Vidaara.orgClass 9 · Mathematics
CodeVID-M09-01-IRR-01
Irrational Numbers & Representation — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
$\sqrt2$ is:
- A.rational
- B.irrational
- C.an integer
- D.zero
2.
An irrational number's decimal is:
- A.terminating
- B.recurring
- C.non-terminating non-recurring
- D.finite
3.
$\sqrt{\text{(perfect square)}}$ is:
- A.irrational
- B.rational
- C.not real
- D.negative
4.
To represent $\sqrt2$ we use the:
- A.distance formula
- B.Pythagoras theorem
- C.section formula
- D.protractor
5.
$\pi$ is:
- A.rational
- B.irrational
- C.terminating
- D.an integer
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
6.
Is $\sqrt7$ rational or irrational?
7.
Is $\sqrt{25}$ irrational?
8.
Classify $2+\sqrt2$.
9.
To represent $\sqrt2$, what are the two legs of the right triangle?
Section C — Short Answer (3 marks)
4 × 3 = 12 marks
10.
Describe how to represent $\sqrt2$ on the number line.
11.
Describe how to represent $\sqrt3$ on the number line.
12.
Is $0.101001000100001\ldots$ rational or irrational?
13.
Give an irrational number between $2$ and $3$.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
14.
Represent $\sqrt5$ on the number line using the Pythagorean construction.
15.
Find an irrational number between $\tfrac17$ and $\tfrac27$.
Answer Key
Section A — Multiple Choice Questions
- (B) irrational
- (C) non-terminating non-recurring
- (B) rational
- (B) Pythagoras theorem
- (B) irrational
Section B — Short Answer (2 marks)
- Irrational.
- No ($=5$).
- Irrational.
- $1$ and $1$.
Section C — Short Answer (3 marks)
- Right triangle with legs $1,1$ gives hypotenuse $\sqrt2$; transfer it with a compass.
- On the $\sqrt2$ segment erect a perpendicular of length $1$; the new hypotenuse is $\sqrt3$.
- Irrational.
- e.g. $\sqrt5$.
Section D — Long Answer (5 marks)
- Right triangle with legs $2$ and $1$ gives hypotenuse $\sqrt5$; transfer with a compass.
- e.g. $0.150150015000\ldots$ (non-terminating, non-recurring).
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