NCERT Solutions for Class 9 Maths Chapter 9 — Exercise Set 9.1#
The converse of “if X then Y” is “if Y then X”.
To show a statement is true, give a reason (a proof) that works in every case. To show it is false, one counterexample is enough.
Questions 1–12: Frame the converse for each of the propositions. Then, determine if each of the two statements is true or not. Justify the true statements and give a counterexample for each false statement.
1. If two lines are parallel, then the corresponding angles formed by a transversal are equal. #
Solution
Proposition — true. When a transversal cuts two parallel lines, the corresponding angles are equal (corresponding angles property of parallel lines).
Converse: If the corresponding angles formed by a transversal with two lines are equal, then the lines are parallel.
Converse — true. If the lines were not parallel, they would meet on one side and form a triangle with the transversal. Then one corresponding angle would be an exterior angle of that triangle, which is greater than the interior opposite angle — so the corresponding angles could not be equal.
Both statements are true.
2. If a quadrilateral is a square, then all its angles are equal. #
Solution
Proposition — true. Every angle of a square is 90°.
Converse: If all the angles of a quadrilateral are equal, then it is a square.
Converse — false. Counterexample: a rectangle of sides 4 cm and 2 cm. All its angles are 90° (each angle is $\dfrac{360°}{4}$), but its sides are not all equal, so it is not a square.
*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 (Fig. 9.1). Proposition: If AB = AC, then IE = IF. #
Proposition — true
If $AB = AC$, then $\angle B = \angle C$, so their halves are equal: $\angle IBC = \angle ICB$.
Step 1. In $\triangle IBC$, the base angles are equal, so $IB = IC$.
Step 2. Compare $\triangle EBC$ and $\triangle FCB$ ($E$ is on $AC$, $F$ is on $AB$):
| Part | Reason |
|---|---|
| $\angle EBC = \angle FCB$ | halves of equal angles |
| $BC = CB$ | common side |
| $\angle ECB = \angle FBC$ | $\angle C = \angle B$ |
By ASA congruence, $\triangle EBC \cong \triangle FCB$, so $BE = CF$.
Step 3. Subtracting the equal parts from Step 1:
$$IE = BE - IB$$$$IF = CF - IC$$
$$\Rightarrow IE = IF$$
Converse — false
Converse: If $IE = IF$, then $AB = AC$.
Why it can fail: $AI$ bisects $\angle A$, so $\triangle AIF$ and $\triangle AIE$ share the side $AI$ and have equal angles at $A$. If also $IF = IE$, the two triangles have two sides and a non-included angle equal — which allows two shapes. Either the triangles are congruent (then $AB = AC$), or the angles at $F$ and $E$ add up to 180°.
Let $\angle B = 2\beta$ and $\angle C = 2\gamma$. Using exterior angles:
$$\angle AFI = \angle B + \tfrac{1}{2}\angle C = 2\beta + \gamma$$$$\angle AEI = \angle C + \tfrac{1}{2}\angle B = 2\gamma + \beta$$
Their sum is 180° when $3(\beta + \gamma) = 180°$, that is, when $\angle B + \angle C = 120°$, or $\angle A = 60°$.
Counterexample: a triangle with $\angle A = 60°$, $\angle B = 80°$, $\angle C = 40°$. Here $IE = IF$ (measuring a careful drawing, or calculating, confirms this), but $\angle B \neq \angle C$, so $AB \neq AC$.
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 — true. Adding the same number $a$ to two equal numbers gives equal results.
Converse: If $a + x = a + y$, then $x = y$.
Converse — true. Subtract $a$ from both sides:
$$a + x - a = a + y - a$$$$\Rightarrow x = y$$
Both statements are true. (This is why we may add or subtract the same number on both sides of an equation.)
5. If a and b are perfect squares, then ab is a perfect square. #
Solution
Proposition — true. Let $a = m^2$ and $b = n^2$. Then:
$$ab = m^2n^2$$$$ab = (mn)^2$$
which is a perfect square.
Converse: If $ab$ is a perfect square, then $a$ and $b$ are perfect squares.
Converse — false. Counterexample: $a = 2$, $b = 8$. Then $ab = 16 = 4^2$ is a perfect square, but neither 2 nor 8 is a perfect square.
6. If x = y, then x² = y². (x and y are real numbers.) #
Solution
Proposition — true. Squaring equal numbers gives equal results.
Converse: If $x^2 = y^2$, then $x = y$.
Converse — false. Counterexample: $x = 3$, $y = -3$. Then $x^2 = 9 = y^2$, but $x \neq y$. (In fact $x^2 = y^2$ only gives $x = y$ or $x = -y$.)
7. If x = y, then x³ = y³. (x and y are real numbers.) #
Solution
Proposition — true. Cubing equal numbers gives equal results.
Converse: If $x^3 = y^3$, then $x = y$.
Converse — true. Suppose $x^3 = y^3$. Then:
$$x^3 - y^3 = 0$$$$\Rightarrow (x - y)(x^2 + xy + y^2) = 0$$
The second factor can be written as
$$x^2 + xy + y^2 = \left(x + \dfrac{y}{2}\right)^2 + \dfrac{3}{4}y^2$$which is a sum of squares, so it is 0 only when $x = y = 0$. In every other case it is positive, so $x - y = 0$. Either way, $x = y$.
Both statements are true. (Unlike squares, a negative number has a negative cube, so cubes never “hide” the sign.)
8. If n is divisible by 24, then it is divisible by both 4 and 6. (n is a positive integer.) #
Solution
Proposition — true. If $n = 24k$, then $n = 4 \times 6k$ and $n = 6 \times 4k$, so $n$ is divisible by 4 and by 6.
Converse: If $n$ is divisible by both 4 and 6, then it is divisible by 24.
Converse — false. Counterexample: $n = 12$ is divisible by 4 and by 6, but not by 24. (4 and 6 share the factor 2, so a number divisible by both need only be divisible by their LCM, 12.)
9. If n is divisible by 60, then it is divisible by both 5 and 12. #
Solution
Proposition — true. If $n = 60k$, then $n = 5 \times 12k$ and $n = 12 \times 5k$.
Converse: If $n$ is divisible by both 5 and 12, then it is divisible by 60.
Converse — true. Let $n = 12k$. Since 5 divides $12k$ and 5 has no common factor with 12, 5 must divide $k$. So $k = 5m$ and:
$$n = 12 \times 5m = 60m$$Both statements are true. (This works because 5 and 12 have no common factor, so their LCM is $5 \times 12 = 60$. Compare Question 8.)
10. If n is the square of a prime number, then it has exactly 3 factors. #
Solution
Proposition — true. If $n = p^2$ with $p$ prime, its factors are $1$, $p$ and $p^2$ — exactly 3. (For example, $49$ has factors 1, 7, 49.)
Converse: If $n$ has exactly 3 factors, then it is the square of a prime number.
Converse — true. Factors come in pairs $(f, \dfrac{n}{f})$. An odd number of factors means one pair repeats, so $n = m^2$ is a perfect square, and its factors include $1$, $m$ and $m^2$. These are already 3, so there are no others. If $m$ had a factor $q$ other than 1 and $m$, then $q$ would also be a factor of $n$ — a fourth factor. So $m$ is prime.
Both statements are true.
11. If n is a product of two unequal prime numbers, then it has exactly 4 divisors. #
Solution
Proposition — true. If $n = pq$ with $p \neq q$ prime, its divisors are $1$, $p$, $q$ and $pq$ — exactly 4. (For example, $15$: 1, 3, 5, 15.)
Converse: If $n$ has exactly 4 divisors, then it is a product of two unequal prime numbers.
Converse — false. Counterexample: $n = 8$ has exactly 4 divisors (1, 2, 4, 8), but $8 = 2^3$ is not a product of two unequal primes. (Every cube of a prime has 4 divisors.)
12. If n and n + 3 have no factors in common, then n is not a multiple of 3. #
Here “no factors in common” means no common factor other than 1.
Solution
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: If $n$ is not a multiple of 3, then $n$ and $n + 3$ have no factors in common.
Converse — true. Any common factor $d$ of $n$ and $n + 3$ also divides their difference:
$$(n + 3) - n = 3$$So $d = 1$ or $d = 3$. But $d = 3$ would mean 3 divides $n$, which is not the case. So $d = 1$.
Both statements are true.
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.
For $n = 1, 2, 3$ we get 5, 17, 37, which are prime, but:
$$n = 4: \quad 4(16) + 1 = 65 = 5 \times 13$$∴ $n = 4$ gives 65, which is not prime.
(ii) All numbers of the form n² + n + 11 are prime.
For $n = 1$ to $9$ the values 13, 17, 23, 31, 41, 53, 67, 83, 101 are all prime, but:
$$n = 10: \quad 100 + 10 + 11 = 121 = 11 \times 11$$∴ $n = 10$ gives 121, which is not prime.
(iii) All numbers of the form 4ⁿ + 3 are prime.
For $n = 1, 2, 3$ we get 7, 19, 67, which are prime, but:
$$n = 4: \quad 256 + 3 = 259 = 7 \times 37$$∴ $n = 4$ gives 259, which is not prime.
14. Find counterexamples to the following statements. #
(i) If n is a prime number, then 2ⁿ − 1 is a prime number.
For $n = 2, 3, 5, 7$ the values 3, 7, 31, 127 are prime, but:
$$n = 11: \quad 2^{11} - 1 = 2047 = 23 \times 89$$∴ $n = 11$ is a counterexample.
(ii) If n is an even number, then 2ⁿ + 1 is a prime number.
For $n = 2$ and $4$ we get 5 and 17, which are prime, but:
$$n = 6: \quad 2^6 + 1 = 65 = 5 \times 13$$∴ $n = 6$ is a counterexample.
15. Consider the statement: ‘If a number is divisible by 8, then it is divisible by both 2 and 4’. #
(i) Justify the statement.
If $n = 8k$, then:
$$n = 2 \times 4k \quad \text{and} \quad n = 4 \times 2k$$So $n$ is divisible by 2 and by 4.
(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?
No. That would be using 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 not by 8.
(Being divisible by 4 already makes a number divisible by 2, so checking 2 adds nothing. To test for 8, check whether the number formed by the last three digits is divisible by 8.)
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 ‘sum of the digits is a multiple of 3’ using ‘If-then’ sentences. #
Solution
The relationship works in both directions:
- If a number is divisible by 3, then the sum of its digits is a multiple of 3.
- 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. We can say both at once: a number is divisible by 3 if and only if the sum of its digits is a multiple of 3.
17. We have identified different types of quadrilaterals — squares, rectangles, parallelograms, rhombi, kites and trapezia. 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 Fig. 9.2). #
(i) Suppose Q satisfies: ‘If a quadrilateral is of type Q, then it has equal-length diagonals.’ (a) Should the two sticks be of equal length? (b) Will it matter how the two sticks are put together?
(a) Yes. Every quadrilateral of type Q has equal diagonals, so the two sticks (the diagonals) must be of equal length. Unequal sticks can never give a type-Q quadrilateral.
(b) Yes, it may matter. This property only says that equal diagonals are necessary; it does not say they are enough. Not every quadrilateral with equal diagonals need be of type Q. For example, if Q is “rectangle”, the equal sticks must also bisect each other; if Q is “square”, they must bisect each other at right angles. So we may have to place the sticks in a particular way.
(ii) Instead, suppose Q satisfies: ‘If a quadrilateral has equal diagonals, then it is of type Q.’ What will be your answers to (a) and (b) now?
This is the converse of the property in (i).
(a) Using two sticks of equal length is enough to guarantee type Q. But it is not required: this property does not rule out type-Q quadrilaterals with unequal diagonals. So equal sticks are a sure way, though not the only possible way.
(b) No, it does not matter. If the sticks are equal, every quadrilateral formed by joining their endpoints has equal diagonals, and so is of type Q — however the sticks are crossed.