NCERT Solutions for Class 9 Maths Chapter 9 Propositions and their Converses – step-by-step answers to every exercise and in-text question of the chapter, as per the latest NCERT syllabus for 2027. Read them online or download the free question-answer PDF.
This first chapter of Part II is about how mathematicians reason with ‘if … then …’ statements. A proposition is a statement that is either true or false, and its converse swaps the ‘if’ part and the ‘then’ part. Using triangles, factors, divisibility and prime numbers, the chapter shows that a true proposition can have a false converse, that both can be true, or that both can be false. You will learn to frame a converse, prove the true statements and knock down the false ones with a single, well-chosen counterexample.
NCERT Solutions for Class 9 Maths Chapter 9 Propositions and their Converses PDF
All the answers are on this page below. The PDF also explains every answer – why it is right and how to check it – so download it for revision.
| Chapter | Download | |
|---|---|---|
| Solutions PDF | Propositions and their Converses | View PDF |
NCERT Solutions for Class 9 Maths Chapter 9 Propositions and their Converses – Questions and Answers
All 28 questions of the chapter (textbook pages Part II, 1 – 7) are answered below, in the order of the book – in-text questions, activities and the end-of-chapter exercises. Each answer is written the way you would write it in the exam.
Think and Reflect — Page 1Statement 1 and its converse, Statement 2
Statement 2: If two angles of a triangle are equal, then the sides opposite the equal angles have equal lengths.
We have proved the first statement in an earlier grade. Is the second statement true? Can you prove it?
(Hint: Draw the altitude from the vertex containing the third angle.)
Yes, Statement 2 is true. Proof:
Draw the altitude AD from A to BC, so ∠ADB = ∠ADC = 90°.
In ΔABD and ΔACD:
∠ABD = ∠ACD (given)
∠ADB = ∠ADC (each 90°)
AD = AD (common side)
So ΔABD ≅ ΔACD (AAS congruence)
Hence AB = AC (corresponding parts of congruent triangles)
So the sides opposite the equal angles are equal. The converse of Statement 1 is also true.
In-text Questions — Page 2Examples 2 and 3, and the Discussion
• Proposition P: If a number is a multiple of 6, then it is a multiple of 3.
• Converse Q: If a number is a multiple of 3, then it is a multiple of 6.
Proposition P is true. Justify why this is so.
P is true because 6 is itself a multiple of 3 (6 = 3 × 2).
n = 6k = (3 × 2) × k = 3 × (2k)
2k is a whole number, so n is a multiple of 3.
For example, 42 = 6 × 7 = 3 × 14, and 90 = 6 × 15 = 3 × 30.
No. The converse Q is false.
Counterexample: 9 is a multiple of 3 (9 = 3 × 3), but it is not a multiple of 6, since 9 ÷ 6 = 1.5 is not a whole number.
Every multiple of 6 is even, so any odd multiple of 3 (3, 9, 15, 21, …) is a counterexample. So P being true does not make its converse true.
• Proposition P: If n is a perfect square, then it has an odd number of factors.
• Converse Q: If n has an odd number of factors, then it is a perfect square.
Which of them are true?
Both P and Q are true.
Factors of n come in factor–partner pairs (d, n ÷ d). Each pair of different numbers adds 2 to the count. A factor is its own partner only when d × d = n, which happens only if n is a perfect square, and then for just one d.
| n | Factors | Number of factors |
| 36 (square) | 1, 2, 3, 4, 6, 9, 12, 18, 36 | 9 (odd) |
| 25 (square) | 1, 5, 25 | 3 (odd) |
| 12 (not square) | 1, 2, 3, 4, 6, 12 | 6 (even) |
So n is a square ⇔ exactly one factor is unpaired ⇔ the number of factors is odd. Each statement implies the other.
If a number has an odd number of factors, there must be a factor-partner pair in which the same number repeats (e.g., the partner factor of 5 in 25). If not, the given number will have an even number of factors since each factor can be paired with its factor pair.
Thus, if a number has an odd number of factors, then it is a perfect square.
It proves Statement Q: if n has an odd number of factors, then n is a perfect square.
The argument starts from “the number has an odd number of factors” and ends at “it is a perfect square”:
⇒ some factor-partner pair repeats, say (f, f)
⇒ n = f × f, a perfect square
It does not prove P. To prove P we would have to start from “n is a perfect square” and reach “odd number of factors”.
In-text Questions — Page 3Example 4: equal area and congruence
• Proposition P: If two triangles have the same area, then they are congruent.
• Converse Q: If two triangles are congruent, then they have the same area.
Determine if these statements are true or not. Justify the true statements and give a counterexample for each false statement.
P is false; Q is true.
P, counterexample: Triangle 1 is right-angled with legs 3 cm and 4 cm (sides 3, 4, 5 cm). Triangle 2 has base 4 cm and height 3 cm, with its top vertex above the middle of the base (sides 4 cm, √13 cm, √13 cm).
The areas are equal but the sides are not, so the triangles are not congruent.
Q, justification: congruent triangles fit exactly on top of each other, so they cover the same region. Their bases are equal and the matching heights are equal, so ½ × base × height is equal.
Think and Reflect — Page 3When a proposition and its converse are both false
Example 1 (n is a positive integer)
• Proposition: If n is a prime number, then n is odd. False: 2 is prime but even.
• Converse: If n is odd, then n is prime. False: 9 is odd but 9 = 3 × 3 is not prime.
Example 2 (geometry)
• Proposition: If a quadrilateral has all four sides equal, then it is a rectangle. False: a rhombus with angles 60° and 120° has equal sides but is not a rectangle.
• Converse: If a quadrilateral is a rectangle, then all four sides are equal. False: a 2 cm × 5 cm rectangle.
In-text Questions — Page 4Example 5: the converse of the Baudhāyana–Pythagoras theorem
Let ΔABC be the triangle with sidelengths a, b, c such that a² + b² = c². Construct another right triangle ΔXYZ whose perpendicular sides YZ and XZ have sidelengths a and b.
(Hint: Use the Baudhāyana–Pythagoras Theorem.)
XY = c.
By the Baudhāyana–Pythagoras theorem,
XY² = YZ² + XZ² = a² + b²
But a² + b² = c² (given for ΔABC)
So XY² = c², and since lengths are positive, XY = c.
So the third side of ΔXYZ has the same length as the third side AB of ΔABC.
(Hint: Use the SSS criterion for congruence.)
ΔABC ≅ ΔXYZ, so ΔABC is right-angled at C.
CA = b = ZX
AB = c = XY (from the previous answer)
So ΔABC ≅ ΔXYZ (SSS), with C ↔ Z, B ↔ Y, A ↔ X
Hence ∠C = ∠Z = 90°
So any triangle whose sides satisfy a² + b² = c² must be right-angled, with the right angle opposite the side c. The converse of the Baudhāyana–Pythagoras theorem is true.
Think and Reflect — Page 5A Note: other ways of saying ‘if X then Y’
| Proposition (true) | Converse (false) | Counterexample |
|---|---|---|
| If a person lives in Chennai, then the person lives in Tamil Nadu. | If a person lives in Tamil Nadu, then the person lives in Chennai. | Someone living in Madurai lives in Tamil Nadu but not in Chennai. |
| If you score 100 out of 100 in a test, then you pass it. | If you pass a test, then you scored 100 out of 100. | A student scoring 65 out of 100 passes. |
| If an animal is a cow, then it has four legs. | If an animal has four legs, then it is a cow. | A dog has four legs but is not a cow. |
| If today is Sunday, then our school is closed. | If our school is closed, then today is Sunday. | Schools are closed on summer-vacation weekdays too. |
| Proposition (true) | Converse (false) | Counterexample |
|---|---|---|
| Geometry | ||
| If a quadrilateral is a square, then its diagonals are equal. | If a quadrilateral has equal diagonals, then it is a square. | A 3 cm × 4 cm rectangle: both diagonals are 5 cm. |
| If a quadrilateral is a rhombus, then its diagonals are perpendicular. | If the diagonals are perpendicular, then it is a rhombus. | A kite with sides 2, 2, 5, 5 cm. |
| Number theory | ||
| If n is divisible by 4, then n is even. | If n is even, then n is divisible by 4. | n = 6 |
| If n is a multiple of 9, then n is a multiple of 3. | If n is a multiple of 3, then n is a multiple of 9. | n = 12 |
Exercise Set 9.1 — Pages 5–7Propositions, converses and counterexamples
If two lines are parallel, then the corresponding angles formed by a transversal are equal.
Converse: If a transversal cuts two lines and the corresponding angles are equal, then the two lines are parallel.
Proposition: true. This is the basic property of parallel lines from earlier grades: if l ∥ m, then ∠1 = ∠2.
Converse: true. Suppose ∠1 = ∠2 but l and m are not parallel. Then they meet at some point R, forming ΔPQR. One of the two angles becomes an exterior angle of ΔPQR and the other equals an interior opposite angle. An exterior angle equals the sum of the two interior opposite angles, so it is greater than either of them, and ∠1 ≠ ∠2. This contradicts ∠1 = ∠2, so l ∥ m.
Converse: If all the angles of a quadrilateral are equal, then it is a square.
Proposition: true. Every angle of a square is 90°, so all four angles are equal.
Converse: false. If all four angles are equal, each is 360° ÷ 4 = 90°, so the quadrilateral is a rectangle, but its sides need not be equal.
Counterexample: a rectangle with sides 2 cm and 4 cm has four angles of 90° but is not a square.
Proposition: If AB = AC, then IE = IF.
Converse: If IE = IF, then AB = AC.
Proposition: true.
⇒ ½∠B = ½∠C, so ∠IBC = ∠ICB and ∠FBI = ∠ECI
⇒ IB = IC (Statement 2, in ΔIBC)
In ΔIBF and ΔICE:
∠FBI = ∠ECI, IB = IC, ∠BIF = ∠CIE (vertically opposite)
⇒ ΔIBF ≅ ΔICE (ASA) ⇒ IF = IE
Converse: false.
Counterexample: AB = 5 cm, AC = 8 cm, ∠A = 60° (so BC = 7 cm). AB ≠ AC, yet IE = IF ≈ 1.76 cm, because IE = IF in every triangle with ∠A = 60°:
In quadrilateral AFIE: ∠AFI + ∠AEI = 360° − 60° − 120° = 180°
So ∠AFI = ∠CEI and ∠BFI = ∠AEI (supplements)
I is the incentre, so IP = IQ (IP ⊥ AB, IQ ⊥ AC)
ΔIPF ≅ ΔIQE (AAS: 90°, equal acute angles at F and E, IP = IQ) ⇒ IF = IE
Converse: If a + x = a + y, then x = y.
Proposition: true. If x and y are the same number, then adding the same number a to each gives the same result.
Converse: true. Add −a (that is, subtract a) to both sides of a + x = a + y. By the proposition itself the results are equal:
x = y
Both statements are true. Example: from 5 + x = 5 + 9 we may conclude x = 9.
Converse: If ab is a perfect square, then a and b are perfect squares.
Proposition: true.
ab = m² × k² = (m × k)², a perfect square.
For example, 4 × 9 = 36 = 6².
Converse: false. Counterexample: a = 2, b = 8. Then ab = 16 = 4² is a perfect square, but neither 2 nor 8 is a perfect square.
If x = y, then x² = y².
Converse: If x² = y², then x = y.
Proposition: true. If x and y are the same number, then squaring the same number gives the same result: x × x = y × y.
Converse: false. Counterexample: x = 3, y = −3.
but 3 ≠ −3
Converse: If x³ = y³, then x = y.
Proposition: true. Cubing the same number gives the same result.
Converse: true.
x² + xy + y² = (x + y/2)² + ¾y²
This is 0 only when y = 0 and x = 0.
If x³ = y³, then (x − y)(x² + xy + y²) = 0,
so x − y = 0, or x = y = 0. Either way, x = y.
Unlike squares, cubes keep the sign: (−3)³ = −27 but 3³ = 27, so different numbers have different cubes.
If n is divisible by 24, then it is divisible by both 4 and 6.
Converse: If n is divisible by both 4 and 6, then it is divisible by 24.
Proposition: true.
n = 24k = 6 × (4k), so 6 divides n
Converse: false. Counterexample: n = 12. It is divisible by 4 (12 = 4 × 3) and by 6 (12 = 6 × 2), but not by 24.
The reason: 4 and 6 share the factor 2, so the smallest number divisible by both is their LCM, 12, not 4 × 6 = 24.
Converse: If n is divisible by both 5 and 12, then it is divisible by 60.
Proposition: true. n = 60k = 5 × (12k) = 12 × (5k).
Converse: true.
5 divides 12k. 5 is prime and does not divide 12,
so 5 must divide k: k = 5m.
n = 12 × 5m = 60m, so 60 divides n.
This works because 5 and 12 have no common factor other than 1, so their LCM is 5 × 12 = 60.
Converse: If n has exactly 3 factors, then n is the square of a prime number.
Proposition: true. If n = p² with p prime, its factors are only 1, p and p², because the only prime in n is p. Example: 49 has factors 1, 7, 49.
Converse: true.
Then 1, m and m² are factors of n.
If m were not prime, it would have a factor d with 1 < d < m,
and d would be a fourth factor of n.
So m is prime, and n = m² is the square of a prime.
Converse: If n has exactly 4 divisors, then n is a product of two unequal prime numbers.
Proposition: true. If n = p × q with p, q different primes, its divisors are exactly 1, p, q and pq. Example: 15 = 3 × 5 has divisors 1, 3, 5, 15.
Converse: false. Counterexample: n = 8. Its divisors are 1, 2, 4, 8: exactly 4. But 8 = 2 × 2 × 2 is not a product of two unequal primes.
Converse: If n is not a multiple of 3, then n and n + 3 have no factors in common.
(“No factors in common” means no common factor other than 1.)
Proposition: true. If n were a multiple of 3, then n + 3 would also be a multiple of 3, and 3 would be a common factor. So if they have no common factor, n is not a multiple of 3.
Converse: true.
Then d divides (n + 3) − n = 3, so d = 1 or d = 3.
d = 3 would mean 3 divides n, which is not so.
Hence d = 1: no common factor.
Example: 10 and 13 share no factor; 9 and 12 share 3.
(i) All numbers of the form 4n² + 1 are prime.
(ii) All numbers of the form n² + n + 11 are prime.
(iii) All numbers of the form 4ⁿ + 3 are prime.
(i) n = 4: 4 × 4² + 1 = 65 = 5 × 13, not prime.
| n | 1 | 2 | 3 | 4 |
| 4n² + 1 | 5 | 17 | 37 | 65 |
(ii) n = 10: 10² + 10 + 11 = 121 = 11 × 11, not prime. (For n = 1 to 9 the values 13, 17, 23, 31, 41, 53, 67, 83, 101 are all prime.)
(iii) n = 4: 4⁴ + 3 = 256 + 3 = 259 = 7 × 37, not prime.
| n | 1 | 2 | 3 | 4 |
| 4ⁿ + 3 | 7 | 19 | 67 | 259 |
(i) If n is a prime number, then 2ⁿ – 1 is a prime number.
(ii) If n is an even number, then 2ⁿ + 1 is a prime number.
(i) n = 11 is prime, but
(The smaller primes 2, 3, 5, 7 give 3, 7, 31, 127, which are all prime, so we must go up to 11.)
(ii) n = 6 is even, but
(n = 2 and n = 4 give 5 and 17, which are prime.)
(i) Justify the statement.
(ii) Recall the divisibility shortcuts that we have studied for different numbers. To check whether a given number is divisible by 8, is it enough to check whether it is divisible by 2 and 4? Why or why not?
(i) If n is divisible by 8, then n = 8k.
n = 8k = 4 × (2k), so 4 divides n
(ii) No, it is not enough. That would need the converse, “if a number is divisible by 2 and 4, then it is divisible by 8”, which is false. Counterexample: 12 is divisible by 2 and by 4, but 12 ÷ 8 = 1.5. Divisibility by 4 already includes divisibility by 2, so the two checks together only guarantee divisibility by 4.
To test for 8, check whether the number formed by the last three digits is divisible by 8. For 1036: 036 = 36 is not, so 1036 is not divisible by 8, even though it passes the tests for 2 and 4.
Proposition: If a number is divisible by 3, then the sum of its digits is a multiple of 3.
Converse: If the sum of the digits of a number is a multiple of 3, then the number is divisible by 3.
Both are true, so each condition implies the other: a number is divisible by 3 exactly when its digit sum is a multiple of 3. That is why the shortcut works both ways.
100a + 10b + c = (99a + 9b) + (a + b + c) = 3(33a + 3b) + (a + b + c)
The number and its digit sum differ by a multiple of 3, so one is a multiple of 3 exactly when the other is. Example: 471 → 4 + 7 + 1 = 12, and 471 = 3 × 157.
Suppose we have identified a category of quadrilaterals called Q, and we have to construct a quadrilateral of this type. For this, we are to use two thin sticks, put them together as diagonals so that the quadrilateral obtained by joining their endpoints is of type Q (see the Fig. 9.2).
If a quadrilateral is of type Q, then it has equal-length diagonals.
(a) Should the two sticks be of equal length? Why or why not?
(b) Will it matter how the two sticks are put together?
(ii) Instead of the property mentioned above, suppose Q satisfies the following property.
If a quadrilateral has equal diagonals, then it is of type Q.
What will be your answers to (a) and (b) now?
(i) Property: type Q ⇒ equal diagonals.
(a) Yes, the sticks must be equal. Every type-Q quadrilateral has equal diagonals. The sticks are the diagonals, so unequal sticks give unequal diagonals, and that quadrilateral cannot be of type Q.
(b) Yes, it may matter. The property says only that equal diagonals are needed; it does not say they are enough (that is the converse). Type Q may also need, say, the sticks to cross at their midpoints, so some ways of placing equal sticks may not give type Q.
(ii) Property: equal diagonals ⇒ type Q.
(a) Equal sticks are enough, so using equal sticks always works. But they are not shown to be necessary: this property does not say every type-Q quadrilateral has equal diagonals.
(b) No, it does not matter. However two equal sticks are crossed, the quadrilateral has equal diagonals and is therefore of type Q.
Chapter 9 of 14 – Maths
- Previous: Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions
- Next: Chapter 10 How Quantities Combine: Understanding Data
- All chapters: NCERT Solutions Class 9 Maths
Also for This Chapter
NCERT Solutions Class 9 Maths Chapter 9 Propositions and their Converses – Overview
| Class | Class 9 |
| Subject | Maths |
| Chapter | 9 – Propositions and their Converses |
| Textbook pages | Part II, 1 – 7 |
| Questions solved | 28 |
| Session | 2027 |
| Format | Full answers on this page + PDF |
NCERT Solutions Class 9 Maths Chapter 9 – FAQs
Where can I download NCERT Solutions Class 9 Maths Chapter 9 Propositions and their Converses PDF?
Read every answer on this page, or download the free PDF from the table at the top.
Is this for the 2027 session?
Yes, it follows the latest NCERT Class 9 Maths book for 2027.
How many questions are solved in NCERT Class 9 Maths Chapter 9?
All 28 questions of the chapter (textbook pages Part II, 1 – 7) – the in-text questions, activities and the end-of-chapter exercises.
How should I use these solutions?
Read the chapter in the NCERT book first, attempt the questions yourself, then check your answers here.