NCERT Solutions Class 9 Maths Chapter 9 Propositions and Their Converses

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C L A S S 9 · M AT H S

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Chapter 9:
Propositions and
their Converses

NCERT Textbook — Ganita Manjari

BOOK PAGES SECTIONS QUESTIONS MEDIUM

Part II, 1 – 7 7 28 English

Every question Step by step Exam ready
Activities, in-text questions The working shown, not Written to match how
and all exercises. just the final answer. marks are awarded.

Solutions, notes, sample papers & more at 35 pages

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

CLASS 9 · MATHS · GANITA MANJARI

NCERT Solutions — Chapter 9: Propositions and their
Converses
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.

TEXTBOOK BOOK PAGES
Ganita Manjari (Class 9) Part II, 1 – 7

SECTIONS QUESTIONS
7 28

MEDIUM
English

Think and Reflect — Page 1
Statement 1 and its converse, Statement 2
THINK AND REFLECT

Q1 Statement 1: If two sides of a triangle are equal, then the angles opposite the equal
sides are equal.
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:

Page 1 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

A

x x
B D C
ΔABC with ∠B = ∠C = x, and the altitude AD drawn from A.

Given: in ΔABC, ∠B = ∠C. To prove: AB = AC.

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.

Page 2 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

Why the altitude from A: A is the vertex with the “third angle”, the one that is
not given equal to another. The altitude from A splits ΔABC into two right
triangles that share the side AD, and each contains one of the equal angles.
That is exactly what a congruence needs.

Tip: if you prefer ASA, use the angle sum. In ΔABD, ∠BAD = 180° − 90° − x =
90° − x, and in ΔACD, ∠CAD = 90° − x too. So ∠BAD = ∠CAD, AD is common
and ∠ADB = ∠ADC, giving ΔABD ≅ ΔACD by ASA.

Does D lie between B and C? Yes. ∠B + ∠C < 180° and ∠B = ∠C, so each is
less than 90°. Both base angles are acute, so the foot of the altitude lands on
BC itself.

Common mistake: using BD = DC to prove the triangles congruent. We do not
know that yet; it is a result of the congruence, not a reason for it.

Check it yourself: draw BC = 6 cm, make angles of 50° at both B and C, and let
the arms meet at A. Measure AB and AC: they come out equal.

In-text Questions — Page 2
Examples 2 and 3, and the Discussion

Q1 Example 2
• 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).

Page 3 of 35

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Class 9 Maths Chapter 9 Propositions and their Conversesg l AglaSem · NCERT Solutions

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e m.
Let n be a multiple of 6. Then n = 6k for some whole number k.
m l as
.co
n = 6k = (3 × 2) × k = 3 × (2k)
m ag
la se
ag
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.
co m
m . ag
la se
ag
m
Why: counting in steps of 6 (6, 12, 18, …) is the same as counting in steps of 3
comultiples
but taking every second number. So every multiple of 6 is on the list .of
e m
of 3. om
.c las ag
se m
a
agl Tip: the general rule is: if a divides b, and b divides n, then a divides n. Here a =
s
3, b = 6.
m a
m .co agl
la se
ag

com of 6.
Q2 Does this now mean that all multiples of 3 are also multiples of 6? This corresponds to
the converse statement Q: If a number is a multiple of 3, then it is a.multiple
m
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co agl
Determine if it is true and if not, give a counterexample.
.
se m
aANSWER
agl No. The converse Q is false.
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Counterexample: 9 is a multiple of 3 (9 =om
g l a
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3 × 3), but it is not a multiple of 6, since 9 ÷ 6 = 1.5 is
m a
aseodd multiple of 3 (3, 9, 15, 21, …) is a counterexample. So P
not a whole number.
Every multiple of 6 is even, soglany
a
being true does not make its converse true.

co m
m .
se
m la
.co ag
a s em Why one example is enough: Q claims something about every multiple of 3. A
agl single multiple of 3 that is not a multiple of 6 shows the claim is false. No further
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argument is needed.
m a s e
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co m
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m as e
.co

ag l Page 4 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

Did you know? Q becomes true if we add a condition: if a number is a multiple
of 3 and is even, then it is a multiple of 6. Multiples of 6 are exactly the even
multiples of 3.

Q3 Example 3 (n is any positive integer)
• 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.

Why only one repeated pair: if f × f = n and g × g = n, then f² = g², and since
both are positive, f = g. A number cannot have two different “square roots”
among its factors. This is the extra fact the book uses on page 3 to prove P.

Check it yourself: 12 pairs as (1, 12), (2, 6), (3, 4): three pairs, six factors. 36
pairs as (1, 36), (2, 18), (3, 12), (4, 9) and (6, 6): four pairs plus the lone 6, nine
factors.

Page 5 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

Q4 Discussion. Consider the following argument.

Each factor of a number has a ‘partner’ factor such that their product yields the
given number, e.g., 5 is a factor of 35, and 5 × 7 = 35. Here, 7 is the partner factor
of 5, and vice-versa. Let us focus on factor-partner pairs of numbers, e.g., (1, 12),
(2, 6), (3, 4) are the pairs for the number 12.
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.

What does this argument prove? Statement P or Q?

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”:

odd number of factors

⇒ 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”.

Why direction matters: an argument proves “if X then Y” only when it assumes
X and arrives at Y. Running the same steps backwards is a different claim, and
each backward step must be checked separately.

Tip: going backwards here fails at one step. Knowing that n = f × f tells us one
pair repeats, but not that only one pair repeats. The book fills this gap: a
number cannot have two repeating pairs, so a square has an odd number of
factors, and P is true as well.

Page 6 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

In-text Questions — Page 3
Example 4: equal area and congruence

Q1 Example 4
• 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).

Triangle 1 Triangle 2

5 cm
3 cm 3 cm

4 cm 4 cm
Two triangles with the same base (4 cm) and the same height (3 cm).

Area of each = ½ × 4 × 3 = 6 cm²

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.

Page 7 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

Why the heights are equal: let ΔABC ≅ ΔPQR, with altitudes AD on BC and PS
on QR. In ΔABD and ΔPQS: ∠B = ∠Q, ∠ADB = ∠PSQ = 90° and AB = PQ, so
ΔABD ≅ ΔPQS (AAS) and AD = PS. With BC = QR as well, ½ × BC × AD = ½ ×
QR × PS.

Tip: area depends only on base and height. Slide the top vertex of a triangle
along a line parallel to the base and you get endless triangles with the same
area, almost none of them congruent to each other.

Check it yourself: the slanting sides of Triangle 2 run 2 cm across and 3 cm up,
so each is √(2² + 3²) = √13 ≈ 3.6 cm. No side of Triangle 1 has this length.

Think and Reflect — Page 3
When a proposition and its converse are both false
THINK AND REFLECT

Q1 It can also happen that both the proposition and converse are false! Can you give an
example?

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.

Page 8 of 35

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Class 9 Maths Chapter 9 Propositions and their Conversesg l AglaSem · NCERT Solutions

co m
m.

o m l ase
Whym g inside the
.c works: look for two groups that overlap but neither asits
se Primes and odd numbers share 3, 5, 7, …, but 2 is only prime and 9 is only
this
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other.
odd. “Equal sides” and “rectangle” share the squares, but a rhombus is only in

com
the first group and a non-square rectangle only in the second. One stray
member in each direction makes both.statements ag
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false.
a
agl
Another one: “If n is a multiple of 4, then n is a multiple of 6” fails for 4, and its
converse fails for 6. m co
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as e
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se m
la
ag
In-text Questions — Page 4
m a s
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co m
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m as e
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ag l Page 9 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

Example 5: the converse of the Baudhāyana–Pythagoras theorem

Q1 Example 5. Converse of the Baudhāyana–Pythagoras theorem: Let a, b, c be the
sidelengths of a triangle. If a² + b² = c², then the triangle is right-angled.
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.

B
Y

a c
a

C b A Z b X
Figures on page 4: ΔABC with BC = a, CA = b, AB = c, and the right triangle ΔXYZ with YZ =
a, ZX = b (not to scale).

What can we say about the length of XY?
(Hint: Use the Baudhāyana–Pythagoras Theorem.)

XY = c.

ΔXYZ is right-angled at Z, so XY is its hypotenuse.
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.

Page 10 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

Why build a new triangle: we cannot apply the theorem to ΔABC, because we
do not yet know that it is right-angled; that is what we want to prove. So we
build ΔXYZ, which we know is right-angled, and use the theorem there.

Common mistake: writing XY = ±c. A length cannot be negative, so only XY = c
is possible.

Q2 (Same figures: ΔABC with BC = a, CA = b, AB = c and a² + b² = c²; right triangle ΔXYZ
with ∠Z = 90°, YZ = a, XZ = b.)

B
Y

a c
a

C b A Z b X
ΔABC and ΔXYZ from page 4 (not to scale).

What can we say about ΔABC and ΔXYZ?
(Hint: Use the SSS criterion for congruence.)

ΔABC ≅ ΔXYZ, so ΔABC is right-angled at C.

Page 11 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

BC = a = YZ
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.

Why the correspondence C ↔ Z: in ΔABC, C is the vertex where sides a and b
meet; in ΔXYZ, Z is the vertex where sides a and b meet. Matching vertices this
way makes all three pairs of sides equal, so the angle at C equals the angle at Z.

Check it yourself: 3² + 4² = 9 + 16 = 25 = 5², and 5² + 12² = 25 + 144 = 169 =
13². Make triangles with sides 3, 4, 5 cm and 5, 12, 13 cm and measure the angle
opposite the longest side: it is 90°.

Did you know? Builders and carpenters still use this converse to check a
corner: if the marks at 3 units and 4 units along two walls are 5 units apart, the
corner is a right angle.

Think and Reflect — Page 5

Page 12 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

A Note: other ways of saying ‘if X then Y’
THINK AND REFLECT

Q1 Identify real-life examples in which a proposition is true but not the converse. Give nice
counterexamples!

PROPOSITION (TRUE) CONVERSE (FALSE) COUNTEREXAMPLE

If a person lives in Chennai, If a person lives in Tamil Someone living in Madurai lives in
then the person lives in Tamil Nadu, then the person lives Tamil Nadu but not in Chennai.
Nadu. in Chennai.

If you score 100 out of 100 in a If you pass a test, then you A student scoring 65 out of 100
test, then you pass it. scored 100 out of 100. passes.

If an animal is a cow, then it If an animal has four legs, A dog has four legs but is not a
has four legs. then it is a cow. cow.

If today is Sunday, then our If our school is closed, then Schools are closed on summer-
school is closed. today is Sunday. vacation weekdays too.

The pattern: in each case the “if” part describes a smaller group inside a bigger
group (Chennai inside Tamil Nadu, cows inside four-legged animals). Going
from small to big is safe; going from big to small is not, because the bigger
group has extra members, and any one of them is a counterexample.

Your own answer: make examples from your own town, district and state, or
from your school timetable, for example “If it is the second period on Monday,
then we have Maths”.

Page 13 of 35

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Class 9 Maths Chapter 9 Propositions and their Converses
ag l AglaSem · NCERT Solutions

. com is
Give more examples from geometry as well as number theory in which a proposition
sem
Q2

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true but not its converse. Give appropriate counterexamples.
a
m . agl
la se
ag

PROPOSITION (TRUE) CONVERSE (FALSE) COUNTEREXAMPLE
co m
m . ag
se
Geometry
la
If a quadrilateral is a square, then aIfga quadrilateral has equal A 3 cm × 4 cm rectangle: both
its diagonals are equal. diagonals, then it is a diagonals are 5 cm.
square.
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m l as
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If a quadrilateral is a rhombus, If the diagonals are A kite with sides 2, 2, 5, 5 cm.
m a
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then its diagonals are perpendicular, then it is a

la perpendicular. rhombus.
ag Number theory
m a s
If n is divisible by 4, then n is
em
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If n is even, then n is n=6
agl
even.
la s divisible by 4.

If n is a multiple of 9, then n is a a g If n is a multiple of 3, then n n = 12

m
multiple of 3. is a multiple of 9.
. co
em
m
co
gl as
. a
a se m
agl Why these work: each “if” part names a special case of the “then” part
(squares are special quadrilaterals with equal diagonals; multiples of 9 are
se m
com wrongly claims that the general case is g l a
.
special multiples of 3). The converse
always the special one. sem a
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More examples: “If two angles are vertically opposite, then they are equal” is
true, but its converse fails for the two base angles of an isosceles. comtriangle
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Check it yourself: the 3 cm × 4 cm rectangle has diagonals √(3² + 4²) = √25 =
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.
all equal. em a s
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co m
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m as e
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ag l Page 14 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

Exercise Set 9.1 — Pages 5–7
Propositions, converses and counterexamples

Q1 Frame the converse for each of the propositions in Questions 1–12. Then, determine if
each of the two statements is true or not. Justify the true statements and give a
counterexample for each false statement.
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.

1
l
P

2
m
Q
t
Transversal t cuts lines l and m at P and Q; ∠1 and ∠2 are corresponding angles.

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.

Page 15 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

The converse in detail: if l and m meet at R to the right of t, then ∠1 is the
exterior angle of ΔPQR at P and ∠2 is its interior angle at Q, so ∠1 = ∠Q + ∠R >
∠2. If they meet to the left, then ∠1 equals the interior angle at P (vertically
opposite) and ∠2 is the exterior angle at Q, so ∠2 = ∠P + ∠R > ∠1. Either way
the angles cannot be equal.

Justifying the proposition: through Q draw the line m′ that makes an angle
equal to ∠1 with t. By the converse just proved, m′ ∥ l. But through Q there is
only one line parallel to l, and m is such a line, so m′ = m and ∠2 = ∠1. Many
books simply take the proposition as a basic property (axiom) of parallel lines.

Result: here the proposition and its converse are both true.

Q2 If a quadrilateral is a square, then all its angles are equal.

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.

Why: a square needs two things: equal angles and equal sides. The converse
gives only the first, so it describes all rectangles, and squares are just the
special rectangles with equal sides.

Page 16 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

Tip: the corrected converse is true: if all angles of a quadrilateral are equal,
then it is a rectangle.

*Q3 Given any ΔABC, let us bisect the angles at B and C. The bisectors meet at the
incentre I of the triangle. Now extend the bisectors beyond I till they meet the
opposite sides at E and F respectively, as shown.
Proposition: If AB = AC, then IE = IF.

A

E
F

I

B C
Fig. 9.1, page 5: bisectors BE and CF of ∠B and ∠C meet at the incentre I.

Converse: If IE = IF, then AB = AC.
Proposition: true.

Page 17 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

AB = AC ⇒ ∠B = ∠C (Statement 1)
⇒ 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°:

∠IBC + ∠ICB = ½(∠B + ∠C) = ½(180° − 60°) = 60°, so ∠FIE = ∠BIC = 120°
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

Page 18 of 35

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Class 9 Maths Chapter 9 Propositions and their Conversesg l AglaSem · NCERT Solutions

co m
e m.
m l as
m .co A ag
la se
ag
60°
o m
5 cm Qsem . c ag
a
agl E 8 cm
F co m
em.
Pm l as
m .co ag
la se I
ag
m a s
em7 cm
.co agl
B a s C
agl
Counterexample: AB = 5 cm, AC = 8 cm, BC = 7 cm. ∠A = 60°. IP ⊥ AB and IQ ⊥ AC (dashed); the
m co
.
red segments IE and IF are equal.
e m
com gl as
. a
em
a s
agl
What the book expects: the book prints no answer. Because the proposition is
se m
m
about an isosceles triangle, manyostudents
c
will guess that the converse is true
g l a
.
motherwise: any non-isosceles triangle with ∠A = a
s
as well. The mathematics says e
a glaThe star marks this as a challenge question, and the
60° breaks the converse.
challenge is most likely to find this counterexample; it fits the chapter’s lesson
that a converse must be checked, not assumed. So write “converse
co m false” and
give the counterexample with its proof. m. se
m la
.co ag
a s em
agl
Why IP = IQ: I lies on the bisector of ∠A, because all three angle bisectors pass
c
through the incentre. In ΔAIP and ΔAIQ: ∠IAP = ∠IAQ, ∠API = ∠AQI = 90° and
m .
AI is common, so ΔAIP ≅ ΔAIQ (AAS) and IP = IQ.
m a s e
. co agl
a sem
agl

co m
m .
m as e
.co

ag l Page 19 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

Why the angles at F and E match: the acute angle between CF and AB is the
smaller of ∠AFI and ∠BFI; the acute angle between BE and AC is the smaller of
∠CEI and ∠AEI. Since ∠AFI = ∠CEI and ∠BFI = ∠AEI, these two acute angles
are equal. (If they are 90°, then F = P and E = Q, and IF = IP = IQ = IE directly.)

Check it yourself: construct ∠A = 60° with arms AB = 5 cm and AC = 8 cm; BC
measures 7 cm (7² = 5² + 8² − 5 × 8 = 49). The bisector of ∠B meets AC at E
with AE = 10/3 cm ≈ 3.33 cm, and the bisector of ∠C meets AB at F with AF =
8/3 cm ≈ 2.67 cm. Measure IE and IF: both are 2√7/3 cm ≈ 1.76 cm.

Did you know? IE = IF happens exactly when AB = AC or ∠A = 60°, so the
converse fails only for the 60° case. A related, famous result, the Steiner–
Lehmus theorem, says that if the whole bisectors BE and CF are equal in length,
then AB = AC. That statement is true, but it is much harder to prove.

Q4 If x = y, then a + x = a + y, where x, y and a are any three numbers. This proposition
and its converse are routinely used while solving equations.

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:

(a + x) − a = (a + y) − a
x=y

Both statements are true. Example: from 5 + x = 5 + 9 we may conclude x = 9.

Page 20 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

Where they are used: the proposition lets us “add the same number to both
sides” of an equation. The converse lets us “cancel” the same number from
both sides. For example, x − 7 = 12 ⇒ x − 7 + 7 = 12 + 7 ⇒ x = 19 uses the
proposition.

Tip: the same is true for multiplication only with care: a × x = a × y gives x = y
only when a ≠ 0 (0 × 2 = 0 × 5, but 2 ≠ 5).

Q5 If a and b are perfect squares, then ab is a perfect square.

Converse: If ab is a perfect square, then a and b are perfect squares.
Proposition: true.

Let a = m² and b = k² (m, k whole numbers).

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.

Why the converse fails: the “missing” factors of one number can be supplied
by the other. 2 = 2 and 8 = 2 × 2 × 2; together they make 2 × 2 × 2 × 2 = (2 ×
2)². Another counterexample: 3 × 12 = 36.

Did you know? The converse does hold if a and b have no common factor other
than 1: then each of them must itself be a perfect square.

Page 21 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

Q6 In Questions 6 and 7, x and y are real numbers.
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.

x² = 3² = 9, y² = (−3)² = 9, so x² = y²
but 3 ≠ −3

What x² = y² really tells us: x² − y² = 0, so (x − y)(x + y) = 0. Hence x = y or x =
−y. Squaring hides the sign, so the converse fails whenever x and y are
opposite non-zero numbers.

Tip: the converse becomes true if we add a condition: if x and y are both non-
negative and x² = y², then x = y. That is why a length found from XY² = c² is
simply XY = c.

Q7 If x = y, then x³ = y³.

Converse: If x³ = y³, then x = y.
Proposition: true. Cubing the same number gives the same result.
Converse: true.

Page 22 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

x³ − y³ = (x − y)(x² + xy + y²)
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.

Another way: cubing keeps the order of numbers. If x < y, then x³ < y³ (check
the cases: both positive, both negative, or x negative and y positive). So two
different numbers can never have the same cube.

Check it yourself: expand (x − y)(x² + xy + y²) = x³ + x²y + xy² − x²y − xy² − y³
= x³ − y³. And (x + y/2)² + ¾y² = x² + xy + ¼y² + ¾y² = x² + xy + y².

Compare with Q6: for squares the converse is false, for cubes it is true. A small
change in a statement can change its truth.

Q8 In Questions 8–12, n is a positive integer.
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 = 4 × (6k), so 4 divides n
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.

Page 23 of 35

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Class 9 Maths Chapter 9 Propositions and their Conversesg l AglaSem · NCERT Solutions

o m
The reason: 4 and 6 share the factor 2, so the smallest number divisible by both is their LCM,
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Without using primes: the last digit of 12k is the last digit of 2k, which is always
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which happens only when k is a multiple of 5. So n = 12 × 5m = 60m.
. co agl
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agl

co m
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m as e
.co

ag l Page 24 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

Compare with Q8: 4 and 6 share the factor 2, so being divisible by both does
not give divisibility by 24. 5 and 12 share nothing, so it does give divisibility by
60.

Q10 If n is the square of a prime number, then it has exactly 3 factors.

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.

3 is odd, so n is a perfect square (Example 3): n = m², with m > 1.
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.

Why m > 1: if m = 1 then n = 1, which has only one factor, not three.

Check it yourself: the numbers below 100 with exactly 3 factors are 4, 9, 25
and 49, which are 2², 3², 5² and 7². A square of a non-prime, such as 36 = 6²,
has more factors: 9 of them.

Q11 If n is a product of two unequal prime numbers, then it has exactly 4 divisors.

Converse: If n has exactly 4 divisors, then n is a product of two unequal prime numbers.

Page 25 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

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.

Why: a prime cubed, p³, also has exactly four divisors: 1, p, p², p³. So 8, 27 (1,
3, 9, 27) and 125 are all counterexamples.

Did you know? The numbers with exactly 4 divisors are precisely the products
of two different primes and the cubes of primes.

Q12 If n and n + 3 have no factors in common, then n is not a multiple of 3.

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.

Let d be a common factor of n and n + 3.
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.

Key step: if a number divides two numbers, it also divides their difference. Here
the difference is always 3, so the only possible common factors are 1 and 3.

Page 26 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

Tip: the proposition was proved by showing “if n is a multiple of 3, then n and n
+ 3 have a common factor”. Proving “if not Y, then not X” is a valid way to prove
“if X, then Y”.

Q13 There are no known ‘neat’ expressions that generate only primes! Find
counterexamples to the following claims.
(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

Lesson: a formula can give primes many times in a row and still fail. Checking
examples can disprove a claim, but it can never prove it for all n.

Tip for (ii): put n = 11. Every term becomes a multiple of 11: 11² + 11 + 11 = 11 × 13
= 143. Choosing n to match the constant term is a quick way to break such
formulas.

Other counterexamples: (i) n = 6 gives 145 = 5 × 29; (iii) n = 5 gives 1027 = 13
× 79. If n = 0 is allowed, (i) gives 1 and (iii) gives 4, neither of which is prime.

Page 27 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

Did you know? Euler’s formula n² + n + 41 gives primes for every n from 0 to
39, but n = 40 gives 1681 = 41 × 41.

Q14 Find counterexamples to the following statements.
(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

2¹¹ − 1 = 2048 − 1 = 2047 = 23 × 89, not prime.

(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

2⁶ + 1 = 64 + 1 = 65 = 5 × 13, not prime.

(n = 2 and n = 4 give 5 and 17, which are prime.)

Why 2⁶ + 1 must factor: 2⁶ + 1 = 4³ + 1, and a³ + 1 is always divisible by a + 1
because a³ + 1 = (a + 1)(a² − a + 1). With a = 4: 65 = 5 × 13. Whenever the even
number n has an odd factor greater than 1 (n = 6, 10, 12, …), 2ⁿ + 1 is composite.
Another counterexample: 2¹⁰ + 1 = 1025 = 25 × 41.

Did you know? Primes of the form 2ⁿ − 1 are called Mersenne primes. The
book’s Fermat example belongs to part (ii): Euler showed 2³² + 1 = 4294967297
= 641 × 6700417.

Page 28 of 35

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Class 9 Maths Chapter 9 Propositions and their Conversesg l AglaSem · NCERT Solutions

om 2 and
Consider the statement: ‘If a number is divisible by 8, then it is divisible by.cboth
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Proposition: If a number is divisible by 3, then the sum of its digits is a multiple of 3.

co m
m .
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.co

ag l Page 29 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

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.

For a 3-digit number with digits a, b, c:
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.

Why it works for any length: 10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1, … Each
place value is one more than a multiple of 9 (and so of 3). So any number = (a
multiple of 3) + (sum of its digits).

Tip: the converse is the one we use to test a number (digit sum is a multiple of 3
⇒ divisible). The proposition tells us that if the digit sum is not a multiple of 3,
the number is not divisible by 3: 472 has digit sum 13, so it is not divisible by 3.

Page 30 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

Q17 We have identified different types of quadrilaterals — squares, rectangles,
parallelograms, rhombi, kites and trapezia. One can identify more types (e.g., we
could create a category of quadrilaterals that have equal-length opposite sides).
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).

Fig. 9.2, page 7: two sticks placed as the diagonals of a quadrilateral.

(i) Suppose Q satisfies the following property.
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.

Page 31 of 35

Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

(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.

Why: in (i) “equal diagonals” is a necessary condition for Q; in (ii) it is a
sufficient condition. A necessary condition must be met but may not finish the
job; a sufficient one finishes the job on its own.

Example for (i)(b): if Q were “rectangle”, equal sticks would have to cross at
their midpoints; if Q were “square”, they would also have to cross at right
angles. Equal sticks crossed anywhere else give a quadrilateral that is neither.

Tip: in every case the sticks must actually cross each other, as in Fig. 9.2, so
that joining their endpoints in order gives a quadrilateral with the sticks as its
diagonals.

Chapter at a glance
A proposition is a statement that is either true or false, for example “If two sides of a
triangle are equal, then the angles opposite them are equal”.
‘If X then Y’ can also be said as ‘X implies Y’ or ‘Y when X’. Its converse is ‘If Y then X’.
A true proposition can have a false converse (multiple of 6 ⇒ multiple of 3, but 9 is a
multiple of 3 and not of 6). Both can be true (equal sides ⇔ equal opposite angles; n is a
perfect square ⇔ n has an odd number of factors). Both can be false (prime ⇒ odd fails at
2; odd ⇒ prime fails at 9).
A counterexample is a single case where the ‘if’ part holds but the ‘then’ part fails; one is
enough to show a proposition is false. Euler disproved Fermat’s claim by showing 2³² + 1 is
composite.
An argument proves ‘if X then Y’ only if it starts from X and reaches Y. The factor-partner
argument proves ‘odd number of factors ⇒ perfect square’; the other direction needs the
extra fact that only one pair can repeat.

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Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

Converse of the Baudhāyana–Pythagoras theorem: if a² + b² = c², the triangle is right-
angled. Proof: build a right triangle with legs a and b; its hypotenuse is c, so the two
triangles are congruent by SSS.
Divisibility: divisible by 4 and 6 does not give divisible by 24 (12 is a counterexample), but
divisible by 5 and 12 does give divisible by 60, because 5 and 12 have no common factor.
x² = y² does not give x = y, but x³ = y³ does.
No simple formula gives only primes: 4n² + 1 fails at n = 4 (65), n² + n + 11 at n = 10 (121),
4ⁿ + 3 at n = 4 (259), 2ⁿ − 1 at n = 11 (2047) and 2ⁿ + 1 at n = 6 (65).

Page 33 of 35

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Class 9 Maths Chapter 9 Propositions and their Converses
ag l AglaSem · NCERT Solutions

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co m
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m as e
.co

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Class 9 Maths Chapter 9 Propositions and their Converses AglaSem · NCERT Solutions

IDEA IN SYMBOLS / WHAT IT SAYS WHERE STUDENTS SLIP
STATEMENT
sticks are enough

Page 35 of 35