SOLUTIONS
PULLOUT
Worksheets
Term 1 (April to September)
Mathematics
Class
10
OSWAAL BOOKS
“Oswaal House” 1/11, Sahitya Kunj, M.G. Road, AGRA-282002
Ph.: 0562-2857671, 2527781, Fax : 0562-2854582, 2527784
email : [email protected], website : www.OswaalBooks.com
CONTENTS
Unit-I : Number System
1.
Real Numbers
Summative Assessment
Ø
Worksheets 1 - 11
3 - 14
Formative Assessment
Ø
Worksheet 12
14 - 14
Unit-II : Algebra
2.
3.
Polynomials
Summative Assessment
Ø
Worksheets 12 - 23
15 - 29
Formative Assessment
Ø
Worksheet 24
29 - 30
Pair of Linear Equations in two Variables
Summative Assessment
Ø
Worksheets 25 - 37
31 - 50
Formative Assessment
Ø
Worksheet 38
50 - 51
Unit-III : Geometry
4.
Triangles
Summative Assessment
Ø
Worksheets 39 - 52
52 - 71
Formative Assessment
Ø
Worksheet 53
72 - 72
Unit-IV : Trigonometry
5.
Introduction to Trigonometry and Trigonometric Identities
Summative Assessment
Ø
Worksheets 54 - 76
73 - 101
Formative Assessment
Ø
Worksheet 77
101 - 102
Unit-V : Statistics
6.
Statistics
Summative Assessment
Ø
Worksheets 78 - 95
103 - 128
Formative Assessment
Ø
Worksheet 96
129 - 129
(v)
CHAPTER
Term-I
1
Summative Assessment
Unit I
Number System
Real Numbers
SUMMATIVE ASSESSMENT
WORKSHEET-1
SECTION A
1. We know that
a × b = L.C.M. × H.C.F.
= 200 × 5 = 1000
1
2. Since
16 = ± 4
So it is a rational number.
1
SECTION B
3. Given
H.C.F.
L.C.M.
First term
Second term
= 18
=?
= 306
= 1314
1
we know that
First × Second term = L.C.M. × H.C.F.
306 × 1314 = L.C.M. × 18
L.C.M. =
306 × 1314
= 22388
18
1
SECTION C
4. First we will find out HCF (1530, 1365)
1530 = 1365 × 1 + 165
1365 = 165 × 8 + 45
165 = 45 × 3 + 30
45 = 30 × 1 + 15
30 = 15 × 2 + 0
Now
HCF (1530, 1365) = 15
Now lets find out HCF (1305, 15)
1305 = 15 × 87 + 0
∴
HCF (1530, 1365, 1305) = 15
1
SECTION D
5. (a) Maximum number of parallel rows of each class
Now,
104
96
Hence
HCF (104, 96)
S O L U T I O N S
= HCF of 104 and 96
= 2 × 2 × 2 × 13
= 2 × 2 × 2 × 12
=8
1
1
P-3
(b) No. of the students of class X in 1 row =
104
= 13
8
96
= 12
8
No. of students of class IX in 1 row =
1
(c) To minimise the tendency of the students to copy and to teach them value of honesty.
SUMMATIVE ASSESSMENT
1
WORKSHEET-2
SECTION A
1. 145 = 29 × 5
Hence, number of prime factors of 145 is 2.
2.
p = a3b2
q = ab3c2
Hence HCF of p and q = ab2.
3. Decimal expansion of
23
1
1
1
will be terminating
23 52
SECTION B
1656 4025 ( 2
–3312
713 1656 ( 2
–1426
230 713 (3
– 690
4.
23 230 (10
–230
0
2
Hence HFC (1656, 4025) = 23.
SECTION C
5.
z = 2 × 17 = 34
½
y = 2 × 34 = 68
½
x = 2 × 68 = 136
½
Yes, value of x can be found without finding value of y or z as x = 2 × 2 × 2 × 17, which are prime
factors of x.
1½
SECTION D
6. Time required by 3 children to complete a card together = LCM of 10, 16, 20
Now
10 = 2 × 5
16 = 2 × 2 × 2 × 2
20 = 2 × 2 × 5
So
required LCM = 2 × 5 × 2 × 2 × 2
These children possess following values :
Caring respect for elders, creative and helpful.
P-4
M A T H E M A T I C S --
X
1
1
1
1
T E R M – 1
SUMMATIVE ASSESSMENT
WORKSHEET-3
SECTION A
1. A rational number can be expressed as a terminating decimal of the denominator has factors 2 or
5 only.
1
2. We know that
a × b = HCF × LCM
1800 = 12 × LCM
LCM =
1800
= 150
12
1
SECTION B
3. If the number 4n, were to end with the digit zero, then it should be divisible by 5.
1
But
4n = (22)n = 22n
⇒ Only prime in the factorization of 4n is 2.
So by fundamental theorem of Arithmetic, there are no other primes in the factorisation of 4n.½
⇒ 4n can never end with the digit zero.
½
SECTION C
4. Let us assume on the contrary that 4 − 3 2 is rational. Then, there exist co-prime possible integers
a and b such that
a
1
4−3 2 = b
a
= 3 2
b
4b − a
= 2
3b
4−
⇒
⇒
⇒
1
[Q a, b are integers
2 is rational
This, contradicts the fact that
4b − a
is a rotational number]
3b
2 is irrational.
So, our assumption is wrong. Hence 4 − 3 2 is an irrational number.
1
SECTION D
5. Let the number of columns be x.
x is the largest number, which should divide both 104 and 96
104 = 96 × 1 + 8
96 = 8 × 12 + 0
∴ HCF of 104 and 96 is 8
Hence, 8 columns are required.
SUMMATIVE ASSESSMENT
1
1
1
1
WORKSHEET-4
SECTION A
3
3
375
3 × 53
= 3 = 3
3 = 103 = 0·375
8
2
2 ×5
1.
2.
6423
4
=
6243 × 2
4
4
=
12486
2 ×5
2 ×5
104
Hence decimal expansion of the rational number will terminate after 4 places of decimal.
S O L U T I O N S
3
1
1
P-5
SECTION B
3
255 867
765
3.
1
2
102 255
204
2
51 102
102
0
∴ Required HCF = 51.
1
SECTION C
4. Let
5 be a rational number.
∴
a
, (a, b are co-prime integers and b ≠ 0.)
b
a =b 5
5 =
a2 = 5b2
⇒ 5 is a factor of
⇒ 5 is a factor of a
Let
a = 5c, (c is some integer)
⇒
25c2 = 5b2
⇒
5c2 = b2
2
⇒ 5 is a factor of b
5 is a factor of b
∴ 5 is a common factor of a, b
a2
1
But this contradicts the fact that a, b are co-primes.
∴
5 is irrational
Let 2 –
∴
2–
⇒
5 =a
2–a =
2 – a is rational, so is
But
1
5 be rational
5
5.
5 is not rational ⇒ contradiction
∴ 2–
1
5 is irrational.
SECTION D
5. Let
∴
2 be a rational number.
a
,
b
a = 2b
2 =
(a, b are co-prime integers and b ≠ 0)
Squaring,
a2 = 2b2
2
⇒ 2 divides a
⇒ 2 divides a
So we can write a = 2c for some integer c, substitute for a, 2b2 = 4c2, b2 = 2c2
This means 2 divides b2, so 2 divides b.
∴ a and b have at least ‘2’ as a common factor.
But this contradicts, that a, b have no common factors other than 1.
P-6
M A T H E M A T I C S --
X
1
1
T E R M – 1
∴ Our supposition is wrong.
Hence,
2 is irrational.
3
Let
2
= a, where a is a rational
a 2 =3
3
2 =
a
1
3
is rational but 2 is not rational.
a
∴ Our supposition is wrong.
∴
3
2
1
is irrational.
SUMMATIVE ASSESSMENT
WORKSHEET - 5
SECTION A
1. According to Euclid’s division Lemma for any positive interger and 3 there exist unique integers q
and r such that a = 3q + r, where r must satisfy 0 ≤ r < 3.
1
2. 1 = 1 × 1,
2 = 1 × 2,
3 = 1 × 3,
4 = 2 × 2,
5 =1×5
6 = 2 × 3,
7 = 1 × 7,
8 = 2 × 2 × 2,
9 = 3 × 3,
10 = 2 × 5.
Hence the least number that, divisible by all the numbers from 1 to 10 i.e., L.C.M.
= 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520.
1
SECTION B
3. By Euclid’s division algorithm
a = bq + r
Take
b =4
∴
r = 0, 1, 2, 3
So,
a = 4q, 4q + 1, 4q + 2, 4q + 3
Clearly, a = 4q, 4q + 2 are even, as they are divisible by 2.
But 4q + 1, 4q + 3 are odd, as they are not divisible by 2.
∴ any positive odd integer is of the form (4q + 1) or (4q + 3).
1
1
SECTION C
4. (i) The number of rooms will be minimum if each room accomodates maximum number of
participants. Since in each room the same number of participants are to be seated and all of them
must be of the same subject. Therefore, the number of paricipants in each room must be the HCF
of 60, 84 and 108. The prime factorisations of 60, 84 and 108 are as under
60 = 22 × 3 × 5
84 = 22 × 3 × 7
108 = 22 × 33
Hence,
HCF = 22 × 3 = 12
Therefore, in each room 12 participants can be seated.
1
∴
Number of rooms required =
=
Total number of participants
12
60 + 84 + 108
252
=
= 21.
12
12
1
(ii) HCF of numbers.
½
(iii) Liberty and equality are the pay marks of democracy.
½
S O L U T I O N S
P-7
SECTION D
a = 4q + r, 0 ≤ r < 4
5. Let
⇒
1
a = 4q, 4q + 1, 4q + 2 or 4q + 3
Case I.
a2 = (4q)2 = 16q2 = 4(4q2) = 4m, m = 4q2
Case II.
a2 = (4q + 1)2 = 16q2 + 8q + 1 = 4(4q2 + 2q) + 1
½
= 4m + 1, where m = 4q2 + 2q
a2
Cases III.
= (4q +
=
a2
Cases IV.
2)2
4(4q2
= (4q +
=
16q2
½
+ 16q + 4
+ 4q + 1) = 4m, where m = 4q2 + 4q + 1
3)2
=
16q2
½
+ 24q + 9
= 4(4q2 + 6q + 2) + 1
= 4m +1, where m = 4q2 + 6q + 2
½
From cases I, II, III and IV, we conclude that the square of any +ve integer is of the form 4m or 4m + 1.
1
SUMMATIVE ASSESSMENT
WORKSHEET-6
SECTION A
1. The smallest prime number = 2
and the smallest composite number = 4.
Hence required HCF (4, 2) = 2
1
3
has terminating decimal expansion.
5
3. If q is some integer, then any positive odd integer is of the form 6q + 1.
2.
1
1
SECTION B
2 × 3381 =
4.
6762
2
½
3381
3
½
1127 = 7 × 161
7
161
7
23
= 161
7
½
∴ Composite number x = 6762.
½
SECTION C
5. Let
Subtracting,
x
10000 x
10 x
9990 x
x =
P-8
1
= 0·3178178178...
= 3178.178178...
= 3.178178...
= 3175
1
3175 635
=
.
9990 1998
M A T H E M A T I C S --
1
X
T E R M – 1
SECTION D
6. Let
2 be a rational number
a
2 = b,
So
⇒
[Q a, b are co-prime integers and b ≠ 0]
a =
a2
2b
= 2b2
Squaring
½
⇒
2 divides a2
⇒
2 divides a
so we can write a = 2c for same integer c, substitute for a, 2b2 = 4c2 ⇒ b2 = 2c2.
½
This means 2 divides b2, so 2 divides b.
∴ a and b have atleast ‘2’ as a common factor but this contradicts, that a, b have no common
factors other than 1.
∴ Our supposition is wrong.
½
Hence
2 is irrational.
p
5 + 3 2 = q [HCF (p, q) = 1]
Let
2 =
2 is irrational but
p − 5q
3q
p − 5q
is a rational.
3q
½
Irrational ≠ rational ⇒ Contradiction.
∴ 5 + 3 2 is an irrational number.
1
SUMMATIVE ASSESSMENT
WORKSHEET - 7
SECTION A
1.
6
6
6 × 23
6 × 23
=
=
=
·
1250 2 × 54 24 × 54
104
Hence decimal expansion of the rational number will terminate after 4 places of decimal.
2. The decimal expansion of
13
15
2 × 510
is terminating.
1
1
SECTION B
3.
⇒
240 = 228 × 1 + 12
228 = 12 × 19 + 0
HCF of 240 and 228 = 12.
1
1
SECTION C
4. (i) In order to arrange the books as required, we have to find the largest number that divides 96,
240 and 336 exactly, clearly, such a number is their HCF.
We have,
96 = 25 × 3
240 = 24 × 3 × 5
336 = 24 × 3× 7
and
∴
HCF of 96, 240 and 336 is 24 × 3 = 48
S O L U T I O N S
P-9
1
So, there must be 48 books in each stack.
∴
Number of stacks of English books =
96
=2
48
Number of stacks of Hindi books =
240
=5
48
Number of stacks of Sociology books =
336
=7
48
1
(ii) HCF of numbers.
½
(iii) Cleanliness has been discussed in this question, it is a good habit that leads to good health.
½
SECTION D
5. Let x be any positive integer, then it is of the form 3q or 3q + 1 or 3q + 2.
Squaring, we get
(3q)2 = 9q2 = 3 × 3q2 = 3m, m = 2q2
(3q + 1) = 9q2 + 6q + 1
=
3(3q2
1
+ 2q) + 1
= 3m + 1, m = 3q2 + 2q
(3q + 2)2 = 9q2 + 12q + 4
=
9q2
1
+ 12q + 3 + 1
= 3(3q2 + 4q + 1) + 1
= 3m + 1, m = 3q2 + 4q + 1
1
⇒ Square of any positive integer is of the form 3m or 3m + 1 for some integer m.
SUMMATIVE ASSESSMENT
1
WORKSHEET-8
SECTION A
1.
189 189 189 × 23 189 × 23
= 3 = 3
=
·
125
5
5 × 23
(10)3
The decimal expansion of
189
will terminate after 3 places of decimal.
125
1
93
93
31
= 2 3
=
2
2
1500
3×5×2 ×5
2 ×5
Hence, decimal representation is terminating.
3. HCF of 33 × 5 and 32 × 52 = 32 × 5 = 9 × 5 = 45.
2.
1
1
SECTION B
4.
90
144
HCF
LCM
= 2 × 32 × 5
= 24 × 32
= 2 × 32 = 18
= 24 × 32 × 5 = 720.
1
1
SECTION C
5. Let n be any +ve integer, then
n = 3q + r, r = 0, 1, 2
n = 3q or 3q + 1 or 3q + 2
P-10
M A T H E M A T I C S --
½
½
X
T E R M – 1
Case I.
When
Case II.
When
n = 3q, which is divisible by 3
n + 2 = 3q + 2, which is not divisible by 3
n + 4 = 3q + 4, which is not divisible by 3
½
n = 3q + 1, which is not divisible by 3
n + 2 = 3q + 1 + 2 = 3q + 3, which is divisible by 3
n + 4 = 3q + 1 + 4 = 3q + 5, which is not divisible by 3
½
Case III.
When
n = 3q + 2, which is not divisble by 3
n + 2 = 3q + 2 + 2 = 3q + 4, which is not divisible by 3
n + 4 = 3q + 2 + 4 = 3q + 6, which is divisible by 3
Case I, II and III ⇒ One and only one out of n, n + 2 or n + 4 is divisible by 3.
½
½
SECTION D
6. The greatest number of cartons is the HCF of 144 and 90
144 = 24 × 32
90 = 2 × 32 × 5
HCF = 2 × 32 = 18
∴ The greatest number of cartons = 18.
SUMMATIVE ASSESSMENT
1
1
1
1
WORKSHEET-9
SECTION A
1. The product of a non-zero rational number and an irrational number is always irrational.
2. The product of two irrational numbers is always a non zero number.
1
1
SECTION B
3.
18018
2
9009
a
3003
3
1001
b
143
c
9009
=3
3003
1001
b =
=7
143
143 = 11 × 13, so c = 11 or 13
d = 13 or 11.
a =
Since
S O L U T I O N S
d
½
½
½
½
P-11
SECTION C
3. (i) Required number of mintues is the LCM of 18 and 12.
18 = 2 × 32
and
12 = 22 × 3
∴
LCM of 18 and 12 = 22 × 32 = 36
Hence, Ravish and Priya will meet again at the starting point after 36 minutes.
(ii) LCM of numbers
(iii) Healthy competition is necessary for personal dvelopment and progress.
1
1
1
1
SECTION D
5. Any positive integer is of the form 2q or 2q + 1, for some integer q.
∴ When
n = 2q
n2 – n = (2q)2 – 2q = 2q(2q – 1)
= 2m, when m = q(2q – 1)
Which is divisible by 2.
When
n = 2q + 1
n2 – n = (2q + 1) (2q + 1 – 1)
= 2q (2q + 1)
= 2m, when m = q(2q + 1)
Which is divisible by 2.
Hence, n2 – n is divisible by 2 for every positive integer n.
SUMMATIVE ASSESSMENT
1
1
1
1
WORKSHEET-10
SECTION A
1. (C) It is clear that 0·102003000..........is non-terminating non-repeating decimal, so it cannot be
expressed in the form of
p
·
q
1
2. (C) If n is an odd number, then (n2 – 1) is divisible by 8.
1
SECTION B
3. No.
15 does not divide 175.
LCM is exactly divisible by their HCF.
∴ Two numbers cannot have their HCF as 15 and LCM as 175.
1
1
SECTION C
4. By Euclid’s division algorithm,
⇒
P-12
510
92
50
42
8
HCF (510, 92)
92
510
LCM
HCF × LCM
Product of 2 numbers
HCF × LCM
= 92 × 5 + 50
= 50 × 1 + 42
= 42 × 1 + 8
=8×5+2
= 2 × 4 + 0.
=2
= 22 × 23
= 2 × 3 × 5 × 17
= 22 × 23 × 3 × 5 × 17 = 23460
= 2 × 23460 = 46920
= 510 × 92
= Product of two numbers.
M A T H E M A T I C S --
1
1
1
X
T E R M – 1
SECTION D
5. Let 3 − 5 is a rational number
p
3− 5 = q, q ≠ 0
∴
p
3q − p
5 =3– q =
q
∴
3q − p
is rational number.
q
1
5 is a irrational number.
Since irrational number cannot be equal to rational number.
∴ Our assumption is wrong.
1
∴ 3 − 5 is an irrational number.
1
But
SUMMATIVE ASSESSMENT
1
WORKSHEET-11
SECTION A
1. Since LCM of a and b = p therefore LCM of 3a and 2b = 3 × 2 × p = 6p.
1
2.
1
3.
5 + 2 − 7 is an irrational number.
3
is the rational number between
2
2 and
3.
1
SECTION B
4.
(7 × 13 × 11) + 11 = 11 (7 × 13 + 1)
= 11 (91 + 1)
= 11 × 92 = 11 × 2 × 2 × 23
which is a composite number (more than one prime factors)
and
(7 × 6 × 5 × 4 × 3 × 2 × 1) + 3 = 3 (7 × 6 × 5 × 4 × 2 × 1 + 1)
= 3 × (1681)
= 3 × 41 × 41
which is a composite number (more than one prime factors)
1
1
SECTION C
5. We have
117 = 65 × 1 + 52
65 = 52 × 1 + 13
52 = 13 × 4 + 0
Hence,
∴
⇒
∴
Now,
S O L U T I O N S
1
HCF = 13
65m – 117 = 13
65m = 117 + 13 = 130
130
=2
65
65 = 13 × 5
117 = 32 × 13
LCM = 13 × 5 × 32 = 585
m =
1
1
P-13
SECTION D
6.
n3 – n = n(n2 – 1) = n(n + 1) (n – 1) = (n – 1) n(n + 1)
= product of three consecutive positive integers.
Now, we have to show that the product of three consecutive positive integers is divisible by 6.
We know that any positive integer a is of the form 3q, 3q + 1 or 3q + 2 for some integer q.
Let a, a + 1, a + 2 be any three consecutive integers.
½
Case I. If
a = 3q.
a(a + 1) (a + 2) = 3q(3q + 1) (3q + 2)
= 3q (even number, say 2r) = 6qr,
(Q Product of two consecutive integers (3q + 1) and (3q + 2) is an even integer)
which is divisible by 6.
1
Case II. If
a = 3q + 1.
∴
a(a + 1) (a + 2) = (3q + 1) (3q + 2) (3q + 3)
= (even number say 2r) (3) (q + 1)
= 6 r (q + 1),
1
which is divisible by 6.
Case III. If
a = 3q + 2.
∴
a(a + 1) (a + 2) = (3q + 2) (3q + 3) (3q + 4)
= multiple of 6 for every q
= 6r (say),
1
which is divisible by 6.
Hence, the product of three consecutive integers is divisible by 6.
½
FORMATIVE ASSESSMENT
WORKSHEET-12
Objective Type Questions
1. (D)
2. (C)
3. (B)
4. (B)
5. (B)
Oral Questions
1323
1323
=
·
(63 × 352 ) 23 × 33 × 52 × 72
In the denominator 3 and 7 prime factors are involved hence the number is non-terminating.
2. Product of numbers = H.C.F. × L.C.M.
1.
H.C.F. =
480 × 672
= 96.
3360
3. H.C.F. of 616 and 32 will be the answer
616 = 2 × 2 × 2 × 11 × 7
32 = 2 × 2 × 2 × 2 × 2
H.C.F. = 2 × 2 × 2 = 8
Fill in the blanks
1. Algorithm
2. Rational
3. Lemma
4. Every composite number can be expressed as a product of primes and this decomposition is unique,
apart from the order in which the prime factors occur.
5. Non-terminating.
6. 2 and 5
7. Unique
8. Euclid’s Division Algorithm
9. Product.
●●
P-14
M A T H E M A T I C S --
X
T E R M – 1
CHAPTER
Unit II
2
Algebra
Polynomials
SUMMATIVE ASSESSMENT
WORKSHEET-13
SECTION A
1. From the graph it is clear that the curve cuts the x-axis at four places, so number of zeroes is 4. 1
2. At the x-axis y = 0, so point on the x-axis is given by 2x – 5 = 0 ⇒ x =
5 
So, required point is  , 0 ·
2 
5
2
1
SECTION B
3. Since a and b are the zeroes of the polynomial, then
α+β =–
Coefficient of x
Coefficient of x2
 −1 
α + β = –
=1
 1 
⇒
Given
α–β = 9
From (1) and (2) α = 5, β = – 4
αβ =
Again
⇒
...(1) 1
...(2)
Constant
Coefficient of x2
αβ = – k
(5) (– 4) = – k
k = 20
1
SECTION C
f(x) = 4x2 + 4x – 3
3.
f
FG 1 IJ
H 2K
= 4
FG 1 IJ + 4FG 1 IJ − 3
H 4 K H 2K
=1+2–3=0
FG 3 IJ
H 2K
f −
= 4
FG 9 IJ + 4FG − 3 IJ − 3
H 4 K H 2K
=9–6–3=0
S O L U T I O N S
P-15
1 3
∴ ,− are zeroes of polynomial 4x2 + 4x – 3
2 2
1
−4
1 3
coeff. of x
=–
− =–1=
4
2 2
coeff. of x 2
Sum of zeroes =
1
æ 1 ö æ 3 ö -3 constant term
=
Product of zeroes = ç
ç ÷÷ çç- ÷÷ =
è2ø è 2 ø 4
coeff. of x 2
∴Relation between zero and coeff. of polynomial is verfied.
1
SECTION D
2 x 2 + 3x + 5
x 2 + 3x − 4 2 x 4 + 3x3 − 12 x 2 + 28 x − 20
5.
2 x 4 + 6 x3 − 8 x2
−
−
+
− 3x − 4 x 2 + 28 x
2
3x3 + 9 x 2 − 12 x
5 x 2 + 16 x − 20
5 x 2 + 15 x − 20
− −
+
x
Hence quotient = 2x2 + 3x + 5; and remainder = x
Hence, x should be substract from f (x) to make it exactly divisible by
1
x2
SUMMATIVE ASSESSMENT
+ 3x – 4.
1
WORKSHEET-14
SECTION A
1. 2x2 + 5x + 1
Since α and β are the zeroes of the polynomial, Therefore
(α + β) + αβ = –
5 1
+
2 2
−5 + 1
−4
=
= – 2.
2
2
α = – 5 and β = 4
Polynomial = (x – α) (x – β)
= (x + 5) (x – 4)
= x2 – 4x + 5x – 20
= x2 + x – 20.
1
=
2. Given
So,
1
SECTION B
3.
Sum of zeroes = a + b = –
⇒
coeff. of x
=–a
coeff. of x 2
1
2a+b =0
constant
=b
coeff. of x 2
a = 1 then b = – 2
Product of zeroes = ab =
⇒
P-16
M A T H E M A T I C S --
1
X
T E R M – 1
SECTION C
p(x) = 2x3 – 11x2 + 17x – 6
p(2) = 2(2)3 – 11(2)2 + 17(2) – 6
4. (i) Now
= 16 – 44 + 34 – 6
= 50 – 50 = 0
1
Hence 2 is the zero of p(x).
p(3) = 2 (3)3 – 11(3)2 + 17(3) – 6
(ii) Again
= 54 – 99 + 51 – 6
1
= 105 – 105 = 0
Hence, 3 is the zero of p(x).
3
2
 1
 1
 1
 1
p   = 2   – 11   + 17   – 6
2
 2
 2
2
(iii) Again
=
1 11 17
–
+
–6
4
4
2
1
=0
Hence,
1
is also the zero of p(x).
2
SECTION D
p(x) = x3 – 10x2 + 31x – 30
5. Let
Since 2 and 3 are zeroes of p(x).
1
So (x – 2) and (x – 3) are the factors of given polynomial.
i.e.,
(x – 2) (x – 3) =
x2
– 5x + 6 is a factor
x–5
2
3
x – 5x + 6) x – 10x2 + 31x – 30
x3 – 5x2 + 6x
– +
–
– 5x2 + 25x – 30
– 5x2 + 25x – 30
+
–
+
0
1
1
Hence other zero is (x – 5) i.e., x = 5.
SUMMATIVE ASSESSMENT
WORKSHEET-15
SECTION A
1. The maximum number of zeroes that a polynomial of degree 3 can have is 3.
2. If 1 is the zero of the polynomial
x2
⇒
(1)2 +
x2
+ kx – 5, then
+ kx – 5 = 0
k(1) – (5) = 0
⇒
1+k–5 =0
⇒
k–4 =0
⇒
S O L U T I O N S
1
k = 4.
1
P-17
SECTION B
3. Let
3 x2 – 8x + 4 3
p(x) =
=
3 x2 – 6x – 2x + 4 3
=
3 x (x – 2 3 ) – 2 (x – 2 3 )
1
1
= ( 3 x – 2) (x – 2 3 )
SECTION C
p(x) = x2 + 7x + 9
4.
q(x) = x + 2
r(x) = – 1
g(x) = ?
1
p(x) = g(x)q(x) + r(x)
x2 + 7x + 9 = g(x) (x + 2) – 1
x 2 + 7 x + 10
x+2
( x + 2)( x + 5)
=
( x + 2)
g(x) = x + 5.
1
g(x) =
1
SECTION D
p(x) = 2x2 + 5x + k
5.
Sum of zeroes = –
coeff. of x
coeff. of x 2
−5
=α+β=
2
Product of zeroes =
1
constant
coeff. of x 2
= αβ =
k
2
According to question,
21
4
21
(α + β)2 – 2αβ + αβ =
4
α2 + β2 + αβ =
⇒
⇒
⇒
∴
Hence,
P-18
FG −5 IJ
H 2K
1
[Q (α + β)2 = α2 + β2 + 2αβ]
2
k
21
=
2
4
25 21
k
−
=
4
4
2
k
1 =
⇒ k=2
2
–
1
1
k =2
M A T H E M A T I C S --
X
T E R M – 1
SUMMATIVE ASSESSMENT
WORKSHEET-16
SECTION A
Given α =
1.
So,
3 and β = – 3
Polynomial = (x – α) (x – β)
= (x – 3 ) (x + 3 )
= x2 – 3.
2. If 3 is the one zero of the polynomial 2x2 + kx – 15, then
2x2 + kx – 15 = 0
2(3)2 + k(3) – 15 = 0
18 + 3k – 15 = 0
3k + 3 = 0
⇒
k+1 =0 ⇒ k=–1
Now,
Polynomial = 2x2 – x – 15
⇒
= 2x2 – 6x + 5x – 15
⇒
= 2x(x – 3) + 5(x – 3)
⇒
= (2x + 5) (x – 3)
5
Hence, other zero is – .
2
1
1
SECTION B
3. Q
Hence, zeroes are 0 and 2.
f (x) = x2 – 2x
= x(x – 2)
= x = 0 or x = 2
1
1
SECTION C
4. According to question,
So,
Sum of zeroes =
21
8
Product of zeroes =
5
16
1
Quadratic polynomial = x2 – (Sum of zeroes) x + Product of zeroes
= x2 –
=
FG 21 IJ x + 5
H 8 K 16
1
(16x2 – 42x + 5)
16
= (16x2 – 42x + 5)
1
.
16
1
1
SECTION D
5. Given the polynomial
f(x) = ax2 – 5x + c
Let the zeroes of f (x) are α and β, then according to question
Sum of zeroes, (α + β) = Product of zeroes, (αβ) = 10
Now,
S O L U T I O N S
α+β =–
FG IJ
H K
−5
coeff. of x
2 = –
a
coeff. of x
1
P-19
⇒
10 =
⇒
a=
and
αβ =
⇒
⇒
10 =
c=
Hence,
a=
+5
a
1
2
constant
c
=
a
coeff. of x 2
2c
5
1
and c = 5.
2
SUMMATIVE ASSESSMENT
1
1
1
WORKSHEET-17
SECTION A
f (x) = ax2 + bx + c
1. (A) Let
Sum of zeros = α + β =
coeff. of x
b
=–
a
coeff. of x 2
But one zero is 0
α+0 =–
i.e.,
b
a
b
·
a
2. (A) If a is added in the polynomial, then
p(x) = x2 – 5x + 4 + a
Given 3 is the zero
∴
p(3) = 0
⇒
(3)2 – 5(3) + 4 + a = 0
⇒
–6+4+a =0
⇒
a =2
1
Hence other zero is –
1
SECTION B
3. Since, – 1 is a zero of polynomial
∴
⇒
⇒
⇒
⇒
Hence,
p(x) = kx2 – 4x + k, then p(–1) = 0
=0
k+4+k =0
2k + 4 = 0
2k = – 4
k =–2
k = – 2.
1
k(–1)2 – 4(–1) + k
1
SECTION C
3
is a zero of the polynomial 2x3 + 9x2 – x – b, then
2
2
2
3
 3
 3
2 −  + 9 −  + − b = 0
2
2
2




4. If (2x + 3) is a factor then –
−
⇒
Hence,
P-20
27 81 6
+
+ −b =0
4
4 4
60
b =
= 15
4
b = 15.
M A T H E M A T I C S --
1
1
1
X
T E R M – 1
SECTION D
5
and –
3
Therefore,
5. Since
5
are two zeroes of f (x).
3

5
5
 2 5 1
2
 x −
  x +
 =  x −  = (3 x − 5) is a factor of f (x)
3
3
3 3


Also, (3x2 – 5) is a factor of f (x)
Let us now divide f (x) by (3x2 – 5)
We have,
x2 + 2x + 1
3x 2 − 5 3x 4 + 6 x 3 − 2 x 2 − 10 x − 5
3x 4 + 0 x3 − 5 x 2
− −
+
6 x 2 + 3x 2 − 10 x − 5
6 x3 + 0 x 2 − 10 x
3x 2 − 5
3x 2 − 5
− +
0
By division alorithm, we have
3x4 + 6x3 – 2x2 – 10x – 5 = (3x2 – 5) (x2 + 2x + 1)
= ( 3 x + 5) ( 3 x − 5) ( x + 1)2
5
,
3
Hence, the zeros of f (x) are –
5
, – 1 and – 1
3
SUMMATIVE ASSESSMENT
WORKSHEET-18
SECTION A
1. Let
Let other zero be k, then
∴
2.
f (x) = x2 – 5x – 6
æ -5 ö
6+k =– ç
ç ÷÷ = 5
è1ø
k =5–6=–1
1
p(x) = ax2 + bx + c
Let zeroes of polynomial are α and − α, then
b
α + (− α) = –
a
⇒
0 =–
b
⇒ b=0
a
1
SECTION B
p(x) = ax2 + bx + c
1
Let α and
be the zeroes of p(x), then
α
1 c
Product of zeroes, α ×
=
α a
So, required condition is c = a.
S O L U T I O N S
3.
1
1
P-21
SECTION C
p(x) = 5x3 + 8x – 4
= 5x2 + 10x – 2x – 4
= 5x(x + 2) – 2(x + 2)
= (x + 2)(5x + 2)
4. Let
Hence, zeroes of the quadratic polynomial 5x2 + 8 x – 4 are – 2 and
Verification :
1
2
·
5
2
−8
=
5
5
 2
−4
Product of zeroes = (– 2) ×   =
5
5
Sum of zeroes = – 2 +
coeff. of x
−8
=
5
coeff. of x 2
constant
−4
Product of zeroes =
2 =
5
coeff. of x
Thus relationship is verified.
1
Again sum of zeroes =
1
SECTION D
3x3 + 4x2 + 5x – 13 = (3x + 10) g(x) + (16x – 43)
5.
⇒
3
1
2
3 x + 4 x − 11 x + 30
= g(x)
3 x + 10
1
3 x + 10 3 x3 + 4 x2 − 11 x + 30 ( x2 − 2 x + 3
3x3 + 10 x2
(–)
(–)
– 6 x 2 − 11 x
– 6 x 2 – 20 x
( + ) (+ )
1
9 x + 30
9 x + 30
(–)
(–)
0
Hence,
g(x) =
x2
1
– 2x + 3.
SUMMATIVE ASSESSMENT
WORKSHEET-19
SECTION A
4x2 – 12x + 9 = 0
1.
4x2
⇒
⇒
⇒
– 6x – 6x + 9 = 0
2x(2x – 3) – 3(2x – 3) = 0
(2x – 3)(2x – 3) = 0
3
3
x =
and
2
2
Hence, zeroes of the polynomial are
P-22
3 3
,
2 2
M A T H E M A T I C S --
X
T E R M – 1
p(x) = 3x2 – kx + 6
coefficient of x
Sum of the zeroes = 3 = –
coefficient of x2
2.
⇒
3 =–
⇒
k = 9.
(– k)
3
SECTION B
3.
Quadratic polynomial = ( x − 15) ( x + 15)
= x 2 − ( 15)2
= x2 – 15.
SECTION C
2x + 2
3 x 2 − 2 x + 1 6 x 3 + 2 x2 − 4 x + 3
4.
6 x3 – 4 x 2 + 2x
–
+
−
6 x2 − 6 x + 3
6 x2 – 4 x + 2
–
+
−
– 2x + 1
Quotient = 2x + 2; Remainder is – 2x + 1
p(x) = g(x) q(x) + r(x)
= (3x2 – 2x + 1) (2x + 2) + (– 2x + 1)
= 6x3 – 4x2 + 2x + 6x2 – 4x + 2 – 2x + 1
=
6x3
+
2x2
– 4x + 3.
1
1
Verified. 1
SECTION D
x2 – 2 5 x – 15
5.
x–
5
) x3 – 3
x3 –
– +
5 x2 – 5x + 15 5
2
5 x2
– 2 5 x2 – 5x
– 2 5 x2 + 10x
+
–
– 15x + 15 5
– 15x + 15 5
+
–
0
Now,
2
x2 – 2 5 x – 15 = x2 – 3 5 x + 5 x – 15
= x (x – 3 5 ) + 5 (x – 3 5 )
= (x + 5 ) (x – 3 5 )
All the zeroes are
5,–
S O L U T I O N S
5,3 5.
2
P-23
SUMMATIVE ASSESSMENT
WORKSHEET-20
SECTION A
f(x) = x2 – 7x – 8
–7
=7
Let other zero be k, then
–1+k =–
1
k =8
1
2. From the graph it is clear that the curve y = f (x) cuts the x-axis at two places between – 2 and 2.
∴ Required number of zeroes = 2.
1
1.
FG IJ
H K
SECTION B
3. Let,
f (x) = 2x2 – 5x – 3
Let the zeroes of polynomial are α and β, then
5
3
, product of zeroes αβ = –
2
2
According to question, zeroes of x2 + px + q are 2α and 2β
Sum of zeroes α + β =
Sum of zeroes = –
coeff. of x
−p
2 =
1
coeff. of x
= 2α + 2β = 2(α + β) = 2 ×
Product of zeroes =
constant
q
2 =
1
coeff. of x
5
=5 ⇒ p=–5
2
1
FG 3 IJ = – 6
H 2K
= 2α × 2β = 4αβ = 4 −
∴
1
p = – 5 and q = – 6.
SECTION C
p(x) = 3x2 – 4x – 7 and α and β are its zeroes.
4.
æ 4ö 4
- ÷=
Sum of zeroes α + β = -ç
è 3ø 3
1
æ 7ö
- ÷
Product of zeroes αβ = ç
è 3ø
For new polynomial
Sum
1 1
a+b
4/3
4
+
==
=
a b
ab
-7 / 3
7
1
1 1
1
1
3
´
==
=
a b
ab
-7 / 3
7
The required polynomial = x2 – (sum of zeroes)x + product of zeroes
product
æ 3ö
2 æ 4ö
- ÷x +ç- ÷
= x -ç
è 7ø è 7ø
=
1
(7 x2 + 4 x - 3)
7
2
= (7 x + 4 x - 3)
P-24
1
7
M A T H E M A T I C S --
½
X
T E R M – 1
SECTION D
x4 – 4x + (8 – k)
5.
x2
– 2x + k) x4 – 6x3 +
x4 – 2x3 +
– +
–
16x2 –
kx2
25x + 10
– 4x3 + (16 – k)x2 –
– 4x3 +
8x2 –
+
–
+
25x + 10
4kx
1
(8 – k)x2 – (25 – 4k)x + 10
(8 – k)x2 – (16 – 2k)x + (8k – k2)
–
+
–
(2k – 9)x + (10 – 8k + k2)
2
Given, remainder = x + a
⇒
2k – 9 = 1 ⇒ k =
10
=5
2
a = 10 – 8k + k2 = 10 – 40 + 25 = – 5.
and
SUMMATIVE ASSESSMENT
1
WORKSHEET-21
SECTION A
1. From the graph it is clear that curve cut the x-axis at three places. So number of zeroes is 3.
2. Let
f (x) = ax2 + bx + c
Now product of zeroes =
1
constant
c
2 =
a
coeff. of x
Sum of zeroes = –
coeff. of x
b
=–
a
coeff. of x 2
1
SECTION B
3. Since α and β are the zeroes of the polynomial
f(x) = x2 – px + q
∴
α + β = p, αβ = q
Now
1 1
α+β p
+
=
=
α β
αβ
q
1
1
SECTION C
4. Let α and β are the zeroes of the polynomial, then as per question
β = 7α
8
∴
α + 7α = 8α = – −
3
1
⇒
α =
3
FG IJ
H K
and
⇒
S O L U T I O N S
½
½
2k + 1
3
2
k
+1
7α2 =
3
α × 7α =
P-25
⇒
F 1I
7 G J
H 3K
⇒
7×
2
=
2k + 1
3
1
1
2k + 1
=
9
3
7
– 1 = 2k
3
2
= k.
3
⇒
⇒
1
SECTION D
2 ) = x2 – 2
2 ) (x –
5. Polynomial g(x) = (x +
x2 – 2 ) x4 – 5x3 + 2x2 + 10x – 8 ( x2 – 5x + 4
–
x4
– 2x2
+
– 5x3 + 4x2 + 10x
– 5x3
+
+ 10x
–
4x2 – 8
4x2 – 8
(–) (+)
2
0
x2 – 5x + 4 = (x – 4) (x – 1)
1
Other zeroes are 4 and 1.
1
SUMMATIVE ASSESSMENT
WORKSHEET-22
SECTION A
1. Given,
p (x) = x2 – 5x + 6
Sum of zeroes = α + β = –
Product of zeros = αβ =
FG −5 IJ = 5
H1K
6
=6
1
Now α+ β – 3αβ = 5 – 3(6)
1
= 5 – 18 = – 13.
2.
Required polynomial = ( x − 2) ( x − 2 2)
= x2 − 2 2 x − 2 x + ( 2) (2 2)
1
= x2 − 3 2 x + 4.
P-26
M A T H E M A T I C S --
X
T E R M – 1
SECTION B
f (x) = 2x2 – 7x + 3
3.
Sum of roots = p + q = –
Product of roots = pq =
coeff. of x
=–
coeff. of x 2
3
constant
=
2
2
coeff. of x
FG −7 IJ = 7
H 2K 2
½
½
We know that
(p + q)2 = p2 + q2 + 2pq
p2 + q2 = (p + q)2 – 2pq
⇒
FG 7 IJ
H 2K
=
2
−3=
49 3 37
− =
.
4 1
4
½
½
SECTION C
4.
2 x 2 +2 x -1
4 x 2 + 3 x - 2 8 x 4 + 14 x3 - 2 x 2 + ax + b
–
8x4 + 6x3 - 4x 2
–
+
8 x3 + 2 x 2 + ax
–
8 x3 + 6 x2 - 4 x
–
+
- 4 x 2 + ( a + 4) x + b
-4 x 2 - 3 x + 2
+
+
–
(a + 7) x + b - 2
For exact division, remainder is zero, then
(a + 7) x + b – 2 = 0
a + 7 = 0, b – 2 = 0
a = – 7, b = 2.
2
1
SECTION D
5.
2 x2 + 2 x − 1
4 x + 3 x − 2 8 x + 14 x − 2 x2 + 8 x − 12
2
4
3
8x4 + 6x3 – 4x2
– –
+
8x3 + 2x2 + 8x
8x3 + 6x2 – 4x
– –
+
– 4x2 + 12x – 12
– 4x2 – 3x + 2
+
+
–
1
1
15x – 14
1
Subtract 15x – 14 or add – 15x + 14 for exact division.
1
S O L U T I O N S
P-27
SUMMATIVE ASSESSMENT
WORKSHEET-23
SECTION A
1. Let α = 3, β = − 3 be the given zeros and γ be the third zero. Then,
 4
α + β + γ = – − 
 1
⇒
Hence third zero is 4.
3− 3+γ =4
x = 4.
1
2. If α, β and γ are the zeros of cubic polynomials f(x)
f (x) = k{x3 – (α + β + γ) x2 + (αβ + βγ + γα) x – αβγ}
Then
Here, α + β + γ = 2, αβ + βγ + γα = – 7 and αβγ = – 14
∴
f(x) = k(x3 – 2x2 – 7x + 14)
1
where k is any non-zero real number.
SECTION B
p(x) = 3x2 + 11x – 4
3. Let
= 3x2 + 12x – x – 4
⇒
= 3x(x + 4) – 1(x + 4)
⇒
= (3x – 1) (x + 4)
So, zeroes are :
Now,
m =
½
1
and n = – 4.
3
FG IJ
H K
½
1
m n
−4
3
+
+
n m = −4
1
3
1
145
= − 12 – 12 = –
.
12
FG IJ
H K
½
½
SECTION C
4.
p(x) = 6y2 – 7y + 2
7
 7
α + β = – −  =
6
 6
αβ =
2
6
1 1
α+β 7/6 7
+
=
=
=
α β
αβ
2/6 2
1
1 1
1
× =
=3
α β
αβ
1
7
y+3
2
= 2[2y2 – 7y + 6]
2
The required polynomial = y −
P-28
M A T H E M A T I C S --
1
X
T E R M – 1
5. Since α and β are the zeroes of polynomial 3x2 + 2x + 1.
2
Hence,
α+β = −
3
1
and
αβ =
3
Now for the new polynomial,
1−α 1−β
+
Sum of the zeroes =
1+α 1+β
(1 − α + β − αβ) + (1 + α − β − αβ)
=
(1 + α)(1 + β)
2
2−
2 − 2αβ
3
=
=
1 + α + β + αβ 1 − 2 + 1
3 3
4
Sum of zeroes = 3 = 2
2
3
(1 − α)(1 − β)
1−α 1–β
Product of zeroes =
=
(1 + α)(1 + β)
1+α 1+β
FG
H
=
IJ FG
KH
1
1
1
IJ
K
1 − α − β + αβ 1 − (α + β) + αβ
=
1 + α + β + αβ 1 + (α + β) + αβ
2 1 6
+
3 3 = 3 =3
Product of zeroes =
2 1 2
1− +
3 3 3
1+
1
Required polynomial = x2 – (Sum of zeroes)x + Product of zeroes
Hence,
= x2 – 2x + 3.
1
FORMATIVE ASSESSMENT
WORKSHEET-24
Objective Type Questions
1. (?)
2. (D)
3. (A)
4. (C)
5. (D)
6. (B)
1×6
Fill in the blanks
1.
S. No.
No. of zeros
Zeroes
1.
1
−3
2.
1
2
3.
3
− 4, − 4, 2
4.
3
− 4, − 2, 2
5.
0
—
1×5
Quiz
1.
S O L U T I O N S
P(x) = 3x3 – 2x2 + 6x – 5
P(2) = 3(2)3 – 2(2)2 + 6(2) – 5
= 3 × 8 – 2 × 4 + 12 – 5
= 24 – 8 + 12 – 5
= 36 – 13
P(2) = 23
1
P-29
α+β =
2.
−b
=–2
a
c
3
=
=3
a
1
Hence the quadratic polynomial = x2 – (α + β) x + αβ
= x2 – 2x + 3
αβ =
3.
1
1 1
+
n m
−b
− b a − b − 11 11
m+n
= a =
× =
=
= ·
c
mn
a c
c
−4
4
a
1
4. For the polynomial t2 – 4t + 3
p+q =–
pq =
b
(− 4)
=–
=4
a
1
....(i)
c
3
=
=3
a
1
( p + q)2 − 4 pq
(p – q) =
=
(4)2 − 4 × 3
=
16 − 12
=
Hence
By (i) + (ii)
Hence
and
p–q
2p
p
q
4 = ±2
=+2
= 6 or 2p = 2
= 3 or 1
= 1 or 3
1 1
14
+ − 2 pq +
=0
p q
3
1
14
+ 1 − 2g3g1 +
=0
3
3
1
14
+1 −6 +
=0
3
3
=
1 + 3 − 18 + 14
18 − 18
=
=0
3
3
1
●●
P-30
M A T H E M A T I C S --
X
T E R M – 1
CHAPTER
3
Pair of Linear Equations
in Two Variables
SUMMATIVE ASSESSMENT
WORKSHEET-25
SECTION A
1. For a unique solution
a1
b1
≠
a2
b2
−2
3
≠
k
2
⇒
–4 ≠k
i.e., all real numbrs except – 4.
2. x = 6, y = 1.
i.e.,
1
1
SECTION B
3.
3x – 2y = 4
2x + y = 5
Putting x = 2 and y = 1 in equation (1), we have
L.H.S. = 3 × 2 – 2 × 1 = 4 = R.H.S.
Putting x = 2 and y = 1 in equation (2), we have
L.H.S. = 2 × 2 + 1 × 1 = 5 = R.H.S.
Thus, x = 2 and y = 1 satisfy both the equations of the given system.
Hence x = 2, y = 1 is solution of the given system.
....(1)
....(2)
½
½
1
SECTION C
4. Let the man can finish the work in x days and the boy can finish the same work in y days.
Work done by one man in one day =
1
x
Now, work done by one boy in one day =
1
y
According to question,
2 7
1
+ =
4
x y
1
4 4
+ =
3
x y
S O L U T I O N S
...(1)
...(2)
P-31
Let
1
1
= a and = b, then
y
x
1
4
1
4a + 7b =
3
Multiply eqn. (3) by 2 and substract from it eqn. (4)
1
4a + 14b =
2
1
4a + 4b =
3
–
–
–
1
10b =
6
1
1
=
=
y
60
⇒
y = 60 days.
1
Put b =
in equation (3), we get
60
1
a=
15
...(3)
2a + 7b =
...(4) 1
1
1
1
=
15
x
x = 15 days.
So
⇒
1
SECTION D
5.
x + 3y = 6
...(1)
x
3
6− x
y =
3
6
y
1
0
⇒
0
2
½
2x – 3y = 12
⇒
y =
...(2)
2 x − 12
3
x
0
6
3
y
−4
0
−2
x+
Putting the above points and drawing a line
joing them, we get the graphs of the equations
x + 3y = 6 and 2x – 3y = 12.
1
Clearly, the two lines intersect at point B (6,
0).
Hence, x = 6 and y = 0 is the solution of the
system.
1
Again ∆ OAB is the region bounded by the line
2x – 3y = 12 and both the co-ordinate axes.
1
P-32
½
y
3y
=6
2
1
x'
(0, 2)
(3, 1)
(6, 0)
–3 –2 –1 0
1
2
3
–1
–2
–3
(0, –4) –4
2
A
3
x–
y=
4
5
6
B
x
(3, –2)
12
y'
M A T H E M A T I C S --
X
T E R M – 1
SUMMATIVE ASSESSMENT
WORKSHEET-26
SECTION A
1. The lines represented by the equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are parallel, if
c
b
a1
= 1 ≠ 1 ·
c2
b2
a2
1
2. The lines represented by the equations a1x + b1x + c1 = 0 and a2x + b2y + c2 = 0 are coincident, if
a1
b
c
= 1 = 1 ·
a2
b2
c2
1
SECTION B
3.
2x – y = 2
y = 2x – 2
x + 3y = 15
Substituting the value of y form (i) in (ii), we get
x + 6x – 6 = 15
7x = 21
x =3
From (1),
y =2×3–2=4
∴
x = 3 and y = 4.
...(i)
...(ii)
1
1
SECTION C
4.
(2m – 1)x + 3y – 5 = 0
On comparing with the eqn.
a1x + b1y + c1 = 0
a1 = 2m – 1, b1 = 3, c1 = – 5
3x + (n – 1)y – 2 = 0
On comparing with the eqn.
a2x + b2y + c2 = 0
a2 = 3, b2 = (n – 1), c2 = – 2
For a pair of equations to have infinite number of solutions.
...(1)
½
...(2)
½
a1
c
b
= 1 = 1
a2
c2
b2
⇒
2m − 1
3
5
=
=
3
n−1 2
2(2m –1) = 15 and 5 (n–1) = 6
17
11
m=
,n=
·
4
5
1
1
SECTION D
5. Let the speed of the boat in still water be ‘x’ km/hr and speed of the stream be ‘y’ km/hr. Speed of
the boat during upstream be (x – y) km/hr and during downstream be (x + y) km/hr.
∴
S O L U T I O N S
30
28
21
21
=7;
=5
+
+
x− y x+ y
x− y x+ y
1
P-33
1
1
= a and
= b, we get
x− y
x+ y
30a + 28b = 7
21a + 21b = 5
Multiplying eqn. (2) by 21 and eqn. (2) by 28 and then subtracting, we get
630a + 558b = 147
588a + 558b = 140
–
– –
Let
...(1)
...(2)
...(3)
...(4)
1
On subtracting,
42a = 7
7 1
⇒
a=
=
42 6
Putting this value of a in eqn, (1), we get
28b = 7 – 30a = 7 – 30 ×
Now,
a=
1
1
=2⇒b=
6
14
1
1
=
⇒x–y=6
x− y 6
...(5) 1
1
1
=
⇒ x + y = 14
x+ y
14
Solving (5) and (6), we get
x = 10, y = 4
Hence, speed of the boat in still water = 10 km/hr and speed of the stream = 4 km/hr.
b=
SUMMATIVE ASSESSMENT
...(6)
WORKSHEET-27
SECTION A
1.
x–y=
and
x+y=
Adding both the equations,
2x =
Put the value of x in the eqn. (2), we get
3+y=
Hence,
a=
2. ad ≠ bc ⇒
2
4
6 ⇒ x=3
...(1)
...(2)
4 ⇒ y=1
3, b = 1.
1
a
b
≠ . Hence, the pair of linear equations has unique solution.
c
d
1
SECTION B
3. Yes, for justification we have For the equation
2x + 3y = 9
a1 = 2, b1 = 3 and c1 = – 9
and for the equation,
4x + 6y = 18
a2 = 4, b2 = 6 and c2 = – 18
½
½
c
–9 1
2 1 b1 3 1
a1
=
= ,
= = and 1 =
=
c2 –18 2
4 2 b2 6 2
a2
b1 c1
a1
=
From above it is clear that
=
b2 c2
a2
Hence system is consistent and dependent.
Here
½
½
SECTION C
4. Given equations are 2x + 3y = 7 and 2αx + (α + β) y = 28.
We know that the condition for a pair of linear equations to be consistent and having infinite
b1 c1
a1
number of solutions is
=
b2 = c2
a2
P-34
M A T H E M A T I C S --
X
T E R M – 1
1
2 = 3 = 7
a + b 28
2α
I
II
III
⇒
2
2α
α
3
α +β
α +β
β
β
β
From I and III,
⇒
From II and III,
⇒
Hence α = 4, and β = 8.
=
=
=
=
=
=
=
7
28
4
7
28
12
12 – α
12 – 4
8
½
½
1
SECTION D
5. 2x + 3y = 12 ⇒ y =
12 − 2 x
3
x
0
6
3
y
4
0
2
x–y=1 ⇒y=x–1
x
0
1
3
y
−1
0
2
Putting the above points and drawing a line joining them, we get the graph of the equations
2x + 3y = 12 and x – y = 1.
y
5
B4
3
(0, 4)
P
M 2
1
(6, 0)
(1, 0)
x'
1 2 3
A (0, – 1)
–5 –4 –3 –2 –1
–1
–2
x–y=1
(3, 2)
4
5
6
7
8
x
2x + 3y = 12
1
–3
–4
y'
Clearly, the two lines interesect at point P (3, 2).
Hence, x = 3 and y = 2 is the solution of the system.
Area of shaded region = Area of ∆ PAB
1
1
=
× base × height =
× AB × PM
2
2
1
=
×5×3
2
= 7.5 square unit.
S O L U T I O N S
1
1
P-35
SUMMATIVE ASSESSMENT
WORKSHEET-28
SECTION A
1. The equation of one line
4x + 3y = 14.
We know that if two lines a1x + b1y + c = 0 and a2x + b2y + c2 = 0 are parallel, then
c1
a1
b
= 1 ≠
c2
a2
b2
1
Hence, second parallel line is – 12x = 9y.
2. Given,
Here,
3x + 4y = 7 and 3x + 4y = 16
c
b
a1
= 1 ≠ 1
c2
b2
a2
1
Hence, pair of linear equations has no solution.
SECTION B
3.
99x + 101y = 499
101x + 99y = 501
Adding equation (1) and (2), we get
200x + 200y = 1000
⇒
x+y=5
Subtracting equation (2) from equation (1)
– 2x + 2y = – 2
⇒
x–y=1
Adding equations (3) and (4)
2x = 6 ⇒ x = 3
Put the value of x in equation (3), we get y = 2.
...(1)
...(2)
...(3)
½
...(4) ½
½
½
SECTION C
4. (i) Let the fixed charge of taxi be Rs. x per km and the running charge be ` y per km.
According to the question,
x + 10y = 75
x + 15y = 110
Subtracting equation (2) from equation (1), we get
– 5y = – 35
⇒
y =7
Putting y = 7 in equation (1), we get x = 5
...(1)
...(2) 1
1
∴ Total charges for travelling a distance of 25 km
= x + 25y
= ` (5 + 25 × 7)
= ` (5 + 175)
1
= ` 180
(ii) Pair of linear equations in two variables.
½
(iii) We should always be justified in our dealings.
½
P-36
M A T H E M A T I C S --
X
T E R M – 1
SECTION D
5. Let the sum of the ages of the 2 children be x and the age of the father be y years.
∴
y = 2x i.e., 2x – y = 0
and
20 + y = x + 40
x – y = – 20
Subtracting (2) from (1), we get
x = 20
From (1),
y = 2x = 2 × 20 = 40
y = 40
Hence, the age of the father = 40 years.
SUMMATIVE ASSESSMENT
...(1) 1
...(2) 1
1
1
WORKSHEET-29
SECTION A
1.
⇒
– 3x + 4y = 7
...(1)
9
21
x – 6y +
=0
2
2
–3x + 4y = 7
...(2)
a1
b1
c1
=
=
a2
b2
c2
From eqns. (1) and (2),
2.
Hence, pair of linear equations has infinite number of solutions.
Put x = 2 and y = 3 in 2x – 3y + a = 0 and 2x + 3y – b + 2 = 0, we get
4–9+a =0⇒a=5
4 + 9 + 2 – b = 0 ⇒ b = 15
From above it is clear that 3a = b.
1
1
SECTION B
3. Given equations are :
4x + py + 8 = 0
2x + 2y + 2 = 0
...(1)
...(2)
The condition of unique solution,
Hence,
∴
a1
b1
≠
a2
b2
4
p
2
p
≠
or
≠
2
2
1
2
p ≠ 4.
1
1
SECTION C
4. (i) Let the monthly rent of the house be ` x and the mess charges per head per month be ` y.
According to the given conditions,
x + 2y = 3900
...(i)
x + 5y = 7500
...(ii) ½
Subtracting equation (ii) from equation (i), we get
– 3y = – 3600
3600
⇒
y =
= 1200
1
3
Putting this value of y in eqn. (i), we get
x + 2400 = 3900
⇒
x = 3900 – 2400 = 1500
Hence, monthly rent = ` 1500
1
S O L U T I O N S
P-37
(ii) Mess charge per head per month = ` 1200
(iii) Pair of linear equations in two variables.
(iv) Monitering is always good to control our of extreme habits.
½
½
½
SECTION D
5.
2(3x – y) = 5xy
...(1)
2(x + 3y) = 5xy
...(2)
6 2
−
=5
y x
...(3)
2 6
+
=5
y x
...(4) 1
Divide eqns. (1) and (2) by xy,
Putting
1
1
= b, then equations (3) and (4) become
= a and
y
x
6a – 2b = 5
...(5)
2a + 6b = 5
....(6) 1
Multiplying eqn. (5) by 3 and then adding with eqn. (6), we get
20a = 20 ⇒ a = 1
Putting this value of a in eqn. (5), we get
b=
1
2
Now
1
= a = 1 ⇒ y =1
y
and
1
1
=b=
⇒ x = 2.
x
2
SUMMATIVE ASSESSMENT
1
1
WORKSHEET-30
SECTION A
1. From the options it is clear that x = 2, y = 3 is a solution of the linear equation 2x + 3y –13 = 0.
2. at y-axis x = 0, then
0–y =8 ⇒ y=–8
Hence, point of interrection is are (0, –8).
1
SECTION B
3. Pair of equations
kx – 4y = 3
6x – 12y = 9
Condition for infinite solutions :
a1
b1 c1
=
=
a2
b2 c2
⇒
P-38
1
k
−4 3
=
=
6
−12 9
k = 2.
M A T H E M A T I C S --
1
X
T E R M – 1
SECTION C
4.
⇒
x – 5y = 6
x = 6 + 5y
y
0
−1
−2
x
6
1
−4
1
2x – 10y = 12 ⇒ x = 5y + 6
y
0
−1
−2
x
6
1
−4
1
y
1
x'
x – 5y = 6
(6, 0)
–5 –4 –3 –2 –1 0
–1
12
0y =
1
–2
–
x
2
2)
–3
(–4, –
1
2
3
4
5
6
x
7
(1, –1)
y'
Since the lines are co-incident, so the system of linear equations is cosistent with infinite solutions.
1
SECTION D
2x + 3y = 12 ⇒
5.
12 − 2 x
3
x–y=1 ⇒ x–1
x
0
6
3
x
0
1
3
y
4
0
2
y
−1
0
2
1
Putting the above points and drawing a line joining them, we get the graphs of equations 2x + 3y
= 12 and x – y – 1 = 0.
y
2x
(0, 4) 4
3
x–y–1=0
+3
y=
C
2
12
(3, 2)
1
x'
(6, 0)
B
A
–3 –2 –1 0
(0, –1) –1
1 2
(1, 0)
–2
3
4
5
6
x
1
–3
y'
Clearly, the two lines interesct at point (3, 2), Hence, x = 3 and y = 2 is the required solution. 1
∆ ABC is the region between the two lines represented by the above equations and the x-axis. 1
S O L U T I O N S
P-39
SUMMATIVE ASSESSMENT
WORKSHEET-31
SECTION A
1. For coincident lines
a1
b1 c1
=
=
a2
b2 c2
k 7
2
=
=
8 14
4
⇒
k = 4.
2.
x+y=0
x+y+7=0
a1
b1
c
¹ 1
Q
a2 = b2
c2
Hence, pair of equations has no solution.
⇒
1
1
SECTION B
3. The condition for no solution,
When
i.e.,
Since
k 3
= , we get k2 = 36
12 k
k=±6
k 3 1
a1 b1 c1
= ≠
=
≠
⇒
12 k 2
a2 b2 c2
½
k ≠ 6, so k = – 6.
½
1
SECTION C
4.
7x – 4y
On comparing with the equation
a1x + b1y
a1
Again,
5x – 6y
On comparing with the equation
a2x + b2y
a2
= 49
...(1)
= 0
= 7, b1 = – 4, c1 = 49
= 57
...(2)
= c2
= 5, b2 = – 6, c2 = 57
a1
b1
7
4
=
and
=
a2
b2
5
6
a1
b1
¹
a2
b2
Since,
1
So, system has unique solution.
Multiply eqn. (1) by 5 and multiply eqn. (2) by 7 and subtract,
35x – 20y = 245
1
35x – 42y = 399
– +
–
22y
⇒
y
Put the value of y in eqn. (1)
5x – 6(–7) = 57 ⇒ 5x
⇒
x
Hence,
x
P-40
= –154
= –7
= 57 – 42 = 15
=3
= 3 and y = –7.
M A T H E M A T I C S --
1
X
T E R M – 1
SECTION D
5
1
+
=2
x −1 y− 2
6
3
−
=1
x −1 y− 2
5.
...(1)
...(2) 1
1
1
= p,
=q
x −1
y−2
Then the equations (1) and (2) will be
5p + q = 2
6p – 3q = 1
Multiplying eqn. (3) by 3 and then adding in eqn. (4), we get
Let
...(3)
...(4) 1
21p = 7
⇒
p=
7 1
=
21 3
1
Putting this value of p in eqn. (3), we get
q = 2 – 5p = 2 – 5 ×
1
1
=
3
3
∴
1
1
= ⇒ x –1 = 3
x −1 3
x =4
and
q=
∴ Solution is
y=5
x = 4 , y = 5.
Now
p=
1
1
y- 2 = 3 ⇒ y – 2 = 3
SUMMATIVE ASSESSMENT
1
WORKSHEET-32
SECTION A
1. Since, line intersect y-axis, then x = 0
⇒
3(0) – 2y = 6
⇒
y = –3
1
Hence, required point is (0,– 3).
2.
4x – 5y = 5
kx + 3y = 3
For the condition of inconsistent
a1
b1
c
¹ 1
a2 = b2
c2
i.e.,
⇒
⇒
S O L U T I O N S
4
−5 5
≠
=
k
3
3
4
−5
=
k
3
−12
k=
5
1
P-41
SECTION B
3. From Fig.,
x+y=
x–y=
Adding (i) and (ii), we get
2x =
or
x =
Put the value of x in equation (i), we get
19 + y =
or
y =
Hence,
x =
22
16
38
19
...(i)
...(ii) ½
½
22
22 – 19 = 3
19 and y = 3.
½
½
SECTION C
4. Let
1
= a, the given equations become
y
4x + 6a = 15
...(1)
...(2) ½
6x – 8a = 14
Multiply eqn (1) by 4 and eqn. (2) by 3 and adding
16x + 24a = 60
18x – 24a = 42
On adding,
34x = 102
⇒
x =
Put the value of x in eqn. (1)
102
=3
34
½
4(3) + 6a = 15
6a = 15 – 12 = 3
3
1
=
6
2
1
1
=
a=
⇒y=2
y
2
x = 3 and y = 2
a=
Q
Hence
Again y = px – 2 ⇒ 2 = p(3) –2 ⇒ 3p = 4 ⇒ p =
1
4
·
3
1
SECTION D
5. Let the speed of the boat be x km/hr and speed of the stream be y km/hr.
∴ Upstream speed = x – y km/hr
and Downstream speed = x + y km/hr
According to the question,
32
36
+
=7
x− y x+ y
and
40
48
+
=9
x− y x+ y
1
1
= A,
= B, we get
x− y
x+ y
32A + 36B = 7
and
40A + 48B = 9
1
1
Solving these equations, we get
A= ,B=
8
2
Let
P-42
M A T H E M A T I C S --
1
X
T E R M – 1
1
1
=
x− y
8
⇒
x–y=8
1
1
=
and
B=
12 x + y
⇒
x + y = 12
Adding equations (1) and (2), we get 2x = 20
⇒
x = 10
Putting this value of x in eqn.(1), we get
y = x – 8 = 10 – 8 = 2
Hence, the speed of the boat = 10 km/hr and speed of the stream = 2 km/hr.
Hence
A=
SUMMATIVE ASSESSMENT
...(1) ½
...(2) ½
1
1
WORKSHEET-33
SECTION A
1.
2x + y = 7
6x – py = 21
For the condition of infinite number of solutions
a1
b
c
= 1 = 1
a2
b2 c2
1
7
2
=
=
−
p
21
6
⇒
1
1
=
−
p
3
⇒
p = – 3.
1
2. If a pair of linear equations is consistent, then the lines represented by these equations will be
intersecting (or) coincident.
1
⇒
SECTION B
3. Condition for unique solution,
a1
a2
k
12
k2
k
b1
b2
3
≠
k
≠ 36
≠ ± 6.
≠
1
1
SECTION C
4. (i) Suppose the person invested ` x at the rate of 12% simple interst and ` y at the rate of 10%
simple interest, then
yearly interest =
12 x 10 y
+
100 100
12 x 10 y
+
= 130
100 100
⇒
12x + 10y = 13000
⇒
6x + 5y = 6500
If the invested amounts are interchanged, then yearly interest increases by ` 4.
∴
10x + 12y = 13400
⇒
5x + 6y = 6700
S O L U T I O N S
∴
..(1) ½
...(2) ½
P-43
Subtracting eqn. (2) from eqn. (1), we get
x – y = – 200
...(3)
Adding equations (1) and (2), we get
11x + 11y = 13200
⇒
x + y = 1200
...(4)
Adding equations (3) and (4), we get
2x = 1000
⇒
x = 500
1
Putting x = 500 in equation (3), we get y = 700
Thus, the person invested ` 500 at the rate of 12% per year and ` 700 at the rate of 10% per year.
(ii) Pair of linear equations in two variables.
½
(iii) Honesty is the best policy.
½
SECTION D
5. Let length = x and breadth = y
Then according to first condition,
(x – 5) (y + 3) = xy – 9
⇒
...(1) 1
3x – 5y = 6
According to second condition,
(x +3) (y +2) = xy + 67
⇒
...(2) 1
2x + 3y = 61
Multiplying eqn. (1) by 3 and eqn. (2) by 5 and then adding, we get
9x – 15y = 18
Aadding,
10x + 15y = 305
323
= 17
19
1
3(17) – 5y = 6 ⇒ 5y = 51 – 6 ⇒ y = 9
1
19x = 323
⇒
x=
Putting this value of x in eqn. (1), we get
Hence, perimeter = 2(x + y) = 2(17 + 9) = 52 units.
SUMMATIVE ASSESSMENT
WORKSHEET-34
SECTION A
1. (B) First line is
For parallel lines,
So, second line is
10x – 8y –7 = 0
c1
a1
b
= 1 ≠
a2
b2 c2
15x –12y –7 = 0.
2. The area of the triangle formed by line
1
x y
+ = 1 with the coordinate axis = ab.
a b
1
SECTION B
3. Suppose my age is x years and my son’s age is y years. Then
x = 3y
Five years later, my age will be (x + 5) years and my son’s age will be (y + 5) years.
5
∴
x + 5 = (y + 5)
2
⇒
2x – 5y – 15 = 0
P-44
M A T H E M A T I C S --
X
...(1) ½
...(2)
T E R M – 1
Put x = 3y in equation (2)
6y – 5y – 15 = 0 ⇒ y = 15
Putting y = 15 in equation (1)
x = 45
Hence, my present age is 45 years and my son’s present age is 15 years.
1
SECTION C
4. For
(2p – 1)x + (p – 1)y – (2p + 1)
a1
and for
3x + y – 1
a2
For the condition of no solution
a1
a2
2p − 1
3
2p − 1
By
3
⇒
3p – 3
⇒
3p – 2p
p
=
=
=
=
0
2p – 1, b1 = p –1 and c1 = – (2p + 1)
0
3, b2 = 1 and c2 = – 1
b1
c1
= b ≠ c
2
2
p −1 2p + 1
≠
=
1
1
p−1
=
1
= 2p – 1
=3–1
= 2.
½
½
1
1
SECTION D
5. Let two digit number is 10x + y.
According to question,
8(x + y) – 5 = 10x + y
⇒
2x – 7y + 5 = 0
and
16(x – y) + 3 = 10x + y
6x – 17y + 3 = 0
Solving eqns. (1) and (2) by cross-multiplication method, we get
x
y
1
=
=
(2)(–7) – (6)(–7)
(–7)(3) – (–17)(5)
(5)(6) – (2)(3)
⇒
...(1)
...(2) 1
1
x
y
1
=
=
−21 + 85
30 − 6 −34 + 42
x
y 1
=
=
64
24 8
⇒
x
y
=
=1
8
3
Hence,
x = 8, y = 3
So required number = 10 × 8 + 3 = 83.
⇒
SUMMATIVE ASSESSMENT
1
1
WORKSHEET-35
SECTION A
1.
3x + ky = 5
2x + y = 16
For unique solution,
S O L U T I O N S
a1
b
≠ 1
a2
b2
1
P-45
3
k
≠
2
1
⇒
⇒
k≠
3
2
SECTION B
2. For equation,
2x + 3y = 4
a1 = 2, b1 = 3, c1 = – 4
For equation,
½
(k + 2) x + 6y = 3k + 2
a2 = k + 2, b2 = 6, c2 = – (3k + 2)
For infinitely many solutions
⇒
½
a1
b1
c1
=
=
a2
b2
c2
2
3
4
= =
k+2
6 3k + 2
k = 2.
1
SECTION C
3.
141x + 93y = 189
...(1)
93x + 141y = 45
...(2)
Adding equations (1) and (2), we get
234x + 234y = 234
x+y=1
...(3)
Subtracting equation (2) from equation (1), we get
48x – 48y = 144
...(4) 1
x–y=3
Adding equations (3) and (4), we get
2x = 4
1
x =2
Put the value of x in equation (3), we get
2+y=1
y =1–2=–1
Hence,
1
x = 2 and y = – 1
SECTION D
4. Let the cost of one pencil be ` x and the cost one chocolate be ` y.
According to question,
2x + 3y = 11
x + 2y = 7
Now,
and
P-46
2x + 3y = 11 ⇒ x =
...(1)
...(2)
11 − 3 y
2
y
1
3
5
x
4
1
−2
1
x + 2y = 7 ⇒ x = 7 – 2y
M A T H E M A T I C S --
X
T E R M – 1
y
0
1
3
x
7
5
1
1
y
8
7
(–2, 5) 6
5
x + 2y = 7
4
3
(1, 3)
2
1
x'
(5, 1)
(4, 1)
–6 –5 –4 –3 –2 –1 0
–1
–2
–3
1
2
3
(7, 0)
4
5
6
7
8
2x + 3y = 11
x
1
–4
–5
–6
–7
y'
Putting the above points and drawing the lines joining them, we get the graph of the above
equations. Clearly, two lines interesect at point (1, 3).
∴ Solution of eqns. (1) and (2) is x = 1 and y = 3
∴ Cost of one pencil = ` 1 and cost of one chocolate = ` 3.
SUMMATIVE ASSESSMENT
1
WORKSHEET-36
SECTION A
1.
4x –3y = 9
2x + ky = 11
c1
a1
b
= 1 ≠
a2
b2 c2
For unique solution,
4
−3
9
=
≠
2
k
11
⇒
⇒
2=
–3
k
⇒
k=
−3
·
2
1
SECTION B
2.
⇒
⇒
x +1 y−1
+
= 9
2
3
3(x + 1) + 2(y –1) = 54
3x + 2y = 53
⇒
S O L U T I O N S
...(1)
x −1 y +1
+
=8
3
2
P-47
⇒
2(x – 1) + 3 (y + 1)
⇒
2x + 3y
Multiply eqn. (1) by (3),
9x + 6y
Multiply eqn. (2) by 2,
4x + 6y
–
–
On subtracting
Put the value of x in eqn. (2),
=
=
=
=
48
47
53 × 3
47 × 2
–
... (2) ½
5x = 65
65
x =
= 13
5
½
2(13) + 3y = 47
Hence
3y = 47 –26 = 21
21
=7
y =
3
x = 13, y = 7.
1
SECTION C
3.
a+b
2
or
2ax + 2by = a + b
and
3x + 5y = 4
Multiplying (i) by 5 and (ii) by 2b, we get
10ax + 10by = 5a + 5b
6bx + 10by = 8b
–
– –
ax + by =
On subtracting,
...(i)
...(ii) 1
1
x(10a – 6b ) = 5a – 3b
5 a − 3b
1
=
x =
2(5a − 3b) 2
1
1
in (1), we get y =
2
2
1
1
x =
and y = ·
2
2
Putting x =
Hence
1
SECTION D
4. Let the speed of bus be x km/hr and the speed of the train be y km/hr.
240 120
+
According to question,
=8
x
y
120 240
+
and
=7
x
y
1
1
Let
= a,
= b, then
y
x
240a + 120b = 8
120a + 240b = 7
Apply [(1) × 2 – (2)], we get
480a + 240b = 16
120a + 240b = 7
–
– –
On subtracting,
⇒
P-48
1
...(1)
...(2)
...(2)
360a = 9
a=
9
1
=
360 40
M A T H E M A T I C S --
1
X
T E R M – 1
Putting this value of a in eqn.(1), we get
1
60
1
1
=
b=
⇒ y = 60
60 y
1
b=
1
1
=
⇒ x = 40
40 x
Hence, speed of bus = 60 km/hr and speed of train = 40 km/hr.
a=
1
SUMMATIVE ASSESSMENT
WORKSHEET-37
SECTION A
1. The area of the triangle formed by the lines x = 3, y = 4 and x = y is =
1
2
sq unit.
1
SECTION B
2.
29x + 41y = 169
41x + 29y = 181
Adding equations (1) and (2), we get
70x + 70y = 350
x+y=5
Subtracting equation (1) from equation (2), we get
– 12x + 12y = – 12
–x+ y = –1
Adding equations (3) and (4), we get
2y = 4
4
=2
2
Putting this value of y in equation (3), we get
x+2=5 ⇒ x=3
⇒ Hence,
x = 3 and y = 2.
⇒
y =
...(1)
...(2)
1
...(3)
...(4)
½
½
SECTION C
3.
For x + 2y = 3
a1 = 1, b1 = 2 , c1 = 3
For
(k – 1) x + (k +1) y = k + 2
a2 = k –1, b2 = k + 1, c2 = k + 2
For no solution,
a1
b1
c1
=
≠
a2
b2
c2
⇒
From I and II,
⇒
S O L U T I O N S
1
2
3
=
≠
k −1
k +1
k+2
I
II
III
1
1
2
=
k −1
k +1
k + 1 = 2k – 2 ⇒ k = 3
P-49
2
3
≠
k +1
k+2
2(k + 2) ≠ 3 (k + 1)
2k + 4 ≠ 3k + 3 ⇒ k ≠ 1
From II and III,
⇒
⇒
½
½
1
3
≠
k–1
k+2
⇒
k + 2 ≠ 3k – 3
5
⇒
k≠–
2
5
Hence, k = 3 but k ≠ 1 and k ≠ – ·
2
From I and III,
½
½
SECTION D
4.
Since BC DE and BE CD with BC ⊥ CD, BCDE is a rectangle.
∴
Opposite sides are equal.
i.e.,
BE = CD ∴ x + y = 5
and
DE = BC = x – y
Since perimeter of ABCDE is 21.
∴
AB + BC + CD + DE + EA = 21
3 + x – y + x + y + x – y + 3 = 21
6 + 3x – y = 21
3x – y = 15
Adding (i) and (ii), we get
4x = 20
x =5
On putting the value of x in (i), we get
y =0
∴
...(i)
1
1
...(ii)
1
x = 5 and y = 0.
FORMATIVE ASSESSMENT
WORKSHEET - 38
Objective Type Questions
1. (C)
2. (B)
3. (B)
4. (B)
1×4=4
Word Problems
1. Let Leela’s present age = x and her daughter’s age = y
Seven years ago :
y – 7 = 7 (x – 7)
y – 7 = 7x – 49
y – 7x = – 42
Three years later :
y + 3 = 3(x + 3)
y + 3 = 3x + 9
y – 3x = 6
2. Let one’s digit is y and ten’s digit is x.
According to the question :
x+y=9
number = 10x + y
On reversing digits
number = 10y + x
Again according to the question,
9(10x + y) = 2(10y + x)
90x + 9y = 20y + 2x
P-50
M A T H E M A T I C S --
...(i) 1
...(ii) 1
...(i) 1
X
T E R M – 1
90x – 2x + 9y – 20y
88x – 11y
8x – y
3. Let the rate of apple’s = Rs. x per kg
The rate of grapes = Rs. y per kg
According to the question,
2x + y
After one month :
4x + 2y
=0
=0
=0
...(ii) 1
= 160
= 300.
...(i) 1
...(ii) 1
Graph :
y
C
4
3
2
E
1
–3
–2
–1
B
A
0
x'
0
–1
–2
1
2
3
4
x
5
6
D
–3
y'
1. Points are A(2, 0) and B(4, 0).
2. Points are C(0, 4) and D(0, – 2).
3. Solution is (3, 1).
2
2
2
4. The area of triangle ABE =
1
× 2 × 1 = 1 unit2
2
2
5. The area of triangle CDE =
1
× 6 × 3 = 9 unit2
2
2
●●
S O L U T I O N S
P-51
CHAPTER
4
Triangles
SUMMATIVE ASSESSMENT
WORKSHEET-39
SECTION A
1. We have
(26)2
169
=
(side of the smaller triangle)2
121
26
13
= side of the smaller triangle
11
⇒
∴
side =
11 × 26
= 22 cm.
13
1
SECTION B
2. In triangles LMK and PNK, we have
∠M = ∠N = 50º
∠K = ∠K
∴
∆LMK ~ ∆PNK
(Given) 1
(Same)
(AA similarity)
LM
KM
=
PN
KN
∴
a
b+ c
=
x
c
b
bc
x=
·
b+ c
∴
½
a
c
½
SECTION C
3. Proof : BA  PQ
⇒ BR  PQ and PR  CA ⇒ PR  CQ
A
R
Q
B
P-52
P
C
D
M A T H E M A T I C S --
X
T E R M – 1
In ∆BRD, BR  PQ
BD
⇒
=
RD
(corr. of BPT) ....(1) 1
QD
PD
In ∆RPD, PR  CQ
RD
PD
=
QD
CD
⇒
(1) and (2)
⇒
BD
PD
=
PD
CD
(corr. of BPT) ....(2) 1
⇒ PD2 = BD × CD
(12)2 = BD × CD
BD × CD = 44 cm
1
SECTION D
4. Given :
2
3
OB = 3 cm
cos α =
and
In ∆AOB
cos α =
Let
1
OA = 2x and AB = 3x
In ∆ AOB
AB2 =
(3x)2 =
9x2 =
5x2 =
Þ
Þ
x=
Hence,
D
2
AO
=
3
AB
AO2 + OB2
(2x)2 + (3)2
4x2 + 9
9
A
O
α
C
1
B
3
9
=
5
5
6
 3 
OA = 2x = 2 
cm
 =
5
 5
 3 
AB = 3x = 3 
 =
 5
9
cm
5
BD = 2 ´ OB = 2 ´ 3 = 6 cm
AC = 2´ AO
So diagonal
and
= 2´
=
1
½
6
5
12
cm
5
SUMMATIVE ASSESSMENT
½
WORKSHEET-40
SECTION A
2
ar (∆DEF)
DE2  DE 
=
=


ar (∆ABC)
AB2  AB 
1.
⇒
44
=
ar (∆ABC)
⇒
ar (∆ABC) =
S O L U T I O N S
FG 2 IJ
H 3K
2
=
4
9
44 × 9
= 11 × 9 = 99 square unit.
4
1
P-53
SECTION B
14 cm
X
Y
3
6 3 cm
cm
2.
60°
3
4.2
cm
P
8.4 cm
70°
Q
7 cm
R
Z
From given figures,
⇒
∴ ∠X = ∠R
PQ 4·2 1 PR 3 2 1 QR
PQ PR QR
7
1
=
= ;
=
= ;
=
= ⇒
=
=
ZY 8·1 2 ZX 6 3 2 YX 14 2
ZY ZX YX
∆PQR ~ ∆ZYX
180º – (60º + 70º) = 50º
1
(SSS)
1
SECTION C
3. We have,
⇒
(Q ∠P is common and
PQR ~ PAB
FG IJ
H K
32
F 4kI
=G J
area ∆PAB H 3k K
area ∆PQR
PQ
=
area ∆PAB
PA
2
P
3k
2
⇒
PA
PB
=
) 1
PQ
PR
4k
1
B
A
k
Q
area ∆PAB = 18 cm2
∴ area of AQRB = area of ∆PQR – area of ∆PAB = 32 – 18 = 14 cm2
R
1
SECTION D
4. Let
BD = DE = EC = x
A
BE = 2x
BC = 3x
AE2 = AB2 + BE2 = AB2 + 4x2
AC2 =
AB2
+
BC2
=
AB2
+
AD2 = AB2 + BD2 = AB2 + x2
Now,
and
...(1) 1
9x2
8AE2 = 8AB2 + 32x2
3(AB2 + 9x2) + 5 (AB2 + x2)
3AC2 + 5AD2 =
B x D x E x
C
1
[Multiply eqn. (1) by 8] ...(2)
1
= 3AB2 + 27x2 + 5AB2 + 5x2
= 8AB2 + 32x2
∴
3AC2 + 5AD2 = 8AE2.
...(3)
[ From eqn. (2) & (3)] Proved. 1
SUMMATIVE ASSESSMENT
WORKSHEET-41
SECTION A
Perimeter of ∆ABC
length of AC
=
length of PR
Perimeter of ∆PQR
1.
⇒
20
AC
=
40
8
⇒
AC =
P-54
8
= 4 cm.
2
M A T H E M A T I C S --
1
X
T E R M – 1
SECTION B
2. Suppose the median AD intersects PQ at E.
Given,
PQ || BC
⇒
∠APE = ∠B and ∠AQE = ∠C
So in ∆APE and ∆ABD,
∠APE = ∠ABD
A
∠PAE = ∠BAD
∴
1
∆APE ~ ∆ABD
PE
AE
=
BD
AD
∆AQE ~ ∆ACD
⇒
Similarly,
P
Q
E
B
...(1) 1
C
D
QE
AE
=
CD
AD
...(2) ½
From eqns. (1) and (2), we have
PE
QE
=
BD
CD
⇒
⇒
Hence, AD bisects PQ.
PE
QE
=
,
BD
BD
PE = QE
(as CD = BD)
Proved. ½
SECTION C
3. Given :
PS
PT
=
SQ
TR
P
∠PST = ∠PRQ
To prove : PQR is isosceles triangle.
Proof :
By converse of B.P.T., we get
PS
PT
=
SQ
TR
ST || QR
S
Q
T
R
1
∴
∠PST = ∠PQR
(Corresponding angles) 1
∴
∠PST = ∠PRQ
(Given)
∠PQR = ∠PRO
So, ∆PQR is isosceles triangle.
Proved. 1
SECTION D
4. Given : ABC is a triangle in which DE || BC.
AD
AE
To prove :
=
BD
CE
Construction : Draw DN ⊥ AE and EM ⊥ AD, Join BE and CD.
1
Proof : In ∆ADE,
area (∆ADE) =
× AE × DN
2
In ∆DEC,
1
area (∆DCE) =
× CE × DN
2
S O L U T I O N S
1
...(i)
...(ii)
P-55
By (i) / (ii)
1
× AE × DN
area (∆ADE)
= 2
1
area (∆DEC)
× CE × DN
2
⇒
AE
area (∆ADE)
=
CE
area (∆DEC)
...(iii) ½
A
1
area (∆ADE) =
× AD × EM
2
Now in ∆ADE,
and in ∆DEB,
area (∆DEB) =
...(iv) ½
1
× EM × BD
2
D
1
× AD × EM
area (∆ADE)
= 2
area (∆DEB)
1
× BD × EM
2
By (iv) / (v),
N
M
...(v)
E
½
C
B
area (∆ADE)
AD
=
·
...(iv) ½
area (∆DEB)
BD
∆DEB and ∆DEC lies on the same base DE and between same parallel lines DE and BC.
∴
area (∆DEB) = area (∆DEC)
⇒
AE
area (∆ADE)
=
CE
area (∆DEB)
From equations (vi) and (vii), we get
From equation (iii),
...(vii)
AE
AD
=
·
CE
BD
Proved. 1
SUMMATIVE ASSESSMENT
WORKSHEET-42
SECTION A
1. Q PQ || MN
So,
⇒
⇒
⇒
KQ
KP
=
QN
PM
KQ
KP
=
KN
− KQ
PM
KQ
4
=
20.4 – KQ
13
4 × 20.4 – 4KQ = 13KQ
⇒
17KQ = 4 ×20·4 ⇒ KQ =
20·4 × 4
= 4·8 cm.
17
1
SECTION B
2. Since ABCD is a rhombus.
So in ∆BOC
A
BC2 = OB2 + OC2
2
 BD 
 AC 
=
+ 
 2 
 2 
BD 2 + AC2
4
4BC2 = AC2 + BD2
BC2 =
⇒
P-56
D
O
2
1
B
M A T H E M A T I C S --
C
1
X
T E R M – 1
3. According to question,
∠QPR = 90º
QR2 = QP2 + PR2
∴
∴
PR =
1
26 2 − 24 2
= 100 = 10 cm
∠PKR = 90º
∴
8 cm
PK = 102 − 82
= 100 − 64
1
= 36 = 6 cm.
SECTION C
4. According to question,
BC
2
BD = CD =
⇒
1
A
BC = 2BD
Using Pythagoras theorem in the right ∆ABC, we have
AC2 = AB2 + BC2
= AB2 + 4BD2
= (AB2 +
BD2)
1
+
3BD2
AC2 = AD2 + 3CD2.
B
D
C
1
SECTION D
5. Given: ABC is right angled at B and D is the mid-point of BC.
∴
BD = DC =
A
B
1
BC
2
½
½
D
C
In ∆ABD,
AD2 = AB2 + BD2
(Pythagoras Theorem) ...(1) ½
In ∆ABC,
AC2 = AB2 + BC2
(Pythagoras Theorem) ...(2) ½
From eqn. (1),
AD2 = AB2 +
⇒
FG BCIJ
H2K
2
4AD2 = 4AB2 + BC2
⇒
BC2 = 4AD2 – 4AB2
Using this in (2), we get
AC2 = AB2 + 4AD2 – 4AB2
AC2 = 4AD2 – 3AB2.
S O L U T I O N S
(D is the mid-point of BC) ½
...(3) ½
Proved. 1
P-57
SUMMATIVE ASSESSMENT
WORKSHEET-43
SECTION A
1.
Given 2AB = DE and BC = 8 cm
Q
∆ABC ~ ∆DEF
So,
AB
DE
=
BC
EF
⇒
AB
2AB
=
8
EF
⇒
A
B
(1)
D
C
E
(2)
F
1
EF = 2 × 8 = 16 cm.
SECTION B
2. Since the ratio of the corresponding sides of similar triangles is same as the ratio of their perimeters.
∴
∆ABC ~ ∆PQR
⇒
AB
BC
AC
36
=
=
=
PQ
QR
PR
24
⇒
AB
36
=
PQ
24
⇒
AB
36
=
10
24
⇒
AB =
1
36 × 10
= 15 cm.
24
1
3. Proof :
In quadrilateral ABCD,
AO
CO
=
BO
DO
⇒
AO
BO
=
CO
DO
In ∆ABD,
...(1)
O
EO | AB
∴
AE
BO
=
ED
DO
From eqns. (1) and (2),
AE
AO
=
ED
CO
In ∆ADC,
AE
AO
=
ED
CO
⇒
E
(Construction)
B
(By BPT) A
...(2) 1
EO | DC
(Converse of BPT) 1
EO | AB
(Construction)
∴
AB | DC
⇒ In quad. ABCD,
AB | DC
1
⇒ ABCD is a trapezium.
P-58
(Given)
C
D
Proved.
M A T H E M A T I C S --
X
T E R M – 1
SECTION C
AP
3.5
1
=
=
AB
10.5
3
1.
A
AQ
3
1
=
=
AC
9
3
⇒
3.5
AP
AQ
=
and ∠A is common
AB
AC
∆APQ ~ ∆ABC
In ∆ABC,
⇒
AP
PQ
=
AB
BC
∴
P
7
1
3
4.5
1
Q
(SAS)
6
C
B
4·5
1
=
⇒ BC = 13·5 cm.
BC
3
1
SECTION D
4. In cyclic quadrilateral ABCD, we have
∠A + ∠C =
and
∠Β + ∠D =
∴
x + 7 + 3y + 23 =
⇒
x + 3y =
and
y + 8 + 4x + 12 =
⇒
4x + y =
Solving equations (1) and (2), we get
y =
∴
∠A =
∠B =
∠C =
∠D =
180º
180º
180
150
180
160
1
..(1) 1
...(2) 1
40º
(x + 7)º = 37º
(y + 8)º = 48º
(3y + 23)º = 143º
(4x + 12)º = 132º
1
SUMMATIVE ASSESSMENT
WORKSHEET-44
SECTION A
1. Given,
Perimeter of ∆ABC = 32 cm
Perimeter of ∆PQR = 48 cm
For similar triangles, we know that
AB
BC
Perimeter of ∆ABC
AC
=
=
=
PQ
QR
Perimeter of ∆PQR
PR
AC
32
=
6
48
6 × 32
AC =
= 4 cm.
48
⇒
⇒
A
B
S O L U T I O N S
P
C
Q
R
P-59
SECTION B
2. According to question, AO = 20 cm, BO = 12 cm, PB = 18 cm
In ∆AQO and ∆BPO,
∠AOP = ∠BOP
∠A = ∠B = 90º
∆AQO ~ ∆BPO
∴
(Vertically opposite angles)
P
1
O
A
AQ
QO
AO
=
=
BP
PO
BO
AQ
20
=
18
12
AQ = 30 cm.
B
Q
1
SECTION C
3. Given : DB⊥BC, DE⊥AB and AC⊥BC.
BE
DE
∠1 + ∠2
∠2 + ∠3
∠1
∠1
∠ACB
∆ABC
To prove :
Proof : In ∆ABC,
But
⇒
In ∆ABC and ∆BDE,
∴
=
=
=
=
=
=
~
AC
BC
90º
90º
∠3
∠3
∠DEB = 90º
∆BDE
A
3
⇒
2
E
B
AC
BE
=
·
BC
DE
(∠C = 90º) 1
(Given)
1
D
1
C
(Proved)
(Given)
(AA) 1
SECTION D
∆ABC ~ ∆PQR,
ar ∆ABC = ar ∆PQR
4. Given :
and
A
B
To prove :
Proof :
⇒
Also
⇒
⇒
∴ By SSS congruency,
P-60
C
Q
R
∆ABC ≅ ∆PQR
∆ABC ~ ∆PQR
AB2 BC2 CA 2
ar (∆ABC)
=
=
=
ar (∆PQR)
PQ2 QR 2 RP 2
ar (∆ABC) = ar (∆PQR)
(Given)
..(1) 1
(Given)
ar (∆ABC)
=1
ar (∆PQR)
From equation (1), we have
⇒
P
1
AB2
BC2
CA 2
=
=
=1
PQ2
QR 2
RP 2
AB
BC
CA
=
=
=1
PQ
QR
RP
AB = PQ
BC = QR
CA = RA
∆ABC ≅ ∆PQR.
M A T H E M A T I C S --
1
1
X
T E R M – 1
SUMMATIVE ASSESSMENT
WORKSHEET-45
SECTION A
1. For similar triangles,
FG perimeter of ∆1 IJ
H perimeter of ∆2 K
F 4 I 2 16 ·
= G J =
H 25 K 625
area of triangle 1
=
area of triangle 2
2
1
SECTION B
2. Since G is mid point of PQ,
∴
PG = GQ
PG
=1
GQ
According to question,
GH | QR
∴
PG
PH
=
GQ
HR
PH
HR
PH = HR.
1=
⇒
Hence, H is mid-point of PR.
P
1
H
G
(BPT)
R
Q
1
SECTION C
3. According to question,
∴
DE | AB
CD
CE
=
AD
EB
(By BPT) ...(1) ½
CF
CE
=
FD
EB
(By BPT) ...(2) ½
Again since FE || DB,
∴
From equations (1) and (2),
CD
CF
=
AD
FD
⇒
AD FD
=
CD CF
⇒
AD
FD
1+
=
+1
CD
CF
CD + AD
FD + FC
=
CD
FC
⇒
S O L U T I O N S
AC
CD
=
CD
FC
2
CD = AC.FC
C
F
D
E
1
A
B
1
P-61
SECTION D
4. Since, ∆ABC is equilateral, and AD || BC
BC
2
AD2 = AB2 – BD2
∴
BD = DC =
In right ∆ADB
= AB2 +
BC2
4
= AB2 –
AB2
4
AD2 =
1
A
B
D
C
3
AB2 = 4AD2 = 3AB2.
4
(Q AB = BC)
SUMMATIVE ASSESSMENT
WORKSHEET-46
SECTION A
1. For similar triangles, we know that
A
D
2
⇒
⇒
BC
area of ∆ABC
=
area of ∆DEF
EF 2
(4) 2 16
80
=
=
area of ∆DEF
(5) 2 25
area of ∆DEF =
80 × 25
= 125 cm2.
16
B
C
4 cm
E
F
5 cm
SECTION B
2. In ∆APB and ∆DPC, we have
∠ A = ∠ D = 90°
and
∠ APB = ∠ DPC
(vertically opposite angles)
Thus by AA criterion of similarity, we have
∆APB ~ ∆DPC
⇒
=
A
D
P
1
B
C
PB
AP
=
PC
DP
AP × PC = PB × DP
1
SECTION C
3. To prove :
O
Construction : Draw AE ⊥BC and DF⊥BC.
Proof :
In ∆AOE and ∆DOF,
⇒
∴
P-62
C
A
ar ( ∆ABC)
AO
=
ar ( ∆DBC)
DO
E
B
∠AOE = ∠DOF
∠AEO = ∠DFO = 90º
∆AOE ~ ∆DOF
½
F
D
½
(Vertically opposite angles)
(Construction)
(AA)
AO
AE
=
DO
DF
M A T H E M A T I C S --
...(1) 1
X
T E R M – 1
1
× BC × AE
AE
ar ( ∆ABC)
= 2
=
1
DF
ar ( ∆DBC)
× BC × DF
2
Now,
=
½
AO
DO
[From equation (1)] ½
SECTION D
4. Proof :
BC = 2 BD
A
B
P
D
C
Q
Given,
AB
BC
AD
=
=
PQ
QR
PM
⇒
AB
2BD
AD
=
=
PQ
2QM
PM
∴
⇒
R
(PM is the median) 1
1
AB
BD
AD
=
=
PQ
PM QM
∆ABD ~ ∆PQM
∠B = ∠Q,
In ∆ABC and ∆PQR,
AB
BC
=
PQ
QR
and
∠B = ∠Q
∴
M
QR = 2QM
and
In triangles ABD and PQM,
(AD is the median)
(SSS Similarty)
(By CPCT) 1
∆ABC ~ ∆PQR.
Proved. 1
SUMMATIVE ASSESSMENT
WORKSHEET-47
SECTION A
1. In the ∆ ABC,
DE || BC
∴
AD
AE
=
DB
EC
⇒
x
x+2
=
x−2
x −1
⇒
A
D
B
x(x – 1) = (x – 2) (x + 2)
⇒
x2 – x = x2 – 4
⇒
x = 4.
S O L U T I O N S
C
E
1
P-63
SECTION B
2. Draw AC intersecting EF at G.
In ∆CAB,
GF || AB
⇒
In ∆ADC,
⇒
From (i) and (ii), we have
B
A
AG
BF
=
CG
FC
EG || DC
AE
AG
=
ED
CG
...(i) 1
E
F
...(ii)
C
D
AE
BF
=
·
ED
FC
1
SECTION C
3. In ∆ABC
PQ || AB
Again in ∆BCD,
D
A
AQ
BP
=
QC
PC
PR || BD
∴
(By BPT) ...(i) 1
BP
DR
=
PC
RC
⇒
AQ
DR
=
QC
RC
QR || AD
From (i) and (ii),
⇒
R
Q
B
P
(By BPT) ...(ii) 1
C
(By converse of BPT) 1
SECTION D
4. Q
⇒
It is given that
⇒
∆FEC
EC
∠1
AE
≅
=
=
=
∆GBD
BD
∠2
AD
...(1) 1
...(2) 1
A
D
1
3
F
B
2
E
4
C
G
From eqns. (1) and (2),
⇒
⇒
Thus in ∆ADE and ∆ABC,
AE
AD
=
EC
BD
DE || BC,
∠ 1 = ∠ 3 and ∠ 2 = ∠ 4
(converse of BPT)
1
∠A = ∠A
∠1 = ∠3
∠2 = ∠4
So by AAA criterion of similarity, we have
∆ADE ~ ∆ABC.
P-64
M A T H E M A T I C S --
Proved. 1
X
T E R M – 1
SUMMATIVE ASSESSMENT
WORKSHEET-48
SECTION A
1. Since
ST || QR
PS
PT
=
QS
TR
⇒
3
PT
=
5
PR − PT
⇒
3
PT
=
5
28 − PT
⇒
⇒
5PT = 84 – 3PT
8PT = 84
⇒
PT =
84
= 10·5 cm
8
1
SECTION B
2. In ∆CAB,
∴
But,
∴
∠A
AC
AD
AC – AD
CD
Dividing (2) by (1), we get
By converse of BPT,
= ∠B
= CB
= BE
= CB – BE
= CE
(Given)
(By isosceles triangle property) 1
(Given) ...(1)
C
...(2)
CD
CE
=
AD
BE
DE | AB.
E
D
1
B
A
SECTION C
3. In ∆ADB,
AB2 = AD2 + BD2
In ∆ADC,
AC2 = AD2 + CD2
Sultracting eqn. (2) from eqn. (1), we get
AB2 – AC2 = BD2 – CD2
(Pythagoras Theorem) ...(1)
(Pythagoras Theorem) ...(2)
A
1
FG 3 BCIJ − FG 1 BCIJ
H4 K H4 K
2
=
∴
∴
2(AB2 – AC2) = BC2
2(AB)2 = 2AC2 + BC2.
2
=
BC2
2
1
B
3x
x
D
C
1
SECTION D
4. Q
∴
In ∆ADB and ∆PDQ,
By A.A. similarity,
∴
S O L U T I O N S
AB
∠ABQ
∠ADB
∠ABQ
∆ADB
|
=
=
=
~
PQ
∠PQD
∠PDQ
∠PQD
∆PDQ
DQ
PQ
=
DB
AB
DQ
z
=
DB
x
A
(Corresponding angles) 1
(Common)
C
x
P
y
1
D
...(i)
z
B
Q
P-65
∆PBQ ~ ∆CBD
Similarly,
z
BQ
=
y
DB
and
...(ii)
Adding (i) and (ii), we get
z z DQ + BQ BD
+ =
=
DB
BD
x y
⇒
z z
+
=1
x y
∴
1 1
1
+
·
=
x y
z
1
1
SUMMATIVE ASSESSMENT
WORKSHEET-49
SECTION A
ar (∆ABC) AC2
=
=
ar (∆PQR) PR 2
1.
⇒
FG AC IJ
H PR K
2
81
 7·2 
= 

169
 PR 
2
⇒
2
 7·2 
 9 

 =  
 PR 
 13 
⇒
PR =
2
⇒
7·2
9
=
PR
13
13 × 7·2
= 10·4 cm.
9
1
SECTION B
2. ABCD is a square. ∆BCE is described on side BC is similar to ∆ACF described on diagonal AC.
Since ABCD is a square.
So AB = BC = CD = DA and AC =
Now
⇒
⇒
⇒
2 BC
∆BCE ~ ∆ACF
Area (∆BCE)
BC2
=
Area (∆ACF)
AC2
Area (∆BCE)
BC2
1
=
=
2
Area (∆ACF)
2
( 2 BC)
Area (∆BCE) =
1
2
F
D
C
1
E
A
Area (∆ACF)
B
1
SECTION C
3. Given : ∆ABC, right angled at A.
BL and CM are medians.
To prove :
4(BL2 + LM2) = 5BC2.
Proof : In ∆ABL,
BL2 = AB2 + AL2
FG AC IJ
H2K
2
F AB IJ
+ G
H2K
2
= AB2 +
In ∆ACM,
1
L
(BL is median)
CM2 = AC2 + AM2
= AC2
P-66
C
A
M
B
(CM is median) 1
,
M A T H E M A T I C S --
X
T E R M – 1
BL2 + CM2 = AB2 + AC2 +
AC2 AB2
+
4
4
4(BL2 + CM2) = 5AB2 + 5AC2
= 5(AB2 + AC2)
= 5BC2.
Proved. 1
SECTION D
4. Given : In ∆ABC and ∆DEF, AP and DQ are medians, such that
A
B
D
P
C
E
AP
AB BC
=
=
DQ
DE EF
Q
...(i)
F
∆ABC ~ ∆DEF
To prove :
AB
=
DE
Proof : From (1),
1
2
1
2
BC
AP
EF = DQ
1
BP
AP
AB
=
=
EQ
DQ
DE
∆ABP ~ ∆DEQ
∠B = ∠E
⇒
⇒
⇒
In ∆ABC and ∆DEF,
and
[SSS similarity]
1
AB
BC
=
DE
EF
∠B = ∠E,
∆ABC ~ ∆DEF.
1
(By SAS criterion)
Proved 1
SUMMATIVE ASSESSMENT
WORKSHEET-50
SECTION A
1.
⇒
⇒
Perimeter of ∆ABC
AB
=
Perimeter of ∆PQR
PQ
60
AB
=
36
9
60 × 9
AB =
= 15 cm.
36
1
SECTION B
1. Construction : Join BD.
Proof :
∴
In ∆ADB,
In ∆BCD,
∴
S O L U T I O N S
AB
AB
AE
ED
BG
GD
AE
ED
|| EF and AB | | CD
|| CD || EF
BG
=
(BPT)
GD
BF
=
(BPT)
FC
BF
=
·
FC
C
D
E
G
F
1
A
B
1
P-67
SECTION C
∆AOB ~ ∆COD
1.
By area theorem,
ar (∆AOB)
AB2
=
ar (∆COD)
DC2
⇒
ar (∆AOB)
(2CD) 2
4
=
=
ar (∆COD)
1
(CD) 2
∴
(A similarity)
C
D
1
ar (∆AOB) : ar (∆COD) = 4 : 1.
O
1
B
A
1
SECTION D
4. Given. The line segment XY is parallel to side AC of ∆ABC.
Proof : In ∆BAC and ∆BXY,
∠B = ∠B
∠BAC = ∆BXY
∴
∆BAC ~ ∆BXY
We have,
FG BA IJ
H BX K
F AB IJ
= G
H BX K
ar (∆BAC)
=
ar (∆BXY)
2 × ar (∆BXY)
ar (∆BXY)
⇒
AB
=
BX
⇒
1
BX
=
2
AB
⇒
⇒
1–
(Common)
(Corresponding angles)
(By AA similarity) 1
2
A
2
1
X
2
B
Y
1
BX
= 1–
2
AB
AB − BX
=
AB
2 −1
AX
=
AB
2 −1
C
1
2
2
1
·
SUMMATIVE ASSESSMENT
WORKSHEET-51
SECTION A
1. We have
Area (∆ABC)
BC2
=
Area (∆DEF)
EF2
⇒
9
(2·1)2
=
16
EF 2
⇒
3
2·1
=
4
EF
⇒
P-68
EF =
4 × 2·1
= 2·8 cm
3
M A T H E M A T I C S --
1
X
T E R M – 1
SECTION B
2. Since,
XY || QR
∴
PX
PY
=
XQ
YR
P
4 cm
1
PY
4
=
=
2
PR − PY
PR − 4
⇒
⇒
X
PR – 4 = 8 ⇒ PR = 12 cm
∴In right ∆PQR,
R
Q
QR2 = PR2 – PQ2
1
Y
= 122 – 62 = 144 –36 = 108
QR = 6 3 cm
1
SECTION C
3. Given : In ∆ABC, AD⊥BC and AD2 = BD × CD.
To prove : ABC is a right triangle.
Proof : In right triangles ABD and ACD, applying Pythagoras theorem
A
AB2 = AD2 + BD2
AC2 = AD2 + CD2
and
∴
AB2 + AC2
=
2AD2
+
BD2
1
+
CD2
= 2(BD × CD) + BD2 +CD2,
B
D
C
given AD2 = BD × CD
= (BD +
i.e.,
AB2 + AC2
=
CD)2
1
=
BC2
BC2
⇒ ∆ABC is a right triangle.
1
SECTION D
4. In trapezium ABCD,
Also,
In trapezium ABCD,
∴
In ∆BGE and ∆BDC,
AB || DC and DC = 2AB.
BE
4
=
EC
3
EF || AB || CD
AF BE
4
=
=
FD BC
3
∠B = ∠B
B
A
F
D
G
E
C
∠BEG = ∠BCD
(Common corresponding angles)
∴
∆BGE ~ ∆BDC
(AA similarity) 1
⇒
EG
BE
=
CD
BC
As,
BE
4
EC
3
=
⇒
=
EC
3
BE
4
⇒
⇒
S O L U T I O N S
EC
3
+1=
+1
BE
4
...(1)
1
EC + BE
7
=
BE
4
P-69
⇒
BC
7
BE
4
=
⇒
=
BE
4
BC
7
⇒
EG
4
=
CD
7
4
CD
7
∆DGF ~ ∆DBA
⇒
EG =
Similarly,
⇒
DF
FG
=
DA
AB
⇒
FG
3
=
AB
7
⇒
FG =
Adding (i) and (ii),
EG + FG =
...(i)
1
LMQ AF = 4 = BE OP
7 BD
MM FD
PP
EC 3
⇒
=
MN BC 7 PQ
3
AB
7
4
3
CD + AB
7
7
b
g
4
3
8
3
11
× 2 AB + AB = AB + AB =
AB
7
7
7
7
7
7EF = 11AB.
⇒
EF =
∴
...(ii)
SUMMATIVE ASSESSMENT
Proved. 1
WORKSHEET-52
SECTION A
Area (∆AOB)
4
Area (∆COD) = 1
1.
⇒
D
C
84
4
Area (∆COD) = 1
Area (∆COD) =
O
84
= 21 cm2.
4
A
B
1
SECTION B
2. Proof : In ∆POQ,
AB | PQ,
(Given)
OB
OA
=
BQ
AP
...(i)
AC | PR,
(Given)
∴
OA
OC
=
AP
CR
...(ii)
From (i) and (ii), we get
OB
OC
=
BQ
CR
∴
In ∆OPR,
∴
P-70
(BPT) 1
(BPT ) 1
BC | QR,
(By converse of BPT) 1
M A T H E M A T I C S --
X
T E R M – 1
SECTION C
3. Since DE || BC.
By BPT, we have
AD
AE
=
DB
EC
⇒
A
DB
EC
=
AD
AE
1
Adding 1 on both sides,
D
DB
EC
+1=
+1
AD
AE
DB + AD
EC + AE
=
AD
AE
B
E
C
1
AB
AC
=
AD
AE
⇒
AD
AE
=
·
AB
AC
1
SECTION D
4. In ∆ABC,
DP || BC
AD
AP
=
,
DB
PC
(BPT) ...(1)
A
D
P
E
B
Similarily, in ∆ABC
Q
EQ || AC
From figure,
BQ
BE
=
QC
EA
EA = AD + DE
= BE + ED
= BD
Then (2) becomes,
BQ
AD
=
QC
BD
⇒
S O L U T I O N S
..(2) 1
(Given, BE = AD)
1
...(3)
1
From (1) and (3), we get
∴ By converse of BPT,
C
BQ
AP
=
QC
PC
PQ || AB.
1
P-71
FORMATIVE ASSESSMENT
WORKSHEET-53
Objective Type Questions
1. (D)
2. (A)
3. (C)
4. (C)
5. (A)
Fill in the blanks
1.
2.
3.
4.
5.
Same
Parallel
Similar
Proportional
SAS
Quiz
1. In given figure B'C'||BC. Find AB.
AB
3
=
3·6
2
3·6 × 3
2
AB = 5·42 cm
2. In a right angled triangle, the square of hypotenuse is equal to the sum of the squares of the other
two sides.
AB =
3.
AB = AD + BD = 1·5 + 3 = 4·5 cm.
According to BPT,
AE
AD
=
AC
AB
or
1·5
1
=
4·5
AC
⇒
1
1
=
3
AC
⇒
AC = 3 cm.
EC = AC – AE = 3 – 1 = 2 cm.
DB
7·2
4
AD
=
=
=
EC
5·4
3
AB
4. According to BPT,
AD
4
=
1·8
3
⇒
AD =
1·8 × 4
= 2·4 cm
3
5. No, because corresponding sides are not in proportion.
A
6
3
2
B
P
4
C
Q
5
4
R
●●
P-72
M A T H E M A T I C S --
X
T E R M – 1
CHAPTER
5
Introduction to
Trigonometry and
Trigonometric Identities
SUMMATIVE ASSESSMENT
WORKSHEET - 54
SECTION A
1. tan is not defined when θ is equal to 90º i.e., tan 90º = ∞.
2.
1
cos 48º cos 42º – sin 48º sin 42º = cos (90º – 42º) cos (90º – 48º) – sin 48º sin 42º
= sin 42º sin 48º – sin 48º sin 42º = 0
1
SECTION B
3.
1
sin 90°
1
1
+
=
+
cos45° cosec 30°
1/ 2
2
=
=
1
1
2 + 2
2 2+1
2
1
SECTION C
4.
L.H.S. =
=
=
S O L U T I O N S
1 − cos θ
1 + cos θ =
1 − cos 60°
1 + cos 60°
1 − 12
1 + 12
1
2
3
2
=
1
2
3
2
=
1
3
1
P-73
R.H.S. =
sin θ
1 + cos θ
=
sin 60°
1 + cos 60°
=
3/2
1
1+
2
=
3/2
3/2
1
=
3
1
1
= L.H.S.
Hence relation is verified for q = 60°.
SECTION D
4
4
2
2
5. 4 sin 30° + cos 60° – 3 cos 45° – sin 90°
(
) (
)
2
éæ
ù
éæ 1 ö 4 æ 1 ö 4 ù
êç 1 ö÷ – 1 2 ú
êç ÷ + ç ÷ ú
÷ ( ) ú
= 4 êç 2 ÷ ç 2 ÷ ú – 3 êç
êè ø è ø ûú
êè 2 ø
ë
ë
ûú
1
é1
é1 ù
1ù
= 4 ê + ú – 3 ê -1ú
ê
ú
ê
ë16 16 û
ë2 úû
1
=
1 3
+
2 2
1
=
4
= 2.
2
1
SUMMATIVE ASSESSMENT
WORKSHEET-55
SECTION A
1.
sin 38° sin 52° – cos 38° cos 52° = sin (90° – 52°) sin (90° – 38°) – cos 38° cos 52°
= cos 52° cos 28° – cos 38° cos 52° = 0.
sin θ = cos θ
2.
⇒
⇒
1
sin θ
cos θ = 1
tan θ = 1 = tan 45°
θ = 45°.
1
SECTION B
3. Q
Squaring on both sides
sin θ – cos θ =
1
2
æ1 ö2
(sin θ – cos) 2 = ç
ç ÷÷
è2 ø
P-74
M A T H E M A T I C S --
X
T E R M – 1
Þ
sin2 θ + cos2 θ – 2 sin θ cos θ =
1
4
1 – 2 sin θ cos θ =
1
4
1
3
=
4
4
(sin θ + cos θ)2 = sin2 θ + cos2 θ + 2 sin θ cos θ
2 sin θ cos θ = 1 –
Again
1
= 1 + 2 sin θ cos θ
3
7
=
4
4
= 1+
sin θ+ cos θ =
Þ
7
7
=
4
2
1
SECTION C
2
2
 3
1
2
5  + 4
 − (1)
2
2
2

2


2

5 cos 60° + 4 cos 30° – tan 45°
2
2
=
1 1
sin 2 30° + cos2 60°
+
   
2 2
5
+3–1
4
= 1 1
+
4 4
5
+2
4
= 1
2
4.
13
26 13
4
= ·
= 1 =
4
2
2
1
1
1
SECTION D
5. Given :
Q
c
sin q =
cos2
2
c +d2
q = 1 – sin2 q
æ
c
ç
= 1– ç 2
2
è c +d
= 1–
=
=
Þ
S O L U T I O N S
cos q =
ö2
÷
÷
ø
1
c2
c2 + d 2
c2 + d2 - c2
c2 + d2
d2
2
c + d2
d
c2 + d 2
1
1
P-75
Again,
tan q =
=
sin q
cos q
c / c2 + d 2
d / c2 + d 2
c
=
d
1
SUMMATIVE ASSESSMENT
WORKSHEET-56
SECTION A
1.
sin 45º + cos 45º =
2.
(sec A + tan A ) (1 – sinA) =
1
2
+
1
2
=
2
2
= 2.
1
FG 1 + sin A IJ (1 – sin A)
H cos A cos A K
 1 + sin A 
=  cos A  (1 – sin A)


=
1 − sin 2 A
cos A
=
cos 2 A
= cos A
cos A
1
SECTION B
3.
L.H.S. =
=
=
=
=
1 − cos A
1 + cos A
1 − cos A 1 − cos A
×
1 + cos A 1 − cos A
1
(1 − cos A)2
(1 − cos2 A)
(1 − cos A)2
sin 2 A
1 − cos A
sin A
cos A
1
= sin A − sin A
Proved. 1
= cosec A – cot A = R.H.S.
SECTION C
4.
L.H.S. = cosec2 q – tan2 (90° –
q)
sin 2 (90° - q)
=
–
cos2 (90° - q)
sin 2 q
1
P-76
M A T H E M A T I C S --
X
T E R M – 1
2
1
sin (90° –q)
–
2
2
sin q
sin q
=
2
1
=
1
–
2
sin q
cos q
2
sin q
2
=
=
=
=
=
=
1 - cos q
2
sin q
sin 2 q
1
sin 2 q
1
sin2 θ + cos2 θ
sin2 θ + sin2 (90° – θ)
R.H.S.
1
SECTION D
cosec2 (90º – θ) = sec2 θ
Sol.
sec2 θ – tan2 θ = 1
1
cos2 40º + cos2 50º = cos2 (90º – 50º) + cos2 50º
sin2 50º + cos2 50º = 1
tan2 30º
F 1 IJ
= G
H 3K
2
=
1
3
1
sec2 52ºsin2 38º = sec2 52º.sin2 (90º – 52º) = sec2 52º.cos2 52º = 1
cosec2 70º – tan2 20º = cosec2 (90º – 20º) – tan2 20º = sec2 20º – tan2 20º = 1
∴
1
Given expression =
−
4
=
1
1
×1
3
3(1)
2×
1 2 9–8 1
− =
=
·
4 9
36
36
SUMMATIVE ASSESSMENT
1
WORKSHEET-57
SECTION A
5 sin q + 4 cos q
5 cos q + 4 5 (1) + 4 9
=
= = 9.
=
5 sin q - 4 cos q
5 cos q - 4 5 (1) - 4 1
1. tan θ = 1,
2.
⇒
Since,
So,
sin2 A
sin A (sin A – 2)
sin A
A
=
=
≠
=
2 sin A
0
2, ∴ sin A = 0
0º.
1
1
SECTION B
2
3.
2 cos ec 2 30° + 3 sin 2 60° −
 3
3 1 
3
2
tan 2 30° = 2(2) + 3 
 − 

2
4
4
 3


= 4+
S O L U T I O N S
1
9
3
−
4 4
P-77
=
16 + 9 − 3
4
=
25 − 3
4
1
SECTION C
4.
L.H.S. =
=
cos 3 θ + sin 3 θ cos 3 θ – sin 3 θ
+
cos θ + sin θ
cos θ – sin θ
(cos θ + sin θ) (cos2 θ + sin 2 θ − sin θ cos θ)
(cos θ + sin θ)
+
(cos θ − sin θ) (cos2 θ + sin 2 θ + sin θ cos θ)
(cos θ − sin θ)
= (1 – sin θ cos θ) + (1 + sin θ cos θ)
= 2 – sin θ cos θ + sin θ cos θ
= 2 = R.H.S.
1
1
1
Proved
SECTION D
5. sin2 30° cos2 45° + 4 tan2 30° +
1
1
sin2 90° – 2 cos2 90° +
2
24
2
2
2
1
1
1  1 
 1 
+ 4
+ (1)2 − 2(0) +
=   ×


2  2 
2
24
 3
=
1 æ1 ö 4 1 1
ç ÷÷ + + +
ç
4 è 2 ø 3 2 24
1
1
=
1 4 1 1
+ + +
8 3 24 2
=
3+32+1+12
24
1
=
48
= 2.
24
1
SUMMATIVE ASSESSMENT
WORKSHEET-58
SECTION A
2
1.
2
 3
 1
2
2
−
sin 60º – sin 30º = 
 

2
 2 
=
2. x = a cos θ, y = b sin θ
Now,
b2x2 + a2y2 – a2b2 =
=
=
=
P-78
3 1
2 1
− =
= ·
4 4
4 2
1
b2(a cos θ)2 + a2 (b sin θ)2 – a2b2
a2b2 cos2 θ + a2b2 sin2 θ – a2b2
a2b2 (sin2 θ + cos2 θ) – a2b2
a2b2 – a2 b2 = 0.
M A T H E M A T I C S --
1
X
T E R M – 1
SECTION B
2 cot2 A – 1 = 2(cosec2 A – 1) – 1
3.
=
=
=
2
sin 2 A
2
æ 3 ö2
ç ÷
ç 2 ÷
è ø
–3
–3
8
-1
-3 =
3
3
1
SECTION C
4. According to question,
1
sin 3θ = cos (θ – 6º)
⇒
cos (90º – 3θ) = cos (θ – 6º)
⇒
90º – 3θ = θ – 6º
⇒
1
4θ = 90º + 6º = 96º
θ=
96º
= 24º
4
1
SECTION D
tan θ =
5.
In ∆ABC,
AC2 = AB2 + BC2
= 1+5=6
⇒
(i)
AB
1
=
BC
5
AC =
cosec 2 θ − sec 2 θ
cosec 2 θ + sec 2 θ
=
1
6
2
 6
( 6)2 − 

 5
 6
( 6) + 

 5
2
2
A
6
5
=
6
6+
5
=
θ
B
24 2
= ·
36 3
F 1 IJ + F 5 I
sin θ + cos θ = G
H 6 K GH 6 JK
2
(ii)
2
5
C
1½
2
2
=
S O L U T I O N S
6
1
6−
1 5
+ = 1.
6 6
½
P-79
SUMMATIVE ASSESSMENT
WORKSHEET-59
SECTION A
cosec θ – cot θ =
1.
1
4
⇒
(cosec q - cot q)(cosec q + cot q)
1
=
(cosec q + cot q)
4
⇒
1
cosec 2 q - cot 2 q
=
4
cosec q + cot q
⇒
⇒
1
1 + cot 2 q - cot 2 q
=
4
cosec q + cot q
cosec θ + cot θ = 4.
11
2.
cot 2 θ
−
1
 sin 2 θ
1 
11 
–
=

2
2
 cos θ cos2 θ 
cos θ
11
 sin 2 θ − 1 

= 11 
2
 cos θ 
 1 − sin 2 θ 

= – 11 
2
 cos θ 
F cos θ I
GH cos θ JK
2
= – 11
2
1
= – 11.
SECTION B
3. Here
3 sin θ – cos θ = 0 and 0º < θ < 90º
⇒
3 sin θ = cos θ
⇒
sin θ
1
=
cos θ
3
⇒
tan θ =
⇒
1
3
1
LMQ tan θ = sin θ OP
cos θ Q
N
= tan 30º
θ = 30º.
1
SECTION C
4. (i) Clearly, distance covered by the artist is equal to the length
of the rope AC. Let AB be the vertical pole of height 12 m.
It is given that ∠ACB = 30º
Thus, in right-angled triangle ABC,
AB
AC
1
12
⇒
=
2
AC
⇒
AC = 24 m.
Hence, the distance covered by the circus artist is 24 m.
A
Rope
12 m
sin 30º =
P-80
M A T H E M A T I C S --
30º
C
B
2
X
T E R M – 1
(ii) Trigonometric ratios of an acute angle of a right angled triangle.
(iii) Single mindedness helps us to gain success in life.
½
½
SECTION D
5.
sec q - 1
sec q + 1
+
sec q + 1
sec q –1
L.H.S. =
=
=
(sec q - 1) + (sec q + 1)
1
(sec q + 1)(sec q - 1)
2 sec q
2
sec q - 1
=
2 sec q
tan q
1
1
cos q
= 2 × cos q ´ sin q
1
= 2 × sin q
1
= 2cosec θ
= R.H.S.
SUMMATIVE ASSESSMENT
WORKSHEET-60
SECTION A
tan θ + cot θ = 5
9.
On squaring both sides
tan2 θ + cot2 θ + 2 tan θ cot θ = 25
⇒
tan2 θ + cot2 θ + 2 tan θ
⇒
1
= 25
tan θ
tan2 θ + cot2 θ = 23.
sec 2A = cosec (A – 27º)
2.
1
We know that
sec 2A = cosec (90º – 2A)
⇒
A – 27º = 90º – 2A
⇒
3A = 117º
⇒
A=
117º
= 39º.
3
1
SECTION B
3.
6cos (90° – 23)º + cosec (90° – 79)º + 3cot (90° – 48)º
6sin 23°+ sec 79°+ 3 tan 48°
=
cosec 11º + 3cot 42º + 6cos 67º
cosec11°+ 3cot 42°+ 6 cos 67°
=
6cos 67º +cosec11º + 3cot 42º
cosec 11º + 3cot 42º + cos 67°
= 1.
S O L U T I O N S
1
1
P-81
SECTION C
4. (i) Let AB be the tree broken at a point C such that the broken part CB takes the position CO and
strikes the ground at O. It is given that OA = 30 m and ∠AOC = 30º.
Let AC = x and CB = y, then CO = y
In ∆OAC, we have
tan 30º =
1
⇒
=
3
⇒
x =
Again in ∆OAC, we have
cos 30º =
AC
OA
B
x
30
y
30
= 10 3
3
y
OA
OC
⇒
x
30º
30 m
O
30
3
=
y
2
60
= 20 3
y =
3
⇒
1
C
A
∴
1
Height of the tree = (x + y) = 10 3 + 20 3 = 30 3
= 30 × 1·732 = 51.96 m
(ii) Trigonometric ratios of an acute angle of a right angled triangle.
(iii) The problem of decreasing ratio of trees and land is discussed here.
½
1
SECTION D
5.
⇒
⇒
Now,
15 tan2 θ + 4 sec2 θ
15 tan2 θ + 4(tan2 θ + 1)
15 tan2 θ + 4 tan2 θ + 4
19 tan2 θ
tan θ
θ
(sec θ + cosec θ)2 – sin2 θ
=
=
=
=
=
=
=
23
23
23
19
1 = tan 45º
45º
(sec 45º + cosec 45º)2 – sin2 45º
e 2 + 2 j − FGH 12 IJK
2 1
= e2 2 j −
2
2
=
= 8−
1
1
2
1
1 15
=
2 2
1
SUMMATIVE ASSESSMENT
WORKSHEET-61
SECTION A
1. Given :
From Fig.
C
a
sin θ =
b
AB =
tan θ =
2.
P-82
a
b2 − a 2
AC
=
AB
a
b2 − a 2
b
·
θ
A
cos2 x + sin2 x = 1
M A T H E M A T I C S --
X
b2 − a2
1
B
1
T E R M – 1
SECTION B
5 cosec θ = 7
3. Given :
⇒
cosec θ =
7
5
⇒
sin θ =
5
7

1 
Q cos ec θ =

sin
θ 1

sin θ + cos2 θ – 1 = sin θ – (1 – cos2 θ)
= sin θ – sin2 θ
=
FG IJ
H K
5
5
−
7
7
2
=
35 – 25 10
=
49
49
1
SECTION C
cos θ + sin θ =
4.
2 cos θ
⇒
sin θ = cos θ ( 2 – 1)
⇒
sin θ =
⇒
sin θ =
cos q ( 2 - 1) ( 2 + 1)
1
( 2 + 1)
cos θ(2 − 1)
2 +1
( 2 + 1) sin θ = cos θ
⇒
⇒
1
2 sin q + sin θ = cos θ
cos θ – sin θ =
2 sin q .
Proved. 1
SECTION D
5.
sin A =
∴
cos A =
sin B =
∴
Now
cos B =
=
cos (A + B) =
S O L U T I O N S
5
1 − sin 2 A
, cos A =
2
F
1 I
1−G
H 5 JK
1
10
1−
=
IJ
10 K
1
2
1
=
5
5
1
1 − sin 2 B
, cos B =
FG
H
1−
2
=
1
=
10
1−
3
1
10
cos ( A + B) = cos Acos B – sin Asin B
=
∴
1
2
5
×
6
50
1
2
A + B = 45º.
−
3
10
–
1
50
1
5
=
×
5
50
1
10
=
5
2 2
=
1
2
1
= cos 45º
1
P-83
SUMMATIVE ASSESSMENT
WORKSHEET-62
SECTION A
1. Q sin θ = cos θ ⇒ tan θ = 1 ⇒ θ = 45°
C
1
cos θ =
Also
2
2 tan θ + cos2 θ = 2(1) +
Now,
=2+
FG 1 IJ
H 2K
2
2
1
1 5
= ·
2 2
θ
A
1
B
1
cosec θ = 2
2.
1
= sin 30º
2
θ = 30º
⇒
sin θ =
⇒
Now
⇒
cot θ =
3p
cot 30º =
3p
3p
⇒
⇒
3 =
p = 1.
1
SECTION B
3.
sec2 (90º – q ) – cot 2 q
2(sin 2 25º +sin 2 65º )
–
2cos2 60º tan 2 28º tan 2 62º
3(sec 2 43º – cot2 47º )
2
2
= (cosec q – cot q ) –
2(sin 2 25º + cos2 25º )
1 1
2× × tan 2 28º×cot2 28º
2 2
3[sec2 43º – tan 2 43º ]
1
1
× tan 2 28º×
1
2
tan 2 28º
=
–
2 × (1)
3
=
1
½
1 1
1
– =
2 6
3
½
SECTION C
4. LHS =
æ cosec A+cot A ö
1
1
1
ç
=
´ç
÷÷ – cosec A
cosec A–cot A sin A cosec A–cot A è cosec A+cot A ø
Now RHS =
1
1
−
sin A cosec A + cot A
=
cos ec A + cot A
– cosec A
cos ec 2 A – cot 2 A
=
cosec A + cot A
– cosec A = cot A
1
P-84
FG
H
½
1
cosec A – cot A
×
cosec A + cot A
cos ec A – cot A
(cosec A–cot A)
= cosec A –
cosec 2 A - cot 2 A
= cosec A –
= cosec A –
∴
1
LHS = RHS
(cosec A - cot A)
= cot A
1
M A T H E M A T I C S --
X
IJ
K
1
½
Proved.
T E R M – 1
SECTION D
5.
2 sin (3x – 15°) =
3
3
2
sin (3x – 15)° = sin 60°
3x – 15 = 60°
3x = 60 + 15 = 75°
sin (3x – 15)° =
⇒
⇒
⇒
Now
1
75
= 25°
3
sin2 (2x + 10)° + tan2 (x + 5)° = sin2 (50 + 10)° + tan2 (25 + 5)°
= sin2 60° + tan2 30°
x =
2
 3  1 
 + 
= 

 2   3
=
3 1
+
4 3
=
9+7
13
·
=
12
12
1
2
1
1
SUMMATIVE ASSESSMENT
WORKSHEET-63
SECTION A
sec θ + tan θ = 7
1. Given :
⇒
(sec θ + tan θ)(sec θ − tan θ)
=7
(sec θ − tan θ)
⇒
sec 2 θ − tan 2 θ
=7
sec θ − tan θ
⇒
1
=7
sec θ − tan θ
⇒
sec θ – tan θ =
1 − sec 2 A
2.
cosec 2 A − 1
=
=
1
·
7
1
1 − (1 + tan 2 A)
1 + cot 2 A − 1
− tan 2 A
cot 2 A
=
– tan 2 A
1 / tan 2 A
= – tan4 A
1
SECTION B
3. Given :
⇒
⇒
Hence,
S O L U T I O N S
2 sin 2θ =
3
3
= sin 60º
2
2θ = 60º
sin 2θ =
cos 2θ = cos 60º =
1
·
2
1
1
P-85
SECTION C
4. (i) Let A be the kite and CA be the string attched to the kite such that its one end is tied to a point
C on the ground. The inclination of the string CA with the ground is 60º.
In ∆ABC, we are given that ∠C = 60º and perpendicular AB = 60 m.
AB
∴
sin C =
AC
AB
⇒
sin 60º =
AC
60
3
⇒
=
AC
2
120
⇒
AC =
= 40 3 m
3
C
A
60 m
60º
B
Hence, the length of the string is 40 3 m.
(ii) Trigonometric ratios of an acute angle of a right angled triangle.
(iii) Playing makes children’s mind and body healthy.
2
½
½
SECTION D
5. Consider an equilateral triangle of sides ‘a’ units
A
30°
a
a
60°
60°
B
∴ BD = CD =
As
∴
D
a
C
a
· . Also ∠BAD = ∠CAD = 30°
2
∠A = ∠B = ∠C = 60°. In the right ∆ADB, ∠ADB = 90°
AD2 = AB2 – BD2
(By Pythagoras Theorem)
⇒
AD2 = a2 –
a2
3a2
=
4
4
3
a
2
BD
tan 30° =
=
AB
1
1
AD =
a/2
3a / 2
=
1
3
·
SUMMATIVE ASSESSMENT
1
WORKSHEET-64
SECTION A
1. The maximum value of sin θ is 1.
2.
1
tan θ =
7
8
1 − sin 2 θ
(1 + sin θ)(1 − sin θ)
1
cos 2 θ
=
=
= cot2 θ =
2
2
(1 + cos θ)(1 − cos θ)
tan 2 θ
1 − cos θ
sin θ
1
64
=
=
·
2
49
(7 / 8)
P-86
M A T H E M A T I C S --
X
T E R M – 1
SECTION B
3.
L.H.S. = – 1 +
sin A sin (90º − A)
cot (90 º − A)
sin A cos A
tan A
½
= – 1 + sin A cos A × cot A
½
cos A
sin A
½
= – 1+
= – 1 + sin Acos A ×
= – 1 + cos2 A = – (1 – cos2 A)
½
= – sin2 A = R.H.S.
SECTION C
4. Let
Q
⇒
Q
∴
⇒
⇒
Hence,
Now
AB
AC – AB
AC
AC2
(x + 1)2
x2 + 2x + 1
=
=
=
=
=
=
x
1
x+1
AB2 + BC2
x2 + (5)2
x2 + 25
24
= 12
2
AB = 12, AC = 13
A
x
x+1
2x = 24 ⇒ x =
sin C =
AB 12
=
AC 13
cos C =
BC 5
=
AC 13
12 25
1+
1 + sin C
13 = 13 = 25 ·
=
5
18 18
1 + cos C
1+
13 13
B
5 cm
C
1
1
1
SECTION D
5.
tan θ + sin θ
tan θ − sin θ
sin q
+ sin q
cos q
= sin q
- sin q
cos q
æ 1
ö
sin q çç
+ 1÷÷
è cos q ø
æ 1
ö
=
sin q çç
–1÷÷
è cos q ø
L.H.S. =
sec q + 1
= sec q - 1 = R.H.S.
S O L U T I O N S
1½
1½
Proved. 1
P-87
SUMMATIVE ASSESSMENT
WORKSHEET-65
SECTION A
5 tan θ = 4
1. Given :
Now
5 sin θ − 3 cos θ
=
5 sin θ + 3 cos θ
=
2. Given :
∴
⇒
∴
4−3
1
1
=
= ·
4+3 4+3 7
1
3
= tan θ
x
3x = sec θ, and
(3x)2 +
FG sin θ IJ − 3
H cos θ K = 5 tan θ − 3
F sin θ IJ + 3 5 tan θ + 3
5G
H cos θ K (Divide Numerator & Denominato by cos θ)
5
FG 3 IJ
H xK
2
= sec2 θ – tan2 θ
1

9  x2 − 2  = 1

x 
x2 –
1
x
=
2
1
9
1
SECTION B
3.
R.H.S. =
=
=
p2 - 1
p2 +1
=
(cosec q + cot q)2 –1
(cosec q + cot q) 2 + 1
cosec2 q + cot 2 q + 2cosec q.cot q –1
cosec2 q + cot 2 q + 2cosec q.cot q + 1
1 + cot 2 q + cot 2 q + 2 cosec q cot q - 1
1
cosec 2 q + cosec2 q - 1 + 2 cosec q cot q + 1
2 cot q (cot q + cosec q)
= 2 cosec q (cosec q + cot q)
=
cos θ
× sin θ = cos θ = LHS.
sin θ
Proved 1
SECTION C
4.
L.H.S. =
=
P-88
cos A
sin A
+
1 – tan A 1 − cot A
cos A
sin A
+
sin A
cos A
1−
1−
cos A
sin A
FG
H
IJ
K
FG
H
IJ
K
=
cos 2 A
sin 2 A
+
cos A − sin A sin A − cos A
=
cos 2 A
sin 2 A
–
cos A − sin A cos A − sin A
M A T H E M A T I C S --
1
1
X
T E R M – 1
=
cos 2 A − sin 2 A
cos A − sin A
=
(cos A − sin A)(cos A + sin A)
(cos A − sin A)
Proved. 1
= cos A + sin A = R.H.S.
SECTION D
5.
L.H.S. =
cosec A
cosec A
+
cosec A–1 cosec A+1
1
cosec2 A+cosec A+cosec2 A–cosec A
=
(cosec A–1)(cosec A+1)
=
=
2cosec2 A
cosec2 A–1
2cosec2 A
1
cot 2 A
2
2
sin
A
=
2
cos A
sin 2 A
=
=
sin 2 A
´
sin 2 A cos 2 A
2
2
cos 2 A
1
= 2 sec2 A = R.H.S.
Proved. 1
SUMMATIVE ASSESSMENT
WORKSHEET-66
SECTION A
1. Given :
tan θ =
So
C
5 tan θ = 12
12
5
13
12
FG IJ
H K
13 12
13 sin θ
=
3 13
3
= 4.
θ
A
5
1
B
sin θ
sin θ(1 − cos θ)
=
1 + cos θ
(1 + cos θ)(1 − cos θ)
2.
=
sin θ(1 − cos θ)
2
1 − cos θ
=
sin θ(1 − cos θ)
2
sin θ
=
1 − cos θ
sin θ
1.
SECTION B
3. Given :
⇒
S O L U T I O N S
4 cos θ = 11sin θ
cos θ =
11
sin θ
4
P-89
11
sin q - 7 sin q
4
11 cos θ − 7 sin θ
=
11
11 cos θ + 7 sin θ
11´ sin q + 7 sin q
4
11´
Now
FG 121 − 7IJ
H4 K
=
F 121 + 7IJ
sin θG
H4 K
1
sin θ
=
121 – 28 93
=
·
121 + 28 149
SECTION C
sec θ + tan θ = λ
4. Let
...(1)
We know that
sec2 θ – tan2 θ = 1
⇒
(sec θ + tan θ) (sec θ – tan θ) = 1
⇒
λ(sec θ – tan θ) = 1
⇒
sec θ – tan θ =
Adding eqns. (1) and (2), we get
1
1
λ
...(2)
1
λ
1
=λ+
λ
2sec θ = λ +
FG
H
IJ
K
1
4x
1
1
⇒
2x +
=λ+
2x
λ
Comparing both sides, we get
⇒
2 x+
1
λ = 2x or λ =
⇒
1
2x
1
·
2x
sec θ + tan θ = 2x or
1
SECTION D
cosec θ =
5. Given :
(i)
cos θ =
2
2/ 5
1/ 5
=2–
1
− 5
sin2 θ
+
cos2 θ
1½
5
F 1 IJ + FG 2 IJ
= G
H 5K H 5K
2
P-90
1 4
=
5 5
5
cos θ
cot θ – cosec θ = sin θ − cosec θ
=
(ii)
5
cos2 θ = 1 – sin2 θ = 1 –
So,
∴
1
5 Þ sin θ =
2
=
1 4
+ = 1.
5 5
M A T H E M A T I C S --
1½
X
T E R M – 1
SUMMATIVE ASSESSMENT
WORKSHEET-67
SECTION A
2 tan 30º
1.
1 + tan 2 30º
=
=
2.
tan
FG A + BIJ
H 2 K
=
=
=
 1 
2
 3 
2
2
3
3
=
=
2
1
4
 1 
1+
1+
3
3
 3 
2 3
× = 3
3 4
2
sin 60º.
æ
Cö
tan ç
ç90º – ÷÷
2ø
è
C
cot
·
2
1
LMQ A + B + C = 90ºOP
N 2 2 2 Q
1
SECTION B
3. In the ∆ABP,
sin 30º =
1
50
=
⇒ AP = 100 cm
2
AP
⇒
In the ∆ AQD,
sin 30º =
AD
AQ
20
1
⇒ AQ = 40 cm
=
AQ
2
⇒
Now,
AB
AP
½
½
½
the length of (AP + AQ) = 100 + 40 = 140 cm.
SECTION C
4.
cos2 (45º + q) + cos2 (45º – q)
+ cosec (75º + θ) – sec (15º – θ)
tan (60º + q) tan (30º – q )
=
cos2 (45º + q) + sin 2 (90º – 45º + q )
+ cosec (75º + θ) – cosec (90º – 15º + θ)
tan (60º + q)cot (90º – 30º + q )
1
=
cos2 (45º + q )+ sin 2 (45º + q )
+ cosec (75º + θ) – cosec (75º + θ)
tan (60º + q).cot (60º + q )
1
=
1
= 1.
1
1
SECTION D
5.
L.H.S. = (1 + cot A – cosec A) (1 + tan A + sec A)
FG cos A − 1 IJ FG1 + sin A + 1 IJ
H sin A sin A K H cos A cos A K
F sin A + cos A − 1 IJ FG sin A + cos A + 1 IJ
= GH
K H cos A K
sin A
= 1+
S O L U T I O N S
1
P-91
=
(sin A + cos A) 2 − 12
sin A cos A
1
=
sin 2 A + cos 2 A + 2 sin A cos A – 1
sin A cos A
1
1 + 2 sin A cos A – 1
sin A cos A
= 2 = R.H.S.
=
1
SUMMATIVE ASSESSMENT
WORKSHEET-68
SECTION A
1.
⇒
⇒
sin (x –20)º =
=
x – 20 =
4x =
⇒
x =
tan θ =
2.
h=
=
cos (3x –10)º
sin [90º – (3x –10)º]
90 – 3x + 10
90 + 10 + 20 = 120
120
= 30.
4
6 h
=
4 28
6 × 28
θ
4
A
42.
C
C
6
4
B
1
h
θ
A
28
B
1
SECTION B
L.H.S. = (cosec θ – cot θ)2
3.
⇒
cos θ 
 1
−


sin θ sin θ 
⇒
 1 − cos θ 


sin θ 
2
2
(1 − cos θ)2
=
1 − cos2 θ
(1 − cos θ)2
(1 − cos θ)(1 + cos θ)
1 − cos θ
= R.H.S.
1 + cos θ
=
=
SECTION C
a2cos2 θ
4. We have
=
+ 2absin θ cos θ + b2sin2 θ
and
n2 = a2sin2 θ – 2absin θ cos θ + b2cos2 θ
Adding (i) and (ii), we get
m2 + n2 = a2(cos2 θ + sin2 θ) + b2(cos2 θ + sin2 θ)
= a2 (1) + b2 (1)
= a2 + b2 = RHS.
m2
...(i) 1
...(ii) 1
Proved. 1
SECTION D
5.
L.H.S. =
tan θ
cot θ
+
1 − cot θ 1 − tan θ
1
tanq
tan
q
+
=
1 1- tanq
1tan q
P-92
M A T H E M A T I C S --
X
T E R M – 1
=
tan 2 q
1
+
tan q - 1 tan q (1- tan q)
=
tan 3 q - 1
(tan q - 1) tan q
1½
1
(tan q - 1)(tan 2 q + tan q + 1) tan 2 q + tan q + 1
=
(tan q - 1)(tan q)
tan q
= tan θ + 1 + cot θ = R.H.S.
1
=
SUMMATIVE ASSESSMENT
½
WORKSHEET-69
SECTION A
1.
tan A =
Now
cosec2 A –1 =
=
8
15
1
sin 2 A
cos 2 A
sin 2 A
1
–1=
1 − sin 2 A
sin 2 A
= cot2 A
1
225
=
tan A (8 / 15) 2
64
tan x = sin 45º cos 45º + sin 30º
=
2.
=
2
=
1
FG 1 IJ FG 1 IJ + 1
H 2 KH 2 K 2
1 1
+ = 1 = tan 45º
2 2
x = 45º.
=
⇒
1
SECTION B
∠C = 90º (Angle in a semi-circle)
3.
∴
∴
AB =
(3) 2 + (2) 2 = 9 + 4 = 13
tan A =
BC 2
=
AC 3
tan B =
AC 3
=
BC 2
tan A tan B =
½
½
½
2 3
. = 1.
3 2
√13 cm
½
SECTION C
4.
cosec θ =
13
12
12
13
cos2 θ = 1 – sin2 θ
sin θ =
 12 
= 1–  
 13 
S O L U T I O N S
2
1
P-93
169 − 144
25
=
169
169
5
cos θ =
13
 12 
 5 
2×  − 3× 
13
 
 13 
2 sin θ − 3 cos θ =
 12 
 5 
4 sin θ − 9 cos θ
4×  − 9× 
 13 
 13 
24 − 15
9
=
=
= 3.
48 − 45
3
=
⇒
Now
1
1
SECTION D
5.
tan θ + sec θ − 1
tan θ − sec θ + 1
tan q + sec q - (sec 2 q - tan 2 q)
=
[Q 1 = sec2 θ – tan2 θ]
tan q - sec q + 1
(tan θ + sec θ) − [(sec θ + tan θ) (sec θ − tan θ)]
1
=
tan θ − sec θ + 1
L.H.S. =
[1 − (sec θ − tan θ)]
tan θ − sec θ + 1
[1 − sec θ + tan θ ]
= (tan θ + sec θ)
[1 − sec θ + tan θ ]
= (tan θ + sec θ)
=
sin θ
1
+
cos θ cos θ
=
1 + sin θ
= R.H.S.
cos θ
1
1
Proved. 1
SUMMATIVE ASSESSMENT
WORKSHEET-70
SECTION A
1
1. Maximum value of sec θ , i.e., cos θ = 1
1
SECTION B
2.
cos 45°
1
+
=
sec 30° sec 60°
=
=
=
1
2 +1
2 2
3
1
2
×
3 1
+
2
2
6 1
+
4
2
1
6 +2
·
4
SECTION C
3.
P-94
L.H.S. =
sin θ − cos θ sin θ + cos θ
+
sin θ + cos θ sin θ − cos θ
M A T H E M A T I C S --
X
T E R M – 1
=
=
=
(sin θ − cos θ) 2 + (sin θ + cos θ) 2
sin 2 θ − cos 2 θ
(sin 2 θ + cos 2 θ) – 2 sin θ cos θ + (sin 2 θ + cos 2 θ) + 2 sin θ cos θ
1+1
1
sin θ − 1 + sin 2 θ
2
=
= R.H.S.
2
2 sin θ − 1
2
sec 41º .sin 49º + cos 29º .cos ec 61º −
4.
1
sin 2 θ − (1 − sin 2 θ)
2
3
Proved. 1
(tan 20º .tan 60º .tan 70º)
3(sin 2 31 º + sin 2 59º )
2
[tan 20º. 3 cot (90º– 70º)]
3
3[sin 2 31º + cos 2 (90º –59º )]
cosec (90º– 41º)sin 49º + cos 29º.sec (90º– 61º) =
2
[tan 20º 3.cot 20º ]
3
3(sin 2 31º + cos 2 31º )
1
cosec 49º.sin 49º + cos 29º.sec 29º =
=
1
1+1− 2
2−2
=
=0
3
3
1
SECTION D
5.
L.H.S. =
=
=
=
=
=
2
θ–
θ – 2 cosec2 θ + cosec4 θ
2 sec2 θ – (sec2 θ)2 – 2 cosec2 θ + (cosec2 θ)2
2 (1 + tan2 θ) – (1 + tan2 θ)2 – 2(1 + cot2 θ) + (1 + cot2 θ)2
2 + 2 tan2 θ – 1 – 2 tan2 θ – tan4 θ – 2 – 2 cot2 θ + 1 + 2 cot2 θ + cot4 θ
cot4 θ – tan4 θ
R.H.S.
sec2
sec4
SUMMATIVE ASSESSMENT
1
1
1
1
WORKSHEET-71
SECTION A
1.
AB =
(AC) 2 − (BC) 2
=
(17) 2 − (8) 2 = 289 − 64
=
225 = 15
15sec A + 8cot A = 15
2.
FG 17 IJ
H 15 K
+8
C
17
8
FG 15 IJ
H 8K
= 17 + 15 = 32.
1
2
sin A =
tan2 45º
2
2
1
1
=
× (1)2 =
2
2
1
sin A =
= sin 45º
2
A
B
15
A
1
FG IJ
H K
⇒
⇒
S O L U T I O N S
A = 45º
1
P-95
SECTION B
3.
cos (A – B) =
3
= cos 30º ⇒ A – B = 30º
2
...(1) ½
sin (A + B) =
3
= sin 60º ⇒ A + B = 60º
2
...(2) ½
2A = 90º ⇒ A = 45º
Adding equations (1) and (2),
From (2),
½
½
B = 60º – A = 60º – 45º = 15º
SECTION C
4. we know that,
sec (90º – θ) = cosec θ, tan (90º – θ) = cot θ, cot (90º – θ) = tan θ, cosec (90º – θ) = sec θ
1
sin q.sec (90º – q ) tan q
tan (90º– q )
sin θ.cosec θ.tan θ cot θ
−
Hence, cosec (90° – q) cos q.cot (90° – q )–
=
cot q
sec θ.cos θ.tan θ
cot θ
1
=
1
× tan θ
sin θ
−1 = 1 – 1 = 0
× cos θ.tan θ
sin θ ×
1
cos θ
1
SECTION D
5. sin θ =
∴
⇒
⇒
3
3
7
⇒ cos θ =
and tan θ =
7
4
4
cosec 2 q-cot 2 q
sec2 q -1
1
2
tan q
+2 cot q =
+ 2´
½+½
7
+ cos q
x
7
7
7
+
=
3
x
4
1
1
2 7
7
7
+
+
=
tan q
3
x
4
⇒
7 2 7
7
7
+
=
3
3
4
x
⇒
7
4 7- 7
=
x
4
1
4
7
3 7
=
⇒x= ·
3
x
4
⇒
SUMMATIVE ASSESSMENT
1
WORKSHEET-72
SECTION A
1.
cos2 17º– sin2 73º = cos2 17º – sin2 ( 90 – 17)º
= cos2 17º – cos2 17º
1
= 0.
2.
P-96
A = 30º, sin 2A = sin 2 (30º) = sin 60º =
3
·
2
M A T H E M A T I C S --
1
X
T E R M – 1
SECTION B
2
2
 2 
1
2
2  + 3
 − 2(1)
2 cos2 60° + 3 sec 2 30° − 2 tan 2 45°
2
 
 3
=
2
2
sin 2 30° + cos2 45°
1  1 
+
  

2  2 
3.
1
2
+4−2
4
= 1 1
+
4 2
=
10
·
3
1
SECTION C
1
= sin 30º
2
A + B – C = 30º
4. We have
sin (A + B – C ) =
∴
...(1) 1
1
= cos 45º
2
∴
B + C – A = 45º
Adding eqs. (1) and (2), we get
2B = 75º
⇒
B = 37.5º
Now subtracting eqn. (2) from eqn. (1), we get
2(A – C) = – 15º ⇒ A – C = – 7·5º
We know that,
A + B + C = 180º
A + C = 142·5º
Adding eqns. (3) and (4), we get
2A = 135º
A = 67·5º
⇒
C = 75º
Hence,
∠A = 67·5º, ∠B = 37·5º, ∠C = 75º.
and
cos ( B + C – A ) =
5.
sec A =
sin A =
L.H.S. =
...(2) 1
...(3)
1
...(4)
1
17
8
⇒ cos A =
8
17
1 − cos2 A =
 8 
1− 
 17 
2
=
15
17
1
3 − 4 sin2 A
4 cos2 A – 2
 225 
3 − 4

 289 
=
 64 
4
−3
 289 
=
867 − 900
256 − 867
1
− 33
33
·
= − 611 =
611
S O L U T I O N S
P-97
R.H.S. =
3 − tan 2 A
1 − 3 tan 2 A
2
 15 
3− 
 8 
=
2
 15 
1 − 3 
 8 
1
192 − 225
33
=
64 − 675
611
L.H.S. = R.H.S.
=
∴
SUMMATIVE ASSESSMENT
1
WORKSHEET-73
SECTION A
1.
cos ( A + B) = cos (180º – C)
=
sin (45º + θ) – cos (45º – θ) =
=
=
2.
1
cos C = cos 90º = 0.
sin [90º – (45º + θ)] – cos (45º – θ)
cos (45º – θ) – cos (45º – θ)
0.
1
SECTION B
3.
3 tan θ = 1
tan θ =
⇒
∴
sin2
θ–
cos2
1
3
1
= tan 30°
θ = 30°
θ = sin2 30° – cos2 30°
2
1  3 

=   − 
 2   2 
=
2
1 3
2
1
− =–
=– ·
4 4
4
2
1
SECTION C
4. Given :
⇒
and
xsin θ = y cos θ
ycos θ
x =
sin θ
x sin3 θ + ycos3 θ = sin θcos θ
...(1) 1
...(2)
Eliminating x from (1) and (2), we get
y cos q
3
3
sin q . sin θ + ycos θ = sin θcos θ
⇒
y cos θ sin2 θ + ycos3 θ = sin θcos θ
y cos θ [sin2 θ + cos2 θ] = sin θ cos θ
⇒
P-98
y = sin θ
M A T H E M A T I C S --
...(3) 1
X
T E R M – 1
Substituting this value of y in (3), we get
x = cos θ
...(4) 1
∴ Squaring and adding (3) and (4), we get
x2 + y2 = cos2 θ + sin2 θ = 1.
Proved.
SECTION D
5.
L.H.S. =
cos 2 θ
sin 3 θ
+
1 − tan θ sin θ − cos θ
cos 2 θ
sin 3 θ
+
sin θ sin θ − cos θ
1−
cos θ
cos 3 θ
sin 3 θ
–
=
cos θ − sin θ cos θ − sin θ
1
=
=
cos 3 θ − sin 3 θ
cos θ − sin θ
=
(cos θ − sin θ)(cos 2 θ + sin 2 θ + sin θ cos θ)
(cos θ − sin θ)
= 1 + sin θ cos θ = R.H.S.
SUMMATIVE ASSESSMENT
1
1
Proved. 1
WORKSHEET-74
SECTION A
tan θ + cot θ
Squaring on both sides,
tan 2 θ + cot2 θ + 2tan θ cot θ
⇒
tan2 θ + cot2 θ
2.
tan 3A
⇒
cot (90º – 3A)
⇒
90º – 3A
1.
⇒
=2
=
=
=
=
=
4
4 – 2 = 2, as tan θ cot θ = 1.
cot (A – 26°)
cot (A – 26º)
A – 26º
1
4A = 90º + 26º = 116º
⇒
Α=
116º
= 29º.
4
1
SECTION C
4.
cos 50º = cos (90° – 40)º = sin 40º
cosec2 59º = cosec2 (90º – 31º) = sec2 31º
1
tan 78º = tan (90° – 12º) = cot 12º
Hence,
cos 50º 4(cosec 2 59º - tan 2 31º ) 2
+
- tan 12º tan 78º. sin 90º
2sin 40º
3
3 tan 2 45º
S O L U T I O N S
1
=
sin 40º
4(sec 2 31º– tan 2 31º ) 2
+
− tan 12º cot 12º × 1
2 sin 40º
3
3 × (1) 2
=
1 4 2 7
+ − =
2 3 3 6
1
P-99
SECTION D
5. Let cot θ = x,
3 cot2 θ – 4cot θ +
3
or
(x –
x2
3 = 0 becomes
– 4x +
1
3 =0
3 ) ( 3 x – 1) = 0
∴
x =
⇒
cot θ =
∴
3 or
1
1
3
3 or cot θ =
1
3
θ = 30º or θ = 60º
If θ = 30º, then
cot2 30º + tan2 30º = ( 3 ) 2 +
If θ = 60º, then
cot2
60º +
tan2
F 1 IJ
60º = G
H 3K
2
FG 1 IJ
H 3K
2
=3+
+ ( 3) 2 =
1 10
=
3
3
1
10
1
+3=
3
3
SUMMATIVE ASSESSMENT
1
WORKSHEET-75
SECTION A
1.
⇒
⇒
cos 2θ = sin 4θ
sin (90º – 2θ) = sin 4θ
90º – 2θ = 4θ ⇒ 6θ = 90º
90º
= 15º.
6
tan x = sin 45º cos 45º + cos 60º
1
1
1
+
=
2
2
2
⇒
θ=
2.
1
FG IJ FG IJ
H KH K
1 1
+ = 1 = tan 45º
2 2
x = 45º.
=
⇒
1
SECTION B
3. cos 68º + tan 76º = cos (90º –22º) + tan (90º – 14º)
1
= sin 22º + cot 14º,
[Q cos (90º – θ) = sin θ and tan (90º – θ) = cot θ ] 1
SECTION C
4.
Given : cos θ + sin θ = p and sec θ + cosec θ = q
∴
L.H.S. = q(p2 – 1) = (sec θ + cosec θ) [(cos θ + sin θ)2 – 1]
= (sec θ + cosec θ) (1 + 2 sin θ cos θ –1)
=
½
FG 1 + 1 IJ (2 sinθ cos θ)
H cos θ sin θ K
1
sin θ + cos θ
× 2 sin θ cos θ
cos θ sin θ
= 2(sin θ +cos θ)
= 2p
= R.H.S.
=
P-100
M A T H E M A T I C S --
1
Proved. ½
X
T E R M – 1
SECTION D
x2 = r2 sin2 A cos2 C
5. Since,
y2 = r2 sin2 A sin2 C
z2 = r2 cos2 A
x2
=
cos2
=
r2 sin2
(cos2 C
=
r2 sin2 A
=
r2 (sin2 A
=
r 2.
Proved. 1
SUMMATIVE ASSESSMENT
WORKSHEET-76
+
z2
1
r2 sin2 A
+
y2
A
+
C+
r2 sin2
A
sin2 C)
+
sin2
+
C+
r2 cos2 A
1
r2 cos2 A
r2cos2 A
+
1
cos2 A)
SECTION A
1. Q
2.
sin 20° = sin (90° – 70°) = cos 70°
3 sin2 20° – 2 tan2 45° + 3 sin2 70° = 3 cos2 70° + sin2 70°) – 2 × (1)2
= 3×1–2=1
2
2
(1 + tan θ) cos θ = sec2 θ cos2 θ
=
1
cos2 θ
= cos2 θ = 1.
1
1
SECTION B
⇒
⇒
⇒
sin (36 +θ)º = cos (16 + θ)º
cos [90º –(36 + θ)º] = cos (16 + θ)º
90 – 36 – θ = 16 + θ
2θ = 90 – 36 – 16 = 38
⇒
θ=
3.
1
38
= 19.
2
SECTION C
4. Given :
(sec A + tan A) (sec B + tan B) (sec C + tan C) = (sec A – tan A) (sec B – tan B) (sec C – tan C)
Multiply both the sides by (sec A – tan A) (sec B –tan B) (sec C – tan C)
(sec A + tan A) (sec B + tan B) (sec C + tan C ) × (sec A – tan A) (sec B – tan B) (sec C – tan C)
= (sec A – tan A)2 (sec B – tan B)2 (sec C –tan C)2
1
2
2
2
2
2
2
⇒ (sec A – tan A) (sec B – tan B) (sec C – tan C)
= (sec A – tan A)2 (sec B – tan B)2 (sec C – tan C)2
⇒ 1 = [ (sec A – tan A) (sec B – tan B) (sec C – tan C)]2
⇒ (sec A – tan A) (sec B – tan B) (sec C – tan C) = ± 1
1
Similarly, multiply both sides by (sec A + tan A) (sec B + tan B) (sec C + tan C), we get
(sec A + tan A) (sec B +tan B) (sec C + tan C) = ± 1.
Proved. 1
5.
Given expression =
S O L U T I O N S
tan q.cos q cot (90º – 50º )
+
– [cos2 20º + cos2 (90º – 20º)]
sin q
tan 50º
1
sin q cos q tan 50º
= cos q . sin q + tan 50º – [cos2 20º + sin2 20º]
1
= 1 + 1 – 1 = 1.
1
P-101
SECTION D
cos 65º
cos 65º
cos 65º
= sin (90º -65º ) = cos 65º
sin 25º
tan(90º–70º ) cot 70º
tan 20º
=
=
cot 70º
cot 70º
cot 70º
sin 90º
tan 5º tan 35º tan 60º tan 55º tan85º
6.
1
=1
=1
=1
= tan (90º – 85º) tan (90º – 55º) tan 55º tan 60º.tan 85º.
= cot 85º.tan 85º.cot 55ºtan 55º.
3= 3
Given expression = 1 – 1 – 1 + 3 =
1
3
1
= 1×1×
∴
3 – 1.
FORMATIVE ASSESSMENT
WORKSHEET-77
Objective Type Questions
1. (C)
2. (B)
3. (A)
4. (D)
5. (A)
6. (C)
7. (C)
Fill in the blanks
1. (I) 25°
(VI) 50°
(II) 65 °
(VII) 1°
(III) 59°
(VIII) 10°
(IV) 51°
(IX) 23°
(V) 55°
(X) 53°
Error’s Correction
2. (I)
sin 2 20º + sin 2 70º
cos 2 20º + cos 2 70 º
=
=
2
(II)
2
sin 20º + sin 70º
cos 2 20º + cos 2 70 º
(III) Correct (No Error)
(IV)
sin 2 20º + sin 2 70º
=
sin 2 (90º −70º ) + sin 2 70º
cos 2 (90º −70º ) + cos 2 70º
cos 2 70º + sin 2 70º
2
2
sin 70 º + cos 70º
=
1
=1
1
sin 2 (90º −70º ) + sin 2 70º
cos 2 (90º −70º ) + cos 2 70º
cos 2 70º + sin 2 70º
1
=
=
=1
2
2
1
sin 70 º + cos 70º
=
sin 2 (90º −70º ) + sin 2 70º
cos 2 (90º −70º ) + cos 2 70º
1
=
=1
2
2
1
sin 70 º + cos 70º
(V) Correct (No error)
cos 2 20º + cos 2 70 º
cos 2 70º + sin 2 70º
A
Answers
5
1. sin C =
13
12 cm
5 cm
B
2.
⇒
⇒
sin A = sin A′
2
B' C'
=
5
20
2 × 20
B'C' =
5
=8
13 cm
C
5
A
C'
20
2
B
C
A'
B'
3. TRUE / FALSE
(I) F, (II) F, (III) T, (IV) F, (V) T, (VI) F.
■
P-102
M A T H E M A T I C S --
X
T E R M – 1
CHAPTER
6
Statistics
SUMMATIVE ASSESSMENT
WORKSHEET - 78
SECTION A
10 + 25 35
=
= 17·5.
1
2
2
2. The abscissa of the point of intersection of the ‘‘less than type’’ and ‘‘more than type’’ cumulative
frequency curve of a grouped data is median.
1
1. Class mark of the class 10 – 25 =
SECTION B
3.
Marks
No. of students
c.f.
0 – 10
5
5
10 – 20
15
20
20 – 30
30
50
30 – 40
8
58
40 – 50
2
60
N = 60
Here
So, median class = 20 – 30
N
60
=
= 30
2
2
l = 20, f = 30, c.f. = 20, h = 10
æN
ö
ç - c.f. ÷
ç2
÷
Median = l + ç
÷ ´ h
f
ç
÷÷
ç
è
ø
1
æ 30 - 20 ö
÷÷ ´ 10
= 20 + ç
ç
è 30 ø
100
= 20 +
30
= 20 +
S O L U T I O N S
10
= 20 + 3.33 = 23.33.
3
1
P-103
SECTION C
4.
Class marks (xi)
fi
fixi
27
4
108
32
14
448
37
22
814
42
16
672
47
6
282
52
5
260
57
3
171
Σf = 70
Σfx = 2755
2
å fi xi
Mean = å f
i
∴
=
½
2755
= 39·36.
70
½
SECTION D
5.
Weight (in Kg)
Cumulative Frequency
More than or equal to 0
120
More than or equal to 10
106
More than or equal to 20
89
More than or equal to 30
67
More than or equal to 40
41
More than or equal to 50
18
Plotting the points
2
y
(0, 120)
120
110
Scale
x-axis 2 cm = 10 units
y-axis 1 cm = 10 units
(10, 106)
100
90
(20, 89)
Cumulative frequency
80
More than ogive
70
(30, 67)
60
50
(40, 41)
40
2
30
20
(50, 18)
10
0
P-104
10
20
30
40
50
Lower limits
60
x
M A T H E M A T I C S --
X
T E R M – 1
SUMMATIVE ASSESSMENT
WORKSHEET-79
SECTION A
1.
The mean of observations x1 , x2 , ......, xn = x then the mean of x1 + a, x2 + a ...... xn + a = x + a. 1
SECTION B
2.
Class Interval
Frequency
0 – 50
8
50 – 100
15
100 – 150
32
150 – 200
26
200 – 250
12
250 – 300
7
Total
100
2
SECTION C
3.
Modal class : 30 – 40,
Here
l = 30, f1 = 45, f2 = 12, f0 = 30, h = 10


f1 − f0
Mode = l +  2 f − f − f  × h
0
2 
 1
1
 45 − 30 
= 30 + 
 × 10
 90 − 30 − 12 
= 30 + 3·125 = 33·125
1
1
SECTION D
4.
Class
f
c.f
0 – 50
2
2
25
50
50 – 100
3
5
75
225
100 – 150
5
10
125
625
150 – 200
6
16
175
1050
200 – 250
5
21
225
1125
250 – 300
3
24
275
825
300 – 350
1
25
325
325
Total
Σf = 25
∴
Mean =
x
fx
Σfx = 4225
2
Σfx 4225
=
= 169
Σf
25


f1 − f0
Mode = l +  2 f − f − f  × h
 1 0 2
1
From table modal class = 150 – 200
so f1 = 6, f0 = 5, f2 = 5, l = 150, h = 50
S O L U T I O N S
P-105
So
Mode = 150 +
(6 − 5)
× 50
12 − 5 − 5
50
2
= 150 + 25 = 175
= 150 +
Now
Median =
=
1
1
2
mode +
mean
3
3
1
2
(175) + × 169
3
3
1
= 58·33 + 112·66 = 170·83
SUMMATIVE ASSESSMENT
WORKSHEET-80
SECTION A
1.
Given frequency distribution = 8·1
Σ fixi = 132 + K
Σ fi = 20
Σ fi xi
132 + K
8·1 = Σ f =
20
i
Q
⇒
1
K=6
SECTION B
2.
Modal class : 20 – 30
Here l = 20, f1 = 40, f0 = 24, f2 = 36, h = 10
Mode = l +
½
( f1 − f 0 )
×h
2 f1 − f 0 − f 2
½
= 20 +
(40 – 24)
× 10
80 – 24 – 36
½
= 20 +
16 × 10
= 28
20
½
SECTION C
3.
P-106
xi
fi
xifi
3
10
30
9
p
9p
15
4
60
21
7
147
27
q
27q
33
4
132
39
1
39
Total
Σfi = 26 + p + q
Σxifi = 408 + 9p + 27q
M A T H E M A T I C S --
X
T E R M – 1
Σ fi = 40,
Given,
⇒
⇒
26 + p + q = 40
p + q = 14
–
∴
Mean, x =
14·7 =
...(i) ½
Σxi f i
Σf i
½
408 + 9 p + 27q
40
1
588 = 408 + 9p + 27q
180 = 9p + 27q
p + 3q = 20
Subtracting eq. (1) from eq. (2), we get
2q = 6
⇒
q=3
Putting this value of q in eq. (1), we get
p = 14 – q = 14 – 3 = 11
...(2)
½
½
SECTION D
4.
Class
Mid values
xi
fi
fixi
c.fi
25 – 35
30
7
210
7
35 – 45
40
31
1240
38
45 – 55
50
33
1650
71
55 – 65
60
17
1020
88
65 – 75
70
11
770
99
75 – 85
80
1
80
100
100
4970
Total
Σ fi xi
4970
Mean x = Σ f =
= 49·7
100
i
2
1
N
100
=
= 50
2
2
Hence median class = 45 – 45
so, l = 45, c.f. = 38, f = 33, h = 10
From table
n
 − c. f
Median = l +  2
 f


×h

 50 − 38 
= 45 + 
 × 10
 33 
= 45 + 3·64 = 48·64.
S O L U T I O N S
1
P-107
SUMMATIVE ASSESSMENT
WORKSHEET-81
SECTION A
1.
According to question,
Mode – Mean = k (Median – Mean)
k=
i.e.,
2.
Mode – Mean
Median – Mean
=
3Median – 2Mean – Mean
Median – Mean
=
3(Median – Mean)
= 3.
(Median – Mean)
1
1
The median of give data is 20·5.
SECTION B
3.
Marks
Cumulative frequency
More than or equal to 0
38
More than or equal to 10
35
More than or equal to 20
27
More than or equal to 30
12
More than or equal to 40
5
2
SECTION C
4.
Classes
15
15
100 – 200
17
32
200 – 300
f
32 + f
300 – 400
12
44 + f
400 – 500
9
53 + f
500 – 600
5
58 + f
600 – 700
2
60 + f
⇒
Since,
∴ Median class is 200 – 300.
P-108
c.f.
0 – 100
From table,
⇒
fi
n = 60 + f
n 60 + f
=
2
2
median = 240
1
½
LM n − c. f .OP
×h
Median = l + M 2
MN f PPQ
LM 60 + f − 32 OP
× 100
240 = 200 + M 2
MN f PPQ
M A T H E M A T I C S --
½
½
X
T E R M – 1
LM 60 + f − 64 OP 100
N 2f Q
⇒
40 =
⇒
8f = 10 f – 40
⇒
2f = 40
∴
½
f = 20.
SECTION D
5.
Class Interval
c.f.
No. of students
0 – 10
7
7
10 – 20
21
14
20 – 30
34
13
30 – 40
46
12
40 – 50
66
20
50 – 60
77
11
60 – 70
92
15
70 – 80
100
8
2
From table, maximum frequency = 20. So modal class = 40 – 50
Now l = 40, f1 = 20, f2 = 11, f0 = 12, h = 10
æ f1 - f 0 ö
÷
Mode = l + ç
ç2 f - f - f ÷ ´ h
è 1 0 2ø
1
æ 20 - 12 ö
ç
= 40 + ç
÷÷ ´ 10
è 40 - 12 - 11 ø
= 40 +
8
´ 10
17
= 40 +
80
= 40 + 4·7 = 44·7.
17
SUMMATIVE ASSESSMENT
1
WORKSHEET-82
SECTION A
1.
The modal class of the distribution is 40 – 50.
1
SECTION B
2.
The multiples of 5, according to the problem are :
5, 15, 25, 35, 45
Mean =
=
S O L U T I O N S
5 + 15 + 25 + 35 + 45
5
125
= 25.
5
½
1
½
P-109
SECTION C
3.
For mean,
Q
Class Interval
xi
fi
f ix i
0 – 10
5
3
15
10 – 20
15
8
120
20 – 30
25
10
250
30 – 40
35
15
525
40 – 50
45
7
315
50 – 60
55
4
220
60 – 70
65
3
195
Σf = 50
Σfx = 1640
1
å fi xi
1640
Mean = å f =
= 32·8
50
i
½
For mode,
Modal class = 30 – 40
and
l = 30, f1 = 15, f2 = 7, f0 = 10, h = 10
f1 - f 0
Mode = l + 2 f - f - f ´ h
1
0
2
= 30 +
15 - 10
´ 10
30 - 10 - 7
= 30 +
5
´ 10
13
= 30 +
50
13
1
= 30 + 3·85
∴
1
Mode = 33·85.
SECTION D
4.
Classes
f
c.f.
5 – 10
2
2
10 – 15
12
14
15 – 20
2
16
20 – 25
4
20
25 – 30
3
23
30 – 35
4
27
35 – 40
3
30
Total
Since,
P-110
2
Σf = 30 = N
N
= 15
2
M A T H E M A T I C S --
X
T E R M – 1
∴
Median class is 15 – 20
æN
ç - c. f.
ç2
Median = l + ç
f
ç
ç
è
l = 15, N = 30,
From table,
Median = 15 +
ö
÷
÷
÷ ×h
÷÷
ø
c.f. = 14, f = 2, h = 5
1
FG 15 − 14 IJ × 5
H 2 K
1
= 15 + 2·5 = 17·5.
SUMMATIVE ASSESSMENT
WORKSHEET-83
SECTION A
1.
According to formula for an even number of 2n terms.
Median = Mean of nth term and (n + 1)th term
1
SECTION B
2.
Age
Number of Patients
Less than 20
Less than 30
Less than 40
Less than 50
Less than 60
Less than 70
60
102
157
227
280
300
2
SECTION C
3.
Sol. :
xi
fi
xifi
3
10
30
9
p
9p
15
4
60
21
7
147
27
q
27q
33
4
132
39
1
39
Total
Σfi = 26 + p + q
Σxifi = 408 + 9p + 27q
Given,
⇒
⇒
∴
S O L U T I O N S
Σfi = 40,
26 + p + q = 40
p + q = 14
Σxi f i
–
Mean, x = Σf
i
408 + 9 p + 27q
14.7 =
40
588 = 408 + 9p + 27q
180 = 9p + 27q
p + 3q = 20
...(1) ½
½
1
...(2)
P-111
Subtracting eq. (1) from eq. (2), we get
2q = 6
⇒
q=3
Putting this value of q in eq. (1), we get
p = 14 – q = 14 – 3 = 11.
½
½
SECTION D
4.
Daily income (classes)
No. of workers (c.f.)
less than 250
10
less than 300
15
less than 350
26
less than 400
34
less than 450
40
less than 500
50
1
y
50
45
Cumulative frequency
40
Scale
x-axis 1 cm = 50 units
y-axis 1 cm = 5 units
35
30
25
20
15
2
10
5
x
250
From graph,
Hence,
300
350 400 450 500
Class limits
N 50
=
= 25
2
2
Median daily income = ` 350.
SUMMATIVE ASSESSMENT
1
WORKSHEET-84
SECTION A
1.
The model class is 10 – 15, than f ≥ 8.
1
SECTION B
2.
According to question
mode = 24·5
and
mean = 29·75
The relationship connecting measures of central tendencies is :
We know that
3 Median = Mode + 2 Mean
3 Median = 24·5 + 2 × 29·75
= 24·5 + 59·50
3 Median = 84·0
84
∴
Median =
= 28
3
M A T H E M A T I C S -P-112
½
½
1
X
T E R M – 1
SECTION C
3.
xi - a
h
fi
f iu i
20
= –2
10
25
–50
40
–40
42
0
10
=1
10
33
33
20
=2
10
10
20
å fi =150
Σfiui = – 37
Class
Class marks
xi
ui =
20 – 30
25
–
30 – 40
35
–
40 – 50
45 = a
50 – 60
55
60 – 70
65
10
= –1
10
0=0
å fi ui
Mean = a + å f ´ h
i
æ -37 ö
÷÷ ´ 10 = 42·5 approx.
= 45 + ç
ç
è 150 ø
1
1
1
SECTION D
4.
xi (class mark)
fi
fixi
0 – 100
50
12
600
100 – 200
150
16
2400
200 – 300
250
6
1500
300 – 400
350
7
2450
400 – 500
450
9
4050
Σfi = 50
Σfixi = 11000
Class
Total
Mean =
∴
Σxi f i 11000
=
50 = 220·00
Σf i
2
1
1
Average daily income = ` 220·00.
SUMMATIVE ASSESSMENT
WORKSHEET-85
SECTION A
1.
Given
mode – median = 24
Using formula
mode = 3 median – 2 Mean
difference between median and mean = a.
SECTION B
2.
The frequency table of the given data is as given below :
xi
14
15
16
18
20
25
x
fi
1
3
1
3
3
4
1
1
It is given that the mode of the given data is 25. So it must have the maximum frequency that is
possible only when xi = 25 hence x = 25.
1
S O L U T I O N S
P-113
SECTION C
3.
Sol.
Classes
Frequency
fi
fixi
Mid points
xi
0 – 20
6
10
60
20 – 40
8
30
240
40 – 60
10
50
500
60 – 80
12
70
840
80 – 100
8
90
720
100 – 120
6
110
660
Σf = 50
Total
Σfx = 3020
Σ fi xi 3020
–
Mean, x = Σ f =
= 60.4
50
i
∴
2
1
SECTION D
4.
Sol.
Class Interval
Frequency
Cumulative frequency
0 –100
2
2
100 – 200
5
7
200 – 300
x
7+x
300 – 400
12
19 + x
400 – 500
17
36 + x
500 – 600
20
56 + x
600 – 700
y
56 + x + y
700 – 800
9
65 + x + y
800 – 900
7
72 + x + y
900 – 1000
4
76 + x + y
N = 100
2
Hence,
76 + x + y = 100
⇒
x + y = 100 – 76 = 24
Median = 525, so median class = 500 – 600
Now,
....(1)
n
- c. f .
median = l + 2
´h
f
é100
ù
- (36 + x) ú
ê
ê 2
ú
525 = 500 + ê
ú ´ 100
20
ê
ú
ê
ë
ûú
25 = (50 – 36 – x) 5
25
=5
5
x = 14 – 5 = 9
Þ
Put the value of x in equation (1), we get
y = 24 – 9 = 15.
Þ
P-114
(14 – x) =
M A T H E M A T I C S --
1
1
X
T E R M – 1
SUMMATIVE ASSESSMENT
WORKSHEET-86
SECTION A
1.
2.
1
The median of the data = 20·5.
2 + 3 + 5 + 7 + 11
The mean of first five prime numbers =
5
28
=
= 5·6
5
1
SECTION B
3.
x
3
9
15
21
27
f
7
5
10
12
6
Σf = 40
fx
21
45
150
252
162
Σfx = 630
Σ fx = 630, Σ f = 40
630
= 15·75.
Mean =
40
∴
1
1
SECTION C
4.
Class
xi (class marks)
fi
fixi
Cum. Frequency
0 – 10
5
8
40
8
10 – 20
15
16
240
24
20 – 30
25
36
900
60
30 – 40
35
34
1190
94
40 – 50
45
6
270
100
Σfi = 100
Σfixi= 2640
Total
å fi xi
2640
Mean = å f =
= 26.4
100
i
median class : 20 – 30
∴
Here,
1
50 − 24
× 10
36
= 20 + 7·22 = 27·22
∴
1
Median = 20 +
1
SECTION D
5.
Class Interval
Frequency
Cumulative Frequency
0 – 10
10 – 20
20 – 30
30 – 40
40 – 50
50 – 60
5
f1
20
15
f2
5
5
5 + f1
25 + f1
40 + f1
40 + f1 + f2
45 + f1 + f2
Total
N = 60
Hence,
and
S O L U T I O N S
2
45 + f1 + f2 = 60
f1 + f2 = 15
N = 60
P-115
Median = 28.5, ∴ Median class is 20 – 30.
N
− c. f
Median = l + 2
×h
f
30 – 5 − f1
28.5 = 20 +
× 10
20
8.5 × 2 = 25 – f1
⇒
⇒
f1 = 25 – 17 = 8
From (1),
f2 = 15 – f1 = 15 – 8 = 7
∴
f1 = 8 and f2 = 7.
...(1)
1
1
SUMMATIVE ASSESSMENT
WORKSHEET-87
SECTION A
1.
x = 27.
2.
 n+1
Median = 

 2 
1
th
 13 + 1 
observation = 

 2 
th
observation = 7th observation.
1
SECTION B
3.
Heights
No. of girls
120 and more
50
130 and more
48
140 and more
40
150 and more
28
160 and more
8
2
SECTION C
4.
(i)
No. of children (xi)
No. of famlies (fi)
fi xi
0
5
0
1
11
11
2
25
50
3
12
36
4
5
20
5
2
10
Total
Σfi = 60
Mean =
=
Σfixi = 127
∑ f i xi
∑ fi
127
= 2·12 approx.
60
1
(ii) Mean of ungrouped data.
(iii) For progress, we should decrease population growth.
P-116
1
M A T H E M A T I C S --
½
½
X
T E R M – 1
SECTION D
5.
y
c.f.
200
More than 80
200
180
More than 100
180
160
More than 120
150
More than 140
130
More than 160
90
Wages
(80, 200)
190
Scale
x-axis 2 cm = 10 units
y-axis 1 cm = 10 units
(100, 180)
170
(120, 150)
150
More than ogive
140
(140, 130)
130
120
2
2
110
100
90
(160, 90)
80
70
60
50
40
30
20
10
0
80
100
120
140
Lower Lts
SUMMATIVE ASSESSMENT
160
x
WORKSHEET-88
SECTION A
1.
From the table it is clear that the frequency is maximum for the class 30 – 40, so modal class is
30 – 40.
1
SECTION B
2.
Modal class : 60 – 80
Here, l = 60, f1 = 61, f0 = 52, f2 = 38, h = 20
Mode = l +
½
f1 − f0
×h
2 f1 − f0 − f2
½
= 60 +
61 − 52
× 20
122 − 52 − 38
½
= 60 +
9
× 20 = 65·62
32
½
SECTION C
3.
Class Interval
Frequency
Cumulative Frequency
135 – 140
4
4
140 – 145
7
11
145 – 150
18
29
150 – 155
11
40
155 – 160
6
46
160 – 165
5
51
Total
S O L U T I O N S
N = 60
3
P-117
SECTION D
4.
Let assumed mean, a = 649·5 and h = 100
Life time
(in hrs)
xi
ui = x i − a
h
fi
f iu i
400 – 499
449·5
–2
24
– 28
500 – 599
549·5
–1
47
– 47
600 – 699
649·5
0
39
0
700 – 799
749·5
1
42
42
800 – 899
849·5
2
34
68
900 – 999
949·5
3
14
42
Σfi = 200
Σfiui = 77
Total
∴
–
Mean, x = a +
FG Σf u
H Σf
i i
i
IJ
K
×h
2
1
77
× 100
200
= 649·5 + 38.5
= 688.
= 649·5 +
1
Average life time is 688 hours.
SUMMATIVE ASSESSMENT
WORKSHEET-89
SECTION A
1.
2.
Median =
1
2
mode +
mean
3
3
=
1
2
(8) + (8)
3
3
=
16 + 8
3
=
24
= 8.
3
1
The median of the data is 42·5.
SECTION B
3.
Modal class : 5 – 7
½
Here l = 5, f1 = 80, f0 = 45, h = 2, f2 = 55
f1 − f0
Mode = l + 2 f − f − f × h
1
0
2
P-118
=5+
80 – 45
×2
160 – 45 – 55
=5+
35 × 2
= 6·17.
60
M A T H E M A T I C S --
½
1
X
T E R M – 1
SECTION C
3.
Let assumed mean, a = 35 and h = 10.
xi − a
h
fi
f iu i
5
–3
5
– 15
15
–2
13
– 26
25
–1
20
– 20
35
0
15
0
45
1
7
7
55
2
5
10
Σfi = 65
Σfiui = – 44
xi
ui =
(Class Marks)
Total
Σfi ui
–
Mean, x = a + Σf × h
i
∴
= 35 +
1½
½
−44
× 10
65
= 35 – 6·76
–
x = 28·24.
1
SECTION D
4.
Draw the ‘less then’ ogive and ‘more than’ ogive simultaneously.
c.f.
Less than
30
40
50
60
70
80
90
c.f.
More than
10
18
30
54
60
85
100
20
30
40
50
60
70
80
100
90
82
70
46
40
15
2
y
(20, 100)
100
90
Less than ogive
(80, 85)
(40, 82)
80
Cumulative frequency
(90, 10)
(30, 90)
More than ogive
70
(50, 70)
(70, 60)
60
(60, 54)
50
(60, 46)
40
(70, 40)
30
2
(50, 30)
20
(40, 18)
10
(80, 15)
(30, 10)
0
10
20
30
40
50
60
70
80
90
x
From The Graph Median = 58.3
S O L U T I O N S
P-119
SUMMATIVE ASSESSMENT
WORKSHEET-90
SECTION A
1.
In the given formula l represents the lower limit of the class with highest frequency.
1
2.
The require formula is 3 Median = Mode + 2 Mean
1
SECTION B
3.
Class
Cumulative frequency
More than 50
60
More than 60
48
More than 70
30
More than 80
20
More than 90
5
2
SECTION C
3.
(i) Here class intervals are not in inclusive form. So, we first convert them in inclusive form by
subtracting 1/2 from the lower limit and adding 1/2 to the upper limit of each cases. where h is
the difference between the lower limit of a class and the upper limit of the preceding class. The
given frequency distribution in inclusive form is as follows :
Age (in years)
No. of cases
4·5 – 14·5
6
14·5 – 24·5
11
24·5 – 34·5
21
34·5 – 44·5
23
44·5 – 54·5
14
54·5 – 64·5
5
we observe that the class 34·5 – 44·5 has the maximum frequency. So, it is the modal class such
that
1
Here, l = 34.5, h = 10, f1 = 23, f0 = 21, f2 = 14
f1 – f0
×h
2 f1 – f 0 – f 2
Now,
Mode = l +
⇒
Mode = 34.5 +
= 34·5 +
23 – 21
× 10
46 – 21 – 14
2
× 10
11
= 34·5 + 1·81
1
= 36·31
(ii) Mode of grouped data.
(iii) If we practise habit of cleanliness we will be able to put disease at on arm’s length.
P-120
M A T H E M A T I C S --
X
½
½
T E R M – 1
SECTION D
4.
We prepare cumulative frequency table :
Height (in cm)
Frequency
Height Less than
Cumulative Frequency
140 – 143
3
143
3
143 – 146
9
146
12
146 – 149
26
149
38
149 – 152
31
152
69
152 – 155
45
155
114
155 – 158
64
155
178
158 – 161
78
161
256
161 – 164
85
164
341
164 – 167
96
167
437
167 – 170
72
170
509
2
Now we mark upper class limits on X-axis and cumulative frequency on Y-axis. We plot
(140, 0), (143, 3), (146, 12), (149, 38), (152, 69), (155, 114), (158, 178), (161, 256),
(164, 341), (167, 437), (170, 509)
These points are joined by line segments to obtain the cumulative frequency polygon as shown
in Figure
Y
550
500
450
400
Cuf
350
300
250
200
150
100
50
X
140
143 146 149 152 155 158 161 164 167 170
Heights
S O L U T I O N S
P-121
SUMMATIVE ASSESSMENT
WORKSHEET-91
SECTION A
1.
In the distribution,
median class = 20 – 25
Hence,
lower limit of median class = 20
and
modal class = 25 – 30
So,
lower limit of modal class = 25
Sum of lower limits of median class and lower limit of modal class = 25 + 20 = 45.
1
SECTION B
xi
2.
fi
xifi
1
1
1
3
2
6
5
1
5
7
5
35
9
6
54
11
2
22
13
3
39
Total
20
162
1
Σfi xi
Mean = Σf
i
½
162
= 8·1
20
∴ Mean number of plants per house is 8·1.
–
x =
SECTION C
xi
(Class marks)
3.
fi
fixi
10
30
50
70
90
12
15
32
p
13
120
450
1600
70p
1170
Total
Σfi = 72 + p
We know that Mean,
–
x =
⇒
53 =
⇒
3340 + 70p =
⇒
3340 + 70p =
⇒
70p – 53p =
⇒
17p =
p=
P-122
Σfixi = 3340 + 70p
1
Σfi xi
Σfi
3340 + 70 p
72 + p
53 (72 + p)
3816 + 53p
3816 – 3340
476
476
= 28.
17
M A T H E M A T I C S --
½
½
1
X
T E R M – 1
SECTION D
4.
Frequency distribution given below :
Age (in years)
Frequency
Age less than
Cumulative Frequency
0·5 – 9·5
5
9·5
5
9·5 – 19·5
15
19·5
20
19·5 – 29·5
20
29·5
40
29·5 – 39·5
23
39·5
63
39·5 – 49·5
17
49·5
80
49·5 – 59·5
11
59·5
91
59·5 – 69·5
9
69·5
100
2
Now we plot points (9·5, 5), (19·5, 20), (29·5, 40), (39·5, 63), (49·5, 80) (59·5, 91) and (69·5, 100) and
join them to obtain the required ogive as shown in figure :
90
80
70
60
50
40
30
20
10
–0.5
9.5
19.5 29.5 39.5 49.5 59.5 69.5
SUMMATIVE ASSESSMENT
WORKSHEET-92
SECTION A
1.
In the given distribution,
Median class = 160 – 165
So
upper limit of Median class = 165
and
So
Modal class = 150 – 155
Lower limit of Modal class = 150
Hence, sum of upper limit of Median class and lower limit of modal class
= 165 + 150 = 315.
1
SECTION B
f1 - f 0
Mode = l + 2 f - f - f × h
1
0
2
2.
1
Here, modal class : 30 – 40,
l = 30, f1 = 25, f0 = 20, f2 = 12, h = 10
S O L U T I O N S
½
P-123
∴
Mode = 30 +
= 30 +
25 – 20
×10
50 – 20 – 12
5 × 10
18
= 30 + 2·77 = 32·77 (Modal age).
½
SECTION C
3.
Class Interval
Frequency
0 – 10
8
10 – 20
12
20 – 30
25
30 – 40
13
40 – 50
12
Total
70
1
Here, modal class 20 – 30
and l = 20, f1 = 25, f2 = 13, f0 = 12, h =10
½
æ f1 - f 0 ö
÷
Mode = l + ç
ç2 f - f - f ÷ × h
è 1 0 2ø
= 20 +
½
FG 25 − 12 IJ × 10
H 50 − 12 − 13 K
½
13
× 10
25
= 20 + 5·2 = 25·2.
= 20 +
SECTION D
4.
xi
ui =
xi - a
h
f iu i
Height (in cm)
Number of girls (fi)
120 – 130
2
125
–2
–4
130 – 140
8
135
–1
–8
140 – 150
12
145
0
0
150 – 160
20
155
1
20
160 – 170
8
165
2
16
Total
Σfi = 50
Σfiui = 24
Let assumed mean, a = 145 and h = 10
∴
Σfi ui
–
Mean, x = a + h × Σf
i
2
24
× 10
50
= 145 + 4·8
–
x = 149·8.
–
x = 145 +
P-124
M A T H E M A T I C S --
1
1
X
T E R M – 1
SUMMATIVE ASSESSMENT
WORKSHEET-93
SECTION A
1.
Time
Frequency
Cumulative frequency
0 – 10
8
8
10 – 20
10
18
20 – 30
12
30
30 – 40
22
52
40 – 50
30
82
50 – 60
18
100
N
= 50
2
Now
Hence, median class is 30 – 40.
1
SECTION B
2.
Here, the modal class : 30 – 40
l = 30, f1 = 25, f0 = 10, f2 = 12, h = 10
½
f1 − f0
Mode = l + 2 f − f − f × h
1
0
2
½
= 30 +
25 − 10
× 10
50 − 10 − 12
½
= 30 +
15
× 10 = 35·35
28
½
SECTION C
3.
(i) Let the assumed mean, A = 1400 and h = 400.
Calculation of Mean
Height (in m) x1
No. of Villages f1
D = xi –1400
200
600
1000
1400
1800
2200
142
265
560
271
89
16
– 1200
– 800
– 400
0
400
800
Total
N= Σfi= 1343
ui =
x i − 1400
400
–3
–2
–1
0
1
2
f iu i
– 426
– 530
– 560
0
89
32
Σfiui = –1395 1
We have A = 1400, h = 400, Σfiui = –1395 and N = 1343.
RS 1 ∑ f u UV
TN
W
F –1395 IJ
= 1400 + 400 × GH
1343 K
Mean = A + h
i i
= 1400 – 415·49 = 984·51
S O L U T I O N S
1
P-125
(ii) Mean by assumed mean method.
½
(iii) Villages are much necessary to keep a balance between the ecological problems.
½
SECTION D
y
4.
More than
c.f.
100
0
100
90
10
90
20
72
30
32
40
12
80
70
60
50
1
40
2
30
20
From graph,
Hence,
N
100
=
= 50
2
2
Median = 25.
10
0
10
20
30
1
SUMMATIVE ASSESSMENT
40
Class
x
WORKSHEET-94
SECTION A
1.
The number of athletes who completed the race in less than 14·4 seconds = 2 + 14 + 16 = 32. 1
SECTION B
2.
From the cumulative frequency distribution
15 + x = 28
⇒
and
1
x = 28 – 15 = 13
43 + 18 = y
1
y = 61
Hence,
x = 13 and y = 61.
SECTION C
3. Modal class : 201 – 202
1
Here l = 201, f1 = 26, f0 = 12, f2 = 6, h = 1
f1 − f0
Mode = l + 2 f − f − f × h
1
0
2
Mode = 201 +
= 201 +
P-126
26 − 12
×1
52 − 12 − 6
1
14
= 201·41.
34
1
M A T H E M A T I C S --
X
T E R M – 1
SECTION D
xi
(Class marks)
4.
fi
fixi
15
12
180
45
21
945
75
x
75x
505
52
5460
135
y
135y
165
11
1815
Total
Σfi = 150
Σfixi = 8400 + 75x + 135y
1
x + y = 54
Σfi xi
–
x = Σf
i
∴
8400 + 75 x + 135 y
150
13650 = 8400 + 75x + 135y
91 =
75x + 135y = 5250 ⇒ 5x + 9y = 350
From table,
...(i) 1
...(ii) 1
96 + x + y = 150
Solving eqs. (i) and (ii), we get x + y = 54
1
x = 34 and y = 20.
SUMMATIVE ASSESSMENT
WORKSHEET-95
SECTION A
1.
Class
Frequency
Cumulative frequency
20 – 40
10
10
40 – 60
12
22
60 – 80
20
42
80 – 100
22
64
N
= 32
2
Q
1
Hence, Median class is 60 – 80.
SECTION B
2.
Modal class : 40 – 60
Also, l = 40, f1 = 28, f2 = 20, f0 = 16, h = 20 1
f1 − f0


Mode = l +  2 f − f − f  × h
 1 0 2
28 − 16
× 20
56 − 16 − 20
= 40 + 12 = 52.
1
½
= 40 +
S O L U T I O N S
½
P-127
SECTION C
3.
From the given table,
12 + a = 25
⇒
½
½
a = 25 – 12 = 13
25 + 10 = b
⇒
b= 35
½
⇒
b + c = 43
½
⇒
c = 43 – b
= 43 – 35 = 8
and
48 + 2 = d
⇒
½
d = 50
SECTION D
4.
Now we draw cumulative frequency distribution table by less than method.
Marks
Number of students
Marks less than
Cumulative
Frequency
0 – 10
6
10
6
10 – 20
25
20
31
20 – 30
48
30
79
30 – 40
72
40
151
40 – 50
116
50
267
50 – 60
63
60
330
2
Now we plot the points (0, 0), (10, 6), (20, 31), (30, 79), (40, 150), (50, 267), (60, 330) we get
following ogive :
Y
350
300
250
200
150
100
50
0
P-128
10
20
30
40
50
60
X
M A T H E M A T I C S --
X
T E R M – 1
FORMATIVE ASSESSMENT
WORKSHEET-96
Objective Type Questions
1.
6.
2. (C)
7. (B)
(B)
(A)
Fill in the blanks
1.
2.
3.
4.
5.
6.
mode
uniform
mode
mode + 2 mean
median
mode
7.
a+b
2
8.
Σ fi xi
Σ fi
9.
x=
10.
10 × 1 = 10
3. (A)
8. (C)
4. (C)
9. (A)
5. (B)
10. (D)
10 × 1 = 10
Σ fi di
Σ fi
mode.
●●
S O L U T I O N S
P-129