries will derive benefits from SAFTA.

Question 10.
India’s neighbours often think that the Indian government tries to dominate and interfere in the domestic affairs of the smaller countries of the region. Is this a correct impression ?
This is not a correct impression. Indian government never thinks to dominate or interferes in the domestic affairs of the smaller countries of the region. India believes in the principles of Panchsheel and Indian government has full respect for the sovereignty and dignity of other states. During the last sixty years India has no1 interfered in the affairs of any state.

## CBSE Class 10th Social Science SST NCERT Solutions in Hindi Medium and English Medium

एनसीईआरटी समाधान कक्षा 10 समाजिक विज्ञान (हिंदी मीडियम): एनसीईआरटी को राष्ट्रीय शिक्षा और अनुसंधान परिषद के रूप में जाना जाता है और यह एक स्वायत्त संगठन है जो पूरे देश में सीबीएसई स्कूलों सहित सभी भारतीय स्कूलों के लिए पाठ्यक्रम निर्धारित करता है। स्कूल शिक्षा के क्षेत्र में कई वर्षों के अनुभव के साथ विशेषज्ञ शिक्षकों के शामिल NCERTSolutions.Guru पर एक उच्च मानक संकाय द्वारा NCERT समाधान कक्षा 10 सामाजिक विज्ञान के लिए हल किया गया है। NCERTSolutions.Guru की NCERT कक्षा 10 सामाजिक विज्ञान नि: शुल्क पीडीएफ डाउनलोड में निम्नलिखित NCERT कक्षा 10 सामाजिक विज्ञान पाठ्य सामग्री में दिए गए प्रश्नों के पीडीएफ समाधान शामिल हैं:

छात्र केवल एनसीईआरटी- समाधान.कॉम पर जाकर और संबंधित पीडीएफ लिंक से डाउनलोड करके इन पीडीएफ का उपयोग कर सकते हैं। प्रत्येक पीडीएफ पुस्तक में सामाजिक विज्ञान की पाठ्यपुस्तकों में निर्धारित कई विषयों के अध्यायवार समाधान शामिल हैं।

### Class 10 Social Science NCERT Solutions in Hindi Medium

एनसीईआरटी समाधान कक्षा 10 समाजिक विज्ञान History: India and The Contemporary World – II (इकाई 1: इतिहास-भारत और समकालीन विश्व-II)

NCERT Solutions for Class 10 Social Science Geography: Contemporary India – II (इकाई 2: भूगोल-समकालीन भारत-II)

NCERT Solutions for Class 10 Social Science Civics (Political Science): Democratic Politics – II (इकाई 3: राजनीति विज्ञान-लोकतांत्रिक राजनीति-II)

NCERT Solutions for Class 10 Social Science Economics: Understanding Economic Development (इकाई 4 अर्थशास्त्र-आर्थिक विकास की समझ)

### Class 10 Social Science NCERT Solutions in English Medium

Here is the list of chapters from History, civics, Geography and Economics with detailsed solutions.

### NCERT Solutions for Class 10 Social Science Economics

#### एसएसटी कक्षा 10 एनसीईआरटी समाधान

सामाजिक विज्ञान ने वास्तविक दुनिया के परिदृश्यों में आर्थिक और सामाजिक रिश्तों की बेहतर समझ के माध्यम से दुनिया में क्रांति ला दी। उपभोक्ता अधिकारों, औद्योगीकरण की आयु और राजनीतिक दलों जैसे विषयों से संबंधित अध्यायवार समाधान सामाजिक-आर्थिक और ऐतिहासिक अवधारणाओं के माध्यम से छात्रों को चलते हैं – सामाजिक विज्ञान के विभिन्न क्षेत्रों से संबंधित हैं। कक्षा 10 सामाजिक विज्ञान के लिए CBSE NCERT समाधान में कक्षा 10 सामाजिक विज्ञान की पाठ्यपुस्तकों से संबंधित विभिन्न विषयों के चरण-दर-चरण समाधान शामिल हैं।

CBSE Class 10 NCERT Solutions को एक विशेषज्ञ संकाय द्वारा NCERTSolutions.Guru पर तैयार किया गया है, जो अवधारणाओं को सरल बनाता है और आसान संशोधन के लिए उन्हें समझना आसान बनाता है। बहुविकल्पीय प्रश्नों से, भूगोल के नक्शों के लिए चिह्नित उत्तर, और व्यक्तिपरक प्रश्नों के विस्तृत उत्तर, इन PDF का अध्ययन करने से छात्र आगामी CBSE कक्षा 10 सामाजिक विज्ञान बोर्ड की परीक्षाओं की पूरी तैयारी कर सकेंगे।

सीबीएसई कक्षा 10 सामाजिक विज्ञान एनसीईआरटी समाधान पीडीएफ फाइलों को उन उपकरणों की एक भीड़ तक पहुँचा जा सकता है जो ई-रीडर समर्थन की सुविधा देते हैं और हमारी वेबसाइट से डाउनलोड किए जा सकते हैं। NCERTSolutions.Guru सामाजिक विज्ञान कक्षा 10 डेमोक्रेटिक पॉलिटिक्स मुक्त पीडीएफ समाधान और हमारे मंच पर दाखिला लेने वाले छात्रों के लिए विशेष पूरक सामग्री भी प्रदान करता है। हालाँकि, PDF सभी के द्वारा डाउनलोड और देखने के लिए स्वतंत्र हैं।

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