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CBSE Mathematics Extra Questions Class 9
With the help of extra questions you can learn and practise the formulas to increase your chances of scoring more in the exam. You can also refer to CBSE sample papers, NCERT books, CBSE revision notes, CBSE past year question papers to prepare for the upcoming examination.
The extra questions with solutions can help to improve your Mathematics skills and thus guide you to move ahead while getting subject clarity.
Importance Of Mathematics Extra Questions For Class 9
The CBSE syllabus is quite vast but with the help of Mathematics Class 9 extra questions, you can get an idea of common question types that will be asked in the exam. Get yourself acquainted with extra questions so you can attempt the paper with more confidence. With CBSE extra questions, you can start practising even the challenging subjects and increase your chances of scoring more marks.
Chapter 1: Number System
 Find five rational numbers between 1 and 2.
Ans:
We have to find five rational numbers between 1 and 2.
So, let us write the numbers with denominator 5 + 1 = 6
Thus, 6/6 = 1, 12/6 = 2
From this, we can write the five rational numbers between 6/6 and 12/6 as:
7/6, 8/6, 9/6, 10/6, 11/6
Chapter 2: Polynomial
 Give an example of a monomial and a binomial having degrees as 82 and 99, respectively.
Ans:
An example of a monomial having a degree of 82 = x⁸⁹
An example of a binomial having a degree of 99 = x⁹⁹ + x
Chapter 3: Coordinate Geometry
Q1. Write the signs convention of the coordinates of a point in the second quadrant.
Ans: (ve, +ve)
Q2. Write the value of the ordinate of all the points lying on the xaxis.
Ans: 0
Chapter 4: Linear Equations in Two Variables
 Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) x – y/5 – 10 = 0
(ii) 2x+3y = 6
(iii) y – 2 = 0
Ans:
(i) The equation xy/510 = 0 can be written as:
(1)x + (1/5) y + (10) = 0
Now compare the above equation with ax + by + c = 0
Thus, we get;
a = 1
b = ⅕
c = 10
(ii) Rearranging the given equation, we get,
–2x + 3y – 6 = 0
The equation –2x + 3y – 6 = 0 can be written as,
(–2)x + 3y +(– 6) = 0
Now comparing (–2)x + 3y +(– 6) = 0 with ax + by + c = 0
We get, a = –2
b = 3
c = 6
(iii) y – 2 = 0
The equation y – 2 = 0 can be written as,
0x + 1y + (–2) = 0
Now comparing 0x + 1y + (–2) = 0with ax + by + c = 0
We get, a = 0
b = 1
c = –2
Chapter 5: Introduction to Euclid’s Geometry
 Define parallel lines.
Ans. Two coplanar lines in a plane that are not intersecting are called parallel lines.
Chapter 6: Lines and Angles
 Find the degree measure of an angle that is half of its complementary angle.
Ans:
Let the required angle be x
∴ Its complement = 90° – x
Now, according to given statement, we obtain
x =
1
2
(90° – x)
⇒ 2x = 90° – x
⇒ 3x = 90°
⇒ x = 30°
Hence, the required angle is 30°.
Chapter 7: Triangles
 Find the measure of each exterior angle of an equilateral triangle.
Ans. Each interior angle of an equilateral triangle is 60°.
So, each exterior angle = 180° – 60° = 120°
Chapter 8: Quadrilaterals
 What is a quadrilateral? Mention 6 types of quadrilaterals.
Ans:
A quadrilateral is a 4 sided polygon having a closed shape. It is a 2dimensional shape.
The 6 types of quadrilaterals include:
 Rectangle
 Square
 Parallelogram
 Rhombus
 Trapezium
 Kite
Chapter 9: Areas of Parallelograms
 If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Ans:
From the question we have the following conditions:
(i) AB and CD are 2 chords which are intersecting at point E.
(ii) PQ is the diameter of the circle.
(iii) AB = CD.
Now, we will have to prove that ∠BEQ = ∠CEQ
Chapter 10: Circles
 In the given figure, ∠ACP = 40° and BPD = 120°, then find ∠CBD.
Ans:
∠BDP = ∠ACP = 40° [angle in same segment]
Now, in ∆BPD, we have
∠PBD + ∠BPD + ∠BDP = 180°
⇒ ∠PBD + 120° + 40° = 180°
⇒ ∠PBD = 180° – 160o = 20°
or ∠CBD = 20°
Chapter 11: Constructions
 Construct an equilateral triangle, given its side as 4cm and justify the construction.
Solution:
Ans: Construction Procedure:
Let’s draw a line segment AB=4 cm.
With A and B as centres, draw two arcs on the line segment AB and note the point as D and E.
With D and E as centres, draw the arcs that intersect the previous arc and form an angle of 60° each.
Now, draw the lines from A and B that are extended to meet each other at point C.
Therefore, ABC is the required triangle.
Chapter 12: Heron’s Formula
 The sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540cm. Find its area.
Ans:
The ratio of the sides of the triangle is given as 12: 17: 25
Now, let the common ratio between the sides of the triangle be “x”
∴ The sides are 12x, 17x and 25x
It is also given that the perimeter of the triangle = 540 cm
12x + 17x + 25x = 540 cm
=> 54x = 540cm
So, x = 10
Now, the sides of the triangle are 120 cm, 170 cm, 250 cm.
So, the semi perimeter of the triangle (s) = 540/2 = 270 cm
Using Heron’s formula,
Area of the triangle
=√[s (sa) (sb) (sc)]
= √[270(270 – 120) (270 – 120) (270 – 250)] cm²
= √[270 × 150 × 100 × 20] cm²
= 9000 cm²
Chapter 13: Surface Areas and Volumes
 The paint in a certain container is sufficient to paint an area equal to 9.375 sq.m. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?
Ans:
Given,
Dimensions of the brick = 22.5 cm × 10 cm × 7.5 cm
Here, l = 22.5 cm, b = 10 cm, h = 7.5 cm
Surface area of 1 brick = 2(lb + bh + hl)
= 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm²
= 2(225 + 75 + 168.75) cm²
= 2 x 468.75 cm²
= 937.5 cm²
Area that can be painted by the container = 9.375 m² (given)
= 9.375 × 10000 cm²
= 93750 cm²
Thus, the required number of bricks = (Area that can be painted by the container)/(Surface area of 1 brick)
= 93750/937.5
= 937500/9375
= 100
Chapter 14: Statistics
 The number of family members in 10 flats of society are
2, 4, 3, 3, 1,0,2,4,1,5.
Find the mean number of family members per flat.
Ans:
Number of family members in 10 flats 2, 4, 3, 3, 1, 0, 2, 4, 1, 5.
So, we get,
Mean = sum of observation/ total no of observations
Mean = (2 + 4+ 3 + 3 + 1 + 0 + 2 + 4 + 1 + 5) / 10
Mean = 25/10 = 2.5
Chapter 15: Probability
 Compute the probability of the occurrence of an event if the probability of the event not occurring is 0.56.
Ans:
Given,
P(not E) = 0.56
We know that,
P(E) + P(not E) = 1
So, P(E) = 1 – P(not E)
P(E) = 1 – 0.56
Or, P(E) = 0.44