Propositions and their Converses Class 9 Solutions Maths Ganita Manjari Chapter 9

Solving questions with the help of Ganita Manjari Class 9 Solutions and Part 2 Class 9 Maths Chapter 9 Propositions and their Converses NCERT Solutions improves confidence.

Ganita Manjari Class 9 Chapter 9 Solutions

Class 9 Ganita Manjari Chapter 9 Solutions

Class 9 Maths Ganita Manjari Chapter 9 Solutions Propositions and their Converses

NCERT Intext Questions (Pages No. 2-4)

Question 1.
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.
Justify why P is true. Determine whether Q is true, and if not, give a counterexample.
Solution:
P is true.
If n is a multiple of 6, then n = 6k for some whole number k.
So n = 3 × (2k), which is 3 times a whole number — so n is a multiple of 3.
Q is false.
Counterexample: 9 is a multiple of 3 (9 = 3 × 3), but it is not a multiple of 6.
(Any odd multiple of 3, such as 3, 15, or 21, works just as well.)

Question 2.
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?
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?
Solution:
The argument proves Q. It starts from “n has an odd number of factors” and ends with “n is a perfect square” — that is exactly the direction of Q.
Factors come in factor-partner pairs; if no pair repeated a number, the factors would split neatly into pairs, and their count would be even.
An odd count forces some pair (f, f), so n = f × f
P is true.
If n = f × f, then (f, f) is a factor partner pair with a repeated number.
There cannot be a second such pair (g, g) because g × g = f × f forces g = f for positive numbers.
So exactly one factor is “unpaired,” and all the others come in distinct pairs — giving an odd total.
For example: 36 has factor pairs (1, 36), (2, 18), (3, 12), (4, 9), (6, 6)
that’s 4 × 2 + 1 = 9 factors.

Question 3.
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.
Solution:
P is false.
Counterexample: a right triangle with legs 3 and 4 (sides 3, 4, 5) has area \(\frac {1}{2}\) × 4 × 3 =6.
A triangle with base 6 and height 2 has \(\frac {1}{2}\) × 6 × 2 = 6 too.
The areas are equal, but the second triangle has a side of length 6, which the first doesn’t have — so they are not congruent.

Q is true.
Congruent triangles can be placed exactly on top of each other, so they cover the same region of the plane — and hence have the same area.
(Equivalently: corresponding sides and altitudes are equal, so \(\frac {1}{2}\) × base × height gives the same value for both.)

Question 4.
Baudhayana-Pythagoras theorem
Let a, b, c be the side lengths of a triangle. If the triangle is right-angled, then a² + b² = c². What is its converse?
Converse of the Baudhayana-Pythagoras Theorem: Let a, b, c be the side lengths of ∆ABC with a² + b² = c².
Construct a right triangle XYZ whose perpendicular sides YZ and XZ have lengths a and b. What can we say about the length of XY? What can we say about ∆ABC and ∆XYZ?
Solution:
Length of XY: ∆XYZ is right-angled at Z, so by the Baudhayana-Pythagoras Theorem: XY² = YZ² + XZ² = a² + b².
But we are given a² + b² = c², so XY² = c² giving XY = c.

Comparing the triangles. Now the three sides match:
BC = YZ, CA = ZX = b, AB = XY = c
So ∆ABC ≅ ∆XYZ by the SSS congruence criterion.
Corresponding angles of congruent triangles are equal, so ∠C = ∠Z = 90°.
Hence ∆ABC is right-angled (at C, the vertex opposite the longest side c), and the converse of the Baudhayana-Pythagoras Theorem is true.

Think and Reflect (NCERT Textbook Page No. 1)

Question 1.
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.)
Solution:
Statement 2: If two angles of a triangle are equal, then the sides opposite the equal angles have equal lengths.

In ∆ABC, ∠B = ∠C, with altitude AD drawn from the third vertex A, perpendicular to BC.
Yes, Statement 2 is true – it is the converse of Statement 1, and in this case both the proposition and its converse hold true.
Given: ∆ABC with ∠B = ∠C.
To prove: AB = AC.
Construction: Following the hint, draw the altitude AD from vertex A (the vertex holding the third angle) to BC, so that AD ⊥ BC.
Proof: Compare ∆ABD and ∆ACD.
∠ABD = ∠ACD (given, ∠B = ∠C)
∠ADB = ∠ADC = 90° (AD is an altitude)
AD = AD (common side)
So ∆ABD ≅ ∆ACD by the AAS congruence criterion, and hence the corresponding sides are equal: AB = AC.

Think and Reflect (NCERT Textbook Page No. 3)

Question 1.
It can also happen that both the proposition and converse are false! Can you give an example?
Solution:
Geometry example.
Proposition P: If a quadrilateral is a rhombus, then it is a rectangle.
False – a rhombus with angles 60° and 120° is not a rectangle.
Converse Q: If a quadrilateral is a rectangle, then it is a rhombus.
False – a 4 cm × 2 cm rectangle does not have all sides equal.
Number example.
Proposition P: If a number is a multiple of 4, then it is a multiple of 6.
This is false – 4 itself is a multiple of 4 but not of 6.
Converse Q: If a number is a multiple of 6, then it is a multiple of 4.
This is also false – 6 is a multiple of 6 but not of 4.

Think and Reflect (NCERT Textbook Page No. 5)

Question 1.
Identify real-life examples in which a proposition is true but not the converse. Give nice counterexamples!
Solution:
Proposition P: If you have the flu, then you have a fever.
Converse Q: If you have a fever, then you have the flu.
Here, Proposition P is true, but its converse may not be true.
You could have a fever because of a different infection, not the flu.
OR
Proposition P: If it is raining, then I carry an umbrella.
Converse Q: If I carry an umbrella, then it is raining.
Here, proposition P is true, but its converse may not be true.
Counterexample: I might carry an umbrella on a sunny day to block the harsh sunlight, even though it isn’t raining.

Question 2.
Give more examples from geometry as well as number theory in which a proposition is true but not its converse. Give appropriate counterexamples.
Solution:
Proposition P: If a quadrilateral is a square, then it has four equal sides.
Converse Q: If a quadrilateral has four equal sides, then it is a square.
Here, Proposition P is true, but its Converse Q may not be true.
Counterexample: A rhombus has four equal sides, but it is not a square because its interior angles are not 90 degrees.

Ex 9.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 9.1 Solutions

Exercise 9.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 9.1 Solutions

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.

Question 1.
If two lines are parallel, then the corresponding angles formed by a transversal are equal.
Solution:
Converse: If the corresponding angles formed by a transversal are equal, then the two lines are parallel.
Checking the proposition: The statement is true.
Justification: Corresponding angles are equal when two parallel lines are cut by a transversal.

Checking the Converse: The statement is true.
Justification: If a pair of corresponding angles are equal, then the two lines are parallel.

Question 2.
If a quadrilateral is a square, then all its angles are equal.
Solution:
Converse: If all the angles of a quadrilateral are equal, then it is a square.
Checking the proposition: The statement is true.
Justification: A square has four right angles, so all its angles are equal.

Checking the Converse: The statement is false.
Counterexample: A rectangle that is not a square has four equal angles of 90°.

Question 3.
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.
Solution:
Converse: If IE = IF, then AB = AC.
Checking the proposition: The statement is true.
Justification: If AB = AC, then ΔABC is an isosceles triangle, so ∠B = ∠C.
Hence, by symmetry, the two angle bisectors BE and CF are equal in the corresponding parts around the incentre I.
Therefore, IE = IF.
Checking the proposition: The statement is false.
Counterexample: If IE = IF, the figure is not necessarily symmetrical.
Hence, we cannot say that AB = AC.

Question 4.
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.
Solution:
Proposition P: If x = y, then a + x = a + y.
Converse Q: If a + x = a + y, then x = y.
Checking the proposition: The proposition is true.
Justification: Since x = y, adding the same number a to both sides gives a + x = a + y.
Checking the converse: The converse is true.
Justification: If a + x = a + y, subtracting a from both sides gives x = y.

Question 5.
If a and b are perfect squares, then ab is a perfect square.
Solution:
Proposition P: If a and b are perfect squares, then ab is a perfect square.
Converse Q: If ab is a perfect square, then a and b are perfect squares.
Checking the proposition: The proposition is true.
Justification: Let a = x² and b = y²
Then, ab = x²y² = (xy)², which is a perfect square.
Checking the converse: The converse is false.
Counterexample: Take a = 3 and b = 12
Then, ab = 3 × 12 = 36, which is a perfect square.
However, neither 3 nor 12 is a perfect square.
Hence, the converse is false.

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

Question 6.
If x = y, then x² = y².
Solution:
Proposition P: If x = y, then x² = y².
Converse Q: If x² = y², then x = y.
Checking the proposition: The proposition is true.
Justification: If x = y, then squaring both sides gives x² = y².
Checking the converse: The converse is false.
Counterexample: Take x = 3, y = -3.
Then, x² = y² = 9, but x ≠ y.
Hence, the converse is false.

Question 7.
If x = y, then x³ = y³.
Solution:
Proposition P: If x = y, then x³ = y³.
Converse Q: If x³ = y³, then x = y.
Checking the proposition: The proposition is true.
Justification: If x = y, then cubing both sides gives x³ = y³.
Checking the converse: The converse is true.
Justification: If x³ = y³, then taking the cube root of both sides gives x = y.

In Questions 8-12, n is a positive integer.

Question 8.
If n is divisible by 24, then it is divisible by both 4 and 6.
Solution:
Converse: If n is divisible by both 4 and 6, then n is divisible by 24.
Checking the proposition: The proposition is true.
Justification: Since 24 = 4 × 6, every multiple of 24 is divisible by both 4 and 6.
Checking the converse: The converse is false.
Counterexample: Take n = 12. It is divisible by both 4 and 6, but 12 is not divisible by 24.

Question 9.
If n is divisible by 60, then it is divisible by both 5 and 12.
Solution:
Converse: If n is divisible by both 5 and 12, then n is divisible by 60.
Checking the proposition: The proposition is true.
Justification: Since 60 = 5 × 12, every multiple of 60 is divisible by both 5 and 12.
Checking the converse: The converse is true.
Justification: If n is divisible by both 5 and 12, then it is divisible by their LCM, 60.
Hence, n is divisible by 60.

Question 10.
If n is the square of a prime number, then it has exactly 3 factors.
Solution:
Converse: If n has exactly 3 factors, then n is the square of a prime number.
Checking the proposition: The proposition is true.
Justification: Let n = m², where m is prime.
Its factors are 1, m, and m².
Thus, n has exactly 3 factors.
For example, let’s take m = 5, a prime number
hence n = m² = 5² = 25
Now factor of n = 1, 5, and 25
i.e., 1, n and m.
Checking the converse: The converse is true.
Justification: A number with exactly 3 factors must be of the form m2, where m is a prime
number.

Question 11.
If n is a product of two unequal prime numbers, then it has exactly 4 divisors.
Solution:
Converse: If n has exactly 4 divisors, then n is a product of two unequal prime numbers.
Checking the proposition: The proposition is true.
Justification: Let n = ab where a and b are unequal primes.
Its divisors are 1, a, b, ab.
Thus, n has exactly 4 divisors.
Checking the converse: The converse is false.
Counterexample: Take n = 8 = 2³.
Its divisors are 1, 2, 4, 8.
So it has exactly 4 divisors.
But 8 is not the product of two unequal prime numbers.

Question 12.
If n and n + 3 have no factors in common, then n is not a multiple of 3.
Solution:
Converse: If n is not a multiple of 3, then n and n + 3 have no factors in common.
Checking the proposition: The proposition is true.
Justification: If n were a multiple of 3, then n + 3 would also be a multiple of 3.
Thus, they would have 3 as a common factor.
Hence, if they have no common factor, n cannot be a multiple of 3.
Checking the converse: The converse is true.
Justification: If n is not a multiple of 3, then n and n + 3 cannot have 3 as a common factor.
Since any common factor of n and n + 3 must divide their difference, (n + 3) – n = 3, their only possible common factor greater than 1 is 3.
Therefore, they have no common factor.

Question 13.
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.
Solution:
(i) Putting n = 4 in 4n² + 1, we get
4n² + 1 = 4(4)² + 1
= 64 + 1
= 65
and 65 = 5 × 13
Hence, 65 is not prime.
The claim is false.
Counterexample: n = 4.

(ii) Putting n = 10 in n² + n + 11, we get
n² + n + 11 = (10)² + 10 + 11
= 100 + 10 + 11
= 121
and 121 = 11 × 11
Hence, 121 is not prime.
The claim is false.
Counterexample: n = 10.

(iii) Putting n = 4 in 4ⁿ + 3, we get
4ⁿ + 3 = 4⁴ + 3 = 256 + 3 = 259
But 259 = 7 × 37
Hence, 259 is not prime.
The claim is false.
Counterexample: n = 4.

Question 14.
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.
Solution:
(i) Putting n = 11 (which is prime) in 2ⁿ – 1, we get
2ⁿ – 1 = 2¹¹ – 1 = 2048 – 1 = 2047
and 2047 = 23 × 89
Hence, 2047 is not a prime number.
The statement is false.
Counterexample: n = 11.

(ii) Putting, n = 6 (even no) in 2ⁿ + 1, we get
2ⁿ + 1 = 2⁶ + 1 = 64 + 1 = 65 and 65 = 5 × 13
Hence, 65 is not a prime number.
The statement is false.
Counterexample: n = 6.

Question 15.
Consider the statement: ‘If a number is divisible by 8, then it is divisible by both 2 and 4.’
(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?
Solution:
(i) Justification: Let the number be n = 8k, where k is an integer.
Then n = 2 × 4k and n = 4 × 2k.
Therefore, n is divisible by both 2 and 4.
Hence, the statement is true.

(ii) No, it is not enough.
For example, 12 is divisible by both 2 and 4, but not by 8.
So, the converse of the given statement is false.
Checking divisibility by both 2 and 4 only guarantees divisibility by 4, since LCM(2, 4) = 4.
The usual shortcut for 8 is to check whether the number formed by the last three digits is divisible by 8.
For a number with fewer than three digits, check the whole number.

Question 16.
Recall that a shortcut to check whether a given number is divisible by 3 is to add the digits of the number and check if the sum is a multiple of 3. Express the relationship between ‘a number is divisible by 3 ’ and ‘the sum of the digits is a multiple of 3’ using ‘If – then’ sentences.
Solution:
Statement 1: If a number is divisible by 3, then the sum of its digits is a multiple of 3.
Statement 2: If the sum of the digits of a number is a multiple of 3, then the number is divisible by 3.
Both statements are true.
Each is the converse of the other.
Example: For 123, the digit sum is 1 + 2 + 3 = 6.
6 is a multiple of 3, and 123 = 3 × 41 is divisible by 3.

Question 17.
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. below).

(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?
Solution:
(i) If a quadrilateral is of type Q, then it has equal-length diagonals.
(a) Yes. The two sticks represent the diagonals of the quadrilateral.
Since every quadrilateral of type Q has equal-length diagonals, the two sticks must be of equal length.

(b) It may matter. The given statement does not tell us that every quadrilateral with equal diagonals is of type Q.
For example, suppose Q means rectangles.
Every rectangle has equal diagonals. But placing two equal sticks in just any way does not always produce a rectangle.
For a rectangle, the diagonals must also bisect each other.
So, we must put the equal sticks together at their midpoints.

Thus, equal lengths are necessary, but equal lengths alone do not guarantee type Q.

(ii) (a) We should choose equal sticks to guarantee a quadrilateral of type Q using this statement.
However, the statement does not say that unequal sticks can never give type Q.
It only guarantees that equal diagonals give type Q.

(b) With equal sticks, any arrangement that forms a proper quadrilateral will give type Q.
The diagonals will be equal, so the given statement applies.
No extra condition about their angle or midpoint is stated.
Here, a proper quadrilateral means that the four endpoints form four vertices of a quadrilateral, rather than a collapsed figure.

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