CBSE Class 9 Maths Ganita Manjari Part 2 Ch 09 Propositions and their Converses Long Answer and Case Study Questions with Answers

Long Answer Questions

Question 1. For each of the following propositions, write the converse, and then determine whether each of the two statements is true. Justify every true statement and give a counterexample for every false one.
(i) If a quadrilateral is a square, then its diagonals are equal.
(ii) If a positive integer n is divisible by 25, then n is divisible by 5.
(iii) If two triangles have equal perimeters, then they are congruent.
Answer:
(i) Converse: If the diagonals of a quadrilateral are equal, then it is a square.
Proposition true: each diagonal splits the square into congruent right triangles with the same legs, so the diagonals are equal.
Converse false: a 6 cm × 4 cm rectangle has both diagonals \( \sqrt{52} \) cm but is not a square.
(ii) Converse: If n is divisible by 5, then n is divisible by 25.
Proposition true: 25k = 5 × 5k. Converse false: 10.
(iii) Converse: If two triangles are congruent, then they have equal perimeters.
Proposition false: triangles 3-4-5 and 4-4-4 both have perimeter 12 cm but are not congruent.
Converse true: congruent triangles have equal corresponding sides, so equal sums.
In simple words: In (i) and (ii) only the original is true; in (iii) only the converse is true.

Teacher's Note:
1. Set out each part as: converse, verdict, reason.
2. Part (iii) shows the converse can be the true one.

Question 2. Let n be a positive integer. Prove that n has an odd number of factors exactly when n is a perfect square. Present your answer as two separate arguments, one for each direction, and state clearly which statement each argument proves.
Answer:
P: If n is a perfect square, then n has an odd number of factors.
Q: If n has an odd number of factors, then n is a perfect square.
Key idea: pair each factor d with its partner n ÷ d.
Proof of P. Let n = \( f^2 \). A pair has equal entries exactly when \( d^2 = n \), i.e. d = f (a positive number has one positive square root). So there is exactly one repeated pair (f, f). All other factors form pairs of two different numbers, giving an even count. Total = even + 1 = odd.
Proof of Q. Let n have an odd number of factors. If every pair had two different numbers, the count would be even, a contradiction. So some pair repeats: (f, f), giving n = f × f, a perfect square.
Since P and Q are both true, the two conditions hold in exactly the same cases.
In simple words: Factors come in pairs; only a perfect square has one pair that is a single repeated number, which makes the count odd.

Teacher's Note:
1. Label clearly which argument proves which direction.
2. Use 36 (9 factors) and 12 (6 factors) as quick checks.

Question 3. State the Baudhāyana–Pythagoras theorem and write its converse. Prove the converse, drawing the figures you use and naming the congruence criterion you apply.
Answer:
Theorem: If a triangle with sides a, b, c is right-angled, with the right angle between a and b, then \( a^2 + b^2 = c^2 \).
Converse: If \( a^2 + b^2 = c^2 \), then the triangle is right-angled, with the right angle between the sides a and b.
Proof. Let △ABC have BC = a, CA = b, AB = c, with \( a^2 + b^2 = c^2 \).
Construct △XYZ with a right angle at Z, ZY = a and ZX = b.
By the theorem, \( XY^2 = a^2 + b^2 = c^2 \), so XY = c.
Now BC = YZ, CA = ZX and AB = XY, so △ABC ≅ △XYZ by the SSS criterion.
Hence ∠C = ∠Z = 90°, and △ABC is right-angled at C.
In simple words: Build a right triangle with the same two shorter sides; its third side matches, so the two triangles are identical and the angle is 90°.

Teacher's Note:
1. The proof uses the theorem itself on a new triangle.
2. Students should draw △ABC and △XYZ side by side.

Question 4. In △ABC, AB = AC. The bisector of ∠B meets AC at E and the bisector of ∠C meets AB at F, as shown.
Prove that BE = CF. Then write the converse of the proposition you have proved, and comment on whether you would expect the converse to be true.
Answer:
AB = AC, so ∠ABC = ∠ACB. Call each 2β.
BE bisects ∠B, so ∠EBC = β; CF bisects ∠C, so ∠FCB = β.
In △EBC and △FCB: ∠EBC = ∠FCB = β, BC = CB (common), ∠ECB = ∠FBC = 2β.
So △EBC ≅ △FCB by ASA, and hence BE = CF (corresponding sides).
Converse: In △ABC, if the bisectors BE and CF are equal, then AB = AC.
This converse is true (the Steiner–Lehmus theorem), but it is much harder to prove than the original; known proofs work indirectly, by contradiction.
In simple words: Equal base angles give two congruent triangles, so the bisectors are equal; the reverse is also true but hard to prove.

Teacher's Note:
1. Mark the equal angles β and 2β on the figure.
2. A converse can be true yet far harder to prove.

Question 5. Examine the statement 'If a number is divisible by 8, then it is divisible by both 2 and 4.'
(i) Prove the statement.
(ii) Write its converse and show by a counterexample that the converse is false.
(iii) Hence explain why testing for divisibility by 2 and by 4 is not a valid test for divisibility by 8, while testing for divisibility by 3 and by 4 is a valid test for divisibility by 12.
Answer:
(i) Let n = 8k. Then n = 2 × 4k and n = 4 × 2k, so n is divisible by 2 and by 4.
(ii) Converse: If a number is divisible by both 2 and 4, then it is divisible by 8.
20 = 2 × 10 = 4 × 5, but 20 ÷ 8 = 2.5. So the converse is false.
(iii) A test must work in both directions. For 8 the converse fails, so the test would wrongly pass 20.
2 and 4 share the factor 2; together they only guarantee divisibility by their LCM, 4.
3 and 4 have no common factor, so LCM(3, 4) = 12. Both directions hold, and checking 3 and 4 is a valid test for 12.
In simple words: Splitting a divisor into two parts works only when the parts share no factor: 12 = 3 × 4 works, 8 = 2 × 4 does not.

Teacher's Note:
1. The rule: split only into coprime factors.
2. Use 20 as the standard counterexample for the 8-test.

Question 6. Write each of the following in 'If … then …' form, write its converse, and decide whether each of the two statements is true.
(i) A number is a multiple of 6 when it is a multiple of 12.
(ii) A triangle is equilateral when all its angles are equal.
(iii) A quadrilateral is a rectangle when all its angles are right angles.
Answer:
(i) If a number is a multiple of 12, then it is a multiple of 6. True (12k = 6 × 2k).
Converse: If a number is a multiple of 6, then it is a multiple of 12. False (6).
(ii) If all the angles of a triangle are equal, then it is equilateral. True: equal angles make opposite sides equal.
Converse: If a triangle is equilateral, then all its angles are equal. True: each angle is 60°.
(iii) If all the angles of a quadrilateral are right angles, then it is a rectangle. True (this is the definition).
Converse: If a quadrilateral is a rectangle, then all its angles are right angles. True.
In (ii) and (iii) both directions hold, so each is a characterisation of the figure.
In simple words: Only (i) has a false converse; (ii) and (iii) work both ways.

Teacher's Note:
1. 'When' introduces the hypothesis.
2. A two-way true pair is called a characterisation.

Question 7. A student writes the following argument about a positive integer n.
Suppose n has an odd number of factors. Pair each factor f with its partner factor n ÷ f. If every such pair contained two different numbers, then n would have an even number of factors. So at least one pair must contain the same number twice, say (f, f), which gives n = f × f. Hence n is a perfect square.
(i) State precisely which one of the two statements — 'If n is a perfect square, then n has an odd number of factors' or its converse — this argument proves.
(ii) Explain why the argument does not prove the other statement.
(iii) Supply a correct argument for the other statement.
Answer:
(i) It proves the converse: 'If n has an odd number of factors, then n is a perfect square.' It starts from 'odd number of factors' and ends at 'perfect square'.
(ii) An argument cannot just be read backwards. Starting from n = f × f gives one repeated pair, but the argument never shows there is only one. Two repeated pairs would make the count even, and this is not ruled out.
(iii) Let n = \( f^2 \). A pair (d, n ÷ d) has equal entries exactly when \( d^2 = n \), i.e. d = f, the only positive square root. So there is exactly one repeated pair.
All other factors form pairs of two different numbers, giving an even count. Total = even + 1 = odd.
In simple words: The given argument runs from 'odd factors' to 'square'; the reverse needs its own proof that there is exactly one repeated pair.

Teacher's Note:
1. The missing idea is uniqueness of the square root.
2. Stress: hypothesis at the start, conclusion at the end.

Question 8. Let x and y be real numbers. Examine each of the following propositions together with its converse, deciding whether each statement is true and giving a counterexample wherever one is needed.
(i) If x = y, then \( x^2 = y^2 \).
(ii) If x = y, then \( x^3 = y^3 \).
(iii) If x = y, then 5x − 2 = 5y − 2.
Then explain why the converse in (i) behaves differently from the converses in (ii) and (iii).
Answer:
(i) Proposition true. Converse 'if \( x^2 = y^2 \) then x = y' false: x = 3, y = −3.
(ii) Proposition true. Converse true: cubing keeps order, so different numbers have different cubes.
(iii) Proposition true. Converse true: add 2 and divide by 5 to get x = y.
Why (i) differs: a step can be reversed only if it never sends two different numbers to the same value. Squaring sends 3 and −3 both to 9, losing the sign. Cubing and x ↦ 5x − 2 never do this, so they can be undone.
In simple words: Squaring hides the sign, so you can't go back; cubing and 5x − 2 can always be undone.

Teacher's Note:
1. Link this to 'reversible steps' when solving equations.
2. 3 and −3 is the standard counterexample.

Question 9. Find a counterexample to each of the following claims about positive integers, showing the full working that justifies your choice.
(i) Every number of the form \( n^2 + n + 17 \) is prime.
(ii) For every odd number n greater than 1, the number \( 2^n - 1 \) is prime.
(iii) Every number of the form \( 6n^2 + 5 \) is prime.
Answer:
(i) For n = 1 to 15 the values (19, 23, 29, …, 257) are all prime. At n = 16: \( 256 + 16 + 17 = 289 = 17 \times 17 \), composite.
Reason: \( n(n + 1) + 17 \) at n = 16 is \( 16 \times 17 + 17 = 17^2 \).
(ii) n = 3, 5, 7 give 7, 31, 127 (prime). At n = 9: \( 2^9 - 1 = 511 = 7 \times 73 \), composite.
(iii) n = 1 to 4 give 11, 29, 59, 101 (prime). At n = 5: \( 6 \times 25 + 5 = 155 = 5 \times 31 \), composite.
In simple words: Each formula looks good for a while and then produces a composite number.

Teacher's Note:
1. Part (i) survives 15 tests and still fails; testing is not proof.
2. Show the factorisation for each counterexample.

Question 10. In the figure, the lines l and m are cut by the transversal t, and the eight angles formed are numbered 1 to 8.
(i) State the proposition that relates parallel lines to equal corresponding angles, and state its converse.
(ii) It is measured that ∠1 = 62° and ∠5 = 62°. What can be concluded, and which of the two statements is used?
(iii) It is given instead that l ∥ m and ∠1 = 118°. Find ∠5, ∠3 and ∠8, naming the statement you use each time.
Answer:
(i) Proposition: If two lines are parallel, then the corresponding angles formed by a transversal are equal.
Converse: If the corresponding angles formed by a transversal are equal, then the two lines are parallel.
(ii) ∠1 and ∠5 are corresponding angles (above the line, left of t, at each crossing). They are equal, so by the converse, l ∥ m. The proposition cannot be used, because it assumes the lines are parallel.
(iii) ∠5 = ∠1 = 118° (corresponding angles; the proposition, since l ∥ m).
∠3 = 180° − 118° = 62° (∠1 and ∠3 form a linear pair).
∠8 = ∠5 = 118° (vertically opposite angles).
In simple words: Equal measured angles prove the lines parallel; once they are parallel, known angles let us find the rest.

Teacher's Note:
1. Ask: is parallelism given, or to be proved?
2. Name the angle relationship at every step.

Question 11. A category Q of quadrilaterals is to be built by placing two thin sticks as the diagonals and joining their four endpoints, as shown in the two arrangements below.
(i) Suppose Q satisfies the property 'If a quadrilateral is of type Q, then it has equal diagonals.' Must the two sticks be of equal length? Give a reason.
(ii) For the property in (i), does it matter how the two sticks are placed against each other? Explain.
(iii) Suppose instead that Q satisfies the property 'If a quadrilateral has equal diagonals, then it is of type Q.' Answer (i) and (ii) again, and explain what difference the change of direction has made.
Answer:
The sticks are the diagonals, so the diagonals always have the two stick lengths.
(i) Yes. Every type-Q figure must have equal diagonals, so the sticks must be equal. Equal sticks are necessary.
(ii) The property cannot tell us, and placement may matter. It says what every type-Q figure must have, not what is enough to make one. Type Q might need more, such as diagonals that bisect each other, which depends on placement.
(iii) Now the sticks need not be equal: unequal-diagonal figures could also be type Q. And placement does not matter: any two equal sticks, however crossed, give equal diagonals and so a type-Q figure.
Difference: in (i)–(ii) the property is a test (a necessary condition: it can reject figures). In (iii) it is a recipe (a sufficient condition: it can build figures). Reversing the arrow turns 'necessary' into 'sufficient'.
In simple words: One direction tells you what every Q-shape must have; the other tells you what is enough to make one.

Teacher's Note:
1. Necessary = test; sufficient = recipe.
2. Link 'if Q then P' to 'P is necessary for Q'.

Question 12. Prove that in a triangle, if two sides are equal then the angles opposite those sides are equal. Then state the converse of this proposition and outline its proof. Draw a clearly labelled figure for each of the two proofs.
Answer:
Proposition. Let △ABC have AB = AC. Draw the bisector of ∠A meeting BC at D.
In △ABD and △ACD: AB = AC (given), ∠BAD = ∠CAD (bisector), AD = AD (common).
So △ABD ≅ △ACD (SAS), giving ∠B = ∠C.
Converse. If two angles of a triangle are equal, then the sides opposite them are equal.
Outline. Let △PQR have ∠Q = ∠R. Draw the bisector PS of ∠P. In △PQS and △PRS: ∠Q = ∠R, ∠QPS = ∠RPS, PS = PS.
So △PQS ≅ △PRS (AAS), giving PQ = PR.
Both are true: a triangle has two equal sides exactly when it has two equal angles.
In simple words: Draw the angle bisector from the top vertex; it splits the triangle into two congruent halves in both proofs.

Teacher's Note:
1. Figures: an isosceles triangle with the bisector from the apex.
2. Note SAS for one proof and AAS for the other.

Question 13. A proposition and its converse can be true or false in four possible combinations. List all four combinations, and for each one give an example from geometry or from number theory, justifying the two truth values briefly in every case.
Answer:
1. Both true. 'Parallelogram ⇒ diagonals bisect each other' and 'diagonals bisect each other ⇒ parallelogram'. Both are standard results.
2. Proposition true, converse false. 'Square ⇒ equal diagonals' (true); 'equal diagonals ⇒ square' (false: a 9 cm × 5 cm rectangle).
3. Proposition false, converse true. 'Multiple of 7 ⇒ multiple of 14' (false: 7); 'multiple of 14 ⇒ multiple of 7' (true: 14k = 7 × 2k).
4. Both false. 'Prime ⇒ odd' (false: 2); 'odd ⇒ prime' (false: 9).
So the truth of one tells us nothing about the other.
In simple words: All four true/false pairings really happen, so each direction must be checked separately.

Teacher's Note:
1. Make students build their own table of four examples.
2. Case 3 surprises many students.

Question 14. The divisibility shortcut for 3 says that a number is divisible by 3 exactly when the sum of its digits is a multiple of 3.
(i) Express this shortcut as two 'If … then …' sentences.
(ii) Verify both of these sentences for the number 4587.
(iii) Explain why the shortcut can be used as a test only because both of the sentences in (i) are true.
Answer:
(i) 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.
(ii) Digit sum: 4 + 5 + 8 + 7 = 24 = 3 × 8. And 4587 = 3 × 1529. Both conditions hold, so both sentences are confirmed.
(iii) When the digit sum passes, the second sentence says the number is divisible by 3. When it fails, the first sentence says the number cannot be divisible by 3 (else the sum would be a multiple of 3). If only one sentence were true, the test would give wrong verdicts in one of the two situations.
In simple words: The test gives the right answer both when it says 'yes' and when it says 'no' only because both directions are true.

Teacher's Note:
1. A reliable test needs both directions.
2. Try a failing number too, e.g. 4588 (digit sum 25).

Question 15. For each of the following propositions about a positive integer n, write the converse and decide whether the converse is true, giving a counterexample where it is false.
(i) If n is divisible by 45, then n is divisible by both 9 and 5.
(ii) If n is divisible by 36, then n is divisible by both 4 and 9.
(iii) If n is divisible by 36, then n is divisible by both 6 and 12.
Then state the general rule that decides when such a converse is true.
Answer:
(i) If n is divisible by 9 and 5, then n is divisible by 45. True: 9 and 5 are coprime, 9 × 5 = 45.
(ii) If n is divisible by 4 and 9, then n is divisible by 36. True: 4 and 9 are coprime, 4 × 9 = 36.
(iii) If n is divisible by 6 and 12, then n is divisible by 36. False: 12 is divisible by 6 and 12 but not by 36.
General rule: the converse is true exactly when the LCM of the two divisors equals the number in the conclusion. LCM(9, 5) = 45 ✓, LCM(4, 9) = 36 ✓, LCM(6, 12) = 12 ≠ 36 ✗.
In simple words: Two divisors together guarantee only their LCM; the converse works when that LCM is the target number.

Teacher's Note:
1. The LCM rule covers all such divisibility converses.
2. Coprime divisors: LCM = product.

Question 16. For each of the following propositions about a quadrilateral ABCD, write the converse and decide whether the converse is true, giving the reason.
(i) If ABCD is a parallelogram, then its opposite angles are equal.
(ii) If ABCD is a rhombus, then its diagonals bisect each other at right angles.
(iii) If ABCD is a square, then its diagonals are equal and bisect each other at right angles.
Explain what it means to call a statement of this kind a characterisation of the figure.
Answer:
(i) If the opposite angles of ABCD are equal, then it is a parallelogram. True: 2∠A + 2∠B = 360° gives ∠A + ∠B = 180°, so AD ∥ BC; similarly AB ∥ DC.
(ii) If the diagonals bisect each other at right angles, then ABCD is a rhombus. True: bisecting diagonals give a parallelogram, and right angles make the four small triangles congruent, so all sides are equal.
(iii) If the diagonals are equal and bisect each other at right angles, then ABCD is a square. True: by (ii) it is a rhombus, and equal bisecting diagonals also make it a rectangle.
Characterisation: when a statement and its converse are both true, the property holds for that figure and no other. It can then serve as an alternative definition: checking the property identifies the shape.
In simple words: All three converses are true, so each property is enough to recognise its shape.

Teacher's Note:
1. A characterisation = a property that works both ways.
2. Useful in proofs to identify special quadrilaterals.

Question 17. Let m and n be positive integers.
(i) Show by a counterexample that 'If mn is a perfect square, then both m and n are perfect squares' is false.
(ii) Prove that the converse of the statement in (i) is true.
(iii) Give one extra condition on m and n under which the statement in (i) does become true, and justify your answer briefly.
Answer:
(i) m = 2, n = 8: mn = 16 = \( 4^2 \), but neither 2 nor 8 is a perfect square. So the statement is false.
(ii) Converse: if m = \( p^2 \) and n = \( q^2 \), then mn = \( p^2 q^2 = (pq)^2 \), a perfect square. True.
(iii) Extra condition: m and n have no common factor other than 1.
In a perfect square every prime appears an even number of times. If m and n share no prime, every copy of a prime p in mn comes from m alone (or n alone). So each prime appears an even number of times in m and in n, making both perfect squares.
(In (i), 2 and 8 share the factor 2, so the condition fails there.)
In simple words: 2 × 8 = 16 breaks the claim, but if m and n share no factor, a square product forces both to be squares.

Teacher's Note:
1. Coprimality again repairs a false statement.
2. Prime-counting argument: even powers make squares.

Question 18. Three students discuss the statement 'If two triangles have the same area, then they are congruent.'
Asha: The statement is true, because congruent triangles always have equal areas.
Bharat: The statement is false, and I can draw two triangles of equal area that are not congruent.
Chetna: Whatever the truth of the statement, its converse is certainly true.
(i) Whose reasoning is faulty, and what exactly is the mistake?
(ii) Give the counterexample that Bharat has in mind, with full working and a labelled figure.
(iii) Prove Chetna's claim.
Answer:
(i) Asha's reasoning is faulty. Her reason, 'congruent ⇒ equal areas', is the converse of the statement being discussed. A true converse is not evidence for the original.
(ii) △LMN right-angled at L with LM = 6 cm, LN = 4 cm: area = \( \frac{1}{2} \times 6 \times 4 = 12 \) cm².
△UVW right-angled at U with UV = 8 cm, UW = 3 cm: area = \( \frac{1}{2} \times 8 \times 3 = 12 \) cm².
Hypotenuses: MN = \( \sqrt{52} \) cm, VW = \( \sqrt{73} \) cm, which are different, so the triangles are not congruent. Bharat is right.
(iii) If △ABC ≅ △DEF, then AB = DE, BC = EF, CA = FD. By Heron's formula the area depends only on the three sides, so the areas are equal. Chetna is right.
In simple words: Equal area does not force congruence (Bharat), but congruence does force equal area (Chetna); Asha mixed the two up.

Teacher's Note:
1. A good discussion task on direction of statements.
2. Heron's formula shows area depends only on the sides.

Question 19. Using the converse of the Baudhāyana–Pythagoras theorem, determine which of the following triangles are right-angled, showing all the arithmetic.
(i) Sides 7 cm, 24 cm and 25 cm.
(ii) Sides 8 cm, 15 cm and 16 cm.
(iii) Sides 20 cm, 21 cm and 29 cm.
Then state what the theorem itself, rather than its converse, allows you to conclude about the triangle in (iii).
Answer:
(i) \( 7^2 + 24^2 = 49 + 576 = 625 = 25^2 \). Right-angled (between 7 cm and 24 cm).
(ii) \( 8^2 + 15^2 = 64 + 225 = 289 \), \( 16^2 = 256 \). 289 ≠ 256, not right-angled.
(iii) \( 20^2 + 21^2 = 400 + 441 = 841 = 29^2 \). Right-angled (between 20 cm and 21 cm).
The theorem itself runs the other way: if we were told the triangle is right-angled with legs 20 cm and 21 cm, it would predict the hypotenuse \( \sqrt{841} = 29 \) cm. The theorem calculates a side; the converse tests for a right angle.
In simple words: Two of the three triangles pass the \( a^2 + b^2 = c^2 \) test; the theorem would instead be used to find a missing side.

Teacher's Note:
1. Always test against the longest side.
2. Note 8-15-17 is right-angled; 8-15-16 is not.

Question 20. Write a short account of the part played by counterexamples in mathematics. Your account must include
(i) what a counterexample is;
(ii) Fermat's claim about numbers of the form \( 2^{2^n} + 1 \) and the way in which it was refuted;
(iii) a counterexample of your own to a false converse taken from geometry;
(iv) an explanation of why counterexamples can never be used to prove a statement true.
Answer:
(i) 'If X then Y' claims Y in every case where X holds. A counterexample is one case where X holds and Y fails. One such case is enough to make the claim false.
(ii) Fermat claimed every \( 2^{2^n} + 1 \) is prime. n = 0 to 4 give 3, 5, 17, 257, 65537, all prime. Euler took n = 5: \( 2^{32} + 1 = 4294967297 = 641 \times 6700417 \), composite. One failure outweighed five successes.
(iii) 'Rhombus ⇒ perpendicular diagonals' is true, but the converse is false: the kite with AB = AD = 3 cm, CB = CD = 7 cm has perpendicular diagonals but unequal sides.
(iv) A general claim covers infinitely many cases; no finite set of examples can check them all. \( n^2 + n + 17 \) is prime for n = 1 to 15 but equals \( 17^2 \) at n = 16. Only a proof that covers every case can establish a claim.
In simple words: Counterexamples destroy false claims in one stroke, but examples alone can never prove a claim true.

Teacher's Note:
1. Organise the answer under the four given headings.
2. The n² + n + 17 example is powerful for part (iv).

Case Study Based Questions

Question 21. The Proposition Board: The mathematics club of a school puts up a board carrying these four sentences.
S1: Every multiple of 8 is a multiple of 4.
S2: Draw a triangle with sides 6 cm, 8 cm and 10 cm.
S3: Every multiple of 4 is a multiple of 8.
S4: 91 is a prime number.
(a) Which of the four sentences are propositions?
(b) Which one of S1 and S3 is the converse of the other?
(c) Which one of S1 and S3 is false? Give a counterexample.
(d) Is S4 true or false? Justify your answer.
Answer:
(a) S1, S3 and S4 are propositions. S2 is an instruction, which cannot be true or false.
(b) Each is the converse of the other: 'multiple of 8 ⇒ multiple of 4' and 'multiple of 4 ⇒ multiple of 8'.
(c) S3 is false: 20 = 4 × 5, but 20 is not a multiple of 8. (S1 is true: 8k = 4 × 2k.)
(d) False: 91 = 7 × 13, so it is composite.
In simple words: Three sentences are claims, S1 and S3 are converses, S3 fails at 20, and 91 is not prime.

Teacher's Note:
1. 91 is a common 'looks prime' trap.
2. Converses come in pairs: each is the converse of the other.

Question 22. The Carpenter's Name-plates: A carpenter makes quadrilateral name-plates. He first fixes two thin metal strips so that they cross each other and act as the diagonals of the plate, and then joins their four endpoints with wood.
(a) If the carpenter uses two strips of equal length, will every plate he makes have equal diagonals? Give a reason.
(b) Write the proposition 'If the plate is a rectangle, then its diagonals are equal' and write its converse.
(c) Is the converse in part (b) true? Support your answer with a counterexample if it is false.
(d) The carpenter now wants every plate to be a rectangle. State the condition the two strips must satisfy.
Answer:
(a) Yes. The strips are the diagonals, so equal strips give equal diagonals, however they are crossed.
(b) Proposition: If the plate is a rectangle, then its diagonals are equal. Converse: If the diagonals of the plate are equal, then it is a rectangle.
(c) False. An isosceles trapezium with parallel sides 10 cm and 6 cm has equal diagonals but no right angle.
(d) The strips must be equal in length and cross at their midpoints (bisect each other). Equal diagonals that bisect each other always give a rectangle.
In simple words: Equal strips give equal diagonals, but only equal strips crossing at their middles guarantee a rectangle.

Teacher's Note:
1. Equal diagonals alone are not enough for a rectangle.
2. Hands-on: try this with two pencils of equal length.

Question 23. Checking a Right Angle on Site: To check whether a corner of a plot is a right angle, a mason marks a point 9 m from the corner along one edge and a point 12 m from the corner along the other edge. He then measures the distance between the two marks and finds it to be 15 m.
(a) Is the mason using the Baudhāyana–Pythagoras theorem or its converse? Give a reason.
(b) Carry out the mason's check and state the conclusion.
(c) At a second corner the same two measurements of 9 m and 12 m give a distance of 16 m between the marks. Is that corner a right angle? Justify your answer.
(d) Suppose the mason already knows that a corner is a right angle and wants to predict the distance between marks made 9 m and 12 m from it. Which statement does he now use?
Answer:
(a) The converse: he starts from three lengths and concludes something about the angle.
(b) \( 9^2 + 12^2 = 81 + 144 = 225 = 15^2 \). The corner is a right angle.
(c) \( 9^2 + 12^2 = 225 \), \( 16^2 = 256 \). 225 ≠ 256, so it is not a right angle.
(d) The theorem itself: it predicts the distance \( \sqrt{225} = 15 \) m.
In simple words: Measuring to test a right angle uses the converse; predicting a length from a known right angle uses the theorem.

Teacher's Note:
1. Real-life use: masons use 3-4-5 multiples like 9-12-15.
2. Ask 'what is given?' to choose theorem or converse.

Question 24. Designing a Divisibility Test: A software team writes a program that tests divisibility. To test whether a number is divisible by 12, the program checks whether it is divisible by 3 and by 4. To test whether a number is divisible by 8, the program checks whether it is divisible by 2 and by 4.
(a) Write, as an 'If … then …' sentence, the claim on which the divisibility-by-12 test depends.
(b) Is the converse of the sentence in part (a) true? Give a reason.
(c) Show by means of one number that the divisibility-by-8 test used by the program is wrong.
(d) Which property of the pair of divisors makes the first test work and the second test fail?
Answer:
(a) If a number is divisible by 3 and by 4, then it is divisible by 12.
(b) Yes, its converse 'If divisible by 12, then divisible by 3 and 4' is true: 12k = 3 × 4k = 4 × 3k.
(c) 20 passes both checks (20 = 2 × 10 = 4 × 5), but 20 ÷ 8 = 2.5.
(d) The divisors must be coprime (no common factor other than 1). 3 and 4 are coprime; 2 and 4 are not.
In simple words: The 12-test is safe because 3 and 4 share no factor; the 8-test fails because 2 and 4 do.

Teacher's Note:
1. The test rests on 'divisible by 3 and 4 ⇒ divisible by 12'.
2. Coprime is the key property.

Question 25. Factor-Partner Pairs: A teacher writes the factor-partner pairs of three numbers on the board.
12: (1, 12), (2, 6), (3, 4)
25: (1, 25), (5, 5)
36: (1, 36), (2, 18), (3, 12), (4, 9), (6, 6)
(a) How many factors does each of 12, 25 and 36 have?
(b) Which of the three numbers are perfect squares?
(c) Complete the sentence: 'If a number has an odd number of factors, then …'
(d) Is the converse of the sentence you completed in part (c) true? Give a reason.
Answer:
(a) A pair of different numbers gives 2 factors; a repeated pair gives 1.
12: 3 × 2 = 6. 25: 2 + 1 = 3. 36: 8 + 1 = 9.
(b) 25 = \( 5^2 \) and 36 = \( 6^2 \). 12 is not a square.
(c) '… then it is a perfect square.'
(d) Yes. If n = \( f^2 \), the pair (f, f) repeats and all other factors form pairs of two, so the count is even + 1 = odd.
In simple words: Only squares have a repeated pair, so only squares have an odd number of factors, and vice versa.

Teacher's Note:
1. Let students list pairs for 16 and 20 to check.
2. Both directions true: this is a characterisation.

Question 26. Markings on a Road: Two edges of a straight road are crossed by a pedestrian crossing. A surveyor measures the two marked angles and finds that each of them is 74°.
(a) Name the pair of angles that the surveyor has measured.
(b) Which statement — the proposition on parallel lines or its converse — allows the surveyor to conclude that the two edges are parallel?
(c) State the statement you named in part (b).
(d) On another road whose two edges are known to be parallel, one of a pair of corresponding angles measures 106°. Find the other angle of that pair.
Answer:
(a) Corresponding angles: each is in the same position at the two crossings.
(b) The converse, because he starts from equal angles and concludes the lines are parallel.
(c) If the corresponding angles formed by a transversal are equal, then the two lines are parallel.
(d) 106°. The edges are known to be parallel, so by the proposition corresponding angles are equal.
In simple words: Equal measured angles prove the edges parallel; known parallel edges give equal angles.

Teacher's Note:
1. Direction of reasoning decides which statement is used.
2. Real-world transversal: the crossing stripe.

Question 27. A Prime-Making Formula: A student works out the value of \( 2n^2 + 5 \) for n = 1, 2, 3 and 4 and obtains 7, 13, 23 and 37. All four values are prime, so the student concludes that every number of the form \( 2n^2 + 5 \) is prime.
(a) Find the value of \( 2n^2 + 5 \) when n = 5.
(b) Is the student's conclusion correct? Justify your answer.
(c) What name is given, in this chapter, to the value you found in part (a)?
(d) Name the famous claim about numbers of the form \( 2^{2^n} + 1 \) that was refuted in the same way, and say who refuted it.
Answer:
(a) \( 2 \times 25 + 5 = 55 \).
(b) No. 55 = 5 × 11 is composite, so the claim 'every such number is prime' fails.
(c) A counterexample.
(d) Fermat's claim that every \( 2^{2^n} + 1 \) is prime. Euler refuted it with n = 5: \( 2^{32} + 1 = 641 \times 6700417 \).
In simple words: Four primes in a row proved nothing; n = 5 gives 55, which is not prime.

Teacher's Note:
1. Connect the student's mistake to Fermat's.
2. Patterns suggest; only proofs establish.

Question 28. Two Triangles in a Design: A designer uses two triangles in a pattern. In △PQR, PQ = PR and ∠Q = 65°. In △XYZ, ∠Y = ∠Z = 58°.
(a) Find ∠R.
(b) Find ∠P.
(c) Which statement — 'the angles opposite two equal sides of a triangle are equal' or its converse — tells the designer that XY = XZ?
(d) State the converse that you named in part (c).
Answer:
(a) PQ = PR, so the angles opposite them are equal: ∠R = ∠Q = 65°.
(b) ∠P = 180° − 65° − 65° = 50°.
(c) The converse: we start from equal angles and conclude equal sides.
(d) If two angles of a triangle are equal, then the sides opposite them are equal. So XZ = XY.
In simple words: Equal sides gave equal angles in △PQR; equal angles give equal sides in △XYZ.

Teacher's Note:
1. Note the side opposite ∠Y is XZ, and opposite ∠Z is XY.
2. Same pair of results used in both directions.

Question 29. Four Statements about a Quadrilateral: A student writes four propositions about a quadrilateral ABCD.
P1: If ABCD is a square, then all its sides are equal.
P2: If all the sides of ABCD are equal, then ABCD is a square.
P3: If ABCD is a square, then all its angles are equal.
P4: If all the angles of ABCD are equal, then ABCD is a square.
(a) Which of P2 and P4 is the converse of P1, and which is the converse of P3?
(b) Which of P1 and P3 are true?
(c) Give a counterexample to P2.
(d) Give a counterexample to P4.
Answer:
(a) P2 is the converse of P1; P4 is the converse of P3.
(b) Both P1 and P3 are true: a square has four equal sides and four 90° angles.
(c) A rhombus of side 5 cm with angles 60° and 120°: equal sides, not a square.
(d) A 7 cm × 4 cm rectangle: equal angles, not a square.
In simple words: A square has equal sides and equal angles, but either one alone gives a rhombus or a rectangle.

Teacher's Note:
1. Rhombus and rectangle are the two 'half-squares'.
2. A square is exactly where the two meet.

Question 30. A Quiz Round on Converses: In the final round of a mathematics quiz, a team is shown the following pair of statements about a positive integer.
P: If a number is a perfect cube, then it is a perfect square.
Q: If a number is a perfect square, then it is a perfect cube.
(a) Is Q the converse of P? Give a reason.
(b) Give a counterexample to P.
(c) Give a counterexample to Q.
(d) What does this pair of statements show about the possible truth values of a proposition and its converse?
Answer:
(a) Yes. Q swaps the hypothesis ('perfect cube') and conclusion ('perfect square') of P.
(b) 8 = \( 2^3 \) is a cube, but it lies between \( 2^2 = 4 \) and \( 3^2 = 9 \), so it is not a square.
(c) 4 = \( 2^2 \) is a square, but it lies between \( 1^3 = 1 \) and \( 2^3 = 8 \), so it is not a cube.
(d) A proposition and its converse can both be false. Each must be tested on its own.
In simple words: Both P and Q fail, showing that a statement and its converse can be false together.

Teacher's Note:
1. Numbers that are both squares and cubes: 1, 64, 729 (sixth powers).
2. Completes the four-combinations picture.