Study Material 9th Maths, 2023-24

केन्द्रीय शवद्यािय संगठन
KENDRIYA VIDYALAYA SANGATHAN
शिक्षा एवं प्रशिक्षण का आं चशिक संस्थान, चंडीगढ़

ZONAL INSTITUTE OF EDUCATION AND TRAINING, CHANDIGARH
अध्ययन सामग्री / STUDY MATERIAL

िैशक्षक सत्र / SESSION – 2023-24
कक्षा / CLASS – नौव ीं / IX

शवषय / SUBJECT – गणित / MATHEMATICS
संकिन - राजीव रं जन, सह-प्रशिक्षक (गशणत)

COMPILATION BY: RAJIV RANJAN, TRAINING ASSOCIATE (MATHS)
शिक्षा एवं प्रशिक्षण का आं चशिक संस्थान, चंडीगढ़

ZONAL INSTITUTE OF EDUCATION AND TRAINING, CHANDIGARH
से क्टर 33-सी, चंडीगढ़ / SECTOR-33 C, CHANDIGARH
वेबसाइट / WEBSITE : zietchandigarh.kvs.gov.in
ई-मेल/ E-MAIL :kvszietchd@gmail.com दू रभाष / Phone : 7102-2921841; 2921994

STUDY MATERIAL FOR CLASS IX MATHEMATICS
INDEX
S.No. NAME OF CHAPTER PAGE
No.
1 NUMBER SYSTEM 2
2 POLYNOMIAL 6
3 COORDINATE GEOMETRY 13
4 LINEAR EQUATION IN TWO 18

VARIABLES
5 EUCLID GEOMETRY 26
6 LINES & ANGLES 31
7 TRIANGLES 45
8 QUADRILATERAL 50
9 CIRCLE 54
10 HERONS FORMULA 59
11 SURFACE AREA AND VOLUME 66
12 STATISTICS 75
1
NUMBER SYSTEM
IMPORTANT CONCEPTS:
 The numbers of the form p/q, where „p‟ and „q‟ are integers and q≠0, are called rational numbers.
 A rational number p/q is said to be in simplest form, if „p‟ and „q‟ are integers having no common
factor other than 1 and q≠0.
 Every rational number can be expressed as decimal. If the decimal expression of p/q terminates, then
it is called a terminating decimal.
 A decimal in which a digit or a group of digits repeats periodically, is called a recurring decimal.
 The decimal expression of a rational number is either terminating or non-terminating recurring.
 The decimal expression of an irrational number is „non-terminating and non-recurring‟.
 All rational and all irrational numbers form the collection of all real numbers.
 The process of converting the irrational denominator of a number by multiplying its numerator and
denominator by a suitable number, is called rationalization.
SOME ILLUSTRATIONS/EXAMPLES:
MCQs
1. 3√6 + 4√6 is equal to:
a) 6√6 b)7√6 c)4√12 d)7√12
Answer: b
3√6 + 4√6 = (3 + 4)√6 = 7√6
2. √6 x √27 is equal to:
a) 9√2 b)3√3 c)2√2 d)9√3
Answer: a
= (3 × 3)√2
= 9√2
3. Which of the following is equal to x3?
a) x6 – x3 b)x6.x3 c)x6/x3 d)(x6)3
6 3 6–3 3
Answer: c x /x = x =x
4. Which of the following is an irrational number?

a) 0.14 b) 0.1416 c) 0. 1416d)0.4014001400014…
Answer: d
0.4014001400014…is an irrational number as it is non-terminating and non-repeating.
5. 2√3+√3 =
a) 6 b)2√6 c)3√3 d)4√6
Answer: c 2√3+√3 = (2+1)√3= 3√3.
SHORT ANSWER TYPE QUESTIONS
1. Add 2√2+ 5√3 and √2 – 3√3.

Solution:
(2√2 + 5√3) + (√2 – 3√3)
= 2√2 + 5√3 + √2 – 3√3
= (2 + 1)√2 + (5 – 3)√3
= 3√2 + 2√3
2. Simplify: (√3+√7) (√3-√7).
Solution:
(√3 + √7)(√3 – √7)
Using the identity (a + b)(a – b) = a2 – b2,
(√3 + √7)(√3 – √7) = (√3)2 – (√7)2
=3–7
2
= -4
3. Rationalize the denominator of 1/[7+3√3].
Solution:
=1/(7 + 3√3)
By rationalizing the denominator,
= [1/(7 + 3√3)] [(7 – 3√3)/(7 – 3√3)]
= (7 – 3√3)/[(7)2 – (3√3)2]
= (7 – 3√3)/(49 – 27)
= (7 – 3√3)/22
PRACTICE QUESTIONS
MCQs
1. The decimal expansion of an irrational number may be:

a) Terminating b)Recurring
c)Either terminating or non- recurring d) non-terminating and non-recurring
2. What would be the denominator after rationalizing 7/(5√3 – 5√2)?
a) 19 b)20 c)25 d)None of these
3. In between two rational number there is/are:
a) Exactly one rational number b)Infinitely many rational number
c) Many irrational numbers d)Only irrational numbers
a) ¼ b)½ c)4 d)1/16

4. √12 X √15 is equal to:
a) 5√6 b)6√5 c)10√5 d)√25
5. Which of the following is irrational?
a) 0.14 b) 0.1416 c) 0. 1416d)0.4014001400014…
6. Which of the following is irrational?
4 12
a) b) c) 5 d) 81
9 3
7. The product of a rational and an irrational numbers is:
a) Always an integer b)Always a rational number c)Always an irrational number
b) Sometimes rational and sometimes irrational
8. √6 x √27 is equal to:
a) 9√2 b)3√3 c)2√2 d)9√3
ASSERTION- REASONING QUESTION
1. Assertion: √5 is an irrational number.
Reason: A number is called irrational, if it cannot be written in the form p/q, where p and q
are integers and q≠0
a) Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
b) Both Assertion and Reason are correct and Reason is not the correct explanation for Assertion.
c) assertion is true but the reason is false.
d) both assertion and reason are false.

1. Simplify: (√5 + √2)2
2. Find the value of √(3)-2.
3. Identify a rational number among the following numbers :
2 + √2, 2√2, 0 and π
4. Evaluate : (√5 + √2)2 + (√8 – √5)2
5. If x = 15√−2 , find the value of x3 – 32 – 5x + 3
6. Find a rational number between 1 and 2.
7. Write a rational number equivalent to 5/9 such that its numerator is 25.
3
8. Find two rational numbers between 0.1 and 0.3
9. Simplify: (4+√3)(4-√3)
10. Simplify: (√3+√2)²
ANSWERS:
MCQs
Q1. d Q2.d, Q3b, Q4.a, Q5. b, Q6. a, Q7. c,
Q8. C, Q9. a, Q10. A
SHORT ANSWER QUESTIONS:

Q1: 7 + 2√10 Q2: 1/3 Q3: 0 is a rational number Q4: 20 – 2√10, Q5. 4,
Q6. 3/2, Q7. 25/45, Q8. 5/30 and 7/30, Q9. 13,
Q10. 5+2√6
4
TEST-1 (MM.20)
1. Find five rational numbers between 1 and 2.
2. Find five rational numbers between 3/5 and 4/5.
3. Locate √3 on the number line.
4. Find the decimal expansions of 10/3, 7/8 and 1/7.
5. Find three different irrational numbers between the rational numbers 5/7 and 9/11.
6. Visualize 3.765 on the number line, using successive magnification.
7. Represent √(9.3) on the number line.
8. Simplify:
(i) 72/3.71/5
(ii) 101/2/101/4
1
9. Express 3 8in the form of decimal.
1
10. Rationalize the denominator of 3− 2
TEST-2 (MM.30)
1. Express in the form p/q
Express 0.4323232… in the form p/q, where p and q are integers and q ≠ 0.
2. Find 6 rational numbers between 6/5 and 7/5.
3. Rationalize the denominator:
1 2 7
a) 9+ 5+ 6 b) 3−1 c) 12− 5
4. Express as Fractions
Express 1.363636... in the form p/q, where p and q are integers and q ≠ 0.
5. Simplify the following:
a) (8+√5)(8-√5)
b) (10+√3)(6+√2)
c) (√3+√11)2 +(√3-√11)2
6. What can the maximum number of digits be in the recurring block of digits in the decimal expansion
of1/17?
7. Classify the following numbers as rational or irrational:
a) 2-√5
b) (3+√23)- √23
c) 1/(√2)
7+3 𝟓
8. Simply by rationalizing denominator: 7−3 𝟓
9. Simplify:
10. Express in the form a/b.
5
POLYNOMIALS
(I) Main Concepts and Results:
Meaning of a Polynomial Degree of a polynomial

Coefficients
Monomials, Binomials etc.
Constant, Linear, Quadratic Polynomials etc.

Value of a polynomial for a given value of the variable Zeroes of a polynomial
Remainder theorem Factor theorem
Factorization of a quadratic polynomial by splitting the middle term Factorization of algebraic expressions by
using the Factor theorem Algebraic identities –
(x + y)2 = x2 + 2xy + y2
(x – y)2 = x2 – 2xy + y2
x2 – y2 = (x + y) (x – y)
(x + a) (x + b) = x2 + (a + b) x + ab
(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
(x + y)3 = x3 + 3x2y + 3xy2 + y3 = x3 + y3 + 3xy (x + y)
(x –y)3 = x3 – 3x2y + 3xy2 – y3 = x3 – y3 – 3xy (x – y)
x3 + y3 = (x + y) (x2 – xy + y2)
x3 – y3 = (x – y) (x2 + xy + y2)
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
EXAMPLES:
1) MCQ‟S-
(i) Which one of the following is a polynomial?

3
𝑥² 2 2 3𝑥 2 𝑥−1
(A) = 𝑥² (B) 2 x  1 (C) x + (D)
2 𝑥 𝑥+1
Answer: (C)
(ii) On factorizing x + 8x + 15, we get :

2
(A) (x + 3) (x – 5) (B) (x – 3) (x + 5) (C) (x + 3) (x + 5 ) (D) (x – 3) (x – 5)

Answer: (C)
(iii) On dividing x – 2x – 15 by (x – 5), the quotient is (x + 3) and

2
remainder is 0. Which of
the following statements is true?
6
(A) x2– 2x – 15 is a multiple of (x – 5)
(B) x2– 2x – 15 is a factor of (x – 5)
(C) (x + 3) is a factor of (x – 5)
(D) (x + 3) is a multiple of (x – 5)
Answer: (A)
(iv) The value of the polynomial 2x2+ 3x- 4 at x = 0 is:
(A) 2 (B) 3 (C) – 4 (D) 4
Answer: (C)
(v) The value of the polynomial 5x – 4x2 + 3, when x = –1 is-
(A) – 6 (B) 6 (C) 2 (D) –2
Answer: (A)
2) SHORT ANSWER QUESTIONS:
(i) Give an example of a monomial and a binomial having degrees of 82 and 99, respectively.
Solution: An example of a monomial having a degree of 82 = x82An example of a binomial having a

degree of 99 = x99 + x.
(ii) Find the value of the polynomial 5x – 4x2 + 3 at x = 2 and x = –1.
Solution: Let the polynomial be f(x) = 5x – 4x2 + 3
Now, for x = 2,
f(2) = 5(2) – 4(2)2 + 3
=> f(2) = 10 – 16 + 3 = –3
Or, the value of the polynomial 5x – 4x2 + 3 at x = 2 is -3.
Similarly, for x = –1,
f(–1) = 5(–1) – 4(–1)2 + 3
=> f(–1) = –5 –4 + 3 = -6
The value of the polynomial 5x – 4x2 + 3 at x = -1 is -6.
(iii) Compute the value of 9x2 + 4y2 if xy = 6 and 3x + 2y = 12.
Solution: Consider the equation 3x + 2y = 12
Now, square both sides:

7
(3x + 2y)2 = 122
=> 9x2 + 12xy + 4y2 = 144
=>9x2 + 4y2 = 144 – 12xy
From the questions, xy = 6 So,
9x2 + 4y2 = 144 – 72
Thus, the value of 9x2 + 4y2 = 72.
3) PRACTICE QUESTIONS:
(A) MCQ‟S QUESTIONS:
Q1. √2 is a polynomial of degree is -
(A) 2 (B) 0 (C) 1 (D) ½
Q2. Degree of the polynomial 4x4 + 0x3 + 0x5 + 5x + 7 is -
(A) 4 (B) 5 (C) 3 (D) 7
Q3. Degree of the zero polynomial is-
(A) 0 (B) 1 (C) Any natural number (D) Not defined
Q4. If p(x)= x2- 2√2x +1, then p(2√ 2) is equal to
(A) 0 (B) 1 (C) 4√2 (D) 8√2+ 1
Q5. The value of the polynomial 5x – 4x2 + 3, when x = –1 is-
(A) – 6 (B) 6 (C) 2 (D) –2
Q6. If p(x) = x + 3, then p(x) + p(–x) is equal to-
(A) 3 (B) 2x (C) 0 (D) 6
Q7. Zero of the polynomial p(x) = 2x + 5 is –
(A) -2/ 5 (B) -5/2 (C) 2/5 (D) 5/2
Q8. The value of 2492 – 2482 is-
(A) 12 (B) 477 (C) 487 (D) 497

1 1
Q9. . If 49x2 – b = {7x + 2 }{7x - 2 }, then the value of b is-
(A) 0 (B) 1/√2 (C) 1/4 (D) ´
ASSERTION & REASON BASED QUESTION:

8
Q10. Assertion: The value of 593 × 607 is 359,951.
Reason: (a + b) (a – b) = a2 – b2
Directions: Choose the correct answer out of the following choices :

(A) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
(B) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
(C) Assertion is correct statement but Reason is wrong statement.
(D) Assertion is wrong statement but Reason is correct statement.
(B) SHORT ANSWER QUESTIONS:
Q11.Calculate the perimeter of a rectangle whose area is 25x2 – 35x + 12.
Q12. Find the value of x3 + y3 + z3 – 3xyz if x2 + y2 + z2 = 83 & x + y + z = 15.
Q13. If 𝑎 + 𝑏 = 15, 𝑎𝑏 = 14 then find 𝑎2 + 𝑏2.
Q14. Check whether (7 + 3x) is a factor of (3x3 + 7x).
Q15. Factorise x2 + 1/x2 + 2 – 2x – 2/x.
Q16. Factorise x2 – 1 – 2a – a2.
Q17. Expand (𝑎 − 2𝑏 + 3𝑐)2
Q18. Expand (3𝑎 – 4b)3
Q19. Classify the following polynomials as polynomials Monomial, Binomial, Trinomial, Polynomial
etc.
(i) 𝑥2 + 𝑥 + 1 (ii) 𝑦3 – 5𝑦 (iii) 𝑥𝑦 (iv) 𝑥2 – 2𝑥𝑦 + 𝑦2 + 1

Q20. Factorise: 𝑥3 – 64
(C) ANSWER OF PRACTICE QUESTIONS:
(i) MCQ QUESTIONS:-
1) (B), 2) (A), 3) (D,) 4) (B), 5) (A), 6) (D), 7) (B), 8) (D) 9), (C) 10) (A)
(ii) SHORT QUESTIONS:

1 1
11) P= (2x – 14), 12) 180, 13) 197, 14) NO, 15) (x+ x ) (x+ 𝑥 − 2),
16) (x-1-a) (x+1+a), 17) (a2 + 4b2 +9 c2 - 4ab - 12bc + 6ca)
18) 27 a3 –64 b3 – 108a2c + 144ac2 ,
19) (i)Trinomial (ii0 Binomial (iii) Monomial (iv) Polynomial
20) (x - 4) (x2 + 4x + 16)
9
CLASS TEST – 1 -
TOPIC- POLYNOMIALS
CLASS-IX
TIME- 45 MIN. M.M. 20
SECTION – A – ( 1 mark for each correct option)

Choose the correct option from the following questions:-
Q1. . On dividing x – 2x – 15 by (x – 5), the quotient is (x + 3) and remainder is 0. Which of the following
2
statements is true?
(a) x2– 2x – 15 is a multiple of (x – 5) (b) x2– 2x – 15 is a factor of (x – 5)
(c) (x + 3) is a factor of (x – 5) (d) (x + 3) is a multiple of (x – 5)
Q2. The value of the polynomial 3x + 2x – 4 at x = 0 is :
2
(a) 2 (b) 3 (c) – 4 (d) 4
Q3. If p(x) = x + 3, then p(x) + p(–x) is equal to :

(a) 3 (b) 2x (c) 0 (d) 6
Q4. If x + kx + 6 = (x + 2) (x + 3) for all x, then the value of k is:
2
(a) 1 (b) – 1 (c) 5 (d) 3

SECTION – B – (2 marks for each correct answer)
2 2 2
Q5. Factorise: 9x + 4y + 16z + 12xy – 16yz – 24xz
Q6. Factorise: 4x2 + 20x + 25
Q7. Verify if 2 and 0 are zeroes of the polynomial x 2 − 2x.
Q8. Evaluate: 993
Q9. Expand: (3a+5b)3.
SECTION – C – (3 marks for each correct answer)
Q10. If ( x − 4) is a factor of the polynomial 2 x 2 + Ax + 12 and ( x − 5) is a factor of the

polynomial x 3 − 7 x 2 + 11 x + B , then what is the value of ( A − 2 B )?
Q11. If (x+3) and(x-3) are factors of ax2+5x+b then show that a=b.
CLASS TEST – 2 -
TOPIC- POLYNOMIALS
CLASS-IX
TIME - 90 MIN. M.M. 30

10
Q1. The value of (5)3 + (7)3 + (−12)3 is:
(a) 1260 (b) -1260 (c) 420 (d) 0
Q2. If 𝑎 + 𝑏 = 7, 𝑎𝑏 = 6 then the value of 𝑎3 + 𝑏3will be:
(a)117 (b) 217 (c) 469 (d) 61
Q3. Product of (2𝑎 + 7) (2𝑎 − 7) is:

(a) 2𝑎2 – 7 (b) 2𝑎2 − 49 (c) 2𝑎2 + 49 (d) 4𝑎2 – 49
Q4. Factors of √2𝑥2 − 𝑥 − 10√2 are:

(a) (√2𝑥 − 5) (𝑥 + 2√2) (b) (√2𝑥 + 5) (𝑥 + 2√2)
(c) (√2𝑥 + 5) (𝑥 − 2√2) (d) (√2𝑥 − 5) (𝑥 − 2√2)
Q5. Factors of 343𝑎3 − 125𝑏3 are:

(a) (7𝑎 + 5𝑏) (49𝑎2 + 35𝑎𝑏 + 25𝑏2)
(b) (7𝑎 − 5𝑏) (49𝑎2 − 35𝑎𝑏 − 25𝑏2)
(c) (7𝑎 − 5𝑏) (7𝑎2 + 35𝑎𝑏 + 5𝑏2)
(d) (7𝑎 − 5𝑏) (49𝑎2 + 35𝑎𝑏 + 25𝑏2)
SECTION – B – (2 marks for each correct answer)

2 2
Q6.If 𝑎 + 𝑏 = 15, 𝑎𝑏 = 14 then find 𝑎 + 𝑏
Q7.Expand (𝑎 − 2𝑏 + 3𝑐)2
Q8.Expand (3𝑎 − 4𝑐)3
Q9. Find the product of (𝑥2 + 3𝑦 + 7) (𝑥 − 2).
Q10.Find the value of 𝑝 (−2√3) if p (𝑦) = (√3𝑦2 − 3𝑦 + 5√3).
SECTION – C – (3 marks for each correct answer)

Q11. If 𝑎 + 𝑏 + 𝑐 = 9 and 𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎 = 26, find 𝑎2 + 𝑏 2 + 𝑐2.
Q12. If 𝑥 + 𝑦 = 12 and 𝑥𝑦 = 27, find the value of 𝑥3 + 𝑦3.
Q13. If (𝑥 − 2) is a factor of (𝑥) = 𝑥4 – 2𝑥3 + 3𝑥2 – 𝑎𝑥 + 3𝑎 – 7
Then find the value of 𝑎.
Q14. Find 𝑘 if 𝑥3 + 6𝑥2 + 11𝑥 + 6 = 𝑘 (𝑥 + 1)
Q15. Factorise: 𝑥3 − 216𝑦3 - 18𝑥2𝑦 + 108𝑥𝑦2.
ANSWERS FOR TEST PAPERS:
TEST 1 :
(i) MCQ QUESTIONS:-
1) (a) 2) (c) 3) (c) 4) (c)
(ii) SHORT ANSWER QUESTIONS-
5) (3x + 2y – 4z)2 , 6) ( 2x + 5 )2, 7)YES, 2 & 0 are zeroes of given polynomial
8) 970299 9)27a3 + 125b3 +135a2b + 225ab2 , 10) A = 11,B = 5 then A-2B =1
11
TEST 2 :
(i) MCQ QUESTIONS:-
1) (b) 2) (b) 3) (d) 4) (a) 5) (d)
(ii) SHORT ANSWER QUESTIONS-
6) 197, 7) a2 + 4b2 +9c2 -4ab – 12bc + 6ac, 8) 9a3 – 64c3 – 108a2c +144ac2
9) x3 + 3xy + 7x – 2x2 -6y -14, 10) 23 3 , 11) 29, 12) 756, 13) a = -5,
14) K= (x + 3) (x + 2), 15) x3 – 216y3 – 18x2y + 108xy2
12
CO-ORDINATE GEOMETRY
(I) Main Concepts and Results:
1. Identify the need of coordinate geometry.
2. Identify Cartesian system.
3. Understand the four quadrants and the nature of signs of points.
4. Identify the quadrant in which a given point lies.
5. Identify the terms - axes and origin.
6. Understand the meaning of coordinates.
7. Acquire skill in plotting points in the Cartesian plane.
8. Find the coordinates of a point plotted in Cartesian plane.
9. Identify the equations of the axes.
10. Understand the nature of coordinates of points on the two axes.
EXAMPLES:
1) MCQ‟S-
1) The name of the horizontal line in the Cartesian plane which determines the position of a point is called:
a. Origin b. X-axis c. Y-axis d. Quadrants
Answer: b
2) The name of the vertical line in the Cartesian plane which determines the position of a point is
called:
Answer: c
3) The section formed by horizontal and vertical lines determining the position of the point in a Cartesian
plane is called:
Answer: d
4) The point of intersection of horizontal and vertical lines determining the position of a point in a Cartesian
plane is called:
13
Answer: a
5) If the coordinates of a point are (0, -4), then it lies in:
a. X-axis b. Y-axis c. At origin d. Between x-axis and y-axis
Answer: b
2) SHORT ANSWER QUESTIONS:
Q6. Which of the following points lie on y-axis?

A(1,1), B(1,0), C(0,1), D(0,0), E(0,-1), F(-1,0), G(0,5), H(-7,0),I(3,3).
Answer: C(0,1), D(0,0), E(0,-1), G(0,5)
Q7. Without plotting the points indicate the quadrant in which they lie, if:
1) Ordinate is -5 and abscissa is 3.

2) abscissa is -5 and ordinate is -3.
Answer: IV & III quadrant.
Q8. Take a rectangle ABCD with A(-6,4), B(-5,2), C(-3,3), D(-,4). Find it‟s mirror image with respect to x-
axis.
Answer: A(-6,-4), B(-6,-2), C(-2,-2), D(-2,-4)
3) PRACTICE QUESTIONS:
(A) MCQ‟S QUESTIONS:
1) If the coordinates of a point are (3, 0), then it lies in:
a. X-axis b. Y-axis c. At origin d. Between x-axis and y-axis
2) If the coordinates of a point are (-3, 4), then it lies in:
a. First quadrant b. Second quadrant c. Third quadrant d. Fourth quadrant
3) Points (1, 2), (-2, -3), (2, -3);
a. First quadrant b. Do not lie in the same quadrant c. Third quadrant d. Fourth quadrant
4) If x coordinate of a point is zero, then the point lies on:
a. First quadrant b. Second quadrant c. X-axis d. Y-axis
ASSERTION & REASON BASED QUESTION:
Q5. Assertion: The abscissa of a point (5, 2) is 5.

Reason: The Ʇ distance of a point from y-axis is called its abscissa.

(A) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
(B) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
14
(C) Assertion is correct statement but Reason is wrong statement.
(D) Assertion is wrong statement but Reason is correct statement.
(B) SHORT ANSWER QUESTIONS:
Q6. Find the coordinates of the point

a) Whose ordinate is -4 and lies on y-axis.
b) Whose abscissa is 5 and lies on x-axis.
Q7. A point lies on x-axis at a distance of 9 units from y-axis. What are itscoordinates? What will be the
coordinates of a point, if it lies on y-axisat a distance of -9 units from x-axis?
Q8. What will be reflections of D(-2, -3) in x-axis and y-axis?
Q9. In the given figure, ABC is an equilateral triangle. The coordinates ofvertices B
and C are (3,0) & (-3,0) respectively. Find the coordinates of its vertex A. Also, find
its area.
Q10. Plot the points ( -1 , -1 ) , ( 2 , 3 ) and ( 8, 11 ) and show that they are collinear.
(C) ANSWER OF PRACTICE QUESTIONS:
(A) MCQ QUESTIONS:-
1) (a), 2) (b), 3) (b) 4) (d), 5) (A),
(B) SHORT QUESTIONS:
6) a) (0,-4), b) (5,0)
7) When the point lies on x axis at a distance of 9 units from y axis then the coordinate of this point is (9,0).
When the point lies on y axis at a distance of -9 units from x axis then the coordinate of this point is (0,-9).
8) On x-axis (-2,3),On y-axis (2,-3)
9) BC = 3+3 = 6 units
Length of altitude OA = √3/2 x BC = √3/2 x 6 =3√3Therefore, coordinate of A are (0, 3√3).
CLASS TEST – 1 -
TOPIC- CO-ORDINATE GEOMETRY
CLASS-IX
TIME- 45 MIN. M.M. 20

Q1. The point (-10, 0) lies in

15
a. Third quadrant b. Fourth quadrant
c. On the negative direction of the x-axis d. On the negative direction of the y-axis
Q2. A quadrant in which both x and y values are negative is
a. First quadrant b. Second quadrant c. Third quadrant d. Fourth quadrant
Q3. Abscissa of all the points on the x-axis is
a. 0 b. 1 c. 2 d. Any number
Q4. Ordinate of all points on the x-axis is
a. -1 b. 0 c. 1 d. Any number
SECTION – B – ( 2 marks for correct answer)
Q5.In which Quadrant abscissa of a point is positive?
Q6. Name the Quadrant in which Quadrant/ on Axis Points (1, 1), (2, -2), (-4, -5), (-3, 4),(0,7), (5,0)
are lying.
Q7. What is the value of abscissa of all the points on the y-axis ?
Q8. On which Axis value of Ordinate of is any number?
Q9. Write the Coordinate of the point which lies on the y-axis at a distance of 5 units in the negative
direction of the y-axis .d. (0, -5).
Q10. . If ( x , y ) = ( y, x ) then find the value of x ,y.
SECTION – C – ( 2 marks for each correct option)
ASSERTION & REASON BASED QUESTIONS

(a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but Reason is correct statement.
Q11. Assertion: Point A (-2, -4) lies on III quadrant

Reason: A point both of whose coordinates are –ve lies in III quadrant.
16
Q12. Assertion: Point (4, -2) lies in IV quadrant.
Reason: The Ʇ distance of a point from y-axis is called its abscissa.
ANSWERS OF TEST 1:-
(A) MCQ QUESTIONS:-
1) (c), 2) (c), 3) (d) 4) (b)
(B) SHORT QUESTIONS:
5) I and IV quadrants , Explanation: In a coordinate plane, x can take positive values in the first and fourth
quadrants. For example, (2, 2) and (2, -4) lie on the first and fourth quadrants, respectively.
6) I,IV,III,II, on Y-Axis & X-Axis ,
7) 0 , Explanation: The abscissa of all the points on the y-axis is 0. We know that the coordinates of any
point on the y-axis is (0, y). Here, the ordinate can take any value and the abscissa is zero.
8) Y-Axis , 9) (0, -5) , 10) x=1,y=1
Q11. (a) Assertion and Reason both are correct statements and Reason is the correct explanation of
Assertion.
Q12. (b) Assertion and Reason both are correct statements but Reason is not the correct explanation of
Assertion.
17
LINEAR EQUATION IN TWO VARIABLE
Main Concepts and Results:

● Any equation which can be written in the form ax + by+ c = 0, where a, b and c are real
numbers a≠0, b≠0 is called a linear equation in two variables.
● An ordered pair ( x ,y)is the solution of linear equation in two variables if this point satisfies the
linear equation ax + by + c = 0.
● Examples of line are equation in two variables -2x ꞱꞱ4y Ʇ1, xꞱ10y ꞱꞱꞱ5, etc.
● A linear equation has a unique solution when there exists only one point which satisfies the
linear equation.
● For example: Solution of 2x+6 =2 is
● 2x+6=2
2x = 2−6
2x = −4
x = −4 ÷ 2
x = −2
In 2x+6=2 has only one variable x therefore x has unique solution. Also, geometrically it
will be a point on rectangular axes whose ordinate will be 0.
● A system of linear equation has unique solution when the system of lines intersects each
other at only one point.
● A linear equation in two variables have infinitely many solutions means there are more
than one ordered pair which satisfy the equation.
● Equation of x-axis is y=0 because in x-axis, y coordinates are always zero and the coordinate
form of any point on x-axis will be (x, 0).
● Equation of y-axis is x =0 because at y-axis x-coordinates are always zero and the coordinate
form of any point on y-axis will be ( 0 , y )
18
ILLUSTRATIONS:
MCQ TYPES OF QUESTIONS:

1. Which points given below satisfy the equation 2x + 3y = 12?
A. (-6, 8) B. (6, -8) C. (3, 2) D.( − 4, 5)
Ans. A
2. Which of the following is a linear equation in one variable?
A. 2x + 3y = 0 B. x2 = 5x + 3 C. 5x = y2 + 3 D. 2x + 5 = 11
Ans. D
3. The cost of book (x) exceeds twice the cost of pen (y) by 10 rupees. This statement can be
expressed as linear equation as:
A. x -2y-10=0 B. 2x-y-10=0 C. 2x+y-10=0 D. x-2y+10=0
Ans. A
4. The linear equation 2x – 5y = 7 has
(A) unique solution (B) Two solutions (C) Infinitely many solutions (D) No solution
Ans. C
5. Assertion: (2,1) is a solution of 2x + 3y = 7
Reason: If Ordered pair (p, q) lies on the line then it is one of the solutions of line ax + by
+ c = 0.
A) Both Assertion and Reason are correct and reason is correct explanation for the assertion.
B) Both Assertion and Reason are false but reason is not correct explanation for assertion.
C) Assertion is correct but reason is false.
D) Both Assertion and reason are false.
Ans. A
SHORT ANSWER TYPE QUESTIONS:
1.Find the points where the graph of the equation 3x + 4y = 12 cuts the x-axis and the y-axis.
Ans. The graph of the linear equation 3x + 4y = 12 cuts the x-axis at the point where y = 0.
On putting y = 0 in the linear equation, we have 3x = 12, which gives x = 4. Thus, the
required point is (4, 0).
5
2. Determine the point on the graph of the equation 2x + 5y = 20 whose x-coordinate is
2
times its ordinate.
19
5 5
Ans. As the x-coordinate of the point is times its ordinate, therefore, x = y. Now putting
2 2
value of x in 2x + 5y = 20, we get, y = 2. Therefore, x = 5. Thus, the required point is (5, 2).
3. At what point does the graph of the linear equation x + y = 5 meet a line which is parallel
to the y-axis, at a distance 2 units from the origin and in the positive direction of x-axis.
Ans. The coordinates of the points lying on the line parallel to the y-axis, at a distance 2 units
from the origin and in the positive direction of the x-axis are of the form (2, a). Putting x = 2,
y = a in the equation x + y = 5, we get a = 3. Thus, the required point is (2, 3).
4. Draw the graph of the equation represented by the straight line which is parallel to the x-
axis and is 4 units above it.
Ans.
5. Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is the
value of y when x = 5.
Ans. This shows that y = 3x
Hence, at x = 5 we get y = 15.
QUESTIONS FOR PRACTICE:

MCQ TYPES OF QUESTIONS:
1. The positive solutions of the equation ax + by + c = 0 always lie in the (A) 1st
quadrant (B) 2nd quadrant (C) 3rd quadrant (D) 4th quadrant
2. x = 5, y = 2 is a solution of the linear equation
(A) x + 2 y = 7 (B) 5x + 2y = 7 (C) x + y = 7 (D) 5 x + y =7
3. How many linear equations in x and y can be satisfied by x = 1 and y = 2?
(A) Only one (B) Two (C) Infinitely many (D) Three
4. Any point on the line y = x is of the form
(A) (a, a) (B) (0, a) (C) (a, 0) (D) (a, – a)
5. Assertion: The graph of y = b is always parallel to x- axis.
Reason: The graph of y = 6 is a line that passes through the origin.
A) Both Assertion and Reason are correct and reason is correct explanation for the assertion.
B) Both Assertion and Reason are false but reason is not correct explanation for assertion.
SHORT ANSWER TYPE QUESTIONS:
1.How many solution(s) of the equation 2x + 1 = x – 3 are there on the:
(i) Number line (ii) Cartesian plane
20
2. Show that the points A (1, 2), B (– 1, – 16) and C (0, – 7) lie on the graph of the linear
equation y = 9x – 7.
3. For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its
solution.
4. The following observed values of x and y are thought to satisfy a linear equation. Write the
linear equation:
x 6 −6
y −2 6
Draw the graph using the values of x, y as given in the above table. At what points the graph
of the linear equation
(i) cuts the x-axis (ii) cuts the y-axis
5. If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a.
ANSWERS:
MCQ QUESTIONS
1. A 2.C 3. C 4. A 5. C

8−𝑥
1.One, infinite 2. yes, all points are solutions. 3. c = 𝑥
4. The graph cuts the x-axis at (3, 0) and the y-axis at (0, 2).
5
5. 3
PRACTICE TEST-1
MARKS: 20
Q NO. QUESTIONS
SECTION - A
1. The equation of x–axis is 1
(a)a = 0 (b)y= 0 (c)x=0 (d)y=k
2. The ordered pair (m , n) satisfies the equation ax + by + c = 0 if 1
(a)am+ bn=0 (b)c =0 (c)am+ bn+ c =0 (d)am+bn–c= 0
3. Which of the following is not a linear equation in two 1

variables?
(a)ax+ by= c (b)ax2+by=c (c)2x+3y=5 (d)3x+2y=6
21
4. A linear equation in two variables has __solutions. 1
(a)no (b)only one (c)only two (d)infinitely many
5. x=5 , y=2is a solution of the linear equation 1
(a)x+2y=7 (b)5x+ 2y=7 (c)x+y=7 (d)5x+y=7

6. The graph of the linear equation 2x+3y =6 is a line which meets 1
the x-axis at the point
(a)(2,0) (b)(0,3) (c)(3,0) (d)(0,2)
7. The point of the form (a , a)always lies on: 1
(a) x–axis (b)y–axis (c)on the line y = x (d)on the line x + y=0
8. The solution of the equation x– 2y = 4 is: 1
(a)(0,2) (b)(4,0) (c)(1,1) (d)(2,0)

9. If (2 , 0) is a solution of the linear equation 2x + 3y = k, then the 1
value of k is
(a)4 (b)6 (c)5 (d)2
10. The equation x = 7, in two variables, can be written as 1
(a)x+0y= 7 b) 0x+ y=7 (c) 0x + 0y= 7 (d)x +y=7
SECTION-B
11. Find the solution of the linear equation x+2y = 8 which 2
represents a point on (i)x-axis (ii)y-axis
12. Solve the equation 2x+1 = x–3, and represent the solution(s) on 2
(i) The number line,

(ii) The Cartesian plane.
13. Let y varies directly as x. If y = 12 when x=4, then write a 2
linear equation. What is the value of y when x= 5?
14. Determine the point on the graph of the equation 2x + 5y = 20 2
5
whose x-coordinate is times its ordinate.
2
15. What is the distance of (2 , 4) from x-axis and y-axis. 2
ANSWERS:
1.b 2. C 3.b 4.d 5. C 6.c 7.c 8.b 9.a 10. a
11. (i) (8,0) (ii) (0,4)
12. x= −4
13.y = 3x, y = 15
14.(5,2)
22
PRACTICE TEST-2
MARKS: 30
Q QUESTION
N
0
.
SECTION - A
1 Which of the following represent a line parallel to x-axis? 1
(A) x + y = 3 (B) 2X + 3 = 7
(C) 2Y – 3 = Y – 7 (D) x + 3= 0
2 The point of the form (a, – a) always lies on the line 1
. (A) x = a (B) y = – a (C) y = x (D) x + y =0
3 If we multiply or divide both sides of a linear equation 1
with a non-zero number, then the solution of the linear
equation:
(A) Changes (B) Remains the same
(C) Changes in case of multiplication only
(D) Changes in case of division only
4 The equation 2x + 5y = 7 has a unique solution, if x, y is: 1
(A) Natural numbers (B) Positive real numbers
(C) Real numbers (D) Rational numbers
5 The linear equation 3x – y = x – 1 has: 1
(A) A unique solution (B) Two solutions
(C) Infinitely many solutions (D) No solution
6 A linear equation in two variables is of the form ax + by + 1
. c = 0, where
(A) a ≠ 0, b ≠ 0 (B) a = 0, b ≠ 0 (C) a ≠ 0, b = 0
(D) a = 0, c = 0
7 Any point on the y-axis is of the form 1
. (A) (x, 0) (B) (x, y) (C) (0, y) (D) ( y, y)
8 The solution of a linear equation in two variables is 1
.
(A) a number which satisfies the given equation
(B) an ordered pair which satisfies the given equation
(C)an ordered pair, whose respective values when substituted for
x and y in the given equation, satisfies it
(D) none of these
9 The graph of ax + by + c= 0 is 1
.
(A)a straight line parallel to x–axis
(B) a straight line parallel to y–axis
(C)a general straight line
(D) a line in the 2nd and 3rd quadrant
1 The ordered pair (m , n) satisfies the equation ax+by+c = 0 1
0 if
.
(A) am+ bn= 0 (B)c = 0
(C)am+ bn+ c =0 (D)am+bn–c= 0
23
1 The linear graph 2x + 3y = 12 cuts y axis at 1
1
. (A) (3, 0) (B) (4, y) (C) (2, 2) (D) (3,2)
1 The graph of the linear equation in two variables y= mx is 1
2
. (A)a line parallel to x–axis
(B)a line parallel to y–axis
(C)a line passing through the origin
(D)not a straight line
1 How many linear equations in x and y can be satisfied by 1
3 x=−1and y=3?
.
(A) Only one (B)two
(B) (C)three (D)infinitely many
1 Point(3,1)lies on the line: 1
4
. (A)x+2y=5 (B)x+ 2y= –6 (C)x+2y=6 (D)x+2y=16
1 The graph of the linear equation x + 2y = 7 passes through 1
5 the point
.
(a)(0, 7) (b)(4, 3) (c)(6,1) (d)(7,0)
1 The graph below is of which linear equation: 1

6
. (A) x + y = 0
(B) x – y = 0
(C) 2x + y = 3
(D) 2x – 3y = 4
1 The following is the graph of which 1
7 linear equation:
.
(A) x – 3 = 0
(B) x = y
(C) 2x +y = 0
(D) x + 2y = 0
1 The graphs of linear equations y = x and y = – x on the same 1
8 cartesian plane. What do you observe?
.
(A) Both lines intersect at origin.
(B) Both lines are parallel to x-axis.
(C) Both lines are parallel to y-axis.
(D) None of these.
1 Assertion: (2,4) is a solution of 2x + 3y = 16 1
9
. Reason: If Ordered pair (p, q) lies on the line then it is one
of the solutions of line ax + by + c = 0.
A) Both Assertion and Reason are correct and reason is

correct explanation for the assertion.
B) Both Assertion and Reason are false but reason is not
correct explanation for assertion.
24
2 The value of k if (3,1) lies on 4x − ky = −2 1
0
. a)10 (b)14 (c)15 (d)12
SECTION - B
2 Form a linear equation whose solutions are represented by 2

1 the points having the sum of the coordinates as 10 units.
.
2 Determine the point on the equation 2x + 5y = 20 whose 2
2 5
. x-coordinate is times its ordinate.
2
2 How many solution(s) of the equation 2

3 3x + 1 = 2x – 3 are there on the:
. (i) Number line (ii) Cartesian plane
2 Find four different solutions of the equation x + 2y = 6. 2
4
.
2 The cost of a notebook is twice the cost of a pen. Write a linear 2
5 equation in two variables to represent this statement.
.
ANSWERS:
1.C 2. B 3.C 4.A 5.C 6.A 7.C 8.C 9.C 10.C 11.D 12.C 13.C 14.A 15.D
5
16.A 17.A 18.A 19.A 20.B 21. x + y = 10 22. 2 23. (i) one (ii) infinite 24. (2,2),
(0,3), (6,0), (4,1)
25. let cost of one pen be x and one notebook be y
y = 2x
y – 2x = 0
25
EUCLID GEOMETRY
CONCEPTS
Points, Line, Plane or surface, Axiom, Postulate and Theorem, The Elements, Shapes
of altars or vedis in ancient India, Equivalent versions of Euclid‟s fifth Postulate,
Postulates
1. A straight line may be drawn from any point to any other point.
2. A terminated line (line segment) can be produced indefinitely.
3. A circle may be described with any centre and any radius.
4. All right angles are equal to one another.
5. If a straight line falling on two straight lines makes the interior angles on the same
side of it, taken together less than two right angles, then the the two straight lines if
produced indefinitely, meet on that side on which the sum of angles is taken together
less than two right angles
Euclid’s axioms
(1) Things which are equal to the same thing are equal to one another.
(2) If equals are added to equals, the wholes are equal.
(3) If equals are subtracted from equals, the remainders are equal.
(4) Things which coincide with one another are equal to one another.
(5) The whole is greater than the part.
(6) Things which are double of the same things are equal to one another.
(7) Things which are halves of the same things are equal to one another.
ILLUSTRATIONS
MCQ
1. Euclid‟s fifth postulate is
(A) The whole is greater than the part.
(B) A circle may be described with any centre and any radius.
26
(C) All right angles are equal to one another.
(D) If a straight line falling on two straight lines makes the interior angles on the same side of
it taken together less than two right angles, then the two straight lines if produced
indefinitely, meet on that side on which the sum of angles is less than two right angles.
Solution : Answer (D)
2: John is of the same age as Mohan. Ram is also of the same age
as Mohan. State the Euclid‟s axiom that illustrates the relative ages of John and Ram
(A) First Axiom (B) Second Axiom
(C) Third Axiom (D) Fourth Axiom
Solution : Answer (A)
3. Euclid divided his famous treatise “The Elements” into :
(A) 13 chapters (B) 12 chapters (C) 11 chapters (D) 9 chapters
ANS A) 13 CHAPTERS
4. The number of dimensions, a solid has :
(A) 1 (B) 2 (C) 3 (D) 0
Short answer type questions
1 .Write whether the following statements are True or False?
Justify your answer
1Euclid‟s fourth axiom says that everything equals itself.
Ans True. Things equal to the same thing are equal.
QUESTION FOR PRACTICE
1.The number of dimensions, a solid has :
(A) 1 (B) 2 (C) 3 (D) 0
2.. It is known that if x + y = 10 then x + y + z = 10 + z. The Euclid‟s axiom that illustrates

this statement is :
(A) First Axiom (B) Second Axiom
(C) Third Axiom (D) Fourth Axiom
3. Which of the following needs a proof ?
(A) Theorem (B) Axiom (C) Definition (D) Postulate
27
4. Euclid belongs to the country :
(A) Babylonia (B) Egypt (C) Greece (D) India
5.Assertion and reason
2) Assertion: Through two distinct points there can be only one line that can be drawn.
Reason: . . From this two point we can draw only one line
a) both Assertion and reason are correct and reason is correct explanation for Assertion
b) both Assertion and reason are correct but reason is not correct explanation for Assertion
c) Assertion is correct but reason is false
d) both Assertion and reason are false
Short answer type
6.Ram and Ravi have the same weight. If they each gain weightby 2 kg, how will their new
weights be compared ?
7 If a point C lies between two points A and B such that AC = BC, then prove that AC =BC‟
then prove that AC=1/2 AB. Explain by drawing the figure.
8.Write postulate 1 in your own words.
9. If A, B and C are three points on a line, and B lies between A and C then prove that AB +
BC = AC.
10. : Solve the equation a – 15 = 25 and state which axiom do you use here.
Answers 1) C 2 .(B) 3(A) 4 (C) 5 (A) 6) Ram and Ravi are equal in weight.
10. Euclid‟s second axiom).or a = 40
Chapter test MM 20
MCQ (1 Mark each)

(A) 1 (B) 2 (C) 3 (D) 0
2. The number of dimensions, a surface has :
(A) 1 (B) 2 (C) 3 (D) 0
3.The number of dimension, a point has :
(A)0 (B) 1 (C) 2 (D) 3
4. The side faces of a pyramid are :
(A)Triangles (B) Squares (C) Polygons (D) Trapeziums
(A)Theorem (B) Axiom (C) Definition (D) Postulate
28
6.It is known that if x + y = 10 then x + y + z = 10 + z. The Euclid‟s axiom that illustrates this
statement is :
7.In Indus Valley Civilisation (about 3000 B.C.), the bricks used for construction work were
having dimensions in the ratio
(A) 1 : 3 : 4 (B) 4 : 2 : 1 (C) 4 : 4 : 1 (D) 4 : 3 : 2 (A) First Axiom (B) Second Axiom
Chapter test
8 „Lines are parallel if they do not intersect‟ is stated in the form of
(A) an axiom (B) a definition (C) a postulate (D) a proof

9. The things which are equal to the same thing are equal to one another in the form of
(A) An axiom B)definition (C) a postulate (D) a proof
10. Boundaries of surfaces are :
(A) surfaces (B) curves (C) lines (D) points
Short answer type (2 marks each)
11. : Solve the equation a – 15 = 25 and state which axiom do you use here.
12. If a point C lies between two points A and B such that AC = BC, then prove
that AC =1/2 AB. Explain by drawing the figure.
13. Write the Euclid s second postulate in your own words.
14 Ram and Ravi have the same weight. If they each gain weight by 2 kg, how will their new
15. Give a definition for each of the following terms.
i)parallel lines ii)line segment
Chapter test MM 30
MCQ(1 MARK)
(A) 1 (B) 2 (C) 3 (D) 0
2. The number of dimensions, a surface has :
(A) 1 (B) 2 (C) 3 (D) 0
3.The number of dimension, a point has :
(A)0 (B) 1 (C) 2 (D) 3
4. The side faces of a pyramid are :
(A)Triangles (B) Squares (C) Polygons (D) Trapeziums
(A)Theorem (B) Axiom (C) Definition (D) Postulate
6.It is known that if x + y = 10 then x + y + z = 10 + z. The Euclid‟s axiom that illustrates this
statement is :
9. The things which are equal to the same thing are equal to one another in the form of (A)An
axiom B)definition (C) a postulate (D) a proof
10. Boundaries of surfaces are :
(A) surfaces (B) curves (C) lines (D) points
29
SHORT ANS TYPE(2 MARKS EACH)
11 .State any two Euclids postulates .
12 .Solve the equation x- 5 =15 and state the axiom that you use here.
13 How many dimensions do boundaries of surfaces have ?
14. What is the difference between axiom and postulates?
15 Which of the following statements are true and which are false? Give reasons for your
16.: Solve the equation a – 15 = 25 and state which axiom do you use here.
17. If a point C lies between two points A and B such that AC = BC, then prove that
AC =1/2 AB. Explain by drawing the figure.
18. Write the Euclid s second postulate in your own words.
19. Ram and Ravi have the same weight. If they each gain weight by 2 kg, how will their new
20.. Give a definition for each of the following terms.
i)parallel lines ii)line segment
30
LINES AND ANGLES
Basic Terms and Definitions on Lines and Angles
Line Segment: A line that has two endpoints is called a line segment.
Ray: A line with one endpoint and the other end of the line extending up to infinity is called a ray.
collinear points: When three or more points lie on the same line, they are said to be collinear.
Non-collinear points: When three or more points do not lie on the same line, they are non-collinear.
Angle: An angle is formed by two rays meeting at a common point (called a vertex), and the rays forming the
angle are called arms of the angle.
Acute Angle: An angle that measures between 0° and 90° is called an acute angle.
Obtuse angle: An angle that measures between 90° and 180° is called an obtuse angle.
Right angle: An angle that is equal to 90° is called a right
angle.
Reflex angle: An angle greater than 180° but less than 360° is
called a reflex angle.
: Types Of Angles
Complementary angles: When sum of two angles is equal to
90°
Supplementary angles: When sum of two angles is equal to 180°.
Adjacent angles: Two angles with a common vertex, a common arm and their
non-common arms on different sides of the common arm.
Linear pairs of angles: When 2 adjacent angles are supplementary, i.e. they form a straight line (add up to
180∘), they are called a linear pair.
Vertically opposite angles: When two lines intersect at a point, they form equal angles that
are vertically opposite to each other.
Intersecting and Non-Intersecting Lines
31
When two lines intersect each other at a common point, they are said to be intersecting lines.
Non-intersecting lines are parallel lines that do not intersect each other at a common point.
Pairs of Angles
Axiom – Linear Pair of Angles
If a ray stands on a line, the sum of two adjacent angles so formed is 180°.
Axiom – Converse of Linear Pair of Angles
If the sum of two adjacent angles is 180°, the non-common arms of the angles form a line.
Theorem – Vertically Opposite Angles
If two lines intersect each other, the vertically opposite angles are equal.
Parallel lines with a transversal
A line that intersects two or more lines is called a transversal.
∠1 = ∠5, ∠2 = ∠6, ∠4 = ∠8 𝑎𝑛𝑑 ∠3 = ∠7(Corresponding angles)
∠3 = ∠5, ∠4 = ∠6 (Alternate interior angles)
∠1 = ∠7, ∠2 = ∠8 (Alternate exterior angles)
Interior angles on the same side of the transversal are referred to as consecutive interior angles, allied angles,
or co-interior angles.
Corresponding angles axiom
If a transversal intersects two parallel lines, then each pair of corresponding angles are equal.
Converse of corresponding angles axiom

If a transversal intersects two lines such that a pair of corresponding angles are equal, then the two lines are
parallel to each other.
Theorem – Alternate interior angles

If a transversal intersects two parallel lines, then each pair of alternate interior angles are equal.
Theorem – Converse of alternate interior angles

If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are
parallel.
Theorem – Sum of co-interior angles is supplementary

If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal
is supplementary.
Theorem – Converse of the sum of co-interior angles is supplementary

If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal are
supplementary, then the two lines are parallel.
32
Lines parallel to the same line
Lines that are parallel to the same line are also parallel to each other.
Theorem – Lines Parallel to the Same Line

Lines which are parallel to the same line are parallel to each other.
Angle Sum Property of a Triangle

Theorem 1: The sum of the angles of a triangle is 180º.
Theorem 2: If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two
interior opposite angles.
∠4=∠1+∠2
M.C.Q. QUESTIONS
Question 1.In a triangle, if the measure of an exterior angle is 105° and its opposite interior angles are equal.
Find the value of these equal angles
a. 72 ½ °
b. 52 ½ °
c. 75°
d. 37°
Answer: b. 52 ½
Explanation: Given, exterior angle = 105°

Let us consider the interior angles as x
Using the exterior angle theorem,
[Sum of the interior opposite angles = exterior angle]
2x = 105°
x = 52 ½ °. Hence, each of the interior opposite angles measure 52 ½ °.
Question2: If the ratio of the angles 2:4:3. The value of the smallest angle will be:
a. 40°
b. 80°
c. 60°
d. 20 °
Answer: a. 40°
Explanation: let us consider 2:4:3 as 2x, 4x and 3x
So, 2x + 4x +3x = 180° [the sum of the interior angles of a triangle is 180°]
9x = 180°
x = 20°
Hence, the value of:
2x = 2(20°) = 40°
4x = 4(20°) = 80°
3x = 3 (20°) = 60°
33
So, the smallest angle is 40°.
Question3: Find the value of x from the given figure, where POQ is a line.
a. 20°
b. 30°
c. 25°
d. 35°
Answer: a. 20°
Explanation: Given POQ is a line, which means POQ = 180°.
40° + 4x + 3x = 180°
40° + 7x = 180°
7x = 180° - 40°
7x = 140°
x = 140°/7
x = 20°
So, x = 20°
Question4: If AOB is a line then the measure of ∠BOC, ∠COD and ∠DOA respectively in the given figure,
are:
a. 36°, 54°, 90°

b. 90°, 54°, 36°
c. 90°, 36°, 54°
d. 36°, 90°, 54°
Answer: a. 36°, 54°, 90°

Explanation: ∠AOD + ∠ DOC + ∠ COR = 180° [sum of the interior angles of a triangle is 180°]
5y + 3y + 2y = 180°
10y =180°
y = 180° / 10
y = 18°
So, the values of 5y, 3yand 2y are:
5y = 5 (18°) = 90°
3y = 3 (18°) = 54°
2y = 2 (18°) = 36°
34
CASE STUDY BASED QUESTIONS
Q1. Read the following and answer the questions given below :
Ramesh singh bought an electric bicycle for his son. He saw the bicycle and felt very happy. After seeing
the bicycle he thought of some geometrical figure:
(i) From the geometrical figure , what is ∠ CBF, if ∠ BCD = 450 and AB ǁ CD?
(a) 900 (b) 450 (c) 750 (d) 300
(ii) In the given figure , ∠AFC = 750, then ∠CFB =
(a) 750 (b) 450 (c) 1050 (d) None of these
(iii) In the given figure, ∠FCB =
(a) 450 (b) 300 (c) 750 (d) None of these
(iv) In the given figure, what is the value of ∠EFB ?
(a) 750 (b) 450 (c) 300 (d) 1050
Answer : (i) We have, AB ǁ CD
∠ BCD = ∠ CBF ( Alternate angles)
0
45 = ∠ CBF
Option (b) is correct
(ii) ∠ AFC + ∠ CFB = 1800 (Linear pair)
750 + ∠ CFB = 1800
∠ CFB = 1800 – 750 =1050
Option © is correct
(iii) Since AB ǁ CD,
∠AFC =∠ FCD (Alternate angles)
0
75 = ∠ FCB + ∠ BCD
750 = ∠ FCB + 450
∠ FCB = 300
Option (b) is correct
(iv)We have, ∠ EFB = ∠ AFC (Vertically opp. Angles)
∠ EFB = 7500
Option (a) is correct.

Q.1: In the figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and
reflex ∠COE.
Solution:
From the given figure, we can see;
∠AOC, ∠BOE, ∠COE and ∠COE, ∠BOD, ∠BOE form a straight line each.
So, ∠AOC + ∠BOE +∠COE = ∠COE +∠BOD + ∠BOE = 180°
35
Now, by substituting the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° we get:
70° +∠COE = 180°
∠COE = 110°
Similarly,
110° + 40° + ∠BOE = 180°
∠BOE = 30°
Q.2: In the Figure, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
Solution:
As we know, the sum of the linear pair is always equal to 180°
So,
∠POY + a + b = 180°
Substituting the value of ∠POY = 90° (as given in the question) we get,
a + b = 90°
Now, it is given that a : b = 2 : 3 so,
Let a be 2x and b be 3x.
∴ 2x + 3x = 90°
Solving this we get
5x = 90°
So, x = 18°
∴ a = 2 × 18° = 36°
Similarly, b can be calculated and the value will be
b = 3 × 18° = 54°
From the diagram, b + c also forms a straight angle so,
b + c = 180°
=> c + 54° = 180°
∴ c = 126°
Q.3: In the Figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays
OP and OR. Prove that ∠ROS = 1/2(∠QOS – ∠POS).
Solution:
In the question, it is given that (OR ⊥ PQ) and ∠POQ = 180°
36
So, ∠POS + ∠ROS + ∠ROQ = 180° (Linear pair of angles)
Now, ∠POS + ∠ROS = 180° – 90° (Since ∠POR = ∠ROQ = 90°)
∴∠POS + ∠ROS = 90°
Now, ∠QOS = ∠ROQ + ∠ROS
It is given that ∠ROQ = 90°,
∴∠QOS = 90° + ∠ROS
Or, ∠QOS – ∠ROS = 90°
As ∠POS + ∠ROS = 90° and ∠QOS – ∠ROS = 90°, we get
∠POS + ∠ROS = ∠QOS – ∠ROS
=>2 ∠ROS + ∠POS = ∠QOS
Or, ∠ROS = ½ (∠QOS – ∠POS) (Hence proved).
LONG QUESTIONS
1. In the Figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
[Hint: Draw a line parallel to ST through point R.]
Solution:
First, construct a line XY parallel to PQ.
The angles on the same side of the transversal are equal to 180°.
So, ∠PQR + ∠QRX = 180°
Or,∠QRX = 180° – 110°
∴∠QRX = 70°
Similarly,
∠RST + ∠SRY = 180°
Or, ∠SRY = 180° – 130°
∴∠SRY = 50°
Now, for the linear pairs on the line XY-
∠QRX + ∠QRS + ∠SRY = 180°
Substituting their respective values we get,
∠QRS = 180° – 70° – 50°
Or, ∠QRS = 60°
37
2. In the figure, if AB || CD || EF, PQ || RS, ∠RQD = 25° and ∠CQP = 60°, then find ∠Q
Solution:
According to the given figure, we have
AB || CD || EF
PQ || RS
∠RQD = 25°
∠CQP = 60°
PQ || RS.
As we know,
If a transversal intersects two parallel lines, then each pair of alternate exterior angles is equal.
Now, since, PQ || RS
⇒∠PQC = ∠BRS
We have ∠PQC = 60°⇒∠BRS = 60° … eq.(i)
We also know that,
If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.
Now again, since, AB || CD
⇒∠DQR = ∠QRA
We have ∠DQR = 25°
⇒∠QRA = 25° … eq.(ii)
Using linear pair axiom,
We get,
∠ARS + ∠BRS = 180°
⇒∠ARS = 180° – ∠BRS
⇒∠ARS = 180° – 60° (From (i), ∠BRS = 60°)
⇒∠ARS = 120° … eq.(iii)
Now, ∠QRS = ∠QRA + ∠ARS
From equations (ii) and (iii), we have,
∠QRA = 25° and ∠ARS = 120°
Hence, the above equation can be written as:
∠QRS = 25° + 120°
⇒∠QRS = 145°
Class 9 Maths Chapter 6 LINES AND ANGLES
PRACTICE QUESTIONS
38
1.The sum of angle of a triangle is
(i) (ii) (iii) (iv) none of these
2. In fig if x= then y=
(i)
(ii)
(iii)
(iv)
3. If two lines intersect each other then

(i) vertically opposite angles are equal
(ii) corresponding angle are equal
(iii) alternate interior angle are equal
(iv) none of these
4. The measure of Complementary angle of is

(a)
(b)
(c)
(d) none of there
5. Question 1 is an ASSERTION REASON type question. This question contains STATEMENT -

1(Assertion) and STATEMENT -2 (Reason) and has following four choices (a), (b) ,(c) and (d) , only one of
which is the correct answer. Mark the correct answer.
(a) Statement -1 and Statement -2 are true ; Statement -2 is a correct explanation for Statement -1.
(b) Statement -1 and Statement -2 are true ; Statement -2 is not a correct explanation for Statement -1.
(c) Statement -1 is True, Statement -2 is False.
(d) Statement -1 is False, Statement -2 is True.
Q1. STATEMENT – 1If one angle of a triangle is equal to the sum of the other two angles, then the triangle
is a right triangle.
STATEMENT – 2 The exterior angle of a triangle is the sum of two interior opposite angles
Q1. Read the following and answer any four questions from (i) to (v)
Ishita loves triangular objects. She want to decorate the wall of her room with some triangular hangings.
When she searched for it she found a number of beautiful options for her room.
(i) The angles of triangle ABC are in the ratio 3 : 4 : 5. The measure of the smallest angle A is
(a) 150 (b) 450 (c) 600 (d) 750
(ii) If the measure of ∠ACE is 1050, then ∠ABC =
(a) 450 (b) 750 (c) 600 (d) 1050
(iii) If AB ǁ DE, then measure of ∠CED is
(b) 600 (b) 450 (c) 1050 (d) 800
(iv) If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
39
(a) An isosceles triangle (b) a right triangle
(c) an obtuse triangle (d) an equilateral triangle
(v) If ∠ABC = 600 and ∠DCE = 400 the measure of ∠BAC is
(a)600 (b) 700 (c) 800 (d) 300

Answers (i) (b) (ii) (c) (iii) (a) (iv) (b) (v) (c)
Q2. In the given figure p ǁ q and t is a transversal.
Based on the above information, answer the following questions
(i) If ∠ 1 = 1000, then ∠ 6 =
(a) 1000 , corresponding angles (b) 1000 exterior alternate angles
(c) 800 , co-interior angles(d)800 , corresponding angles
(ii) If ∠ 2 = 800 then ∠ 7 =
(a) 800 , corresponding angles (b) 800 exterior alternate angles
(c) 800 , interior alternate angles (d) 1000 , exterior alternate angles
(iii) If ∠ 3 = 1280, then ∠ 5 =
(a) 560 , corresponding angles (b) 1280co-interior angles
(c) 520 , co-interior angles (d) 640 , interior alternate angles
(iv) If ∠ 1 : ∠ 2 = 9 : 4, then ∠ 7 : ∠ 8 =
(a) 3 : 2 (b) 9 : 4
(c) 4 : 9 (d) 2 : 3
0
(v) If ∠ 3 + ∠ 6 = 200 , then ∠ 7 =
(a) 500 (b) 1000 (c) 1200 (d) 800
Answer (i) (a) (ii) (b) (iii) (c) (iv) (c) (v) (d)
Q3. PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B,
the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD.
(i) Draw the figure from above statement.

(ii) Which is the incident ray and which one is the reflected ray.
(iii) Prove that AB || CD.
[iv]Short Answer Type Question (10 questions)
Q1.If two lines intersect, prove that the vertically opposite angles are equal.
Q2.Bisectors of interior ∠B and exterior ∠ACD of a Δ ABC intersect at the point T.Prove that ∠ BTC = ½ ∠
BAC.
Q3.A transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding angles so
formed are parallel.
Q4. In Fig. , ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
40
Q5. Can a triangle have all angles less than 60°? Give a reason for your answer.
Q6. If the measures of two supplementary angles are ( 3x + 15)0 and (2x + 5)0, then find the value of x.
Q7. Can a triangle have two obtuse angles? Give the reason for your answer.
Q8. How many triangles can be drawn having its angles as 45°, 64° and 72°? Give the reason for your
answer.
Q9. How many triangles can be drawn having its angles as 53°, 64° and 63°? Give the reason for your
answer.
Q10.If the difference between two supplementary angles is 800, then find the angles.
[v]Long Answer Type Question (5 questions)

Q1. If two parallel lines are intersected by a transversal, prove that the bisectors of the two pairs of
interior angles enclose a rectangle.
Q2. The angles of a triangle are arranged in ascending order of magnitude. If the difference between
two consecutive angles is 10°, find all the three angles.
Q3.It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If
ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.
Test Paper 1
SUBJECT –MATHEMATICS
TIME : 30 min. CLASS – IX MAX. MARKS: 20
General Instruction
(1) This question paper contains 3 Sections.
(2) Section A contains 3 questions of 2 marks each.
(3) Section B contains 2 questions of 3 marks each.
(4) Section C contains 2 questions of 4 marks each.
Section A
Q1. In the given figure, AOC is a line, find x.
Q2. In the given figure, line PQ and line MN intersect at O.

(a) Determine y, when x = 60°.
(b) Determine x, when y = 40°.
41
Q3. In the given figure, lines AB, CD and EF intersect at O.
Find the measure of ∠ AOC, ∠ COF.
Section B
Q4. The exterior angles obtained on producing the base of a triangle both ways are 100° and 120°.
Find all the angles.
Q5. ΔABC is right angled at A and AL ┴ BC. Prove that ∠ BAL = ∠ ACD.
Section C
Q.6: It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given
information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.
Q7. In the Figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Test Paper 2
SUBJECT –MATHEMATICS
TIME : 45 min. CLASS –IX MAX. MARKS: 30
General Instruction
(1) This question paper contains 3 Sections.
(2) Section A contains 5 questions of 2 marks each.
(3) Section B contains 4 questions of 3 marks each.
(4) Section C contains 2 questions of 4 marks each.
Section A
Q1. How many triangles can be drawn having its angles as 45°, 64° and 72°? Give reason for your
answer.
Q3. ΔABC is right angled at A and AL ┴ BC. Prove that ∠ BAL = ∠ ACD.
42
Q5.ΔABC is right angled at A and AL ┴ BC. Prove that ∠BAL = ∠ACD
Section B
Q 6. In the figure, OD is the bisector of ∠AOC, OE is the bisector of ∠BOC and OD ⊥ OE. Show that
the points A, O and B are collinear.
Q7. In Fig. , ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of Δ
XYZ, find ∠OZY and ∠YOZ.
Q8. In Fig. , lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
Q9. In Fig. 6.16, if x+y = w+z, then prove that AOB is a line.
Section C
Q10. If two parallel lines are intersected by a transversal, prove that the bisectors of the two pairs of
interior angles enclose a rectangle.
43
Q11. In Fig. , the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at
point T, then prove that ∠QTR = ½ ∠QPR.
44
TRIANGLES
Important Concepts
• Two figures are congruent if they are of the same shape and of the same size.
• Two circles of the same radii are congruent.
• Two squares of the same sides are congruent.
• Two triangles are congruent if their corresponding parts are congruent.
• If two triangles ABC and PQR are congruent under the correspondence A ↔ P, B ↔ Q and C ↔ R, then
symbolically, it is expressed as Δ ABC ≅ Δ PQR.
Some congruence rules are SAS(Side-Angle-Side)
Congruence Rule: Two triangles are congruent if two sides and the included angle of one triangle are equal to the
sides and the included angle of the other triangle. ASA(Angle-Side-Side) Congruence Rule: Two triangles are
congruent if two angles and the included side of one triangle are equal to two angles and the included side of the other
triangle. AAS Congruence Rule: Two triangles are congruent if any two pairs of angles and one pair of corresponding
sides are equal.
NCERT SOLUTIONS
Question 1
In quadrilateral ACBD, AC = AD and AB bisects ∠A (See the given figure). Show that ΔABC ≅ ΔABD. What can
you say about BC and BD?
ANSWER:
In ΔABC and ΔABD,
AC = AD (Given)
∠CAB = ∠DAB (AB bisects ∠A)
AB = AB (Common)
∴ ΔABC ≅ ΔABD (By SAS congruence rule)
∴ BC = BD (By CPCT)
Therefore, BC and BD are of equal lengths.
Question 2:
AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB
45
ANSWER:
In ΔBOC and ΔAOD,
∠BOC = ∠AOD (Vertically opposite angles)
∠CBO = ∠DAO (Each 90º)

BC = AD (Given)
∴ ΔBOC ≅ ΔAOD (AAS congruence rule)
∴ BO = AO (By CPCT)
⇒ CD bisects AB.
MCQ
Q1. The exterior angle of a triangle is equal to the
(a) sum of the two interior opposite angles.
(b) sum of the three interior angles.
(c) difference of two interior angles.
(d) opposite of the interior angle.
Q2.In two right-angled triangle ABC and triangle DEF, the measurement of hypotenuse
and one side is given. Check if they are congruent or not? If yes, by which rule.
a) SAS b)ASA c) SSS d) RHS

Q3.The angles of a triangle are in the ratio 3 : 4: 2. Find all the angles of the triangle.
(a) 110°,40° ,30° b) 60°,80° ,40° c) 60°,50° ,70° d) 110°,50° ,20°
Q4. In SAS congruence rule
a) The angle should be included
b) The angle should not be included
c) Any two sides and one angle
d) None of the above
Q5. Assertion: In Δ ABC, BC = AB and B = 80° ,Then , A = 50°
Reason: In a triangle, angles opposite to two equal sides are equal
a)Both Assertion and Reason are correct and Reason is correct explanation for Assertion.
b) Both Assertion and Reason are correct and Reason is not correct explanation for Assertion.
c) Both Assertion is true but and Reason is false.
d) Both Assertion and Reason are false.
SECTION B SHORT ANSWERS TYPE
Q1.The angle of triangle are (x + 10° ),(2x - 30° ) and x°.Find the value of x.
Q2. ∆ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D .Prove
that AC + AD = CD.
46
Q3.D, E, F are the midpoints of the sides BC,CA and AB respectively of ΔABC , then ΔDEF is congruent
to triangle ΔAEF
Q4. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join
A to O. Show that:
i) OB = OC (ii) AO bisects ∠A
Q5. ABC is a right angled triangle in which ∠A = 90º and AB = AC. Find ∠B and ∠C.
Q6. ABC is an isosceles triangle with AB = AC. Drawn AP ⊥ BC to show that ∠B = ∠C.
Q7.BE and CF are two equal altitude of a triangle ABC. Using RHS congruence rule , Prove that the triangle
ABC is isosceles
Q8. Prove that the Perimeter of a triangle is greater than the sum of three median
Q9. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle
ABC is isosceles.
Q10. In the given figure ∠CPD =∠ BPD and AD is the bisector of ∠BAC.
Prove that Δ BAP≅ Δ CAP and hence BP = CP.
--------------------------------------------------------------------------------------------------
LONG ANSWER TYPE
Q1. In the given figure BA is perpendicular to AC , DE is perpendicular to DF such that
BA = DE and BF = EC. Show that Δ ABC ≅ ΔDEF.
Q2. In an isosceles triangle ABC with AB = AC, D and E are points on BC such that
BE = CD. Show that AD = AE
47
Q3. In the given figure ∠ BCD = ∠ACD and ∠ ACB = ∠BDA.
Prove that AD = DB and ∠A=∠B.
Q4. Prove that if two angles of a triangle are equal then sides opposite to them are also equal.
Q5. PQR is a triangle in which PQ = PR and S is any point on the side PQ. Through S, aline is drawn parallel to
QR and intersecting PR at T. Prove that PS = PT.
ANSWERS (MCQ) Q1 (a) , Q2(d), Q3(b), Q4(a) ,Q5(a)
TEST 1
(20Marks : Q1– 2Marks, Q2 TO Q7 are of 3 Marks each)
Q1.ABC is a isosceles triangle with AB + AC. Draw AP = BC to show that ∠B = ∠C.
Q2.ABCD is quadrilateral such that AB = AD and CB = CD . prove that AC is the perpendicular bicector of
BD.
Q3. ABC is an isosceles triangle with AB = AC and BD and CE are two medians.
Show that BD = CE.
Q4.BE and CF are two equal altitude of a triangle ABC. Using RHS congruence rule ,
Prove that the triangle ABC is isosceles.
Q5. .If two isosceles triangles have a common base , prove that the line joining the vertex bisect the base at
right angle..
Q6.ABC is right angled triangle in which ∠A= 900 and AB = AC .Find ∠B and∠C.
Q7.Prove that the Perimeter of a triangle is greater than the sum of the three median.
TEST 2
(30 Marks: Each Question is of 5 Marks each)
Q1.Prove that the medians of an equilateral triangle are equal.
Q2.Prove that “angle opposite to equal sides of a triangle are equal.
Q3. Prove that,”A triangle is isosceles if and only if any two altitude are equal.
Q4.AD is an altitude of a isosceles triangle ABC in which AC =AB..

48
Show that (i)AD bisects BC (ii)AD bisects ∠A.
Q5.ΔABC is an isosceles triangle in which AB = AC .Side BA is produced to D such that AD = AB .Show

that ∆BCD is right angle.
Q5.ABC and DBC are triangles on the same base BC such that A and D lie on the opposite side of BC,AB
=AC and DB = DC .show that AD is the perpendicular bisectors of BC.
OR
.Line segment joining the mid-points. M and N of parallel sides. AB and DC , respectively of a trapezium
ABCD is perpendicular to both the sides AB and DC. Prove that AD = DC
49
QUADRILATERALS
MULTIPLE CHOICE QUESTIONS:

1. In a ΔABC, P,Q,R, are the midpoints of the sides BC, CA and AB respectively. If AC = 21cm, BC=
29cm, AB=30cm.Find the perimeter of quadrilateral ARPQ.
(A)20 cm. (B) 52cm (C) 51cm (D) 80cm
2. The quadrilateral formed by joining the midpoints of the sides of the quadrilateral PQRS taken in
order, is a rectangle if diagonals of
(A) PQRS are at right angles (B) PQRS is rectangle
(C) PQRS is a parallelogram (D) none of these
3. The diagonal of a rectangle is inclined to one side of the rectangle at 25° .The acute angle between
the diagonals is
(A)55° (B) 50° (C) 40° (D) none of these
4. ABCD is rhombus such that ∠𝐴𝐶𝐵 = 40° then ∠𝐴𝐷𝐵 is
(A)40° (B) 45° (C) 50° (D) 60°
ASSERTION AND REASON:
Direction: Each of these questions contains an assertion followed by reason. Read them carefully
and answer the questions on the basis of following options, select the one that best describes the two
statements.
(a) If both assertion and reason are correct and reason is the correct explanation of assertion.
(b) If both assertion and reason are correct but reason is not the correct explanation of assertion.
(c) If assertion is correct but reason is incorrect.
(d) If assertion is incorrect but reason is correct
5. ASSERTION: The line segment joining the mid points of any two sides of a triangle is parallel to
the third side and equal to half of it.
REASON: Diagonal of a parallelogram divides it into two congruent triangles.
SHORT ANSWER QUESTIONS:

6. ABCD is a rectangle in which diagonals BD bisects ∠B. show that ABCD is a square.
7. If in a parallelogram ABCD, AB= x+5 and BC= x+11 and perimeter is 40cm. Find x.
8. If in a parallelogram ABCD, AC is a diagonal. If the area of ABCD is 180cm2. Find the area of
∆ABC.
9. Show that each angle of rectangle is a right angle.
10. The perimeter of parallelogram is 32cm. If the longer side is 9.5cm, then find the measure of shorter
side.
11. In a trapezium ABCD, AB||CD, if ∠A =55°, ∠B=70°, find ∠C and ∠D.
12. The diagonals of rectangle ABCD intersect at a point O. If ∠COD is 78°, then find ∠OAB.
13. The angles of a quadrilateral are in the ratio 2:3:4:6. Find the angles of quadrilateral.
14. In a parallelogram PQRS, If angle P = (3x -5) and angle Q = (2x + 15). Find the value of x
15. The adjacent angles of a parallelogram are (3x + 10) and (5x -30). Find the value of x
50
LONG ANSWER QUESTIONS:
16. ABCD is a quadrilateral in which P, Q, R and S are mid-points of sides AB, BC, CD and DA
respectively. AC is the diagonal. Show that:
(i) SR ǁ AC and SR = (1/ 2) AC
(ii) PQ = SR
(iii) PQRS is a parallelogram
17. In ABCD is parallelogram, AE is perpendicular to DC and CF is perpendicular to AD. If AB =12
cm, AE =5 cm, CF =8 cm find AD.
18. Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is rectangle.
19. E and F are respectively the mid points of the non-parallel sides AD and BC of a trapezium ABCD.
1
Prove that EF ǁ AB and EF = 2 (AB + CD)
20. O is any point on the diagonal PR of parallelogram PQRS. Prove that ar(PSO) = ar(PQO).
21. In Figure given, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD.
If AB = CD, then show that:
(i) 𝑎𝑟 (DOC) = 𝑎𝑟(AOB)
(ii) 𝑎𝑟 (DCB) = 𝑎𝑟 (ACB)
(iii) DA || CB
22. In figure given, prove that the quadrilateral EFGH formed by the
internal angle bisectors of the quadrilateral ABCD is cyclic.
23. There is a Holi celebration in the KV school Rishikesh. Girls

are asked to prepare Rangoli in a triangular shape. They made a rangoli in the
shape of triangle ABC. Dimensions of △ABC are 26 cm, 28 cm, 25 cm.
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1. In fig, R is mid-point of AB and RQ || BC then AQ is equal to
a. QC b. RB c. BC d. AD
2. In fig R and Q are mid-points of AB and AC respectively. The length of RQ is:
a. 13 b. 14 c. 12.5 d. 13.5
3. If Garland is to be placed along the side of △QPR which is formed by joining
midpoint, what is the length of garland?
a. 39.5 cm b. 49.5 cm c. 35 cm d. 79.5 cm
24. During Math Lab Activity each student was given four broomsticks of lengths
10cm, 10cm, 6cm, 6cm to make different types of quadrilaterals.
1. How many quadrilaterals can be formed using these sticks?

a. Only one type of quadrilateral can be formed
b. Two types of quadrilaterals can be formed.
c. Three types of quadrilaterals can be formed.
d. Four types of quadrilaterals can be formed.
2. Name the types of quadrilaterals formed?
a. Rectangle, Square, Parallelogram
b. Kite , Trapezium, parallelogram
c. Rectangle, Square, Kite
d. Rectangle, Kite, Parallelogram
3. Which of the following is not true for a parallelogram?
a. opposite sides are equal
b. opposite angles are equal
c. opposite angles are bisected by the diagonals
d. diagonals do not bisect each other.
25. There was four plants in Rama‟s field. rama named their bases as P, Q, R, S. He joined PQ, QR, RS
and SP. His teacher told him that the quadrilateral PQRS was a parallelogram. He asked him to find
the measure of all the angles of the parallelogram, provided that the measure of anyone interior angle
of PQRS.
P Q
52
S R
(i) Obtain all the angles of the paralellogram PQRS if ㄥ R=80°.

(ii) Which mathematical concept is used in the above problem?
(ii) If PQ is 8cm then SR is_______.
ANSWERS:
1. (C) 51cm
2. (A) PQRS are at right angles
3. (B) 50°
4. (C) 50°
5. (A)
6. Solve
7. 2cm
8. 90cm2
9. Solve
10. 6.5cm
11. ∠C = 1100 and ∠D = 1250
12. 78°
13. 480, 720, 960, 1440
14. 340
15. 200
16. Solve
17. AD = 7.5CM
18. Solve
19. Solve
20. Solve
21. Solve
22. Solve
23. 1) a. QC 2) b. 14 3) a. 39.5 cm
24. 1) c 2) d 3) d
0 0 0
25. 1) 80 , 100 , 100 2) Opposite angles of a parallelogram are equal
3) 8cm (Opposite sides are equal in a parallelogram)
53
CIRCLE
EXPECTED LEARNING OUTCOMES
1. Recall and review the definition and basic terms related to Circle.
2. Revise statements of basic theorems on Circles.
3. To appreciate the theorems
a. Equal chords of a circle subtend equal angles at the centre.
b.If the angles subtended by the chords of a circle at the centre are equal, then the chords
are equal.
c. The perpendicular from the centre of a circle to a chord bisects the chord.
d.The line drawn through the centre of a circle to bisect a chord is perpendicular to the
chord
e. There is one and only one circle passing through three given non-collinear points
f. Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).
g. Chords equidistant from the centre of a circle are equal in length.
h.The angle subtended by an arc at the centre is double the angle subtended by it at any
point on the remaining part of the circle.
i. Angles in the same segment of a circle are equal.
j. If a line segment joining two points subtends equal angles at two other points lying on the
same side of the line containing the line segment, the four points lie on a circle (i.e. they are
concyclic).
k.The sum of either pair of opposite angles of a cyclic quadrilateral is 180º.
3. If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic.
4. Apply Knowledge gained on the topic „Circles‟ to solve problems.
MCQ:-
Q1. In Fig. ,A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC=130°
And ∠ ECD = 20°. Value of ∠ BAC is.

(a) 50° (b) 40° (c) 90° (d) 110°
Answer:- (d)
Short answer type question:-
Q1. In Fig, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠ DBC = 55°
and ∠BAC = 45°, find ∠ BCD.
Solution : ∠ CAD = ∠ DBC = 55°
(Angles in the same segment) Therefore,

54
∠ DAB = ∠ CAD + ∠ BAC
= 55° + 45° = 100° But ∠ DAB + ∠ BCD = 180°
(Opposite angles of a cyclic quadrilateral) So,
∠ BCD = 180° – 100° = 80°
Long answer type question:-
Q.1 The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
Let AB be a chord of a circle with centre O and O is joined to the mid-point M of AB. You have to prove
that OM ⊥ AB. Join OA and OB . In triangles OAM and OBM,
OA = OB (Radii)
AM = BM (given)
OM = OM (Common)
Therefore, ∆ OAM ≅ ∆ OBM (By SSS rule)
This gives ∠ OMA = ∠ OMB = 90° (CPCT)
Case Study question :- Q 1.
Ankit visited in a mall with his father. He sees that three shops are situated at P, Q, R as shown in the figure
from where they have to purchase things according to their need. Distance between shop P and Q is 8 m, that
of between shop Q and R is 10 m and between shop P and R is 6 m.
(i) Find the radius of the circle.

(a) 5 m (b) 7 m (c) 14 m (d) 8 m
(ii) (ii) Measure of ∠QPR is
(a) 60° (b) 90° (c) 120° (d) 180°
(iii) (iv) Length of the longest chord of the circle is
(a) 6 m (b) 8 m (c) 10 m (d) 24 m
(iv) (v) In figure, PSQP is known as
(a) Major segment (b) Minor segment (c) Major sector (d) Minor sector
Answer:- (i) (a) (ii) (a) (iii) (c) (iv) (b)
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MCQ:-
1. In the figure, if ∠ACB = 50°, then ∠OAB what is
(a) 50° (b) 40° (c) 90° (d) 100°
2. In the figure, quadrilateral PQRS is cyclic. If ∠P= 80°, then
what is the value of ∠R ?
(a) 30° (b) 40° (c) 100° (d) 60°
3. In the given figure, O is the centre of the circle. If OA = 5 cm and

OC = 3 cm, then find the length of AB .
(a) 7 cm (b) 9 cm (c) 8 cm (d) 10 cm
4. Two concentric circles with centre O have A, B, C and D as
points of intersection with a line l as shown in the figure. If
AD =12 cm and BC = 8 cm, find the length of AB and CD.
(a) 4 cm (b) 6 cm (c) 10 cm (d) 2 cm
5. In the figure, ΔABC is equilateral. Find ∠BDC and ∠BEC
(a) 600, 1100 (b) 500, 1200 (c) 600, 1200 d) 700, 1300
6. Assertion: A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
Reason: As we know that any chord whose length is twice as long as the radius of the circle always
passes through the centre of the circle
d) both Assertion and reason are false
1. In the figure, if AB is the diameter of the circle, then find the value of x.
2. A chord of a circle is equal to the radius of the circle. Find the angle subtended
by the chord at a point on the minor arc and also at a point on the major arc.
3. If two non- parallel sides of a trapezium are equal, prove that it is cyclic.
4. Prove that a cyclic parallelogram is a rectangle.
5. Show that two circles cannot intersect at more than two points.
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Long answer type question:-
1. In the given figure, OD is perpendicular to the chord AB of a circle

whose centre is O. If BC is a diameter, show that CA = 2OD.
2. A circular park of radius 20 m is situated in a village. Three girls Rita,
Sita and Gita are sitting at equal distance on its boundary each having a toy
telephone in their hands to talk to each other. Find the length of the string of each phone
3. If two circles intersect at two points, prove that their centers lies on the perpendicular bisector of the
common chord.
4. Find the length of a chord of a circle which is at a distance of 4 cm from the centre of the circle with
radius 5 cm.
5. Prove that if chords of congruent circles subtend equal angles at their centres, then they are equal.
MCQ
ANSWER:- 1. (b) 2. (c) 3. (c) 4. (d) 5. (c) 6. (a)
ANSWER:- 1. 50° 2. 300, 1500
CHAPTER-TEST (20 Marks)

SECTION – A (2 marks for each question)
1. In the given figure, if O is the centre of circle and ∠OBA=300 ,
determine ∠APB.
2. In the figure, if ∠ACB = 50°, then what is the measure of ∠OAB ?
3. Can we have a cyclic quadrilateral ABCD such that ∠A = 90°,
∠B = 70°, ∠C = 95° and ∠ D = 105° ?

4. In the figure, O is the centre of the circle and ∠ABC = 55°, then
what is ∠ADC ?
5. Two concentric circles with centre O have A, B, C and D as
points of intersection with a line l as shown in the figure. If AD =12 cm
and BC = 8 cm, find the length of AB and CD.
6. In the given figure, O is the centre of the circle. If OA = 5 cm and
OC = 3 cm, then find the length of AB .
SECTION – B (4 marks for each question)

7. Prove that , The angle subtended by an arc at the centre is double the angle subtended by it at any
point on the remaining part of the circle.
8. In the given figure, O is the centre of the circle. Find ∠BAO,
∠AOB, ∠BOD,∠ODB, if ∠AOC = 1300 and ∠OCD = 300.
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CHAPTER-TEST (30 Marks)
SECTION – A (2 marks for each question)

1. AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of
AB from the centre of the circle.
2. If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the
radius of the circle passing through the points A, B and C is
3. In Fig. 10.6, if ∠ OAB = 40º, then find the ∠ ACB.
4. ABCD is a cyclic quadrilateral such that AB is a diameter of the circle

circumscribing it and
∠ ADC = 140º, then find the ∠ BAC.
5. AB and AC are two equal chords of a circle. Prove that the bisector of the angle BAC passes through
the centre of the circle.
6. If a line segment joining mid-points of two chords of a circle passes through the centre of the circle,
prove that the two chords are parallel.
SECTION – B (3 marks for each question)

7. On a common hypotenuse AB, two right triangles ACB and ADB are situated on
opposite sides. Prove that ∠ BAC = ∠ BDC.
8. If non-parallel sides of a trapezium are equal, prove that it is cyclic.
SECTION – C (4 marks for each question)

9. In the given figure, OD is perpendicular to the chord AB of a circle
whose centre is O. If BC is a diameter, show that CA = 2OD.
10. A circular park of radius 20 m is situated in a village. Three girls Rita,
Sita and Gita are sitting at equal distance on its boundary each having a toy
telephone in their hands to talk to each other. Find the length of the string of each phone
11. If two circles intersect at two points, prove that their centers lies on the perpendicular bisector of the
common chord.
58
HERONS FORMULA
Multiple choice Questions
1 If the perimeter of an equilateral triangle is 180cm.Then its area will be:
a.900cm2 b.900√3cm2 c.300√3cm2 d.600√3cm2
2 Heron’s formula to find the area of an equilateral triangle of side ‘a' is given by:
2
a. 𝑎2 𝑠 2 b. [𝑠(𝑠 − 𝑎)(𝑠 − 𝑏) c. 𝑠 𝑠 − 𝑎 d. 𝑠(𝑠 − 𝑎)3
3 Find the area of a regular hexagon of side a.
a.3√3a2/2cm2 b.√3a2cm2 c.3√3a2cm2 d.4cm2
4 The area of triangle with given two sides18cm and 10cm respectively and perimeter equal to 42cm is:
a.20√11cm2 b.19√11cm2 c.22√11cm2 d.21√11cm2
5 Directions: In the following questions, a statement of assertion (A) is followed by a statement of

reason (R). Mark the correct choice as:
Assertion: Area of a triangle with sides 3cm, 4cm and 5cm is 6 cm2.
Reason: Heron‟s formula for area of a triangle is ½ base × height.

(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion
(A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of
assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answers: 1(b)900√3cm2
2 (d)√s(s-a)3
3 (a) 3√3a2/2cm2
4 (d)21√11cm2
5 (c)Assertion(A)is true and Reason(R)isfalse.
Multiple choice Questions (For Practise)
1 Directions: In the following questions, a statement of assertion (A) is followed by a statement of

reason (R). Mark the correct choice as:
Assertion: Area of a scalene triangle is calculated by Heron’s formula
59
Reason: Area of a quadrilateral whose sides and one diagonal are given, can be calculated by
dividing the quadrilateral into two triangles using Heron‟s formula.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion
(A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of
assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
2 The area of an equilateral triangle with side 2cm is
a. √6cm2 b.√3cm2 c.√8cm2 d.4cm2

3 The sides of a triangular plot are in the ratio of 3:5:7 and its perimeter is 300m.Find its area
a.4√30 b.1500√3 c.12√30 d.16√30

4 The equal sides of isosceles triangle are12cm and perimeter is 300cm.The area of this triangle
is:
a.9√15 sq.cm b.6√15sq.cm c.3√15 sq.cm d.√15 sq.cm
5 The area of an equilateral triangle having side length equal to√3/4cmis:

a.2/27sq.cm b.2/15sq.cm c.3√3/64sq.cm d.3/14sq.cm
Answer 2
Multiple choice Questions (For Practise)
1 (b)Both Assertion (A)and Reason (R)are the true,
but Reason(R) is not a correct explanation of Assertion(A).
2 (b)√3cm2
3 (b)1500√3sq.cm
4 (a)9√15sq.cm
5 © 3√3/64sq.cm
CASE STUDY (solved)Question 1:
A farmer has a triangular plot of land, and he intends to divide it equally among his three children. The sides
of the plot are 50m, 60m, and 70m. Each child will receive a triangular piece of land with a common point.
Find the dimensions of the triangular pieces resulting from the division and calculate their areas.
Answer: Step 1: Find the area of the triangular plot using Heron's formula.
𝑠 = (𝑎 + 𝑏 + 𝑐) / 2
𝑠 = (50 + 60 + 70) / 2
𝑠 = 180 / 2
𝑠 = 90
60
Area = 𝑠 𝑠 − 𝑎 𝑠 − 𝑏 𝑠 − 𝑐
𝐴𝑟𝑒𝑎 = 90 90 − 50 90 − 60 90 − 70
𝐴𝑟𝑒𝑎 = 90 ∗ 40 ∗ 30 ∗ 20
𝐴𝑟𝑒𝑎 = 3600 𝑠𝑞. 𝑚
Step 2: Since the farmer wants to divide the land equally, each child will get a triangular piece with
one-third of the area.
Each child's area = 3600 / 3
Each child's area = 1200 sq.m
Step 3: Apply the formula to find the lengths of the triangle's sides:
𝑆𝑖𝑑𝑒1 = 2 × (𝐴𝑟𝑒𝑎 / 𝐵𝑎𝑠𝑒1) = 2400 / 50 = 48𝑚
𝑆𝑖𝑑𝑒2 = 2 × (𝐴𝑟𝑒𝑎 / 𝐵𝑎𝑠𝑒2) = 2400 / 60 = 40𝑚
𝑆𝑖𝑑𝑒3 = 2 × (𝐴𝑟𝑒𝑎 / 𝐵𝑎𝑠𝑒3) = 2400 / 70 = 34.29𝑚
Step 4: Calculate dimensions of triangular pieces:
Child 1 gets a triangular piece with sides 50m, 48m, and an adjacent side to the common point.
Child 2 gets a triangular piece with sides 60m, 40m, and an adjacent side to the common point.
Child 3 gets a triangular piece with sides 70m, 34.29m, and an adjacent side to the common point.
(Case study for practice)
Question 1:
A triangular park has sides measuring 45m, 60m, and 75m. Due to increased pollution in the city, the local
government decides to double the size of the park while maintaining the shape of the triangle. Calculate the
new dimensions of the park and find the increase in area.
Question 2: The students in XYZ School decided to set up a triangular garden with a tiled pathway around
it. They chose the dimensions of the triangle to be 15 meters, 30 meters, and 35 meters for the sides. The
width of the pathway is 1.5 meters.
a) Calculate the area of the triangular garden.
b) What will be the new dimensions of the triangle if we include the pathway?
c) Calculate the area of the triangular garden including the pathway.
d) Determine the area covered by tiles for the pathway.
Question 3:
In a triangular park, the lengths of the sides are 15 m, 22 m, and 25 m. A smaller triangular flower bed is to
be made inside the park with midpoints of each side of the park as vertices. Find the area of the smaller
triangular flower bed and the remaining area of the park outside the flower bed.
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Solutions case study 1
Step 1: Calculate the current area of the triangular park = 1350 sq.m
New dimensions: (45*√2)m, (60*√2)m, (75*√2)m
New Area = 2700 sq.m
Increase in area = 1350 sq.m
Solution case study 2:
𝑠 = 40,
𝐴 = 500 𝑠𝑞𝑢𝑎𝑟𝑒 𝑚𝑒𝑡𝑒𝑟𝑠
b) To determine the new dimensions of the triangle including the pathway, add the width of the
pathway (1.5m) to each side.
A': = 16.5m
B': = 31.5m
C': = 36.5m
c) Calculate the area of the new triangular garden including the pathway:
Let a' = 16.5, b' = 31.5, and c' = 36.5.
A' ≈ 609.83 square meters
d) To find the area covered by tiles for the pathway, subtract the area of the original triangle from the
new area (A' - A).
Area of tiled pathway = A' - A
Area of tiled pathway = 609.83 - 500
Area of tiled pathway ≈ 109.83 square meters
Solution case study 3:
Area of park = 528 m².
Sides of smaller triangle are half the sides of larger triangle,
a1 = 15/2 = 7.5 m, b1 = 22/2 = 11 m, and c1 = 25/2 = 12.5 m.
Area of flower bed = 132 m².
Remaining area= 396 m².
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Short Answer Type Questions (Solved)
1 The. perimeter of an isosceles triangle is 32cm. The ratio of the equal side to base is 3:2. Find the area of the
triangle.
Answer: The ratio of the equal side to the base is 3: 2.
Let the sides be 3x, 2x. Let the third = 3x
Given, perimeter = 32
We know that the perimeter is equal to the sum of the sides. Thus,
⇒ 𝟑𝐱 + 𝟐𝐱 + 𝟑𝐱 = 𝟑𝟐
⇒ 𝟖𝐱 = 𝟑𝟐
⇒𝐱=𝟒
⇒ 𝐬𝐞𝐦𝐢 𝐩𝐞𝐫𝐢𝐦𝐞𝐭𝐞𝐫 𝐬, 𝟑𝟐/𝟐 = 𝟏𝟔
𝐓𝐡𝐮𝐬, 𝐭𝐡𝐞 𝐬𝐢𝐝𝐞𝐬 𝐚𝐫𝐞 𝟏𝟐 𝐜𝐦, 𝟖 𝐜𝐦, 𝟏𝟐 𝐜𝐦
𝐓𝐡𝐮𝐬, 𝐀𝐫𝐞𝐚 𝐨𝐟 𝐭𝐡𝐞 𝐭𝐫𝐢𝐚𝐧𝐠𝐥𝐞 = 𝟑𝟐/𝟐(𝟏𝟔 − 𝟏𝟐)(𝟏𝟔 − 𝟖)(𝟏𝟔 − 𝟏𝟐)
= 𝟏𝟔 × 𝟒 × 𝟖 × 𝟒
= 𝟑𝟐 𝟐𝐜𝐦𝟐
2 Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs
7 per m2.Also find the cost of fencing the field at the rate of Rs 9 per m2
Answer: Sides of the triangle are a=50m,b=65m,c=65m
Area of triangle, by Heron's formula =s(s-a)(s-b)(s-c)
where,
s=2a+b+c
s=250+65+65
s=90
Area of triangle = 90(40)(25)(25)
Area of triangle = 1500m2
Cost of laying grass = Area ×7

Cost of laying grass =1500×7
Cost of laying grass = Rs 10500
Short Answer Type Questions (For practice)
1 The perimeter of a triangular field is 240 m with two sides 78m and 50m.Now,calculate the length
of the altitude on the side of 50m length from its opposite vertex.
2 The side of a triangle are in the ratio of 25:14:12and its perimeter is510m.Find the greatest side
of the triangle and area of given triangle.
3 In the figure given below, ABCD is a rectangle and DEC is an equilateral triangle. Find the area of
ΔDEC.
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4 Each side of an equilateral triangle is 2xcm. If x√3 = √48,then find its area.
5 The sides of a triangle are in the ratio of 3:5:7 and its perimeter is300cm. Find its area.
Solution:
1. 67.5m
2
2. 4800m
2
3. 48-9√3cm
2
4. 16√3cm
2
5. 4500cm
TEST (20)
1. Sides of a triangle are in the ratio of 3: 5 :7 and its perimeter is 300cm. Its area will be: (1)
a.1000√3 sq.cm b.1500√3sq.cm c.1700√3 sq.cm d.1900√3sq.cm
2. The length of altitude of an equilateral triangle of side a unit is (2)
2 2
a.√3/2a b.√3/4a c.√3/2a d. none of these
3. Find the area of a triangle having the length of sides as 3,4 and 5 units respectively. (2)
4. The length of the sides of a triangle is 5x, 5x and 8x. Find the area of the triangle. (3)
5. Find the area of the triangle having sides1 m, 2m and 2 m. (3)
6. The sides of a triangular flower bed are 5m,8 m and11m.Find the area of the flower bed. (4)
2
7. An isosceles right triangle has area 8cm . Find the length of its hypotenuse. (5)
Ans: (TEST 1)
1. (b)1500√3sq.cm
2. c.√3/2a
3. 6cm2
4. 12x2cm2
5. 0.25√15m2
6. 4√21m2
7. 4√2cm
64
Test (30)
1. If the area of an equilateral triangle is 36√3cm2,then its perimeter is (1)
a.64 cm b.60cm c.36 cm d. None of these
2. What is the length of each side of an equilateral triangle having an area of 4√3? (1)
a.4cm b.5cm c.5cm d.6cm
2
3. The area of a triangle is 150cm and its sides are in the ratio 3:4:5.What is its perimeter? (2)
a.10 cm b.30cm c.45cm d.60 cm
4. The sides of a parallelogram are 100 m each and length of the longest diagonal is
160m. Find the area of the parallelogram. (2)
5. The sides of a quadrilateral ABCD are 6cm, 8cm,12cm and 14cm respectively. The angle between
the first two sides is a right angle. Find its area. (3)
6. A rhombus-shaped sheet with perimeter 40cm and one diagonal l12cm, is painted on both sides at
the rate of Rs.5 per m2. Find the cost of painting. (3)
7. Find the area of the quadrilateral shown in thefigure. (3)
8. The hypotenuse of a right-angled triangle is 41cm and the area of the triangle is180 sq.cm, then
find the difference between the lengths of the legs of the triangles. (5)
9. Find the area of a trapezium, the length of whose parallel sides is given as 22cm and 12cm and
the length of other sides is 14 cm each. (5)
10. A rhombus-shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30
m and its longer diagonal is 48 m, how much area of grass field will each cow be getting? (5)
Answers (TEST 2)
1 c.36 cm
2 (a) 4
3 d.60 cm
4 9600m2
5 546 cm2
6 82.8or24(√6+1)cm2
7 Rs 960
8 31 cm
9 51√19cm2
10 48m2
65
SURFACE AREAS AND VOLUMES
Gist of the lesson
1. Spheres (including hemispheres)

2. Right circular cones.
ILLUSTRATIVE EXAMPLES
MULTIPLE CHOICE QUESTIONS

66
1. Total surface area of a hemisphere is 4158 cm², the diameter of the hemisphere is equal to
__________ cm. (Take π = 22/7)
a) 40 cm b) 20 cm c) 21 cm d) 42 cm
Ans: d) 42 cm
2. If the surface area of a sphere of radius “R” is equal to the curved surface area of a hemisphere
of radius “r”, what is the ratio of R/r?
a) ½ b) 1/√2 c) 2 d) √2
Ans: 1/√2
3. If a right circular cone has radius 4 cm and slant height 5 cm then what is its volume?
(a) 16 π cm³ (b) 14 π cm³ (c) 12 π cm³ (d) 18 π cm³
Ans: 16 π cm³
4. Two right circular cones of equal curved surface areas have slant heights in the ratio of 3 : 5.
Find the ratio of their radii.
(a) 4 : 1 (b) 3 : 5 (c) 5 : 3 (d) 4 : 5
Ans: 5 : 3
5. Assertion: If the diameter of a sphere is decreased by 25%, then its curved surface area is
decreased by 43.75%.
Reason : Curved surface area is increased when diameter decreases
d) both Assertions and reason are false
Ans: c) Assertion is correct but reason is false
CASE BASED QUESTION:-
6. Sangita had a hemispherical bowl of radius r. She made a conical vessel of radius r with a tin
sheet.
(i) find the height of the conical vessel so that it can hold the water same as that of the
hemispherical bowl.
(ii) if the radius of the cone formed in the above part is 14 cm, then find how much sheet is used?
(iii) if the height of the conical vessel is doubled, how much more water can it hold than the
hemispherical bowl?
67
Ans: (i) since, volume of conical vessel = volume of hemispherical bowl
1 2
⇒ 𝝅 r²h = 𝝅 r3
3 3
1 2
⇒ 3 𝝅 r²h - 3 𝝅 r3 = 0
⇒ h = 2r
The height is 2r
(ii) since, radius = r = 14 cm
Height = 28 cm
2
𝑙 = ℎ2+ 𝑟 2
⇒𝑙 2
= 282 + 142
⇒ 𝑙 = 14 5 cm
⇒ area of sheet required = 𝝅 r𝑙

2
= 1377.41 𝑐𝑚
1
3
𝝅𝑟 ²ℎ
(iii) 2 = 2:1
3
𝝅𝑟 3
it can hold twice the volume of the hemisphere.
11 SHORT ANSWER TYPE QUESTIONS

7. How many square metres of canvas is required for a conical tent whose height is3.5 m and the
radius of the base is 12 m?
2
Ans: 𝑙 = ℎ2+ 𝑟 2
𝑙 2
= 3. 52 + 122
⇒ 𝑙 = 12.25 m
total canvas required = πrl
π×12×12.5 =471 𝑚2
8. A shopkeeper has one spherical laddoo of radius 5cm. With the same amount of material, how
many ladoos of radius 2.5 cm can be made?
Ans: Given, radius of the spherical laddu, r = 5 cm

4 4 500
∴ Volume of a spherical laddu = 3 𝝅𝑟 3 = 3 𝝅 53 = 𝝅 𝑐𝑚 3
3
Now, radius of small laddu = 2.5 cm
68
4 4 62.5
Volume of a small laddu = 3 𝝅𝑟 3 = 3 𝝅 2.53 = 𝝅 𝑐𝑚 3
3
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎 𝑠𝑝 ℎ 𝑒𝑟𝑖𝑐𝑎𝑙 𝑙𝑎𝑑𝑑𝑢

∴ Number of laddoos =
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎 𝑠𝑚𝑎𝑙𝑙 𝑙𝑎𝑑𝑑𝑢
500 3
𝝅 𝑐𝑚
= 3
62.5
𝝅 𝑐𝑚 3
=8
3
So, he can make 8 ladoos.
9. A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm.Find the
volume and the curved surface of the solid so formed.
Ans: When a right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8
cm, then solid formed is a cone whose height, h = 8 cm.
The radius of the cone, r = 6 cm.
Slant height of the cone, l = 10 cm

1 1 22
∴ Volume of the cone = 3 𝝅 r²h = 3 ∗ ∗6² * 8
7
= 301.7 cm3
Curved surface area of the cone = πrl

22
= ∗6 * 10 = 188.5cm2
7
Hence, the volume and surface area of the cone are 301.7 cm3 and 188.5 cm2 respectively.
LONG ANSWER TYPE QUESTIONS
10. A semi-circular sheet of metal of diameter 28 cm is bent to form an open conicalcup. Find the
capacity of the cup.
Ans: Circumference of a semicircle = πrWhereas, r is the radius of circle
Diameter of circular sheet =28 cm
∴ Radius of circular sheet = 14 cm
Therefore,
Circumference of circular sheet =14π
69
When a semi-circular sheet is bent to form an open conical cup, the radius of the sheet becomes
the slant height of the cup and the circumference of the sheet becomes the circumference of the
base of the cone.
Slant height of cup (l)= Radius of circular sheet =14cm
Circumference of the base of cone = circumference of circular sheet =14π
Let r be the radius of the base of cone
∴ 2πr = 14 ⇒ r = 7cm
Let h be the height of the cup.
2
Therefore, 𝑙 = ℎ2+ 𝑟 2
2
14 = 72 + ℎ 2
⇒ h = 7 2 cm
Now,Capacity of cup = Volume of cone
As we know that, volume of cone is given as-

1
V = 𝝅 r²h
3
1
Therefore, Capacity of cup = 3 𝝅 7² * 7 2
= 622.4cm3
Thus the capacity of the cup is 622.4cm3
11. Two solid spheres made of the same metal have weights 5920 g and 740 g,
respectively.Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm.
Ans: Mass is directly proportional to volume for same metal (Density)
Let Mass of Solid 1 be M1, Volume be V1, Mass of Solid 2 be M2 and Volume be V2
𝑀1 𝑉 1
=
𝑀2 𝑉 2
Volume of sphere is directly proportional to R3
𝑀1 𝑉 1 (𝑅 1)3
= =
𝑀2 𝑉 2 (𝑅 2)3
5920 (𝑅 1)3
=
740 (𝑅 2)3
70
(𝑅 1)3
(𝑅 2)3
=8
𝑅1
=2
𝑅2
𝑅 1 = 2 * 2.5 = 5 cm
12. A corn cob shaped somewhat like a cone, has the radius of its broadest end as 2.1 cm and
length (height) as 20 cm. If each 1 cm2 of the surface of the cob carries an average of four grains,
find how many grains you would find on the entire cob.
Ans: We know the curved surface area of cone cob = πrl.
Given, r = 2.1cm , h = 20 cm.
2
𝑙 = ℎ2+ 𝑟 2
𝑙 2
= 2.12 + 202
l = 20.11 cm
22
∴ Curved surface area of corn cab = ∗ 2.1 ∗ 20.11 = 132.73 cm2
7
Since, the number of grains on 1 cm2 of the surface corn cob =4,
∴ Number of grain on 132.73 cm2 of the surface of corn cab =132.73×4=530.92≈531.
QUESTIONS FOR PRACTICE:-

MULTIPLE CHOICE QUESTIONS
1. The diameter of the moon is approximately one-fourth of the diameter of the earth. What
fraction of the volume of the earth is the volume of the moon?
1 1 1 1
a. b. c. d
64 32 48 16
2. A hemispheric dome of radius 3.5m is to be painted at a rate of ₹600/m2. What is the cost of
painting it? (Take π = 22/7)
a) ₹46200 b) ₹45000 c) ₹47260 d) ₹48375
3. The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is beingpumped into

it. The ratios of the surface areas of the balloon in the two cases is
(A) 1 : 4 (B) 1 : 3 (C) 2 : 3 (D) 2 : 1
4. What is the total surface area of a cone of radius 7cm and height 24cm? (Take π = 22/7)
a) 710 cm² b) 704 cm² c) 700 cm² d) 725 cm²
5. Assertion: if a ball is in the shape of a sphere has a surface area of 221.76cm² then it‟s
diameter is 8.4 cm
71
Reason: if the radius of the sphere be r then the surface area, S=4πr²
d) both Assertions and reason are false
CASE BASED QUESTION:-
6. In a grinding mill, 5 types of mills were installed. These mills used spherical shaped steel
balls of radius 5 mm, 7 mm, 10 mm, 14 mm, 16 mm respectively. For repairing purposes the mill
needs 10 balls of radius 7 mm and 20 balls of radius 3.5mm. The workshop had 20000mm³ steel
which was melted and 10 balls of radius 7mm and 20 balls of radius 3.5 m were made and the
remaining steel was stored for future use.
i) What was the volume of 10 balls of radius 7mm?
ii) How much steel was kept for future use?
iii) What was the surface area of one ball of radius 7mm?
7. A class teacher brings some clay in the classroom to teach the topic mensuration. First she
forms a cone of radius 10 cm and height 5 cm and then she moulds that cone into a sphere.
(i) Find the volume of the conical shape.
(ii) Find the radius of the sphere.
(iii) Find the volume of the sphere the teacher made.
8. Monica has a piece of canvas whose area is 551 m². She uses it to have a conical tent made,
with a base radius of 7 m. Assuming that all the stitching margins and the wastage incurred while
cutting, amounts to approximately 1 m²
(i) find the slant height of the conical tent so formed.
(ii) Find the height of the conical tent so formed.
(iii) find the volume of the conical tent?
9.The surface area of a sphere of radius 5 cm is five times the area of the curved surface of a cone
of radius 4 cm. Find the height and the volume of the cone (taking=22/7)
10.A dome of a building is in the form of a hemisphere. From inside, it was whitewashed at the
cost of ₹498.96. If the rate of whitewashing is ₹4 per
square metre, find the :
(i) Inside surface area of the dome
72
(ii) Volume of the air inside the dome.
11. A metallic sphere is of radius 4.9 cm. If the density of the metal is 7.8 g/cm³, find the mass of
the sphere (π = 22/7).
12. The height of a cone is 16 cm and its base radius is 12 cm. Find the curved surface area and
the total surface area of the cone (Use = 3.14).
13. A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a
sphere. Find the radius of the sphere so formed.
14. A joker‟s cap is in the form of a right circular cone with a base radius of 7 cm and a height of
24 cm. Find the area of the sheet required to make 10 such caps.
15. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm.
Find the outer curved surface area of the bowl.
16. If the radius of a sphere is doubled then what is the ratio of their volumes?
17. Find the capacity in litres of a conical vessel whose diameter is 14 cm and slant height is 25
cm.
18. The area of the circular base of a right circular cone is 78.5 cm². If its height is 12 cm then
find its volume.
LONG ANSWER TYPE QUESTIONS
19. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made
of recycled cardboard. Each cone has a base diameter of 40 cm and height 1m. If the outer side of
the cone is to be painted and the cost of painting is Rs 12/m². What will be the cost? ( Take π=
3.14 and take 1.04 = 1.02)
20. To maintain the beauty of the monument, the students of the school cleaned and painted the
dome of the monument. The monument is in the form of a hemisphere. From inside, it was white
washed by the students whose area is 249.48 m². Find the volume of the air inside the dome.
21. A right triangle of hypotenuse 13 cm and one of its sides 12 cm is made to revolve taking side
12 cm as its axis. Find the volume and curved surface area of the solid so formed.
22. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm. Find
the volume of the solid so obtained. If it is now revolved about the side 12 cm, then what would
be the ratio of the volumes of the two solids obtained in two cases ?
23. A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how
much syrup would be found in 45 gulab jamuns each shaped like a sphere of diameter 2.8 cm .
ANSWERS OF PRACTICE QUESTIONS
1
1.
64
2. ₹46200
3. 1 : 4
73
4. 704 cm²
5. Ans: a) both Assertion and reason are correct and reason is correct explanation for Assertion
6. (i) 14373.3 mm³
(ii) 2033.4mm³
(iii) 616 mm²

500π
7. (i) cm3
3
(ii) 5 cm
500π
(iii) cm3
3
8. (i) 25m
(ii) 24m
(iii) 1232 m³
9. h= 3 cm
V = 50.29 cm³
10.the inner surface area of the dome is 249.48 m² and the volume of the air inside the dome is
523.9 m³
11. 3845.44g=3.85kg (nearly)
12. CSA= 753.6 cm²
TSA = 1205.76 cm²
13. 2.1 cm
14. Curved surface area of 10 jokers cap =5500 cm2
15. 173.25 cm2
16. 1:8
17. 1.232 litres
18. V = 314cm3
19. ₹384.34
20. V = 523.908 m3
21. CSA = 65π cm2 VOL = 100π cm3
22. VOLUME = 240π cm3
RATIO IS 5:12
23. 1552.32 cm3

74
STATISTICS
1 The marks obtained by 17 students in a mathematics test (out of100) are given below :
98, 82, 100, 100, 96, 65, 82, 76, 79, 90, 41, 64, 72, 68, 66, 48, 49.
The range of the data is :
(A) 59 (B) 54 (C) 90 (D) 100
2 The class-mark of the class 120-160 is :
(A) 130 (B) 135 (C) 140 (D) 145
3 In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The
lower limit of the class is :
(A) 6 (B) 7 (C) 8 (D) 12
4 In the class intervals 10-20, 20-30, 30-40 , the number 30 is included in :
(A) 30-40 (B) 20-30 (C) both the intervals (D) none of these
ASSERTION- REASONING
5 DIRECTION : In each of the following questions,a statement of Assertion is given followed

by acorresponding statement of Reason just below it.Of the statements, mark the correctanswer as
(a) Both assertion and reason are true andreason is the correct explanation of assertion.
(b) Both assertion and reason are true butreason is not the correct explanation of Assertion.
(c) Assertion is true but reason is false.
(d) Assertion is false but reason is true.
1. Assertion : If the class mark of a class interval (10- X ) is 20 then upper limit X = 30
Reason : (Upper limit + Lower limit) = Class mark / 2
CASE STUDY PROBLEMS
6 The Class teacher of Class X preparing result analysis of a student. She compares the
marks of a student obtained in Class IX (2018-19) and Class X (2019-20) using the
75
double bar graph as shown below
(I) In which subject has the performance improved the most?

(a) Maths (b) Social Science
(c) Science (d) English
(II) In which subject has the performance deteriorated?

(a) Maths (b) Social Science
(III) In which subject is the performance at par?

(a) Hindi (b) Maths
(IV) What is the difference in Maths Subject?

(a) 5 (b) 30
(c) 0 (d) 10
(V) What is the percentage of marks obtained by a student in Class X (2019-20)?

(a) 60% (b) 55%
(c) 54% (d) 65%
7 A Mathematics teacher asks students to collect the marks of Mathematics in Half yearly
exam. She instructed to all the students to prepare frequency disctribution table using the data
collected. Ram collected the following marks (out of 50) obtained in Mathematics by students of
Class IX
21, 10, 30, 22, 33, 5, 37, 12, 25, 42, 15, 39, 26, 32, 18, 27, 28, 19, 29, 35, 31, 24, 36, 18, 38,
22, 44, 16, 24, 10, 27, 39, 28, 49, 29, 32, 23, 31, 21, 34, 22, 23, 36, 24, 36, 33, 47, 48, 50,
39, 20, 7, 16, 36, 45, 47, 30, 22, 20, 60,17.
76
(I)How many students scored more than 20 but less than 30?
(a) 20 (b) 21
(c) 22 (d) 23
(II) How many students scored less than 20 marks?
(a) 10 (b) 11
(d) 14
(c) 12
(III) How many students scored more than 50% marks?
(a) 1 (b) 2
(d) 3
(c) 26
(IV) What is the class size of the classes?
(a) 10 (b) 5
(d) 20
(c) 15
(V) What is the class mark of the class interval 30 – 40?
(a) 30 (b) 35
(c) 40 (d) 70
8 The COVID-19 pandemic, also known as the coronavirus pandemic, is an ongoing pandemic
of coronavirus disease 2019 (COVID-19) caused by severe acute respiratory syndrome
coronavirus 2 (SARS-CoV-2). It was first identified in December 2019 in Wuhan, China.
During survey, the ages of 80 patients infected by COVID and admitted in the one of the City
hospital were recorded and the collected data is represented in the less than cumulative
frequency distribution table.
Based on the information, answer the following questions:
Age
No. of
(in
patients
yrs)
77
5-
6
15
15 -
11
25
25 -
21
35
35 -
23
45
45 -
14
55
55 -
5
65
(A) The class interval with highest frequency is:
(i) (ii) (iii) (iv)
45- 35- 25- 15-
55 45 35 25
(B) Which age group was affected the least?
(i) 35-45 (ii) 25-35
(iii) 55-65 (iv) 45-55
(C) Which are group was affected the most?
(i) 35-45 (ii) 25-35
(iii) 15-25 (iv) 45-55
(D) How many patients of the age 45 years and above were admitted?
(iv)
(i) (ii) (iii)
23
61 19 14
(E) How many patients of the age 35 years and less were admitted?
(i) (ii) (iii) (iv)
17 38 61 41
SHORT ANSWER TYPE
9 A grouped frequency table with class intervals of equal sizes using 250-270 (270 not included in this
interval) as one of the class interval is constructed for the following data : 268, 220, 368, 258, 242, 310,
272, 342, 310, 290, 300, 320, 319, 304, 402, 318, 406, 292, 354, 278, 210, 240, 330, 316, 406, 215, 258,
236. Write the frequency of the class 310-330 .
10 From the given frequency table ,

Write the total
number of Age of 1-2 2-3 3-4 4-5 5-6
cats that are cats
(years)
above the age
of 3 years No. of 6 4 7 6 17
cats
From the
given frequency table ,
Age of 25- 30- 40- 50- 60-

78
labour 30 40 50 60 70
Write the (years)
total number No. of 6 4 7 6 17
of Labours labour
that are
BELOW the age of 50 years
12 In a histogram, the areas of the rectangles are proportional to the frequencies. Can we say
that the lengths of the rectangles are also proportional to the frequencies ?
13 The Histogram is drawn between which of the two things of the data ?
14The frequency polygon is drawn between which of the two things of the data ?
15 To draw a frequency polygon , a point for a class interval ( 40-60) with corresponding
frequency 8. What is coordinate of that point of the frequency polygon?
16 The class marks of a continuous distribution are : 1.04, 1.14, 1.24, 1.34, 1.44, 1.54 and
1.64 Is it correct to say that the last interval will be 1.55 - 1.73? Justify your answer.
17 Write the difference between Bar graph and Histogram .
18 The class intervals which are 5-7 , 8-10 & 11-13 are not continuous (Inclusive way).
Write these class interval with true limits.
LONG ANSWER TYPE PROBLEMS

19 Prepare a frequency polygon for the given frequency table .
20 In the
given Fig. Class 0-5 5-10 10- 15- 20-
, there is a interval 15 20 25
histogram frequency 3 6 11 5 3
depicting
daily wages of workers in a factory. Construct the frequency distribution table. ( Along x- axis
members age( years) and y- axis number of villages
21 The expenditure of two families on different heads in a month is given below:

Head Food Education cloth Medicine House
rent
Expenditure 6000 1500 2000 700 900

of family A
79
Expenditure 8000 2000 1500 1200 950
of family B
Draw the double Bar graph

22 Draw the Histogram for the given data
Class interval 1-4 4-6 6-8 8-12 12-20
Frequency 6 30 44 16 4
23 Draw the frequenct for the given data

Class interval 1-4 4-6 6-8 8-12 12-20
Frequency 6 30 44 16 4
24 The following are the marks (out of 100) of 60 students in mathematics. 16, 13, 5, 80, 86,
7, 51, 48, 24, 56, 70, 19, 61, 17, 16, 36, 34, 42, 34, 35, 72, 55, 75, 31, 52, 28,72, 97, 74, 45,
62, 68, 86, 35, 85, 36, 81, 75, 55, 26, 95, 31, 7, 78, 92, 62, 52, 56, 15, 63,25, 36, 54, 44, 47, 27,
72, 17, 4, 30. Construct a grouped frequency distribution table with width 10 of each class
starting from 0 – 10 and Draw the Histogram and frequency polygon.
80

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केन्द्रीय शवद्यािय संगठन

KENDRIYA VIDYALAYA SANGATHAN

शिक्षा एवं प्रशिक्षण का आं चशिक संस्थान, चंडीगढ़

अध्ययन सामग्री / STUDY MATERIAL

कक्षा / CLASS – नौव ीं / IX

संकिन - राजीव रं जन, सह-प्रशिक्षक (गशणत)

शिक्षा एवं प्रशिक्षण का आं चशिक संस्थान, चंडीगढ़

से क्टर 33-सी, चंडीगढ़ / SECTOR-33 C, CHANDIGARH

वेबसाइट / WEBSITE : zietchandigarh.kvs.gov.in

ई-मेल/ E-MAIL :kvszietchd@gmail.com दू रभाष / Phone : 7102-2921841; 2921994

4 LINEAR EQUATION IN TWO 18

6 LINES & ANGLES 31

11 SURFACE AREA AND VOLUME 66

4. Which of the following is an irrational number?

1. Add 2√2+ 5√3 and √2 – 3√3.

1. The decimal expansion of an irrational number may be:

a) ¼ b)½ c)4 d)1/16

SHORT ANSWER TYPE QUESTIONS

SHORT ANSWER QUESTIONS:

10. Express in the form a/b.

Meaning of a Polynomial Degree of a polynomial

Monomials, Binomials etc.

Constant, Linear, Quadratic Polynomials etc.

(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

(x + y)3 = x3 + 3x2y + 3xy2 + y3 = x3 + y3 + 3xy (x + y)

(x –y)3 = x3 – 3x2y + 3xy2 – y3 = x3 – y3 – 3xy (x – y)

x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)

(i) Which one of the following is a polynomial?

(ii) On factorizing x + 8x + 15, we get :

(A) (x + 3) (x – 5) (B) (x – 3) (x + 5) (C) (x + 3) (x + 5 ) (D) (x – 3) (x – 5)

(iii) On dividing x – 2x – 15 by (x – 5), the quotient is (x + 3) and

(iv) The value of the polynomial 2x2+ 3x- 4 at x = 0 is:

(A) 2 (B) 3 (C) – 4 (D) 4

(v) The value of the polynomial 5x – 4x2 + 3, when x = –1 is-

(A) – 6 (B) 6 (C) 2 (D) –2

2) SHORT ANSWER QUESTIONS:

Solution: An example of a monomial having a degree of 82 = x82An example of a binomial having a

(ii) Find the value of the polynomial 5x – 4x2 + 3 at x = 2 and x = –1.

Solution: Let the polynomial be f(x) = 5x – 4x2 + 3

f(2) = 5(2) – 4(2)2 + 3

Or, the value of the polynomial 5x – 4x2 + 3 at x = 2 is -3.

Similarly, for x = –1,

f(–1) = 5(–1) – 4(–1)2 + 3

The value of the polynomial 5x – 4x2 + 3 at x = -1 is -6.

(iii) Compute the value of 9x2 + 4y2 if xy = 6 and 3x + 2y = 12.

Solution: Consider the equation 3x + 2y = 12

Now, square both sides:

=> 9x2 + 12xy + 4y2 = 144

=>9x2 + 4y2 = 144 – 12xy

From the questions, xy = 6 So,

9x2 + 4y2 = 144 – 72

Thus, the value of 9x2 + 4y2 = 72.

(A) MCQ‟S QUESTIONS:

Q1. √2 is a polynomial of degree is -

(A) 2 (B) 0 (C) 1 (D) ½

Q2. Degree of the polynomial 4x4 + 0x3 + 0x5 + 5x + 7 is -

(A) 4 (B) 5 (C) 3 (D) 7

Q3. Degree of the zero polynomial is-

(A) 0 (B) 1 (C) Any natural number (D) Not defined

Q4. If p(x)= x2- 2√2x +1, then p(2√ 2) is equal to