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

Class 9 Ganita Manjari Chapter 9 Quick Links:

• Exercise Set 9.1

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The NCERT Class 9 Maths Chapter 9 solutions help students understand these concepts and approach Exercise 9.1 systematically. The exercise contains 17 questions covering logical reasoning, mathematical proofs and counterexamples. Working through these questions helps students develop the ability to analyse statements, identify incorrect assumptions and present mathematically valid explanations.

(Hint: Draw the altitude from the vertex containing the third angle.)
Answer:
Yes, the statement is true.

In △ABC, let ∠ABC = ∠BCA. Draw AD perpendicular to BC.
In △ABD and △ACD:
∠ADB = ∠ADC = 90°.
∠BAD = 90° − ∠ABC.
∠CAD = 90° − ∠BCA.
Therefore, ∠BAD = ∠CAD.
Also, AD is common to both triangles.
So, △ABD ≅ △ACD by ASA congruence.
Therefore, AB = AC, because corresponding sides of congruent triangles are equal.
Thus, the sides opposite the equal angles are equal.

Answer:
Yes, both can be false.

Proposition: If a positive integer is even, then it is a multiple of 3.
This is false. For example, 2 is even, but it is not a multiple of 3.

Converse: If a positive integer is a multiple of 3, then it is even.
This is also false. For example, 3 is a multiple of 3, but it is not even.

Answer:
Example 1: Rain and a wet road

• Proposition: If it rains on an uncovered road, then the road becomes wet.
  This is true in the situation described in the chapter.
• Converse: If the road is wet, then it has rained.
  This is false. A tanker may have spilled water on the road.

Example 2: Buying notebooks
Suppose a shop sells each notebook for ₹10.

• Proposition: If I buy three notebooks, then I spend ₹30.
  This is true because 3 × ₹10 = ₹30.
• Converse: If I spend ₹30 at the shop, then I buy three notebooks.
  This is false. I could buy one pen costing ₹30 instead.

Answer:
Geometry example 1

• Proposition: If a quadrilateral is a square, then all its angles are equal.
  This is true because every angle of a square is 90°.
• Converse: If all the angles of a quadrilateral are equal, then it is a square.
  This is false. A rectangle of length 6 cm and breadth 4 cm has four equal angles but is not a square.

Geometry example 2

• Proposition: If two triangles are congruent, then they have equal areas.
  This is true because congruent triangles have the same shape and size.
• Converse: If two triangles have equal areas, then they are congruent.
  This is false. Consider two right triangles with perpendicular sides 3 cm and 4 cm, and 2 cm and 6 cm.
  Their areas are ½ × 3 × 4 = 6 cm² and ½ × 2 × 6 = 6 cm².
  But their side lengths are different, so they are not congruent.

Number theory example 1

• Proposition: If a number is a multiple of 6, then it is a multiple of 3.
  This is true because 6k = 3 × 2k.
• Converse: If a number is a multiple of 3, then it is a multiple of 6.
  This is false. For example, 9 is a multiple of 3 but not of 6.

Number theory example 2

• Proposition: If a positive integer is divisible by 24, then it is divisible by both 4 and 6.
  This is true because 24 is divisible by both 4 and 6.
• Converse: If a positive integer is divisible by both 4 and 6, then it is divisible by 24.
  This is false. For example, 12 is divisible by both 4 and 6 but not by 24.

Class 9 Maths Ganita Manjari Chapter 9 Exercise Set 9.1 Question Answers and 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.

Answer:
Proposition: True.
When a transversal cuts two parallel lines, the corresponding angles are equal. This is the corresponding-angles property of parallel lines.

Converse: If the corresponding angles formed by a transversal with two lines are equal, then the two lines are parallel.
Converse: True.
Equal corresponding angles are a condition for the two lines to be parallel.

Therefore, we can say that: l ∥ m implies α = β, and α = β implies l ∥ m.

Answer:
Proposition: True.
Every angle of a square is 90°. Therefore, all its angles are equal.

Converse: If all the angles of a quadrilateral are equal, then it is a square.
Converse: False.
Counterexample: A rectangle of length 6 cm and breadth 4 cm has four angles of 90°, but its four sides are not equal. So, it is not a square.

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

Answer:
Proposition: True.
Since AB = AC, ∠ABC = ∠BCA.
Since BI and CI bisect these equal angles, ∠IBC = ∠BCI.
Therefore, BI = CI in △BIC.

Now compare △CIE and △BIF:
∠ICE = ∠IBF, because each is half of an equal base angle.
∠CIE = ∠BIF, because they are vertically opposite angles.
CI = BI, as proved above.
Therefore, △CIE ≅ △BIF by ASA congruence.
Hence, IE = IF.

Converse: In the same construction, if IE = IF, then AB = AC.
Converse: False.
Counterexample: Take △ABC with ∠A = 60°, ∠B = 90° and ∠C = 30°. Draw the angle bisectors and label I, E and F as in the question.
In this triangle, IE = IF even though AB ≠ AC.

Here is a proof of the equal lengths in this counterexample:

1. Join AI. Since I is the incentre, AI also bisects ∠A.
   Therefore, ∠EAI = ∠IAF = 30°.
   Also, ∠ABE = 45° and ∠ACF = 15°.
2. In △ABE:
   ∠AEI = ∠AEB = 180° − 60° − 45° = 75°.
3. In △ACF:
   ∠AFI = ∠AFC = 180° − 60° − 15° = 105°.
4. Mark G on the ray AB so that AG = AE, as shown in the diagram. Join IG.
   In △AEI and △AGI:
   AE = AG, AI is common, and ∠EAI = ∠GAI = 30°.
   Therefore, △AEI ≅ △AGI by SAS congruence.
   Hence, IE = IG and ∠AGI = 75°.
5. G cannot lie between A and F, because △IFG would then have two angles of 105°, which is impossible. G cannot equal F either, because ∠AGI = 75° but ∠AFI = 105°.
   Therefore, F lies between A and G.
   ∠IGF = ∠IGA = 75°.
   ∠IFG = 180° − ∠IFA = 180° − 105° = 75°.
   Thus, △IFG has equal base angles, so IF = IG.
6. Therefore, IE = IG = IF.
   But ∠B ≠ ∠C, so AB ≠ AC.
   This is a counterexample to the converse.

This proposition and its converse are routinely used while solving equations.
Answer:
Proposition: True.
Adding the same number a to equal numbers gives equal results.

Converse: If a + x = a + y, then x = y.
Converse: True.

Subtract a from both sides:
a + x − a = a + y − a.
Therefore, x = y.

Answer:
Proposition: True.
Let a = m² and b = n², where m and n are integers.
Then ab = m² × n² = (mn)².
Since mn is an integer, ab is a perfect square.

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

Counterexample: Let a = 2 and b = 8.
Then ab = 16 = 4², which is a perfect square.
But neither 2 nor 8 is a perfect square.

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

Answer:
Proposition: True.
Equal numbers have equal squares.

Converse: If x² = y², then x = y.
Converse: False.

Counterexample: Let x = 2 and y = −2.
Then x² = 4 and y² = 4, but 2 ≠ −2.

Answer:
Proposition: True.
Equal numbers have equal cubes.

Converse: If x³ = y³, then x = y.
Converse: True.
Taking the real cube root of both sides gives x = y. Each real number has only one real cube root.

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

Answer:
Proposition: True.
Let n = 24k, where k is a positive integer.
Then n = 4 × 6k and n = 6 × 4k.
Therefore, n is divisible by both 4 and 6.

Converse: If n is divisible by both 4 and 6, then it is divisible by 24.
Converse: False.
Counterexample: n = 12.
12 is divisible by both 4 and 6, but it is not divisible by 24.

Answer:
Proposition: True.
Let n = 60k, where k is a positive integer.
Then n = 5 × 12k and n = 12 × 5k.
Therefore, n is divisible by both 5 and 12.

Converse: If n is divisible by both 5 and 12, then it is divisible by 60.
Converse: True.
LCM(5, 12) = 60.
Every number divisible by both 5 and 12 is a multiple of their LCM, 60.

Answer:
Proposition: True.
Let n = p², where p is prime.
Its only positive factors are 1, p and p².
Therefore, n has exactly 3 factors.

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

Let the three factors be 1, p and n, where 1 < p < n.
Since p divides n, n ÷ p is also a factor of n.
This factor lies between 1 and n, so it must be p.
Therefore, n ÷ p = p, giving n = p².
Also, p must be prime. If p were composite, one of its factors between 1 and p would give an extra factor of n.
Thus, n is the square of a prime number.

Answer:
Proposition: True.
Let n = pq, where p and q are unequal prime numbers.
Its positive divisors are 1, p, q and pq.
These are four different divisors and there are no others.

Converse: If n has exactly 4 divisors, then it is a product of two unequal prime numbers.
Converse: False.
Counterexample: n = 8.
Its divisors are 1, 2, 4 and 8, so it has exactly 4 divisors.
But 8 = 2 × 2 × 2. It is not a product of two unequal prime numbers.

Answer:
Proposition: True.
If n were a multiple of 3, then n + 3 would also be a multiple of 3.
Then 3 would be a common factor, which contradicts the given condition.
Therefore, n is not a multiple of 3.

Converse: If n is not a multiple of 3, then n and n + 3 have no common factor other than 1.
Converse: True.
Any common factor of n and n + 3 must also divide their difference:
(n + 3) − n = 3.
The only positive factors of 3 are 1 and 3.
Since n is not divisible by 3, the only possible common factor is 1.

(i) All numbers of the form 4n² + 1 are prime.
Answer:
Take n = 4.
4n² + 1 = 4 × 4² + 1
= 64 + 1
= 65
= 5 × 13.
So, 65 is composite. The claim is false.

(ii) All numbers of the form n² + n + 11 are prime.
Answer:
Take n = 10.
n² + n + 11 = 10² + 10 + 11
= 100 + 10 + 11
= 121
= 11 × 11.
So, 121 is composite. The claim is false.

(iii) All numbers of the form 4ⁿ + 3 are prime.
Answer:
Take n = 4.
4ⁿ + 3 = 4⁴ + 3
= 256 + 3
= 259
= 7 × 37.
So, 259 is composite. The claim is false.

(i) If n is a prime number, then 2ⁿ − 1 is a prime number.
Answer:
Take n = 11, which is prime.
2¹¹ − 1 = 2048 − 1
= 2047
= 23 × 89.
So, 2¹¹ − 1 is composite even though 11 is prime. The statement is false.

(ii) If n is an even number, then 2ⁿ + 1 is a prime number.
Answer:
Take n = 6, which is even.
2⁶ + 1 = 64 + 1
= 65
= 5 × 13.
So, 2⁶ + 1 is composite even though 6 is even. The statement is false.

(i) Justify the statement.
Answer:
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.
So, the statement is true.

(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?
Answer:
No.
For example, 12 is divisible by both 2 and 4, but it is not divisible 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.

Answer:
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.

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).

(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?
Answer:
Yes.
The sticks are the diagonals. Every quadrilateral of type Q has equal diagonals, so the two sticks must have equal lengths.
Unequal sticks cannot give a quadrilateral of type Q under this condition.

(b) Will it matter how the two sticks are put together?
Answer:
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) 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?
(a) Answer:
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) Answer:
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.