Class 9th

Class 9 Maths Exercise 9.1 – Proposition and Converse

Class 9 Maths – Exercise 9.1

Proposition and Converse – Complete Solutions

📘 For Students: This page contains simple, step-by-step solutions of Exercise 9.1. Important propositions, converses, examples and counterexamples are explained in an easy-to-understand way.
Question 1. If two lines are parallel, then corresponding angles formed by a transversal are equal.
Given: Two parallel lines are cut by a transversal.

To prove: Corresponding angles are equal.

Solution:
When two parallel lines are cut by a transversal, the corresponding angles are equal.
Therefore, corresponding angles are equal.
Converse:
If corresponding angles are equal, then the two lines are parallel.
Question 2. If a quadrilateral is a square, then all its angles are equal.

A square has four right angles.

∠A = ∠B = ∠C = ∠D = 90°
Therefore, all angles of a square are equal.
Converse:
If all four angles of a quadrilateral are equal, it need not necessarily be a square.
Example: A rectangle has all four angles equal, but its four sides need not be equal.
Hence, the converse is not always true.
Question 3. If AB = AC, then IE = IF.
Given:
AB = AC

Therefore, △ABC is an isosceles triangle.

∠ABC = ∠BCA

Since BI and CI are angle bisectors:

∠IBC = ∠BCI

Thus, in △BIC:

IB = IC
Also, because AB = AC, the triangle is symmetrical about the angle bisector through A. Hence the corresponding points E and F are at equal distances from I.
Therefore, IE = IF. Hence proved.
Question 4. If x = y, then a + x = a + y.
Given:
x = y
Adding a to both sides:
a + x = a + y
Converse:
If a + x = a + y, subtract a from both sides:
x = y
Hence, both the proposition and its converse are true.
Question 5. If a and b are perfect squares, then ab is a perfect square.

Let:

a = m²,    b = n²

Then:

ab = m²n²
ab = (mn)²
Therefore, ab is a perfect square.
Example:
a = 16 = 4²
b = 25 = 5²

ab = 16 × 25 = 400 = 20²
Question 6. If x = y, then x² = y².
Given:
x = y

Squaring both sides:

x² = y²
Converse:
If x² = y², then:
x = y   or   x = −y
Counterexample:
x = 3, y = −3
x² = y² = 9, but x ≠ y.
Therefore, the converse is not always true.
Question 7. If x = y, then x³ = y³.
Given:
x = y

Cubing both sides:

x³ = y³
Converse:
If x³ = y³, then, for real numbers:
x = y
Therefore, the converse is also true.

Questions 8–12

Question 8. If n is divisible by 24, then it is divisible by both 4 and 6.
24 = 4 × 6

If 24 divides n, then n must be divisible by both 4 and 6.

Example:
n = 120

120 ÷ 4 = 30
120 ÷ 6 = 20
Therefore, the proposition is true.
Question 9. If n is divisible by 60, then it is divisible by both 5 and 12.
60 = 5 × 12

If 60 divides n, then 5 divides n and 12 divides n.

Example:
n = 180

180 ÷ 5 = 36
180 ÷ 12 = 15
Therefore, the proposition is true.
Question 10. If n is the square of a prime number, then it has exactly 3 factors.

Let p be a prime number.

n = p²

The factors of p² are:

1, p, p²

So there are exactly 3 factors.

Example:
7² = 49
Factors of 49 = 1, 7, 49
Therefore, the proposition is true.
Question 11. If n is a product of two unequal prime numbers, then it has exactly 4 divisors.

Let p and q be two unequal prime numbers.

n = pq

Its divisors are:

1, p, q, pq
Example:
n = 3 × 5 = 15

Divisors = 1, 3, 5, 15
Therefore, it has exactly 4 divisors.
Question 12. If n and n + 3 have no factors in common, then n is not a multiple of 3.

Suppose n were a multiple of 3.

3 | n

Let:

n = 3k

Then:

n + 3 = 3k + 3 = 3(k + 1)

Therefore, both n and n + 3 would have 3 as a common factor.

This contradicts the given condition that n and n + 3 have no factors in common.
Hence, n is not a multiple of 3.
Question 13. Find counterexamples.

(i) All numbers of the form 4n² + 1 are prime.

Take:

n = 6

Then:

4(6)² + 1 = 4(36) + 1 = 145
145 = 5 × 29
Therefore, 145 is not prime. Counterexample: n = 6.

(ii) All numbers of the form n² + n + 11 are prime.

Take:

n = 10
10² + 10 + 11 = 100 + 10 + 11 = 121
121 = 11 × 11
Therefore, 121 is not prime. Counterexample: n = 10.

(iii) All numbers of the form 4n + 3 are prime.

Take:

n = 3
4(3) + 3 = 15
15 = 3 × 5
Therefore, 15 is not prime. Counterexample: n = 3.
Question 14. Find counterexamples.

(i) If n is a prime number, then 2ⁿ − 1 is a prime number.

Take the prime number:

n = 11
2¹¹ − 1 = 2048 − 1 = 2047
2047 = 23 × 89
Therefore, 2047 is not prime. Counterexample: n = 11.

(ii) If n is an even number, then 2ⁿ + 1 is a prime number.

Take:

n = 6
2⁶ + 1 = 64 + 1 = 65
65 = 5 × 13
Therefore, 65 is not prime. Counterexample: n = 6.
Question 15. Divisibility by 8

(i) Justify the statement.

If a number is divisible by 8, then it is divisible by both 2 and 4.

8 = 2 × 4

Therefore, every number divisible by 8 is automatically divisible by 2 and 4.

Example:
24 ÷ 8 = 3
24 ÷ 2 = 12
24 ÷ 4 = 6
Therefore, the statement is true.

(ii) Is checking divisibility by 2 and 4 enough to check divisibility by 8?

No.

Consider the number 12.

12 ÷ 2 = 6
12 ÷ 4 = 3

So 12 is divisible by both 2 and 4.

But:

12 ÷ 8 = 1.5
Therefore, divisibility by 2 and 4 is not enough to prove divisibility by 8.
Question 16. Divisibility by 3

Statement 1

If a number is divisible by 3, then the sum of its digits is a multiple of 3.

Example:
372 ÷ 3 = 124

Sum of digits = 3 + 7 + 2 = 12
12 is a multiple of 3.

Statement 2 – Converse

If the sum of the digits of a number is a multiple of 3, then the number is divisible by 3.

Example:
Number = 528

Sum = 5 + 2 + 8 = 15
15 is a multiple of 3.

528 ÷ 3 = 176
Both statements are true and are converses of each other.
Question 17. Quadrilateral Q

(i) Property:

If a quadrilateral is of type Q, then it has equal-length diagonals.

(a) Should the two sticks be of equal length?

Yes.

The two sticks represent the diagonals. Since a Q-type quadrilateral must have equal diagonals, the two sticks should have equal lengths.

Yes, the sticks should be of equal length.

(b) Will it matter how the two sticks are put together?

No.

As long as the two sticks have equal lengths, they represent equal diagonals. Their angle of intersection does not affect the fact that the diagonals have equal lengths.

No, it will not matter how they are put together.

(ii) Property:

If a quadrilateral has equal diagonals, then it is of type Q.

(a) Should the two sticks be of equal length?

Yes.

The sticks represent the diagonals. To make the diagonals equal, the two sticks must have equal lengths.

Yes, the two sticks must have equal lengths.

(b) Will it matter how the two sticks are put together?

No.

According to the given property, any quadrilateral having equal diagonals is of type Q. Therefore, once the sticks have equal lengths, their arrangement does not matter, provided they form a valid quadrilateral.

No, it will not matter how they are put together.
⭐ Quick Revision – Answers at a Glance
Q Answer
1 Corresponding angles are equal. Converse: equal corresponding angles imply parallel lines.
2 All angles of a square are 90°. Converse is not always true.
3 AB = AC ⇒ IE = IF.
4 x = y ⇒ a + x = a + y. Converse is also true.
5 If a and b are perfect squares, ab is a perfect square.
6 x = y ⇒ x² = y². Conver

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