ARITHMETIC PROGRESSIONS 57
Now, looking at the pattern formed above, can you find her monthly salary for
the 6th year? The 15th year? And, assuming that she will still be working in the job,
what about the monthly salary for the 25th year? You would calculate this by adding
= 500 each time to the salary of the previous year to give the answer. Can we make
this process shorter? Let us see. You may have already got some idea from the way
we have obtained the salaries above. Salary for the 15th year
= Salary for the 14th year + = 500 "
500 + 500 +500 + ...+500
= «| s000 + ----_- : = 500
13 times
= % [8000 + 14 x 500] . :
= = [8000 + (15-1) x 500) =% 15000. . Le., First salary + (15 - 1) x Annual increment. ~
In the same way, her monthly salary for the 25th year would be
= (8000 + (25 - 1) x 500] = % 20000
= First salary + (25 - 1) x Annual increment
This example would have given you some idea about how to write the 15th term,
or the 25th term, and more generally, the nth term of the AP.


be = (8000 + (25 - 1) x 500] = % 20000 = First salary + (25 - 1) x Annual increment This example would have given you some idea about how to write the 15th term, or the 25th term, and more generally, the nth term of the AP. Let a,, a,, a,,... be an AP whose first term a, is a and the common
difference is d. Then, -
the second term a, = a+d=a+(2-I1)d
thethirdterm a, = a, +d=(at+d)+d=a+2d=a+(3-1)d
the fourthterm a,= a,+d=(a+2d)+d=a+3d=a+(4-I)d
Looking at the pattern, we can say that the nth term a, =a + (n--1)d. So, the nth term a, of the AP with first term a and common difference d is
given by a, =a + (n -1)d. Reprint 2025-26


58 MATHEMATICS a, is also called the general term of the AP. If there are m terms in the AP, then
a,, tepresents the last term which is sometimes also denoted by /. Let us consider some examples. Example 3 : Find the 10th term of the AP: 2,7, 12,... Solution : Here,@=2, d=7-2=5 and n= 10. We have a,=at(n-Il1)d
So, a,=2+(0-1)x5=2+45=47
Therefore, the 10th term of the given AP is 47. Example 4 : Which term of the AP: 21, 18, 15,...is -81? Also, is any term 0? Give
reason for your answer. -
Solution : Here, a= 21, d= 18-21 =-3 and a, = - 81, and we have to find n. As a,=a+(n-1)d, po
wehave -81=21+(n-1)-3) > _ ~81=24-3n ss
-~105=-3n : | |
So, n= 35
Therefore, the 35th term of the given AP is- 81. Next, we want to know if there is any n for which a, = 0. If such an v7 is there, then
214 (n- 1) 3) = 0,
1.¢., 3(n - 1) = 21
Le., n= 8
So, the eighth term is 0. . Example 5 : Determine the AP whose 3rd term is 5 and the 7th term is 9.


which a, = 0. If such an v7 is there, then 214 (n- 1) 3) = 0, 1.¢., 3(n - 1) = 21 Le., n= 8 So, the eighth term is 0. . Example 5 : Determine the AP whose 3rd term is 5 and the 7th term is 9. Solution : We have
° a,=at+@GB-I)d=a+2d=5 (1)
and : a,=at+(7-l)d=a+6d=9 (2)
Solving the pair of linear equations (1) and (2), we get
a= 3, d=1
Hence, the required AP is 3, 4,5,6,7,... Reprint 2025-26


ARITHMETIC PROGRESSIONS 59
Example 6 : Check whether 301 is a term of the list of numbers 5, 11, 17, 23,... Solution : We have : a,-a,=11-5=6, a,-a,=17-11=6, a,-a,=23-17=6
As a, ,,- 4, is the same for k = 1, 2, 3, etc., the given list of numbers is an AP. Now, a=5 and d=6. Let 301 be a term, say, the mth term of this AP. We know that a,=a+(n-l1)d . So, 301=54+@-1)x6
ie. 301= 6n-1 | 4
302 151 . So, n= 6 3 a |
But n should be a positive integer (Why?). So, 301 is not a term of the given list of
numbers. oe -
Example 7 : How many two-digit numbers are divisible by 3? Solution : The list of two-digit numbers divisible by 3 is:
12, 15, 18,..., 99
Is this an AP? Yes it is. Here, a=12, d=3, a,=99. As NA a, =at(n--l)d,
we have 99s 12+ (2-1) x3
i.e., 87 = (n-1)x3
i.e. gets Lao
, . ; 3
1.¢., n= 29+1=30
So, there are 30 two-digit numbers divisible by 3. Example 8 : Find the 11th term from the last term (towards the first term) of the
AP: 10,7, 4,...,-62. Solution : Here, a= 10, d=7-10=-3, /=- 62,
where l=a+(n-l)d
Reprint 2025-26


60 MATHEMATICS To find the | 1th term from the last term, we will find the total number of terms in
the AP. So, -62=10+(@- 1)(-3)
Le., -72 = (n - 1)(-3) Le., n--1=24 or n= 25 So, there are 25 terms in the given AP. The 11th term from the last term will be the 15th term. (Note that it will not be
the 14th term. Why?)
So, a,, = 10+ (15 - 1)(-3) = 10-42 =- 32
i.e., the 11th term from the last term is - 32, " Alternative Solution : : . . _ If we write the given AP in the reverse order, then a= _ 62 and d = 3 (Why?) So, the question now becomes finding the 11th term with these a and d. So, a,, = -62+ (11-1) x3=-62+30=-32
So, the 11th term, which is now the required term, is - 32. . Example 9: A sum of = 1000 is invested at 8% simple interest per year. Calculate the
interest at the end of each year. Do these interests form an AP? If so, find the interest
at the end of 30 years making use of this fact. Solution : We know that the formula to calculate simple interest is given by
PxRxT


simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years making use of this fact. Solution : We know that the formula to calculate simple interest is given by PxRxT Simple Interest = 100
So, the interest at the end of the Ist year = & "a = = 80
The interest at the end of the 2nd year= 2 Se =% 160
The interest at the end of the 3rd year= % - = = 240
Similarly, we can obtain the interest at the end of the 4th year, 5th year, and so on. So, the interest (in %) at the end of the Ist, 2nd, 3rd, .. . years, respectively are 80, 160, 240,... Reprint 2025-26


ARITHMETIC PROGRESSIONS 61
It is an AP as the difference between the consecutive terms in the list is 80, i.e.,
d= 80. Also, a = 80. So, to find the interest at the end of 30 years, we shall find a,,. Now, a,, = a+ (30-1) d=80 + 29 x 80 = 2400
So, the interest at the end of 30 years will be = 2400. Example 10 : In a flower bed, there are 23 rose plants in the first row, 21 in the
second, 19 in the third, and so on. There are 5 rose plants in the last row. How many
rows are there in the flower bed? Solution ; The number of rose plants in the 1st, 2nd, 3rd, .. ., rows are : .
23,21, 19,...,5 / ,
It forms an AP (Why?). Let the number of rows in the flower bed be n. -
Then a=23, d=21-23=-2,4,55 0 9 |
As, a,=at(n-1)d oo -
We have, 5= 23 +(n-1)(-2) °
ie., ~18 = (n-1)(-2)- oe
i.e., n= 10 :
So, there are 10 rows in the flower bed. | © ,
(- + .


number of rows in the flower bed be n. - Then a=23, d=21-23=-2,4,55 0 9 | As, a,=at(n-1)d oo - We have, 5= 23 +(n-1)(-2) ° ie., ~18 = (n-1)(-2)- oe i.e., n= 10 : So, there are 10 rows in the flower bed. | © , (- + . Ne EXERCISE 5.2
1, Fill in the blanks in the following table, given that a is the first term, d the common
difference and a, the nth term of the AP:
rr ee ee
(i) 7 3 8 vee
(ii) -18 vee 10 0
Gi) |... 3 18 5
(iv) | -18.9 2.5 vee 3.6
(v) 3.5 0 105 vee
Reprint 2025-26


62 MATHEMATICS
2. Choose the correct choice in the following and justify :
(i) 30th term of the AP: 10, 7,4,..., is
(A) 97 (B) 77 (C) -77 (D) -87
(i) 11th term of the AP: - 3, 5 »2,+++,18
1
(A) 28 (B) 22 (C) -38 (D) -48 5
3. In the following APs, find the missing terms in the boxes :
02 [ |, 2
[| [| 3 |
] .
o OO |
(iii) > | SO
wa OOO. |
2 A
4. Which term of the AP:3,8,13,18,...,is 78?
5. Find the number of terms in each of the following APs :
s ' 1]
(i) 7,13, 19,...,205 . - (i) 18, 15-,13,...,-47
rN |
6. Check whether - 150 is aterm of the AP: 11, 8,5,2...
7, Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
8. AnAP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th
term. ,
9. Ifthe 3rd and the 9th terms of an AP are 4 and - 8 respectively, which term of this AP is
zero? , >
10. The 17th term of an AP exceeds its 10th term by 7.


12 and the last term is 106. Find the 29th term. , 9. Ifthe 3rd and the 9th terms of an AP are 4 and - 8 respectively, which term of this AP is zero? , > 10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
11, Which term of the AP: 3, 15, 27, 39, ... will be 132 more than its 54th term?
12. Two APs have the same common difference. The difference between their 100th terms is
100, what is the difference between their 1000th terms?
13. How many three-digit numbers are divisible by 7?
14. How many multiples of 4 lie between 10 and 250?
15. For what value of n, are the nth terms of two APs: 63, 65, 67, ...and3, 10, 17,... equal?
16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12. Reprint 2025-26


ARITHMETIC PROGRESSIONS 63
17. Find the 20th term from the last term of the AP: 3, 8, 13,..., 253.
18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is
44. Find the first three terms of the AP.
19. Subba Rao started work in 1995 at an annual salary of € 5000 and received an increment
of % 200 each year. In which year did his income reach = 7000?
20. Ramkali saved & 5 in the first week of a year and then increased her weekly savings by
= 1.75. If in the nth week, her weekly savings become 20.75, find n.
5.4 Sum of First n Terms of an AP
Let us consider the situation again
given in Section 5.1 in which Shakila , ed N
. : V7 VV N .
put = 100 into her daughter's money TWAIIV-V aa .
box when she was one year old, Pm ip :
~ 150 on her second birthday, ye SO "
% 200 on her third birthday and will VEING eS
continue in the same way. How much A KX) =p 7
money will be collected in the money ' 4 (XN.
box by the time her daughter is 21 . .


~ 150 on her second birthday, ye SO " % 200 on her third birthday and will VEING eS continue in the same way. How much A KX) =p 7 money will be collected in the money ' 4 (XN. box by the time her daughter is 21 . . Uf: J?
years old? . "es Here, the amount of money (in ) put in the money box on her first, second, third,
fourth .. . birthday were respectively 100, 150, 200, 250, ... till her 21st birthday. To
find the total amount in the money box on her 21st birthday, we will have to write each
of the 21 numbers in the list above and then add them up. Don't you think it would be
a tedious and time consuming process? Can we make the process shorter? This would
be possible if we can find a method for getting this sum. Let us see. We consider the problem given to Gauss (about whom you read in
Chapter 1), to solve when he was just 10 years old. He was asked to find the sum of
the positive integers from 1 to 100. He immediately replied that the sum is 5050. Can
you guess how did he do? He wrote :


Gauss (about whom you read in Chapter 1), to solve when he was just 10 years old. He was asked to find the sum of the positive integers from 1 to 100. He immediately replied that the sum is 5050. Can you guess how did he do? He wrote : S=14+24+3+...+99+4 100
And then, reversed the numbers to write
- $= 1004+994+...434+24+1
Adding these two, he got
28S = (100+ 1) + (994+ 2)+...4+(3 + 98) 4+ (24+ 99) + (1 + 100)
= 101+101+...+101+4101 (100 times)
100 x 101
So, S= > 5050 , i.e., the sum = 5050. Reprint 2025-26


64 MATHEMATICS We will now use the same technique to find the sum of the first n terms of an AP : a,a+d,at+2d,... The ath term of this APis a + (n - 1) d. Let S denote the sum of the first terms
of the AP. We have S=at+(atd)+(at+2d)+...+[a+(m---1)d] (1)
Rewriting the terms in reverse order, we have S=[a+(n--1)d]+flat+(n---2)d]+...+(a+d)+a (2)
On adding (1) and (2), term-wise. we get ° [2a+ (vn --Dd)+[2a+ (--)d)+...4+ [24+ (n -- 1d) +[2at+ @ -- 1d]
- ee _
28 = n times .
o, 2S=n[2a+(n--1)d] (Since, there are m terms) . _
Or, S= 5 [2a + (m-- 1) d] . . . -
So, the sum of the first 1 terms of an AP is given by
n
S= 3 [2a + (2 --1) d]
. . 7 . n :
We can also write this *) S= 2 [a+ a+(n-1)d]
n
1.e., S= 3 (a+a,) (3)
Now, if there are only n terms in an AP, then a, = 1, the last term. From (3), we see that ~
. n
7 S=- (+l) (4)
. . 2 This form of the result is useful when the first and the last terms of an AP are
given and the common difference is not given.


an AP, then a, = 1, the last term. From (3), we see that ~ . n 7 S=- (+l) (4) . . 2 This form of the result is useful when the first and the last terms of an AP are given and the common difference is not given. Now we return to the question that was posed to us in the beginning. The amount
of money (in Rs) in the money box of Shakila's daughter on 1st, 2nd, 3rd, 4th birthday,
..., were 100, 150, 200, 250, . . ., respectively. This is an AP. We have to find the total money collected on her 21st birthday, i.e.,
the sum of the first 21 terms of this AP. Reprint 2025-26


ARITHMETIC PROGRESSIONS 65
Here, a= 100, d=50 and n =21. Using the formula : n S= gat @-D d], 21 21
we have S= 3 [2100+ 21-1) x50) = 5 [200 + 1000] 21 = 5 x 1200 = 12600 So, the amount of money collected on her 21st birthday is = 12600. Hasn't the use of the formula made it much easier to solve the problem? We also use Sin place of S to denote the sum of first n terms of the AP. We
write S,, to denote the sum of the first 20 terms of an AP. The formula for the sum of
the first n terms involves four quantities S, a, d and n. If we know any three of them,
we can find the fourth. - . Remark : The nth term of an AP is the difference of the sum to first n terms and the
sum to first (n - 1) terms of it,ie.,a,=S,-S, oj." Let us consider some examples. ~ . Example 11 : Find the sum of the first 22 terms of the AP: 8, 3, -2,... Solution : Here, a= 8, d=3-8=-5, n= 22.
f ;
We know that )
MN n .


to first (n - 1) terms of it,ie.,a,=S,-S, oj." Let us consider some examples. ~ . Example 11 : Find the sum of the first 22 terms of the AP: 8, 3, -2,... Solution : Here, a= 8, d=3-8=-5, n= 22. f ; We know that ) MN n . S = at (nda
22
Therefore, S=> [16 + 21(-5)] = 11116 - 105) = 11(-89) = - 979
So, the sum of the first 22 terms of the AP is - 979. Example 12 : If the sum of the first 14 terms of an AP is 1050 and its first term is 10,
find the 20th term. Solution : Here, S,, = 1050, n= 14, a=10.
n
As S.= 724+ (n-Dd],
14
so, 1050= 7 120+ 134] = 140+ 91d
Reprint 2025-26


66 MATHEMATICS
Le., 910= 91d
Or, d= 10
Therefore, Ay) = 10 + (20-1) x 10 = 200, i.e. 20th term is 200. Example 13 : How many terms of the AP: 24, 21, 18, .. . must be taken so that their
sum is 78? Solution : Here, a = 24, d= 21 --24=-3, S = 78. We need to find n.
n
We know that S= i244 (n--Nd] :
n n ;
So, 78= 5[48+(2-D-3)] = 5[51-37] |
or 3n2 -51n + 156 = 0 . : .
or nw -17In+52= 0 . -
or (n--4)(n - 13) = 0 - _
or n=4 or 13. . / :
Both values of n are admissible. So, the number of terms is either 4 or 13. Remarks: 1. In this case, the sum of the first 4 terms = the sum of the first 13 terms = 78. 2. Two answers are possible because the sum of the terms from 5th to 13th will be
zero. This is because a is positive and d is negative, so that some terms will be
positive and some others negative, and will cancel out each other.


first 13 terms = 78. 2. Two answers are possible because the sum of the terms from 5th to 13th will be zero. This is because a is positive and d is negative, so that some terms will be positive and some others negative, and will cancel out each other. Example 14 : Find the sum of :
(i) the first 1000 positive integers (ii) the first n positive integers
Solution : 4
(i) LetS=1+2+3+...+ 1000
. n
Using the formula S_ = 54 +1) for the sum of the first n terms of an AP, we
have
° 1000
: Si000 = > (1+ 1000) = 500 x 1001 = 500500
So, the sum of the first 1000 positive integers is 500500. (i) LetS =1+2+4+3+...4+n
Here a = | and the last term / is n. Reprint 2025-26


ARITHMETIC PROGRESSIONS 67 n(l+n) n(iat+l Therefore, S.= > or S = a So, the sum of first n positive integers is given by n(n +1)
S= 32 Example 15 : Find the sum of first 24 terms of the list of numbers whose nth term is given by .
a,=3+2n Solution : ; . As a,=342n, So, a,=34+2=5 , So
a=342x2=57
a,=3+2x3=9 cs _ List of numbers becomes 5, 7,9, 11,... . Here, 7-5=9-7=11-9=2 and s0 on. So, it forms an AP with common difference d = 2. To find S,,, we have ne 24, a=5, d=2.. SS . Therefore, S,,= = [2x5 +(24-1)x2] = 12[10+ 46] = 672 So, sum of first 24 terms of the list of numbers is 672. Example 16 : A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find :
(i) the production in the Ist year (ii) the production in the 10th year (iii) the total production in first 7 years Solution : (i) Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in Ist, 2nd, 3rd, ..., years will form an AP.


production in the Ist year (ii) the production in the 10th year (iii) the total production in first 7 years Solution : (i) Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in Ist, 2nd, 3rd, ..., years will form an AP. Let us denote the number of TV sets manufactured in the nth year by a... Then, a,= 600 and a, = 700 Reprint 2025-26


68 MATHEMATICS
or, a+ 2d = 600
and a+ 6d = 700
Solving these equations, we get d=25 and a=550. Therefore, production of TV sets in the first year is 550. Gi) Now Gy = 4+ 9d =5504+9 x 25=775
So, production of TV sets in the 10th year is 775.
- 7
(iii) Also, S,= 5 [2x 550+ (7-1) x25] .
7 :
= 3 [1100 + 150] =4375 . Thus, the total production of TV sets in first 7 years is4375. -
EXERCISE 5.3. - |
1. Find the sum of the following APs: o ; :
(i) 2,7, 12,...,to 10 terms. ' di) -37,-33,-29,..., to 12 terms. (ui) 0.6, 1.7, 2.8,..., to 100 terms. (iv) 1 i I ,--., to 11 terms.
. 15 12 10
2. Find the sums given below: ~
f :
1: ,
a) 7+ 105 hiss) asa . (i) 344324+304+...4+10
(iii) -5 +(-8)+(-11) +... + (230)
3. InanAP: ,
@) givena=5,d=3,a,=50, findnandS . (ii) given a =7,a,,=35, find dand §,,. (ii) given a,, = 37, d=3, find a and S.,. (iv) givena,=15,S,,= 125, findd anda,,. (v) givend=5,S,=75, find a and a,. (vi) givena=2, d=8, S =90, findn anda,
(vil) givena=8,a,=62,S =210, find n and d. (viii) givena =4,d=2, S =-14, find n and a. (x) givena=3,n=8,S = 192, findd.


§,,. (ii) given a,, = 37, d=3, find a and S.,. (iv) givena,=15,S,,= 125, findd anda,,. (v) givend=5,S,=75, find a and a,. (vi) givena=2, d=8, S =90, findn anda, (vil) givena=8,a,=62,S =210, find n and d. (viii) givena =4,d=2, S =-14, find n and a. (x) givena=3,n=8,S = 192, findd. (x) given/=28, S = 144, and there are total 9 terms. Find a. Reprint 2025-26


ARITHMETIC PROGRESSIONS 69 4, How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636? 5, The first term of an APis 5, the last term is 45 and the sum is 400. Find the number of terms
and the common difference. 6. 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? 7. Find the sum of first 22 terms of an AP in which d=7 and 22nd term is 149. 8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18
respectively. 9, If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of
first n terms. 10. Show that a,,a,,...,a,,...formanAP where a, is defined as below : @) a, =3+44n (i) a,=9-5n ; :
Also find the sum of the first 15 terms in each case. . , 11. If the sum of the first n terms of an AP is 4n - n?, what is the first term (that is S,)? What
is the sum of first two terms? What is the second term?


find the sum of the first 15 terms in each case. . , 11. If the sum of the first n terms of an AP is 4n - n?, what is the first term (that is S,)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and
the nth terms. , ° - 12. Find the sum of the first 40 positive integers divisible by 6. 13. Find the sum of the first 15 multiples of 8. 14. Find the sum of the odd numbers between 0 and 50. 15. 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? ° 16. 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. 17.


work by 30 days? ° 16. 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. 17. Ina school, students thought of planting trees in and around the school to reduce air
pollution. It was decided that the number of trees, that each section of each class will
plant, will be the same as the class, in which they are studying, e.g., a section of Class I
will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are
three sections of each class. How many trees will be planted by the students? 18. 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
Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive . 22
semicircles? (Take n = 7? Reprint 2025-26


70 MATHEMATICS
I, ,
I, :
Fig. 5.4 ;
[Hint : Length of successive semicircles is Lt, L, Lye. with centres at A, B, A, B,...,
respectively. ] . 19. 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 Fig. 5.5). In how many rows are the 200 logs placed
and how many logs are in the top row? a " APIS TASA SAS LALA LLL J
Lip ipl iphiPh ipl flip iphip_iphiPhiplhip ipl phi php
Aofofofofofofofofofofofofofeofofofofofe)
-_~ " Fig. 5.5 20. Ina 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 the
line (see Fig. 5.6). 5m a) m 3m e e . ° e °
Fig. 5.6 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?


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? [Hint : To pick up the first potato and the second potato, the total distance (in metres)
run by acompetitor is 2x5+2x(5+3)] Reprint 2025-26


ARITHMETIC PROGRESSIONS 71
EXERCISE 5.4 (Optional)*
1. Which term of the AP: 121, 117, 113,..., is
its first negative term?
{Hint : Find n for a, < 0] 7S cn
2. The sum of the third and the seventh terms ' I
of an AP is 6 and their product is 8. Find ! \
the sum of first sixteen terms of the AP. ! \
! 1
3. A ladder has rungs 25 cm apart. ' 1
(see Fig. 5.7). The rungs decrease ! ' ot m :
uniformly in length from 45 cm at the 2
bottom to 25 cm at the top. If the top and
1] , S
the bottom rungs are 25 m apart, what is cope
the length of the wood required for the . . ; 4 cm
rungs? " 45 cm cee
{Hint : Number of rungs = ~ +1] / Fig. 5.7
4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value
of x such that the sum of the numbers of the houses preceding the house numbered x is
equal to the sum of the numbers of the houses following it. Find this value of x.
[Hint:S,_,=S,,7 S.J . °
5.


from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x. [Hint:S,_,=S,,7 S.J . ° 5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and
built of solid concrete." .
1 1
Each step has a rise of qm and a tread of 5m. (see Fig. 5.8). Calculate the total volume
of concrete required to build the terrace.
. 1 1
{Hint : Volume of concrete required to build the first step = 4 x 5 x 50m' ]
7 5
1 an oF
q m~__y, go
Fig. 5.8
* These exercises are not from the examination point of view. Reprint 2025-26


72 MATHEMATICS
5.5 Summary
In this chapter, you have studied the following points : 1. An arithmetic progression (AP) is a list of numbers in which each term is obtained by
adding a fixed number d to the preceding term, except the first term. The fixed number d
is called the common difference. The general form of anAPisa, a+d, a+2d, a+3d,... 2. A given list of numbers a,, a,, a,, .. . is an AP, if the differences a, - a,, a, - a,
A,- y+. give the same value, 1.e., if 4,,,-4 is the same for different values of k. 3. In an AP with first term a and common difference d, the nth term (or the general term) is
givenby a, =a+(n-l)d. 4, The sum of the first m terms of an AP is given by : S= 5l24 + (nd) .
5. If J is the last term of the finite AP, say the nth term, then the sum of all terms of the AP
is given by: - ' ° -
S= (atl
= 5 (a +1)
- A NOTE TO THE READER
. a + c . » . If a, b, c are in AP, then b = 3 and b is called the arithmetic
mean of a and c. Reprint 2025-26


a . il
talk : arf
1062CHO6
6.1 Introduction . You are familiar with triangles and many of their properties from your earlier classes. In Class IX, you have studied congruence of triangles in detail. Recall that two figures
are said to be congruent, if they have the same shape and the same size. In this
chapter, we shall study about those figures which have the same shape but not necessarily
the same size. Two figures having the same shape (and not necessarily the same size)
are called similar figures. In particular, we shall discuss the similarity of triangles and
apply this knowledge in giving a simple proof of Pythagoras Theorem learnt earlier. Can you guess how heights of mountains (say Mount Everest) or distances of
some long distant objects (say moon) have been found out? Do you think these have
_- - WAH WAHT =
AAH! Soon Iwill nS
It is so easy reach the moon,
to measure the} |
height of the Sf
mountain, Uh
ry
coN Aas
BNL? Et)
. S9 ooo - tn
¥ . a: ----
"Os - J [st vie
Va br pe
ey, V Chey s TW,
eS | | rT
eT i
Reprint 2025-26


74 MATHEMATICS
been measured directly with the help of a measuring tape? In fact, all these heights
and distances have been found out using the idea of indirect measurements, which is
based on the principle of similarity of figures (see Example 7, Q.15 of Exercise 6.3
and also Chapters 8 and 9 of this book). 6.2 Similar Figures In Class IX, you have seen that all circles with the same radii are congruent, all
squares with the same side lengths are congruent and all equilateral triangles with the
same side lengths are congruent. " Now consider any two (or more)
circles [see Fig. 6.1 (i)]. Are they
congruent? Since all of them do not
have the same radius, they are not
congruent to each other. Note that . Ce .
some are congruent and some are not, - @
but all of them have the same shape. _ °
So they all are, what we call, similar. - : Two similar figures have the same_ |. , shape but not necessarily the same size. Therefore, all circles are similar. . . | |
What about two (or more) squares or two (or more) equilateral triangles : (ii)
[see Fig. 6.1 (ii) and (iii)]? As observed + in the case of circles, here also all squares are similar and all equilateral


necessarily the same size. Therefore, all circles are similar. . . | | What about two (or more) squares or two (or more) equilateral triangles : (ii) [see Fig. 6.1 (ii) and (iii)]? As observed + in the case of circles, here also all squares are similar and all equilateral triangles are similar. /\ From the above, we can say
that all congruent figures are (iii)
similar but the similar figures need
not be congruent. Fig. 6.1 Can a circle and a square be P S
similar? Can a triangle and a square A D
be similar? These questions can be | \
answered by just looking at the
figures (see Fig. 6.1). Evidently B c Q
these figures are not similar. (Why?) R Fig. 6.2
Reprint 2025-26


TRIANGLES 75 What can you say about the two quadrilaterals ABCD and PQRS
(see Fig 6.2)?Are they similar? These figures appear to be similar but we cannot be
certain about it.Therefore, we must have some definition of similarity of figures and
based on this definition some rules to decide whether the two given figures are similar
or not. For this, let us look at the photographs given in Fig. 6.3: « Lan at a . Lo .
nef a} | | aod sh, Ay
ie Crh) [alee |jpd owl
CATR NOC acre ytty | bBiaseernty
Fig. 6.3 : You will at once say that they are the photographs of the same monument
(Taj Mahal) but are in different sizes. Would you say that the three photographs are
similar? Yes,they are. 7 ce a What can you say about the two photographs of the same size of the same
person one at the age of 10 years and the other at the age of 40 years? Are these
photographs similar? These photographs are of the same size but certainly they are
not of the same shape. So, they are not similar. What does the photographer do when she prints photographs of different sizes
from the same negative? You must have heard about the stamp size, passport size and
postcard size photographs.


photographs are of the same size but certainly they are not of the same shape. So, they are not similar. What does the photographer do when she prints photographs of different sizes from the same negative? You must have heard about the stamp size, passport size and postcard size photographs. She generally takes a photograph on a small size film, say
of 35mm size and then enlarges it into a bigger size, say 45mm (or 55mm). Thus, if we
consider any line segment in the smaller photograph (figure), its corresponding line 45 55 segment in the bigger photograph (figure) will be 35 [0 =) of that of the line segment. This really means that every line segment of the smaller photograph is enlarged
(increased) in the ratio 35:45 (or 35:55), It can also be said that every line segment
of the bigger photograph is reduced (decreased) in the ratio 45:35 (or 55:35). Further,
if you consider inclinations (or angles) between any pair of corresponding line segments
in the two photographs of different sizes, you shall see that these inclinations(or angles)
are always equal. This is the essence of the similarity of two figures and in particular
of two polygons. We say that:


Further, if you consider inclinations (or angles) between any pair of corresponding line segments in the two photographs of different sizes, you shall see that these inclinations(or angles) are always equal. This is the essence of the similarity of two figures and in particular of two polygons. We say that: Two polygons of the same number of sides are similar, if (i) their
corresponding angles are equal and (ii) their corresponding sides are in the
same ratio (or proportion). Reprint 2025-26


76 MATHEMATICS Note that the same ratio of the corresponding sides is referred to as the scale
factor (or the Representative Fraction) for the polygons. You must have heard that
world maps (i.e., global maps) and blue prints for the construction of a building are
prepared using a suitable scale factor and observing certain conventions. In order to understand similarity of figures more clearly, let us perform the following
activity: Activity 1: Place a lighted bulb at a Oo point O on the ceiling and directly below it a table in your classroom. Let us cut a Ss R= : polygon, say a quadrilateral ABCD, from "Zs - x a plane cardboard and place this AN , cardboard parallel to the ground between i (i . the lighted bulb and the table. Then a a if '\ shadow of ABCD is cast on the table. hiss Hf WW Mark the outline of this shadow as Lak A ij mee \ B A'B'C'D' (see Fig.6.4). ; As. - Note that the quadrilateral A'B'C'D' is ° aK io oc iN
so : ,) ; i a / - BY » an enlargement (or magnification) of the & j Yy LN. 2 quadrilateral ABCD. This is because of: | V yy LY Jaa the property of light that light propogates v {n.


the quadrilateral A'B'C'D' is ° aK io oc iN so : ,) ; i a / - BY » an enlargement (or magnification) of the & j Yy LN. 2 quadrilateral ABCD. This is because of: | V yy LY Jaa the property of light that light propogates v {n. TRA TN are <= | in a straight line. You may also note that. oe Fig. 6.4 A' lies on ray OA, B' lies on ray OB, C' » . lies on OC and D' lies on OD. Thus, quadrilaterals A'B'C'D' and ABCD are of the same shape but of different sizes. . So, quadrilateral A'B'C'D' is similiar to quadrilateral ABCD. We can also say
that quadrilateral ABCD is similar to the quadrilateral A'B'C'D'. Here, you can also note that vertex A' corresponds to vertex A, vertex B'
corresponds to vertex B, vertex C' corresponds to vertex C and vertex D' corresponds
to vertex D. Symbolically, these correspondences are represented as A' > A, B' OB,
C' © C and D' © D. By actually measuring the angles and the sides of the two
quadrilaterals, you may verify that @) ZA=ZA',ZB=2ZB',Z2C=2C,ZD=ZD' and (ii) AB BC CD DA ny - = --_ = --_ = --_
AB BC CD' DA'


represented as A' > A, B' OB, C' © C and D' © D. By actually measuring the angles and the sides of the two quadrilaterals, you may verify that @) ZA=ZA',ZB=2ZB',Z2C=2C,ZD=ZD' and (ii) AB BC CD DA ny - = --_ = --_ = --_ AB BC CD' DA' This again emphasises that two polygons of the same number of sides are
similar, if (@) all the corresponding angles are equal and (ii) all the corresponding
sides are in the same ratio (or proportion). Reprint 2025-26


TRIANGLES 77
From the above, you can easily say that quadrilaterals ABCD and PQRS of
Fig. 6.5 are similar.
4.8 cm | R
5 70°
\/ss°
2.4 Cc
D = 70°) 5.0 cm
35° 4.2 cm "
2.5 cm
2.1 cm 105° 3
105° 100 100 .
1.5 cm P. 3.0 cm ~Q.. Fig.6.5 0 | -
Remark : You can verify that if one polygon is similar to another polygon and this
second polygon is similar to a third polygon, then the first polygon is similar to the third
polygon. You may note that in the two quadrilaterals (a square and a rectangle) of
Fig. 6.6, corresponding angles are equal, but their corresponding sides are not in the
same ratio. fc ) . D 3 cm Cc . 3.5 cm
[J [J S R
3 cm "ot 3 cm 3cm 3 cm
7 : TI P
A 3 cm B 3.5 cm Q
_ Fig. 6.6
So, the two quadrilaterals are not similar. Similarly, you may note that in the two
quadrilaterals (a square and a rhombus) of Fig. 6.7, corresponding sides are in the
same ratio, but their corresponding angles are not equal. Again, the two polygons
(quadrilaterals) are not similar. Reprint 2025-26


78 MATHEMATICS
S
4.2 cm
D 2.1em_¢ 4.2 cm R
2.1..em 2.1.em
P 4.2 cm
4.2 cm
A Ziem ®
Q -
Fig. 6.7 ;
Thus, either of the above two conditions (i) and (ii) of similarity of two
polygons is not sufficient for them to be similar. :
EXERCISE 6.1. Do
1. Fill in the blanks using the correct word given in brackets : °
(i) All circles are _______. (congruent, similar) . (i) All squares are _______. (similar, congruent) (ai) All_____ triangles are similar. (isosceles, equilateral) (iv) Two polygons of the same number of sides are similar, if (a) their corresponding
angles are <_~____ and (b) their corresponding sides are _______. (equal,
proportional) ) , 2. Give two different examples of pair of ,
(i) similar figures. (ii) non-similar figures.
3. State whether the following quadrilaterals are similar or not:
a ; D 3cm C
S_15cm_R 3cm 3cm
1.5 of fs cm
P 1.5 cm Q A 3cm B
Fig. 6.8
Reprint 2025-26


TRIANGLES 79
6.3 Similarity of Triangles
What can you say about the similarity of two triangles? You may recall that triangle is also a polygon. So, we can state the same conditions
for the similarity of two triangles. That is: Two triangles are similiar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). Note that if corresponding angles of two
triangles are equal, then they are known as
equiangular triangles. A famous Greek Ke
mathematician Thales gave an important truth relating iG a .
to two equiangular triangles which is as follows: . at oY, The ratio of any two corresponding sides in Ces ff ;
two equiangular triangles is always the same. - J Nn wy It is believed that he had used a result called : , "Dp
the Basic Proportionality Theorem (now known as | -
the Thales Theorem) for the same. ; , 4 mares c 640 - 546 B.C. To understand the Basic Proportionality ( )
Theorem, let us perform the following activity: . Activity 2 : Draw any angle XAY and on its one
arm AX, mark points (say five points) P, Q, D, R and A E Cc ¥
B such that AP = PQ = QD = DR = RB- P


( ) Theorem, let us perform the following activity: . Activity 2 : Draw any angle XAY and on its one arm AX, mark points (say five points) P, Q, D, R and A E Cc ¥ B such that AP = PQ = QD = DR = RB- P Now, through B, draw any line intersecting arm Q D
AY at C (see Fig. 6.9). R 3 Also, through the point D, draw a line parallel x
to BC to intersect AC at E. Do you observe from tructions that "> ==?M AE and neo
your constructions that > = ? Measure an
EC. What about =? Ob hat A= is al Ito =. Th h
. What about EC. serve that EC 18 also equa to 7; Thus, you can see that
; AD_AE.._., ue ,
in A ABC, DE|| BC and DB = EC' Is it a coincidence? No, it is due to the following
theorem (known as the Basic Proportionality Theorem):
Reprint 2025-26


80 MATHEMATICS
Theorem 6.1 : [f a line is drawn parallel to one side of a triangle to intersect the
other two sides in distinct points, the other two sides are divided in the same
ratio. A
Proof : We are given a triangle ABC in which a line M
parallel to side BC intersects other two sides AB and N,
AC at D and E respectively (see Fig. 6.10). D E
Ww dt that AD = AE
e need to prove that D> EC' B
Cc
Let us join BE and CD and then draw DM L AC and
EN 1 AB. Fig. 6.10
1 ; 1 :
Now, area of A ADE (= 2 base x height) = 2 AD x EN. . Recall from Class IX, that area of A ADE is denoted as ar(ADE). .
1 a
So, ar(ADE) = 5 AD x EN * ;
_. 1 . Similarly, ar(BDE) = 2 DB x EN, _
ar(ADE) = 5 AE x DM and ar(DEC) = 5 EC x DM.
!


Recall from Class IX, that area of A ADE is denoted as ar(ADE). . 1 a So, ar(ADE) = 5 AD x EN * ; _. 1 . Similarly, ar(BDE) = 2 DB x EN, _ ar(ADE) = 5 AE x DM and ar(DEC) = 5 EC x DM. ! AD x EN
Theref ar(ADE) 5" * AD a)
erefore, -r OE OT Er
ar(BDE) 1 pgxEN PB
ao 2
I AE x DM
i ar(ADE) 5 /*~* AE Q)
an 2
» ar(DEC) 1 ECxpM EC
2
Note that A BDE and DEC are on the same base DE and between the same parallels
BC and DE. So, ar(BDE) = ar(DEC) (3)
Reprint 2025-26


TRIANGLES 81
Therefore, from (1), (2) and (3), we have : AD _ AE DB EC Is the converse of this theorem also true (For the meaning of converse, see
Appendix 1)? To examine this, let us perform the following activity: Activity 3: Draw an angle XAY on your B, A notebook and on ray AX, mark points B,, B,, B B, C, BoB and such that AB, = B,B, = B,B, = B, 2 C,
374 7 "4" B C; : Similarly, on ray AY, mark points X Cc,
C,, C,, C,, C, and C such that AC, = C,C, = : Cc
CC, = C,C, = C,C. Then join B,C, and BC © ; Se
(see Fig. 6.11). , . oe Y . - Fig. 6.11
AB, AC, . 1 :
Note that B,B = Gc: (Each equal to 4
You can also see that lines B,C, and BC are parallel to each other, i.c.,
BC, || BC. (1)
Similarly, by joining B,C,, B,C, and B,C,, you can see that: f . BB cc \ 3) mM BGI ABs AC, - 3) d BY, || BC 3 BB Gc. 2) @™ BGI @)
~ AB, _ ACS (=) and 3,¢, BC ) B,B. c,c (1) O° Pas


BC. (1) Similarly, by joining B,C,, B,C, and B,C,, you can see that: f . BB cc \ 3) mM BGI ABs AC, - 3) d BY, || BC 3 BB Gc. 2) @™ BGI @) ~ AB, _ ACS (=) and 3,¢, BC ) B,B. c,c (1) O° Pas From (1), (2), (3) and (4), it can be observed that if a line divides two sides of a
triangle in the same ratio, then the line is parallel to the third side. You can repeat this activity by drawing any angle XAY of different measure and
taking any number of equal parts on arms AX and AY . Each time, you will arrive at
the same result. Thus, we obtain the following theorem, which is the converse of
Theorem 6.1: Reprint 2025-26


82 MATHEMATICS
Theorem 6.2 : /f a line divides any two sides of a A
triangle in the same ratio, then the line is parallel
to the third side. This theorem can be proved by taking a line DE such D E',
= E
h AD _ AE d ing that DEi Hel .
that DB. EC 22 assuming that DE is not paralle
to BC (see Fig. 6.12). B
Cc
If DE is not parallel to BC, draw a line DE' . ;
parallel to BC. Fig. 6.12
S AD _ AE Why ?
0, DB _ E'C ( y 2) ;
Theref AE AE why? | ot
erefore, FC" EC ( y ?) / a . Adding 1 to both sides of above, you can see that E and E' must coincide. (Why ?) oo -
Let us take some examples to illustrate the use of the above theorems. Example 1 : Ifa line intersects sides AB and AC of a A ABC at D and E respectively
d is parallel to BC, prove that "2 = "© (see Fig. 6.13
and is parallel to BC, Prove at 7B = AC (see Fig. 6.13). MN . Solution : DE || BC (Given)
S ADL AE Thi 6.1
0, DB = kc (Theorem 6.1)
.


D and E respectively d is parallel to BC, prove that "2 = "© (see Fig. 6.13 and is parallel to BC, Prove at 7B = AC (see Fig. 6.13). MN . Solution : DE || BC (Given) S ADL AE Thi 6.1 0, DB = kc (Theorem 6.1) . A
DB _ EC
" - AD AE
DB 1 EC 1
Or, . "AD + = AE + D E
. AB _ AC
° AD AE
B
5 AD _ AE c
° AB AC Fig. 6.13
Reprint 2025-26


TRIANGLES 83
Example 2 : ABCD is a trapezium with AB || DC. E A B
and F are points on non-parallel sides AD and BC
respectively such that EF is parallel to AB E F
Fig. 6.14). Show th AE. BF
(see Fig. 6.14). Show that ED FC' 5
Cc
Solution : Let us join AC to intersect EF at G . (see Fig. 6.15). Fig. 6.14
AB || DC and EF || AB (Given) '
So, EF||DC (Lines parallel to the same line are ;
parallel to each other)
A B
EG|| DC (As EF || DC) ok F
go, "B= AS 6.1 1
0, ED = GC (Theorem 6.1) CO)
OD Cc
Similarly, from A CAB, . -
cG CF Fig. 6.15
AG BF...
| C) AG _ BF 2
rE \/ GC FC P 2)
Therefore, from (1) and (2), ©
AE _ BF
-_ ED FC
PS) pT s v
Example 3 : In Fig. 6.16, Re) = TR and Z PST =
Z PRQ. Prove that PQR is an isosceles triangle. Q R
: PS _ PT Fi
eo =~ tf, g. 6.16
Solution : It is given that SQ TR
So, ST || QR (Theorem 6.2)
Therefore, Z PST = ZPQR (Corresponding angles) (1)
Reprint 2025-26


84 MATHEMATICS
Also, it is given that
ZPST= Z PRQ Q2)
So, Z PRQ = Z PQR [From (1) and (2)]
Therefore, PQ = PR (Sides opposite the equal angles)
i.e., PQR is an isosceles triangle. EXERCISE 6.2
1. InFig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). . A , x
A
15m Aj em a \us cm
D E 7.2m E:
3cm ,
B 5.4m
B C . . (i) | Gi) c
Fig. 6.17 | B
2. E and F are points on the sides PQ and PR M
respectively of a A PQR. For each of the following A Cc
cases, state whether EF || QR : x
(i) PE=3.9cm, EQ=3 cm, PF=3.6cmand FR=2.4cm D
(i) PE=4cm, QE=4.5 cm, PF=8 cmand RF=9cm Fig. 6.18
(iit) PQ=1.28 cm, PR = 2.56 cm, PE =0.18 cm and PF=0.36cm
3. In Fig. 6.18, if LM || CB and LN || CD, prove that
A
AM _ AN'
AB AD D
4, In Fig. 6.19, DE || AC and DF || AE. Prove that
BF BE
aa. B Cc
FE EC FUE
Fig. 6.19
Reprint 2025-26


TRIANGLES 85 5. In Fig. 6.20, DE || OQ and DF || OR. Show that P
EF] QR. 6. In Fig. 6.21, A, B and C are points on OP, OQ and E F
OR respectively such that AB || PQ and AC || PR. Show that BC |] QR.
7. Using Theorem 6.1, prove that a line drawn through
. : . . Q R
the mid-point of one side of a triangle parallel to .
another side bisects the third side. (Recall that you Fig. 6.20
have proved it in Class IX). 8. Using Theorem 6.2, prove that the line joining the P '
mid-points of any two sides of a triangle is parallel A
to the third side. (Recall that you have done it in
Class IX). . 9. ABCD is a trapezium in which AB || DC and its LO
diagonals intersect each other at the point O. Show 6B nat 22 = £2. fe
that BO. DO . . Fig. 6.21
10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that
AO 5, hat ABCD i seri
BO.


|| DC and its LO diagonals intersect each other at the point O. Show 6B nat 22 = £2. fe that BO. DO . . Fig. 6.21 10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO 5, hat ABCD i seri BO. DO ow that is a trapezium. |
6.4 Criteria for Similarity of Triangles
In the previous section, we Stated that two triangles are similar, if (1) their corresponding
angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). That is, in A ABC and A DEF, if
G@) ZA=2D,ZB=ZE,2C=Z Fand
(ii) AB _BC _CA> hen th iangl imil Fig. 6.22
ll) .-- = 2 = 1" then the two triangles are similar (See Fig. 6. . DE EF FD 6 (see Fig. 6.22)
A
- L™. |
B Cc E F
Fig. 6.22
Reprint 2025-26


86 MATHEMATICS
Here, you can see that A corresponds to D, B corresponds to E and C
corresponds to F. Symbolically, we write the similarity of these two triangles as
'A ABC ~ A DEF' and read it as 'triangle ABC is similar to triangle DEF'. The
symbol '~' stands for 'is similar to'. Recall that you have used the symbol '=' for
"is congruent to' in Class IX. It must be noted that as done in the case of congruency of two triangles, the
similarity of two triangles should also be expressed symbolically, using correct
correspondence of their vertices. For example, for the triangles ABC and DEF of
Fig. 6.22, we cannot write A ABC ~ A EDF or A ABC ~ A FED. However, we
can write A BAC ~ A EDF. Now a natural question arises : For checking the similarity of two triangles, say
ABC and DEF, should we always look for all the equality relations of their corresponding
angles (ZA=2D, ZB=ZE, ZC= Z FP and all the equality relations of the ratios | - (AB = BE- CA), stusexamine. Y ith
of their corresponding sides DE EF ED).


similarity of two triangles, say ABC and DEF, should we always look for all the equality relations of their corresponding angles (ZA=2D, ZB=ZE, ZC= Z FP and all the equality relations of the ratios | - (AB = BE- CA), stusexamine. Y ith of their corresponding sides DE EF ED). Let us examine. 'ou may recall that
in Class IX, you have obtained some criteria for congruency of two triangles involving
only three pairs of corresponding parts (or elements) of the two triangles. Here also,
let us make an attempt to arrive at certain criteria for similarity of two triangles involving
relationship between less number of pairs of corresponding parts of the two triangles,
instead of all the six pairs of corresponding parts. For this, let us perform the following
activity: i ) . ML . Activity 4: Draw two line segments BC and EF of two different lengths, say 3 cm
and 5 cm respectively. Then, at the points B and C respectively, construct angles PBC
and QCB of some measures, say, 60° and 40°. Also, at the points E and F, construct
angles REF and SFE of 60° and 40° respectively (see Fig. 6.23).
- * R
5 S D
Q P
NA
poe 407\¢ EA 40(\F
3cm 5 cm
Fig. 6.23
Reprint 2025-26


TRIANGLES 87
Let rays BP and CQ intersect each other at A and rays ER and FS intersect each
other at D. In the two triangles ABC and DEF, you can see that
ZB=ZE,2C=ZFand Z A= Z D. That is, corresponding angles of these two
triangles are equal. What can you say about their corresponding sides ? Note that
BC - 2-06 What ab , 0B dro ing AB, DE, CA and FD
[Fs . a ve DE 2nd Ep ? On measuring AB, DE, an , you
will find that DE and Ep we also equal to 0.6 (or nearly equal to 0.6, if there is some
AB BC CA ..
error in the measurement). Thus, DE EF ED You can repeat this activity by
constructing several pairs of triangles having their corresponding angles equal. Every
time, you will find that their corresponding sides are in the same ratio (or proportion). This activity leads us to the following criterion for similarity of two triangles. .


Thus, DE EF ED You can repeat this activity by constructing several pairs of triangles having their corresponding angles equal. Every time, you will find that their corresponding sides are in the same ratio (or proportion). This activity leads us to the following criterion for similarity of two triangles. . Theorem 6.3 : /f in two triangles, corresponding angles are equal, then their
corresponding sides are in the same ratio (or proportion) and hence the two
triangles are similar. : D
This criterion is referred to as the AAA" _ oe o
(Angle-Angle-Angle) criterion of _ A °
similarity of two triangles. . :
This theorem can be proved by taking two P Q
triangles ABC and DEF such that F
ZA=ZD,ZB=ZEandZC=ZF. B cr
(see Fig. 6.24) f ) . Fig. 6.24
Cut DP = AB and DQ = AC and join PQ.


criterion of _ A ° similarity of two triangles. . : This theorem can be proved by taking two P Q triangles ABC and DEF such that F ZA=ZD,ZB=ZEandZC=ZF. B cr (see Fig. 6.24) f ) . Fig. 6.24 Cut DP = AB and DQ = AC and join PQ. So, 'A ABC = A DPQ (Why 2)
This gives ZB=ZP =ZE and PQ|| EF (How?)
therel ~ DP DQ Why?
erefore, 7 PE = OF (Why?)
| AB _ AC Why?
ie., | | DE DE (Why?)
simitary, OE = B andeo AB = BC-AC
mary DE" BF ""'S° DE EF ODF'
Remark : If two angles of a triangle are respectively equal to two angles of another
triangle, then by the angle sum property of a triangle their third angles will also be
equal. Therefore, AAA similarity criterion can also be stated as follows:
Reprint 2025-26


88 MATHEMATICS If two angles of one triangle are respectively equal to two angles of another
triangle, then the two triangles are similar. This may be referred to as the AA similarity criterion for two triangles. You have seen above that if the three angles of one triangle are respectively
equal to the three angles of another triangle, then their corresponding sides are
proportional (i.e., in the same ratio). What about the converse of this statement? Is the
converse true? In other words, if the sides of a triangle are respectively proportional to
the sides of another triangle, is it true that their corresponding angles are equal? Let us
examine it through an activity : '
Activity 5 : Draw two triangles ABC and DEF such that AB = 3 cm, BC = 6 cm,
CA =8 cm, DE = 4.5 cm, EF = 9 cm and FD = 12 cm (see Fig. 6.25). D ; :
A a |
8 em 4.5 cm 12 om -
3 em . B 6cm c E 9 cm F
Fig. 6.25 .
\ AB BC_ CA 2
So,youhave: (== =-~=-- (each equal to =
o, you have ( y DE EF FD (each equal to 3 ) Now measure Z A, Z B, ZC, Z D, Z E and Z F.


. B 6cm c E 9 cm F Fig. 6.25 . \ AB BC_ CA 2 So,youhave: (== =-~=-- (each equal to = o, you have ( y DE EF FD (each equal to 3 ) Now measure Z A, Z B, ZC, Z D, Z E and Z F. You will observe that
ZA=2D,ZB=ZEand Z C= ZF ie,, the corresponding angles of the two
triangles are equal. You can repeat this activity by drawing several such triangles (having their sides
in the same ratio). Everytime you shall see that their corresponding angles are equal. It is due to the following criterion of similarity of two triangles: Theorem 6.4 : /f in two triangles, sides of one triangle are proportional to
(i.e., in the same ratio of ) the sides of the other triangle, then their corresponding
angles are equal and hence the two triangles are similiar. This criterion is referred to as the SSS (Side-Side-Side) similarity criterion for
two triangles. This theorem can be proved by taking two triangles ABC and DEF such that
AB _ BC _©A ey (see Fig. 6.26): DE EF ED (< 1) (see Fig. 6.26):
Reprint 2025-26


TRIANGLES 89
D
A
| P Q
B Cc E F
Fig. 6.26
Cut DP = AB and DQ = AC and join PQ. It can b h Pe OS and PQ IEF (How?)
t can be seen that PE = OF an Q || (How?) ;
So, ZP=ZE and. ZQ=ZF °
DP DQ PQ °
Theref - =a SE
"Ns DE DF EF
5 DP _DQ_BC
° 'DE DF EF WY?)
So, Cc) BC = PQ (Why?)
Thus, A ABC = A DPQ (Why ?)
So, ZA=ZD, ZB=ZE and ZC=ZF _ (How?)
Remark : You may recall that either of the two conditions namely, (i) corresponding
angles are equal and (11) corresponding sides are in the same ratio is not sufficient for
two polygons to be similar. However, on the basis of Theorems 6.3 and 6.4, you can
now say that in case of similarity of the two triangles, it is not necessary to check both
the conditions as one condition implies the other. Let us now recall the various criteria for congruency of two triangles learnt in
Class IX. You may observe that SSS similarity criterion can be compared with the SSS
congruency criterion. This suggests us to look for a similarity criterion comparable to
SAS congruency criterion of triangles. For this, let us perform an activity.


the various criteria for congruency of two triangles learnt in Class IX. You may observe that SSS similarity criterion can be compared with the SSS congruency criterion. This suggests us to look for a similarity criterion comparable to SAS congruency criterion of triangles. For this, let us perform an activity. Activity 6 : Draw two triangles ABC and DEF such that AB = 2 cm, Z A = 50°,
AC =4cm, DE=3 cm, Z D = 50° and DF = 6 cm (see Fig.6.27). Reprint 2025-26


90 MATHEMATICS
D
A DS
\>
4cm 6 cm
2cem 3cm
B Cc E F
Fig. 6.27 . H bserve that ~~ = "© (cach equal to ~) and Z A (included
ere, you may observe that DE ~ DE (each equal to 3 ) an (ine ude
between the sides AB and AC) = Z D (included between the sides DE and DF). That
is, one angle of a triangle is equal to one angle of another triangle and sides including
these angles are in the same ratio (i.e., proportion). Now let us measure 2 B, 2 C,
ZEand ZF ; oe . You will find that 7B = 2 Eand 2 C= ZF Thatis, ZA=ZD,ZB=ZEand
ZC=ZFE So, by AAA similarity criterion, A ABC ~ A DEF. You may repeat this
activity by drawing several pairs of such triangles with one angle of a triangle equal to
one angle of another triangle and the sides including these angles are proportional. Everytime, you will find that the triangles are similar. It is due to the following criterion
of similarity of triangles: " y & / -_ ;
Theorem 6.5 : /f 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.


similar. It is due to the following criterion of similarity of triangles: " y & / -_ ; Theorem 6.5 : /f 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. This criterion is referred to as
the SAS (Side-Angle-Side) D
similarity criterion for two
triangles. ' ; A As before, this theorem can
be proved by taking two triangles Q
ABC and DEF such that P
AB _ AC B CE F
- == 1 ZA=ZD
DE DE (< 1) and
(see Fig. 6.28). Cut DP = AB, DQ Fig. 6.28
= AC and join PQ. Reprint 2025-26


TRIANGLES 91
Now, PQ || EF and A ABC = A DPQ (How ?)
So, ZA=ZD, ZB=ZPandZC=ZQ
Therefore, A ABC ~ A DEF (Why?)
We now take some examples to illustrate the use of these criteria. Example 4 : In Fig. 6.29, if PQ || RS, prove that A POQ ~ A SOR. R
P '
oO , 5
Q So
Fig. 6.29 - oo
Solution : PQRS = - / : (Given)
So, ZP=ZS - ° (Alternate angles)
and ZQ=ZR
Also, Z POQ = ZSOR |. (Vertically opposite angles)
Therefore, vA POQ ~ ASOR (AAA similarity criterion)
Pa) . Example 5 : Observe Fig. 6.30 and then find Z P. R
A.
. 613 7.6
8
yee
Ast -
B ; Cc P a Q
Fig. 6.30
Solution ; In A ABC and A PQR,
Reprint 2025-26


92 MATHEMATICS
AB_38_1 BC_6_1 = CA_3V3_1
RQ 76 2 QP 12 2° be 63 2
_ --AB_BC_CA
That is, RQ QP PR
So, A ABC ~ A RQP (SSS similarity)
Therefore, ZC=ZP (Corresponding angles of similar triangles)
But ZC= 180°-ZA-ZB (Angle sum property)
= 180° - 80° - 60° = 40° a
So, ZP= 40° . : CC
Example 6 : In Fig. 6.31, . ",
A. OA .OB= OC. OD. Show that ZA=ZCandZB=ZD.. | is
Solution: OA.OB=OC.OD~° (Given) .P |
OA OD : Fig. 6.31 3
-_ = -__ ] . . So. OC ~ OB moe
Also, we have io ZAOD= Z COB (Vertically opposite angles) (2)
Therefore, from (1) and (2), AAOD~ ACOB (SAS similarity criterion)
So, ZA=ZCandZD=2B
. (Corresponding angles of similar triangles)
Example 7: A girl of height 90 cmis A
walking away from the base of a rs
lamp-post at a speed of 1.2 m/s. If the lamp ay
is 3.6 m above the ground, find the length "
of her shadow after 4 seconds. Solution : Let AB denote the lamp-post Cc
and CD the girl after walking for4seconds J} , on
away from the lamp-post (see Fig. 6.32). HE E
From the figure, you can see that DE is the B ; D
shadow of the girl. Let DE be x metres.


seconds. Solution : Let AB denote the lamp-post Cc and CD the girl after walking for4seconds J} , on away from the lamp-post (see Fig. 6.32). HE E From the figure, you can see that DE is the B ; D shadow of the girl. Let DE be x metres. Fig. 6.32
Reprint 2025-26


TRIANGLES 93
Now, BD = 1.22mx4=4.8 m. Note that in A ABE and A CDE, ZB=ZD_~- (Each is of 90° because lamp-post
as well as the girl are standing
vertical to the ground) and ZE=ZE (Same angle)
So, A ABE ~ A CDE (AA similarity criterion)
Theref BE _ AB
erefore, DE = CD |
; 48+x 3.6 90 90 0 9
1.¢., x. = 09 ; ( cm = 700 m= 0. m)
ie, AS+x=4x - a |
i.e., 3x= 4.8 a - _
i.e., x= 1.6 - -
So, the shadow of the girl after walking for 4 seconds is 1.6 m long. Example 8 : In Fig. 6.33, CM and RN are : Q N P
respectively the medians of A ABC and -
A PQR. If A ABC ~ A PQR, prove that :" A
(i) AAMC ~ APNR ) :
.. €M _ AB
@) RN PQ Cc
B R
(iit) A CMB ~ ARNQ
Solution: G) |. AABC~ A PQR ig. 6.33 (Given)
"AB BC_CA
So, PQ = QR. RP (1)
and - ZA=ZP, ZB=ZQandZC=ZR (2)
But AB = 2 AM and PQ = 2 PN
(As CM and RN are medians)
2AM CA
So, fi 1), = =
0, from (1) PN RP
Reprint 2025-26


94 MATHEMATICS
| AM _ CA ;
i.e., PN = RP (3)
Also, ZMAC= Z NPR [From (2)] (4)
So, from (3) and (4),
A AMC ~ A PNR (SAS similarity) (5)
li) Fi 5 Me 6
(ii) From (5), aN = RP 6)
B CA _ AB Fi 1
ut RP = PQ [From (1)] (7)
CM _ AB
Therefore, RN = PQ [From (6) and (7) (8)
Gii) Again, PQ = OR . . ; [From (1)]
Theref CM _ BOO [From (8)] (9
erefore, RN = OR rom (8)] (9)
CM AB 2BM
Also, = =
- RN °PQ 2QN
. C) CM .BM 10
Le., - RN = ON (10)
| CM _ BC _BM . ;
Le., 4 RN = OR ON [From (9) and (10)]
Therefore, "OO A CMB ~ A RNQ (SSS similarity)
[Note ; You can also prove part (iii) by following the same method as used for proving
part (i).]
EXERCISE 6.3
1. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by
you for answering the question and also write the pairs of similar triangles in the symbolic
form :
Reprint 2025-26


TRIANGLES 95
P
P
A a A g 5
D>
sf ad =
BO 40/\\c¢ Qh 40-NR Bz Q R
(i) (ii)
P :
D ;
L M
4 g DP 5 °
"TL \ 2
[\
M7 P E 5 F N -L Q - 10 R
iv . (ii > ey) .
: dD P
a
7
sy, : iw 80 39
B CE - t F.
3 (v) > ) . (vi)
: Fig. 6.34
2. In Fig. 6.35, AODC ~ AOBA, Z BOC = 125°
and ZCDO=70°. Find ZDOC, ZDCO and -
ZOAB. + '
OX )125°
3. Diagonals AC and BD of a trapezium ABCD / x
" A B
with AB || DC intersect each other at the
point O. Using a similarity criterion for two Fig. 6.35
anoles. chow that 2A = 2B.
triangles, show that OC OD
Reprint 2025-26


96 MATHEMATICS
T
4. InFi 636, = 2 dZ1=22.Sh K
» InFig. 6.36, O¢ PR 20 = Z 2. Show
that A PQS ~ ATQR.
5. S and T are points on sides PR and QR of
A PQR such that Z P= 2 RTS. Show that Q S R
ARPQ~ ARTS. Fig. 6.36
6. In Fig. 6.37, if A ABE = A ACD, show that A
A ADE ~ A ABC. °
7. InFig. 6.38, altitudes AD and CE of AABC E .
intersect each other at the point P. Show D 4 .
that: . . (i) AAEP ~ ACDP . (ii) AABD ~ ACBE °
es: . " B Cc
(iii) AAEP~ AADB mo
(iv) APDC~ ABEC , | _ Fig. 6.37
8. Eis a point on the side AD produced of a Cc
parallelogram ABCD and BE intersects CD »D
at F. Show that A ABE ~ A CFB. _ .
9. In Fig. 6.39, ABC and AMP are two right :
triangles, right angled at B and M
respectively. Prove that:
A o B
(i) AABC ~ AAMP E
CA BC' Fig. 6.38
© PA MP
: C
10.


at F. Show that A ABE ~ A CFB. _ . 9. In Fig. 6.39, ABC and AMP are two right : triangles, right angled at B and M respectively. Prove that: A o B (i) AABC ~ AAMP E CA BC' Fig. 6.38 © PA MP : C 10. CD and GH are respectively the bisectors
of Z ACB and Z EGF such that D and H lie M
on sides AB and FE of AABC and A EFG
respectively. If AABC ~ A FEG, show that:
. CD AC
© GH FG
A P
(ii) ADCB ~ AHGE B
(iii) ADCA~ AHGF Fig. 6.39
Reprint 2025-26


TRIANGLES 97
11. In Fig. 6.40, E is a point on side CB A
produced of an isosceles triangle ABC
with AB =AC. If AD 1 BC and EF 1 AC,
prove that AABD ~ A ECF, F
12. Sides AB and BC and median AD of a
triangle ABC are respectively propor- E-=- -§ D Cc
tional to sides PQ and QR and median
PM of A PQR (see Fig. 6.41). Show that Fig. 6.40
A ABC ~ APQR.
13. Disa point on the side BC of a triangle A P
ABC such that Z ADC = Z BAC. Show :
that CA? =CB.CD.
14. Sides AB and AC and median AD of a
triangle ABC are respectively B D C QM M R
proportional to sides PQ and PR ands. oe ,
median PM of another triangle PQR. - Fig. 6.41
Show that AABC ~ A PQR. 5 . - _
15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time
a tower casts a shadow 28 m long. Find the height of the tower.
16.


- Fig. 6.41 Show that AABC ~ A PQR. 5 . - _ 15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower. 16. If AD and PM are medians of triangles ABC and PQR, respectively where
AB_AD
AABC~A wey PQ PM
6.5 Summary "- :
In this chapter you have studied the following points :
1. Two figures having the same shape but not necessarily the same size are called similar
figures. .
2. All the congruent figures are similar but the converse is not true.
3. Two polygons of the same number of sides are similar, if (1) their corresponding angles
are equal and (11) their corresponding sides are in the same ratio (i.e., proportion).
4. Ifa line is drawn parallel to one side of a triangle to intersect the other two sides in
distinct points, then the other two sides are divided in the same ratio.
5. Ifa line divides any two sides of a triangle in the same ratio, then the line is parallel to the
third side.
6.


to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio. 5. Ifa line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. 6. Ifintwo triangles, corresponding angles are equal, then their corresponding sides are in
the same ratio and hence the two triangles are similar (AAA similarity criterion).
7. Ifin two triangles, two angles of one triangle are respectively equal to the two angles of
the other triangle, then the two triangles are similar (AA similarity criterion). Reprint 2025-26


98 MATHEMATICS 8. If in two triangles, corresponding sides are in the same ratio, then their corresponding
angles are equal and hence the triangles are similar (SSS similarity criterion). 9. If one angle of a triangle is equal to one angle of another triangle and the sides including
these angles are in the same ratio (proportional), then the triangles are similar
(SAS similarity criterion). A NOTE TO THE READER If in two right triangles, hypotenuse and one side of one triangle are
proportional to the hypotenuse and one side of the other triangle,
then the two triangles are similar. This may be referred to as the
RHS Similarity Criterion. If you use this criterion in Example 2, Chapter 8, the proof will become
simpler. ' OO. Reprint 2025-26


br, | |
am! [pr
2 fH3" er
1. er
a)! ate ai
loe2cHo? COORDINATE GEOMETRY 7
7.1 Introduction . In Class IX, you have studied that to locate the position of a point on a plane, we
require a pair of coordinate axes. The distance of a point from the y-axis is called its
x-coordinate, or abscissa. The distance of a point from the x-axis is called its
y-coordinate, or ordinate. The coordinates of a point on the x-axis are of the form
(x, 0), and of a point on the y-axis are of the form (0, y). _ Here is a play for you. Draw a set of a pair of perpendicular axes on a graph
paper. Now plot the following points and join them as directed: Join the point A(4, 8) to
BG, 9) to C(3, 8) to D(1, 6) to E(1, 5) to F(3, 3) to G(6, 3) to H(8, 5) to 1(8, 6) to
J(6, 8) to K(6, 9) to L(5, 8) to A. Then join the points P(3.5, 7), Q (3, 6) and R(4, 6) to
form a triangle. Also join the points X(5.5, 7), Y(5, 6) and Z(6, 6) to form a triangle. Now join S(4, 5), T(4.5, 4) and U(5, 5) to form a triangle.


9) to L(5, 8) to A. Then join the points P(3.5, 7), Q (3, 6) and R(4, 6) to form a triangle. Also join the points X(5.5, 7), Y(5, 6) and Z(6, 6) to form a triangle. Now join S(4, 5), T(4.5, 4) and U(5, 5) to form a triangle. Lastly join S to the points
(0, 5) and (0, 6) and join U to the points (9, 5) and (9, 6). What picture have you got? Also, you have seen that a linear equation in two variables of the form
ax + by + c = 0, (a, 6 are not simultaneously zero), when represented graphically,
gives a straight line. Further, in Chapter 2, you have seen the graph of
y=ax?+bx+c(a#0), isa parabola. In fact, coordinate geometry has been developed
as an algebraic tool for studying geometry of figures. It helps us to study geometry
using algebra, and understand algebra with the help of geometry. Because of this,
coordinate geometry is widely applied in various fields such as physics, engineering,
navigation, seismology and art! In this chapter, you will learn how to find the distance between the two points
whose coordinates are given. You will also study how to find the coordinates of the
point which divides a line segment joining two given points in a given ratio.


physics, engineering, navigation, seismology and art! In this chapter, you will learn how to find the distance between the two points whose coordinates are given. You will also study how to find the coordinates of the point which divides a line segment joining two given points in a given ratio. Reprint 2025-26


100 MATHEMATICS
7.2 Distance Formula North : . Y
Let us consider the following situation: A town B is located 36 km east and 15
km north of the town A. How would you find
the distance from town A to town B without B
actually measuring it. Let us see. This situation
can be represented graphically as shown in 15 km
Fig. 7.1. You may use the Pythagoras Theorem A East x . oy 36 km ;
to calculate this distance. Now, suppose two points lie on the x-axis. Fig. 7.1 ;
Can we find the distance between them? For Y
instance, consider two points A(4, 0) and B(6, 0) 0.8 -
in Fig. 7.2. The points A and B lie on the x-axis. - 8 D(0, 8) . From the figure you can see that OA = 4 7 ~ .
units and OB = 6 units, - 6 . Therefore, the distance of B from A, i.e., °
AB =OB-OA=6-4=2 units. - 4 So, if two points lie on the x-axis, we can ;
easily find the distance between them. . 1 Now, suppose we take two points lying on * B x
the y-axis. Can you find the distance between O| 1 2 3/4 5 oN '0 . - : ,
them.


So, if two points lie on the x-axis, we can ; easily find the distance between them. . 1 Now, suppose we take two points lying on * B x the y-axis. Can you find the distance between O| 1 2 3/4 5 oN '0 . - : , them. If the points C(0, 3) and D(O, 8) lie on the (4, 0) (6, 0)
y-axis, similarly we find that CD = 8 --3 =5 units Fig. 7.2
(see Fig. 7.2). Next, can you find the distance of A from C (in Fig. 7.2)? Since OA = 4 units and
OC =3 units, the distance of A from C, i.c., AC= [3+ 4? =5 units, Similarly, you can
find the distance of B from D = BD = 10 units. Now, if we consider two points not lying on coordinate axis, can we find the
distance between them? Yes! We shall use Pythagoras theorem to do so. Let us see
anexample. ~ In Fig. 7.3, the points P(4, 6) and Q(6, 8) lie in the first quadrant. How do we use
Pythagoras theorem to find the distance between them? Let us draw PR and QS
perpendicular to the x-axis from P and Q respectively. Also, draw a perpendicular
from P on QS to meet QS at T.


P(4, 6) and Q(6, 8) lie in the first quadrant. How do we use Pythagoras theorem to find the distance between them? Let us draw PR and QS perpendicular to the x-axis from P and Q respectively. Also, draw a perpendicular from P on QS to meet QS at T. Then the coordinates of R and S are (4, 0) and (6, 0),
respectively. So, RS = 2 units. Also, QS = 8 units and TS = PR = 6 units. Reprint 2025-26


CoorDINATE GEOMETRY 101
Therefore, QT = 2 units and PT = RS = 2 units. Y
Now, using the Pythagoras theorem, we 8 Q(6, 8)
have 7
PQ = PT? + QT' 6 / T
= 242728 | 46)
So, PQ = 2,/2 units 3
How will we find the distance between two 2 ,
points in two different quadrants? 1 R R
Consider the points P(6, 4) and Q(-5, -3) ~ Of 1 2 374 5 ORG 0)
(see Fig. 7.4). Draw QS perpendicular to the (4,0) ,
x-axis. Also draw a perpendicular PT from the * Fig. 7.3.
point P on QS (extended) to meet y-axis at the a,
point R. ; a _
8
7
" "Ir 0,4 .
\ ') 3
- 2
5 1
"A [0123456 >
5,0) ~
oo 72
65-3)
, " Fig. 7.4
Then PT = 11 units and QT = 7 units. (Why?)
Using the Pythagoras Theorem to the right triangle PTQ, we get
PQ= J1? +7? = J170 units. Reprint 2025-26


102 MATHEMATICS
Let us now find the distance between any two Y
points P(x,, y,) and Q(x,, y,). Draw PR and QS Qe V2)
perpendicular to the x-axis. A perpendicular from the r
point P on QS is drawn to meet it at the point A
T (see Fig. 7.5). (py)
Then, OR=x,,OS=x,. So, RS =x, -x, = PT. Also, SQ=y,, ST=PR=y,. So, QT=y,-y,. Now, applying the Pythagoras theorem in A PTQ, we get *6 R S x
PQ? = PT? + QT? Fis. 7.5
ig. 7.
= (x,- x, + 0,-y,)
Therefore, PQ=, I(x» -X y +(y.- y,) . . Note that since distance is always non-negative, we take only the positive square
root. So, the distance between the points P(,, y,) and Q(x,, y,) is."
2-7, 2 . PQ = y(x2- 4) +(%- IH) >
which is called the distance formula. ° . Remarks : ,
1. In particular, the distance of a point P(x, y) from the origin O(0, 0) is given by
co" .
\ OP = yx? +y"
2. We can also write, PQ = ./ (x, - x, y +(y, - yp y . (Why?)
Example 1 : Do the points (3, 2), (-2, -3) and (2, 3) form a triangle?


y) from the origin O(0, 0) is given by co" . \ OP = yx? +y" 2. We can also write, PQ = ./ (x, - x, y +(y, - yp y . (Why?) Example 1 : Do the points (3, 2), (-2, -3) and (2, 3) form a triangle? If so, name the
type of triangle formed. |
Solution : Let us apply the distance formula to find the distances PQ, QR and PR,
where P(3, 2), Q(-2, -3) and RQ, 3) are the given points. We have
PQ = (342) + (2+3) =/5? +5? = V50 = 7.07 (approx.)
QR = (-2-2)? +3 -3)? =fC4)? + C6)? = V52 = 7.21 (approx.)
PR = (3 - 2)? +(2-3) =VI + (1? = V2 = 1.41 (approx.)
Since the sum of any two of these distances is greater than the third distance, therefore,
the points P, Q and R form a triangle. Reprint 2025-26


CoorDINATE GEOMETRY 103
Also, PQ? + PR? = QR', by the converse of Pythagoras theorem, we have Z P = 90°. Therefore, PQR is a right triangle. Example 2 : Show that the points (1, 7), (4, 2), (-1, -1) and (- 4, 4) are the vertices
of a square. Solution : Let AC, 7), B(4, 2), C(-1, -1) and D(- 4, 4) be the given points. One way
of showing that ABCD is a square is to use the property that all its sides should be
equal and both its digonals should also be equal. Now, AB =,/(1- 4)?+ (7-2) = 9 + 25 = V34 . BC =,(44+1)?+ (241)? = /2549 = V34 CD =,/(-14+ 4)? + C1- 4)? =/9 + 25 = V34 . DA=/(1+4)? + (7-4)? = f2549 = V34 CO AC = (+1? + (7+? = (4+ 64=V68 . BD = (4+ 4) +(2-4) = 644+4=J68
Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral
ABCD are equal and its diagonals AC and BD are also equal. Thereore, ABCD is a
square. . . Alternative Solution : We find 10
the four sides and one diagonal, say, 9 P| ttt tt tt
AC as above. Here AD? + DC? = .


sides of the quadrilateral ABCD are equal and its diagonals AC and BD are also equal. Thereore, ABCD is a square. . . Alternative Solution : We find 10 the four sides and one diagonal, say, 9 P| ttt tt tt AC as above. Here AD? + DC? = . PT} | fy | tty
34 + 34 = 68 = AC'. Therefore, by 8
the converse of Pythagoras. 7 Pt tt tt Tt
theorem, Z D=90°. A quadrilateral 6 BRR
with all four sides equal and one
mcs asa Sonsed ®™ spt
is a square. . . 4 3 S| st le |
Example 3 : Fig. 7.6 shows the | | ff fy} toi
arrangement of desks in a 2 |
classroom. Ashima, Bharti and 1 ttt ttt
Camella are seated at A(3, 1), | | EL ELL] Ly
B(6, 4) and C(8, 6) respectively. 123 45 6 7 8 9 10
Do you think they are seated in a Columns
line? Give reasons for your Fig. 7.6
answer. Reprint 2025-26


104 MATHEMATICS
Solution : Using the distance formula, we have AB = (6-3)? +(4-D? =/9+9 = Vi8 =3V2 BC = (8 - 6)? +(6 - 4)? =./4 +4 = v8 = 22 AC = (8 - 3)? + (6-1) = 25 + 25 = V50 = 5V2
Since, AB + BC = 32 + 2/2 = 5/2 = AC, we can say that the points A, B and C
are collinear. Therefore, they are seated in a line. |
Example 4 : Find a relation between x and y such that the point (x , y) is equidistant
from the points (7, 1) and (3, 5). Solution ; Let P(x, y) be equidistant from the points A(7, 1) and BG, 5). We are given that AP = BP. So, AP? = BP? | Ce
1.¢., (x-7+(y-1P = («-3P +(y-5P
i.e., x -14x+49+y--2y+1=x-6x4+9+y? - 10y + 25
Le., X-y=2 ° . | :
which is the required relation. - . . Y
Remark : Note that the graph of the equation (8
x-y=2 isa line. From your earlier studies, 7
you know that a point which is equidistant ~ 6 BG, 5)
from A and B lies on the perpendicular 5
bisector of AB.


is the required relation. - . . Y Remark : Note that the graph of the equation (8 x-y=2 isa line. From your earlier studies, 7 you know that a point which is equidistant ~ 6 BG, 5) from A and B lies on the perpendicular 5 bisector of AB. Therefore, the graph of 4
x-y=2 is the perpendicular bisector of AB 3
(see Fig. 7.7). 2
Example 5: Finda point on the y-axis which 1 A(7, 1)
. . . . xX
is equidistant from the points A(6, 5) and O 134567
B(- 4, 3). . \
Solution : We know that a point on the 9
y-axis is of the form (0, y). So, let the point +7 . P(O, y) be equidistant from A and B. Then Fig. 7.7
(6-0 +G6-yY=C4-0P + B-yy
1.¢., 364+25+y -10y= 164+9+y-6y
1.¢., 4y = 36
Le., y=9
Reprint 2025-26


CoorDINATE GEOMETRY 105
So, the required point is (0, 9). Let us check our solution: AP = ,/(6 - 0)? + (5-9)? =./36+16 = V52
BP = (- 4-0)? + 3-9) = /16 + 36 = V52
Note : Using the remark above, we see that (0, 9) is the intersection of the y-axis and
the perpendicular bisector of AB. EXERCISE 7.1
1. Find the distance between the following pairs of points :
(i) (2,3), 4, D (ii) (-5,7), 1, 3) (iii) (a, b), (-a,-b)
2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance
between the two towns A and B discussed in Section 7.2. . "~~
3. Determine if the points (1, 5), (2, 3) and (-2, - 11) are collinear. _
4, Check whether (5, - 2), (6, 4) and (7, - 2) are the vertices of an isosceles triangle.
5. In a classroom, 4 friends are 10 :
seated at the points A, B, C and F | tT ] | tt fy
Das shown in Fig. 7.8.


collinear. _ 4, Check whether (5, - 2), (6, 4) and (7, - 2) are the vertices of an isosceles triangle. 5. In a classroom, 4 friends are 10 : seated at the points A, B, C and F | tT ] | tt fy Das shown in Fig. 7.8. Champa 9 PT} fT | fd pd]
and Chameli walk into the class 8 =
and after observing for a few 7 | | { [ [plea | [|
minutes ChampaasksChamei, 7} {Tt | LTT tt
ors gow tink ABCP 8 # owe SFT ETT ET]
square?" Chameli disagrees. 5
Using distance formula, find . 4 | {| fal | [| [| ce
whichofem comet Cele
a 3
6 Nanetienpotamaitent 37 TPT TT
ormed, if any, by the following '
points, and give reasons for 1 tT TT | ty TT
your answer PET TT TT Tt tl
() C1,-2),(1,0),1,2,¢3,0 9 1 ? 3 _ . 67 8 9 10
.. olumns
(i) (-3, 5), 3, 1), (0, 3), (-l, -4)
(ii) (4,5), (7,6), 4,3), (1,2) Fig. 7.8
7, Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
8. Find the values of y for which the distance between the points P(2, - 3) and Q(10, y) is
10 units. Reprint 2025-26


106 MATHEMATICS
9, If Q(O, 1) is equidistant from P(5, -3) and R(, 6), find the values of x. Also find the
distances QR and PR.
10. Find a relation between x and y such that the point (x, y) is equidistant from the point
(3, 6) and (-3, 4).
7.3 Section Formula
Let us recall the situation in Section 7.2. Suppose a telephone company wants to Y
position a relay tower at P between A and B
is such a way that the distance of the tower B(36, 15)
from B is twice its distance from A. If P lies (x, y) P
on AB, it will divide AB in the ratio | : 2 lL» Cc. (see Fig. 7.9). If we take A as the origin O, A X
and 1 km as one unit on both the axis, the 0 Ds. E
coordinates of B will be (36, 15). In orderto - a
know the position of the tower, we must know _ Fig. 19
the coordinates of P. How do we find these _ ° "
coordinates? . oS °
Let the coordinates of P be (x, y).. Draw perpendiculars from P and B to the
x-axis, meeting it in D and E, respectively. Draw PC perpendicular to BE.


must know _ Fig. 19 the coordinates of P. How do we find these _ ° " coordinates? . oS ° Let the coordinates of P be (x, y).. Draw perpendiculars from P and B to the x-axis, meeting it in D and E, respectively. Draw PC perpendicular to BE. Then, by
the AA similarity criterion, studied in Chapter 6, APOD and A BPC are similar. Theref OD _ OPS 1 geo oP}
ererer'> PC PB 2?" BC. PB 2
~ a s
- 1
So, _x _! and _y Lt,
36-x 2 Is-y 2 Y
These equations give x = 12 and y=5S. Bey ¥2)
You can check that P(12, 5) meets the m
condition that OP : PB = 1 : 2. OVP C
Now let us use the understanding that my 4
you may have developed through this A
example to obtain the general formula. (x )
p71
Consider any two points A(x,, y,) and
B(x,, y,) and assume that P (x, y) divides
AB internally in the ratio m, : m,, i.e., 0 R S T xX
PA m .
- = - (see Fig. 7.10). i
PB m, ( g ) Fig. 7.10
Reprint 2025-26


CoorDINATE GEOMETRY 107
Draw AR, PS and BT perpendicular to the x-axis. Draw AQ and PC parallel to
the x-axis. Then, by the AA similarity criterion,
A PAQ ~ A BPC
PA AQ PQ
Theref -= == 1
errr BP PC BC ()
Now, AQ =RS = OS - OR=x-x,
PC =ST=OT-OS=x,-x
PQ = PS-- QS = PS-- AR=y-y, |
BC = BT- CT = BT-PS=y,-y
Substituting these values in (1), we get . Mm *7KH_ ITM -
My Xy-X YY - . _
m x-xXx . MX» + MND X,
Taking - = -- we get x= 4 -*+ . My Xy- X . . m +m, _
m - me +
Similarly, taking et Fe get y = me FN
my y27y m, +m,
So, the coordinates of the point P(x, y) which divides the line segment joining the
points AG, yp) and BG,, Vo)s internally, in the ratio m, :m, are
(~ MX, + M4% , mh Y2 +m, y, (2)
SY im +m + m+m
This is known as the section formula. This can also be derived by drawing perpendiculars from A, P and B on the
y-axis and proceeding as above.


Vo)s internally, in the ratio m, :m, are (~ MX, + M4% , mh Y2 +m, y, (2) SY im +m + m+m This is known as the section formula. This can also be derived by drawing perpendiculars from A, P and B on the y-axis and proceeding as above. If the ratio in which P divides AB is k : 1, then the coordinates of the point P will be
. . key t+ 1 kyn +
k+1 k+1
Special Case : The mid-point of a line segment divides the line segment in the ratio
1: 1. Therefore, the coordinates of the mid-point P of the join of the points A(,, y,)
and B(x,, y,) is
I-x,+1-x, ly, +]l-y,)_ (4 +x. y+.
1+1 1+1 2 2 . Let us solve a few examples based on the section formula. Reprint 2025-26


108 MATHEMATICS
Example 6 : Find the coordinates of the point which divides the line segment joining
the points (4, - 3) and (8, 5) in the ratio 3 : 1 internally. Solution : Let P(x, y) be the required point. Using the section formula, we get
3(8) + 1(4) 3(5)+ 1(-3)
x= ----=7, y= -- 83
34+1 34+1
Therefore, (7, 3) is the required point. Example 7 : In what ratio does the point (- 4, 6) divide the line segment joining the
points A(- 6, 10) and B(3, - 8)? Solution : Let (- 4, 6) divide AB internally in the ratio m, : m,.


34+1 34+1 Therefore, (7, 3) is the required point. Example 7 : In what ratio does the point (- 4, 6) divide the line segment joining the points A(- 6, 10) and B(3, - 8)? Solution : Let (- 4, 6) divide AB internally in the ratio m, : m,. Using the section
formula, we get .
3m,--6m, -8m, +10 .
ca,= [MES 2, ™) | ()
m+m, - mtM po
Recall that if (x, y) = (a, b) thenx=aandy=b. | " -
3my- 6m, . _, 8m, +10
So, 4-200" and 6 =m
m +m, m+ mM,
N 4 3m, - 6m, .
ow, a = m, + m, gives us
- 4m, - 4m, = 3m, - 6m,
1e., 7m, = 2m,
Le, m,:m,=2:7
You should verify that the ratio satisfies the y-coordinate also.
"em + 10 -8"1 +10
= + m
Now, a ee =, (Dividing throughout by m,)
m, +My), Mm +]
-8 x 2 +10
= -- =6
-+1
7
Reprint 2025-26


CoorDINATE GEOMETRY 109
Therefore, the point (- 4, 6) divides the line segment joining the points A(- 6, 10) and
B(3, - 8) in the ratio 2 : 7. Alternatively : The ratio m, : m, can also be written as ma :1, ork: 1. Let (4, 6)
My
divide AB internally in the ratio k : 1. Using the section formula, we get
C4,6)= 3k --6 -8k+10 (2)
kL ok +1 /
3k --6
So, -4= -
k+1 .
i.e., -4k--4= 3k-6 © :
i.e., Tk=2 ° ce
i.e., kil=2:7 o
You can check for the y-coordinate also. Soe "
So, the point (- 4, 6) divides the line segment joining the points A(- 6, 10) and
B(3, - 8) in the ratio 2: 7. Note : You can also find this ratio by calculating the distances PA and PB and taking
their ratios provided you know that A, P and B are collinear.
pO Y -
Example 8 : Find the coordinates of the points of trisection (i.e., points dividing in
three equal parts) of the line segment joining the points A(2, - 2) and B(- 7, 4). Solution ; Let P and Q be the points of A P Q B
trisection of AB i.e., AP = PQ = QB (2, -2) (-7, 4)
(see Fig. 7.11).


points dividing in three equal parts) of the line segment joining the points A(2, - 2) and B(- 7, 4). Solution ; Let P and Q be the points of A P Q B trisection of AB i.e., AP = PQ = QB (2, -2) (-7, 4) (see Fig. 7.11). Fig. 7.11
Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordinates of P, by
applying the section formula, are
\(-7) + 2(2) 14) + 2(-2) |
L T42) 142 debe C10)
Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are
2(-7) +122) 2(4) + 1-2) | |
241 241. Shes 4,2)
Reprint 2025-26


110 MATHEMATICS
Therefore, the coordinates of the points of trisection of the line segment joining A and
B are (-1, 0) and (- 4, 2). Note : We could also have obtained Q by noting that it is the mid-point of PB. So, we
could have obtained its coordinates using the mid-point formula. Example 9 : Find the ratio in which the y-axis divides the line segment joining the
points (5, - 6) and (-1, - 4). Also find the point of intersection. Solution ; Let the ratio be k : 1. Then by the section formula, the coordinates of the
point which divides AB in the ratio k : 1 are Th+ 5 Ak 6),
k+1 k+l
This point lies on the y-axis, and we know that on the y-axis the abscissa is 0. Theref ktS gg a
erefore, ke] =. :
So, k=5 0
That is, the ratio is 5 : 1. Putting the value of k = 5, we get the point of intersection as
|
(o =) | oo.
3 a -
Example 10 : If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a
parallelogram, taken in order, find the value of p. Solution ; We know that diagonals of a parallelogram bisect each other.


as | (o =) | oo. 3 a - Example 10 : If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p. Solution ; We know that diagonals of a parallelogram bisect each other. So, the coordinates of the mid-point of AC = coordinates of the mid-point of BD
. 7 6+9 144) (8+p 2+3
vee 2° 2) 2° 2
: 15 5 8+p 5
Le, - Bale l ta US
2 2 2 2
15 8+p
aa 27 2
1.¢., p=7
Reprint 2025-26


CoorDINATE GEOMETRY 111
EXERCISE 7.2
1. Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the
ratio 2:3.
2. Find the coordinates of the points of trisection of the line segment joining (4, -1)
and (-2, -3).
3. To conduct Sports Day activities,in p Cc
your rectangular shaped school
ground ABCD, lines have been
drawn with chalk powder at a /
distance of Imeach. 100 flower pots ¥4
have been placed at a distance of 1m al
from each other along AD, as shown NG
LJ
1 .
in Fig. 7.12. Niharika runs a th the Ne . WV .
tl :
distance AD on the 2nd line and us i; *
1 WG I; : : jj Me
posts a green flag. Preet runs 5 th a oo fe
the distance AD on the eighth line v - ,
and posts a red flag. What is the -~ Uy 2
distance between both the flags? If ¥7 1
Rashmi has to post a blue flag exactly A B
halfway between the line segment. 123 45 678 9 10
joining the two flags, where should :
she post her flag? ) Fig. 7.12
4.


flag. What is the -~ Uy 2 distance between both the flags? If ¥7 1 Rashmi has to post a blue flag exactly A B halfway between the line segment. 123 45 678 9 10 joining the two flags, where should : she post her flag? ) Fig. 7.12 4. Find the ratio in which the line segment joining the points (- 3, 10) and (6, - 8) is divided
by © 1,6).
5. Find the ratio in which the line segment joining A(1, - 5) and B(- 4, 5) is divided by the
x-axis. Also find the coordinates of the point of division.
6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find
xandy. ~~
7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is
(2,-- 3) and B is (1, 4).
8. If A and B are (- 2, -2) and (2, - 4), respectively, find the coordinates of P such that
3
AP= 7 AB and P lies on the line segment AB.
9. Find the coordinates of the points which divide the line segment joining A(- 2, 2) and
B(2, 8) into four equal parts.
10.


2, -2) and (2, - 4), respectively, find the coordinates of P such that 3 AP= 7 AB and P lies on the line segment AB. 9. Find the coordinates of the points which divide the line segment joining A(- 2, 2) and B(2, 8) into four equal parts. 10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (- 1, 4) and (2, - 1) taken in
1
order. [Hint : Area of a rhombus = 5 (product of its diagonals)]
Reprint 2025-26


112 MATHEMATICS
7.4 Summary
In this chapter, you have studied the following points :
1. The distance between P(x,, y,) and Q(x,, y,) is J, - 4) +O. - y,)'.
2. The distance of a point P(x, y) from the origin is x? + y'.
3. The coordinates of the point P(@, y) which divides the line segment joining the
points A(x,, y,) and B(x,, y,) internally in the ratio m, : m, are
MyXy + MX MY. +MY Y,
m, + m, m, +m, ,
4. The mid-point of the line segment joining the points P(x,, y,) and Q(x,, y,) is
[Se Ase) .
2 2 ) ;
A NOTETO THE READER
Section 7.3 discusses the Section Formula for the coordinates (x, y) of a
point P which divides internally the line segment joining the points
A(x,, y,) and B(x,, y,) in the ratio m, : m, as follows :
MX, + MX ~_ My, +My,
x= > ys
. m+m m +m,
Note that, here, PA: PB =m, : m,. However, if P does not lie between A and B but lies on the line AB,
outside the line segment AB, and PA: PB =m, : m,, we say that P divides
externally the line segment joining the points A and B.


Note that, here, PA: PB =m, : m,. However, if P does not lie between A and B but lies on the line AB, outside the line segment AB, and PA: PB =m, : m,, we say that P divides externally the line segment joining the points A and B. You will study
Section Formula for such case in higher classes. Reprint 2025-26


[a] es Le]
7 a
(a) ee
1062CHO8
There is perhaps nothing which so occupies the i
middle position of mathematics as trigonometry.
- J.F. Herbart (1890)
8.1 Introduction . . eT
You have already studied about triangles, and in particular, right triangles, in your
earlier classes. Let us take some examples from our surroundings where right triangles
can be imagined to be formed. For instance : , f
1. Suppose the students of a school are , a
visiting Qutub Minar. Now, if a student 4 we i
is looking at the top of the Minar, a right a E.
triangle can be imagined to be made, © wn -
as shown in Fig 8.1. Can the student we ;
find out the height of the Minar, without aw | :
actually measuringit? (ore If
2. Suppose a girl is sitting on the balcony verre
of her house located on the bank of a Fig. 8.1
river. She is looking down at a flower
pot placed ona stairofatemple situated
nearby on the other bank of the river. f-5. A right triangle is imagined to be made > f Pa KA
in this situation as shown in Fig.8.2.


house located on the bank of a Fig. 8.1 river. She is looking down at a flower pot placed ona stairofatemple situated nearby on the other bank of the river. f-5. A right triangle is imagined to be made > f Pa KA in this situation as shown in Fig.8.2. If -_Na i me awa
you know the height at which the © 7] ma Arif
person is sitting, can you find the width os seas leet |
TE |
iver?
of the river? Fig. 8.2
Reprint 2025-26


114 MATHEMATICS
3. Suppose a hot air balloon is flying in the air. A girl happens to spot the balloon in the sky and runs to her @, @) mother to tell her about it. Her mother a at rushes out of the house to look at the o weg balloon.Now when the girl had spotted Obes the balloon intially it was at point A. oC ae When both the mother and daughter fe came out to see it, it had already Ags A travelled to another point B. Can you . ' find the altitude of B from the ground? Fig. 8.3 In all the situations given above, the distances or heights can be found by using
some mathematical techniques, which come under a branch of mathematics called
'trigonometry'. The word 'trigonometry' is derived from the Greek words 'tri'
(meaning three), 'gon' (meaning sides) and ."metron' (meaning measure). In fact,
trigonometry is the study of relationships between the sides and angles of a triangle. The earliest known work on trigonometry was recorded in Egypt and Babylon. Early
astronomers used it to find out the distances of the stars and planets from the Earth. Even today, most of the technologically advanced methods used in Engineering and
Physical Sciences are based on trigonometrical concepts.


a triangle. The earliest known work on trigonometry was recorded in Egypt and Babylon. Early astronomers used it to find out the distances of the stars and planets from the Earth. Even today, most of the technologically advanced methods used in Engineering and Physical Sciences are based on trigonometrical concepts. In this chapter, we will study some ratios of the sides of a right triangle with
respect to its acute angles, called trigonometric ratios of the angle. We will restrict
our discussion to acute angles only. However, these ratios can be extended to other
angles also. We will also define the trigonometric ratios for angles of measure 0° and
90°. We will calculate trigonometric ratios for some specific angles and establish
some identities involving these ratios, called trigonometric identities. 8.2 Trigonometric Ratios Cc
In Section 8.1, you have seen some right triangles <
imagined to be formed in different situations. z Let us take a right triangle ABC as shown Hypotenuse £
in Fig. 8.4, = Here, Z CAB (or, in brief, angle A) is an 5
acute angle. Note the position of the side BC
with respect to angle A. It faces Z A. We call it =
the side opposite to angle A.


right triangle ABC as shown Hypotenuse £ in Fig. 8.4, = Here, Z CAB (or, in brief, angle A) is an 5 acute angle. Note the position of the side BC with respect to angle A. It faces Z A. We call it = the side opposite to angle A. AC is the A Side adjacent to B
hypotenuse of the right triangle and the side AB angle A
is a part of Z A. So, we call it the side .
adjacent to angle A. Fig. 8.4 Reprint 2025-26


INTRODUCTION TO TRIGONOMETRY 115
Note that the position of sides change Cc
when you consider angle C in place of A -
(see Fig. 8.5). <
You have studied the concept of 'ratio' in Ht 3
: . . ypotenuse &
your earlier classes.


INTRODUCTION TO TRIGONOMETRY 115 Note that the position of sides change Cc when you consider angle C in place of A - (see Fig. 8.5). < You have studied the concept of 'ratio' in Ht 3 : . . ypotenuse & your earlier classes. We now define certain ratios 3 Y
involving the sides of a right triangle, and call xz Sb
them trigonometric ratios. wa
The trigonometric ratios of the angle A A B
in right triangle ABC (see Fig. 8.4) are defined Side opposite to -
as follows : angle C
. side opposite to angle A | BC Fig. 8.5
sine of Z A = = -
hypotenuse AC
. side adjacent to angle A_- AB oo .
cosine of Z A= --W___--_ = -- :
hypotenuse _ AC _ -
side opposite to angle A BCtangent of Z A = ---_________ = -_
side adjacent to angle A = AB
1 hypot AC
cosecant of Z A= poise
sine of ZA side opposite to angle A BC
"oN 1 oh A
secant of ZA= - 1 ___- hypotenuse _ AC
'cosine of ZA side adjacent to angle A AB
4 ide adjacent to angle A AB
cotangent of Z A= -----_ = = em eo mee
tangent of Z A side opposite to angle ABC
The ratios defined above are abbreviated as sin A, cos A, tan A, cosec A, sec A
and cot A respectively.


angle A AB 4 ide adjacent to angle A AB cotangent of Z A= -----_ = = em eo mee tangent of Z A side opposite to angle ABC The ratios defined above are abbreviated as sin A, cos A, tan A, cosec A, sec A and cot A respectively. Note that the ratios cosec A, sec A and cot A are respectively,
the reciprocals of the ratios sin A, cos A and tan A. BC
Also, observe that tan A= P= = AC = MA ag cota = SA
so, observe that tan A= 7 'AB cos A and cot A= sinA
AC
So, the trigonometric ratios of an acute angle in a right triangle express the
relationship between the angle and the length of its sides. Why don't you try to define the trigonometric ratios for angle C in the right
triangle? (See Fig. 8.5)
Reprint 2025-26


116 MATHEMATICS
The first use of the idea of 'sine' in the way we use
it today was in the work Aryabhatiyam by Aryabhata, he (Ray
in A.D. 500. Aryabhata used the word ardha-jya oH) SN
for the half-chord, which was shortened to jya or 4 ta
jiva in due course. When the Aryabhatiyam was
translated into Arabic, the word jiva was retained as f 7
itis. The word jiva was translated into sinus, which j
means curve, when the Arabic version was translated
into Latin. Soon the word sinus, also used as sine, -
became common in mathematical texts throughout UJ
Europe. An English Professor of astronomy Edmund
Gunter (1581-1626), first used the abbreviated Aryabhata
notation 'sin'. C.E. 476 - 550
The origin of the terms 'cosine' and 'tangent' was much later. The cosine function
arose from the need to compute the sine of the complementary angle. Aryabhatta
called it kotijya. The name coesinus originated with Edmund Gunter. In 1674, the
English Mathematician Sir Jonas Moore first used the abbreviated notation 'cos'. Remark : Note that the symbol sin A is used as an
abbreviation for 'the sine of the angle A'. sin A is not Q
the product of 'sin' and A. 'sin' separated from A c "
has no meaning.


Mathematician Sir Jonas Moore first used the abbreviated notation 'cos'. Remark : Note that the symbol sin A is used as an abbreviation for 'the sine of the angle A'. sin A is not Q the product of 'sin' and A. 'sin' separated from A c " has no meaning. Similarly, cos A is not the product of .
'cos' and A. Similar interpretations follow for other P :
trigonometric ratios also. 7 . Now, if we take a point P on the hypotenuse Hypotenuse :
AC or a point Q on AC extended, of the right triangle
ABC and draw PM perpendicular to AB and QN A ot
perpendicular to AB extended (see Fig. 8.6), how M BN
will the trigonometric ratios of Z A in A PAM differ
from those of Z Ain A CAB or from those of Z Ain Fig. 8.6
A QAN? To answer this, first look at these triangles. Is A PAM similar to A CAB? From
Chapter 6, recall the AA similarity criterion. Using the criterion, you will see that the
triangles PAM and CAB are similar. Therefore, by the property of similar triangles,
the corresponding sides of the triangles are proportional. AM AP _ MP
So, we have AB = AC BC
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INTRODUCTION TO TRIGONOMETRY 117
F hi find MP = BE- sin A
rom this, we fin AP = AC .
. AM AB MP BC
Similarly, -- = - = cos A, -- =- =tan A and so on. AP AC AM AB
This shows that the trigonometric ratios of angle A in A PAM not differ from
those of angle A in A CAB. In the same way, you should check that the value of sin A (and also of other
trigonometric ratios) remains the same in A QAN also. |
From our observations, it is now clear that the values of the trigonometric
ratios of an angle do not vary with the lengths of the sides of the triangle, if
the angle remains the same. . Note : For the sake of convenience, we may write sin'A, cos?A, etc., in place of
(sin A)', (cos A)', etc., respectively. But cosec A = (sin A)! ¥ sin"! A (it is called sine
inverse A). sin"! A has a different meaning, which will be discussed in higher classes. Similar conventions hold for the other trigonometric ratios as well. Sometimes, the
Greek letter 6 (theta) is also used to denote an angle. . We have defined six trigonometric ratios of an acute angle.


sine inverse A). sin"! A has a different meaning, which will be discussed in higher classes. Similar conventions hold for the other trigonometric ratios as well. Sometimes, the Greek letter 6 (theta) is also used to denote an angle. . We have defined six trigonometric ratios of an acute angle. If we know any one
of the ratios, can we obtain the other ratios? Let us see.
1 , Cc
If in a right triangle ABC, sin A= 3° .
then thi th cp Bell i.e., th 3 k
en this means that 7 (=, ie., the .
lengths of the sides BC and AC of the triangle
ABC are in the ratio 1 : 3 (see Fig. 8.7). Soif A B
BC is equal to k, then AC will be 3k, where Fig. 8.7
kis any positive number. To determine other
trigonometric ratios for the angle A, we need to find the length of the third side
AB. Do you remember the Pythagoras theorem? Let us use it to determine the
required length AB. ,
AB? = AC? - BC? = (3k)? - (KY = 82 = (2V2 ky
Therefore, .


trigonometric ratios for the angle A, we need to find the length of the third side AB. Do you remember the Pythagoras theorem? Let us use it to determine the required length AB. , AB? = AC? - BC? = (3k)? - (KY = 82 = (2V2 ky Therefore, . AB= + 2/2 k
So, we get AB= 2V2k (Why is AB not --2V2k2)
x ae ABE 22k _ 2/2
om oS AC 3k
Similarly, you can obtain the other trigonometric ratios of the angle A. Reprint 2025-26


118 MATHEMATICS
Remark : Since the hypotenuse is the longest side in a right triangle, the value of
sin A or cos A is always less than 1 (or, in particular, equal to 1). Let us consider some examples.
4 Cc
Example 1 : Given tan A = 3° find the other
trigonometric ratios of the angle A. Solution : Let us first draw a right A ABC 4k
(see Fig 8.8). N k h A Bef ,
ow, we know that tan A= AB 3°
Therefore, if BC = 4k, then AB = 3k, where k is a A 3k B
positive number. . ,
j ~ Fig. 8.8
Now, by using the Pythagoras Theorem, we have OO
AC' = AB* + BC? = (4k) + (3k)? = 25k? So, AC=5k . sO
Now, we can write all the trigonometric ratios using their definitions.
mae BOM 4
me=AC 5k 5
y= AB 3k 3
EON AC SKS
Th f tA _1 23 cosee Az lL => d A I 5
ePETOREs COLIN tanA 4 sinA 4 anesee A= cosA 3
Los P
Example 2: If Z B and Z Q are
acute angles such that sin B = sin Q, A
then prove that Z B = Z Q.


tA _1 23 cosee Az lL => d A I 5 ePETOREs COLIN tanA 4 sinA 4 anesee A= cosA 3 Los P Example 2: If Z B and Z Q are acute angles such that sin B = sin Q, A then prove that Z B = Z Q. Solution : Let us consider two right
triangles ABC and PQR where
ae .; Cc BOR Q
sin B = sin Q (see Fig. 8.9). Fig. 8.9
. AC
We have sin B= AB
; now Ee
an sin Q= PQ
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