1. Think and Reflect (Page 1): Prove Statement 2: "If two angles of a triangle are equal, then the sides opposite the equal angles have equal lengths." (Hint: Draw the altitude from the vertex containing the third angle.) 2. Example 5 (Page 4): Complete the proof of the converse of the Baudhāyana-Pythagoras theorem: "Let a, b, c be the sidelengths of a triangle. If a2 + b2 = c2, then the triangle is right-angled."
Left: Altitude AD in ΔABC • Right: Comparing ΔABC with constructed right triangle ΔXYZ
Part 1: Proof of Statement 2 (Converse of Isosceles Triangle Theorem)
Given: In ΔABC, ∠B = ∠C.Construction: Draw altitude AD ⊥ BC from vertex A to side BC.In right triangles ΔADB and ΔADC:
∠B = ∠C (Given), ∠ADB = ∠ADC = 90° (Construction), AD = AD (Common Side)
⇒ ΔADB ≅ ΔADC (by AAS Congruence Criterion) ⇒ AB = AC (by CPCT)
Part 2: Proof of Converse of Baudhāyana-Pythagoras Theorem (Example 5)
Given: In ΔABC, sidelengths BC = a, CA = b, AB = c satisfy a2 + b2 = c2.Step 1: Construct a right-angled triangle ΔXYZ with ∠Z = 90°, YZ = a, and XZ = b.By the Baudhāyana-Pythagoras theorem on ΔXYZ:
XY2 = YZ2 + XZ2 = a2 + b2
Step 2: Since a2 + b2 = c2 (given), we have XY2 = c2 ⇒ XY = c = AB.Comparing ΔABC and ΔXYZ:
BC = YZ = a, CA = XZ = b, AB = XY = c ⇒ ΔABC ≅ ΔXYZ (by SSS Criterion)
⇒ ∠C = ∠Z = 90° (Therefore, ΔABC is a right-angled triangle)
Part 3: Solutions to Examples 2, 4 & Think and Reflect (Pages 2–3)
Example 2 (Multiples of 6 and 3): Proposition P (If n is a multiple of 6, then n is a multiple of 3) is TRUE because n = 6k = 3(2k).Converse Q (If n is a multiple of 3, then n is a multiple of 6) is FALSE.Counterexample:n = 9 (or n = 3, 15) is a multiple of 3, but not a multiple of 6.Example 4 (Area and Congruence of Triangles): Proposition P (If two triangles have the same area, then they are congruent) is FALSE.Counterexample: A triangle with base 6 cm, height 4 cm (Area = 12 cm2) and a triangle with base 8 cm, height 3 cm (Area = 12 cm2) have the same area but are not congruent.Converse Q (If two triangles are congruent, then they have the same area) is TRUE because congruent figures can be superimposed exactly.Think & Reflect (Both Proposition and Converse False):Proposition P: "If a quadrilateral is a rectangle, then it is a rhombus." (FALSE — a 4 cm × 2 cm rectangle is not a rhombus).Converse Q: "If a quadrilateral is a rhombus, then it is a rectangle." (FALSE — a rhombus with 60° and 120° angles is not a rectangle).
Exercise Set 9.1 — Questions 1 to 5 (Pages 5–6)
Exercise 9.1 — Question 1
Parallel Lines & Corresponding Angles
Question 1: Frame the converse, determine if each of the two statements is true or not, justify the true statements, and give a counterexample for each false statement: "If two lines are parallel, then the corresponding angles formed by a transversal are equal."
Parallel lines l || m cut by transversal t with corresponding angles ∠1 = ∠2
Step 1: Frame the Converse Statement
Converse: "If the corresponding angles formed by a transversal with two lines are equal, then the two lines are parallel."
Step 2: Truth Value & Justification of Original Proposition
Original Proposition: TRUE
Justification: By the Corresponding Angles Axiom (and parallel line properties studied in Euclidean geometry), whenever a transversal intersects two parallel lines l || m, the corresponding angles make the exact same inclination with the transversal and are therefore equal (∠1 = ∠2).
Step 3: Truth Value & Justification of Converse Statement
Converse Statement: TRUE
Justification: Suppose corresponding angles ∠1 = ∠2, but lines l and m intersect at some point P to form a triangle with the transversal.Then ∠1 would be an exterior angle of that triangle and ∠2 would be a non-adjacent interior angle, forcing ∠1 > ∠2 (by the Exterior Angle Theorem), which contradicts ∠1 = ∠2.Hence lines l and m cannot intersect and must be parallel (l || m).
Exercise 9.1 — Question 2
Squares & Equal Angles
Question 2:"If a quadrilateral is a square, then all its angles are equal."
A rectangle with unequal adjacent sides has all four angles equal to 90° but is not a square
Step 1: Frame the Converse Statement
Converse: "If all the angles of a quadrilateral are equal, then the quadrilateral is a square."
Step 2: Truth Value & Justification of Original Proposition
Original Proposition: TRUE
Justification: By definition, a square is a quadrilateral with all four sides equal and each of its four interior angles equal to a right angle (∠A = ∠B = ∠C = ∠D = 90°). Therefore, all its angles are equal.
Step 3: Truth Value & Counterexample for Converse Statement
Converse Statement: FALSE
Counterexample: Consider a rectangle with length AB = CD = 6 cm and breadth BC = DA = 3 cm.All four of its angles are equal (each is 90°), but since its adjacent sides are unequal (6 cm ≠ 3 cm), it is not a square.
Exercise 9.1 — Question 3 (*Starred Problem)
Incentre & Angle Bisector Segments
*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 (Fig. 9.1). Proposition:"If AB = AC, then IE = IF."
Fig. 9.1: Angle bisectors BE and CF intersecting at incentre I in ΔABC
Step 1: Frame the Converse Statement
Converse: "In ΔABC (with angle bisectors BE and CF meeting at I), if IE = IF, then AB = AC."
Step 2: Proof that the Original Proposition is TRUE
Original Proposition: TRUE
Step-by-Step Geometric Proof:1. Since AB = AC, angles opposite to equal sides in ΔABC are equal: ∠ABC = ∠ACB.2. Since BE bisects ∠ABC and CF bisects ∠ACB, their halves are also equal:
∠IBC = ∠ABE = ½∠ABC = ½∠ACB = ∠ICB = ∠ACF
3. In ΔIBC, since ∠IBC = ∠ICB, the sides opposite to these equal angles are equal: IB = IC.4. Now consider ΔIFB and ΔIEC:
5. By the ASA congruence criterion, ΔIFB ≅ ΔIEC. Therefore, by CPCT:
IE = IF
Step 3: Truth Value & Justification of Converse Statement
Converse Statement: TRUE
Justification: Suppose AB ≠ AC, say AB < AC, so ∠C < ∠B. Let ∠B = 2β and ∠C = 2γ with γ < β.In ΔIBC, since ∠IBC = β > γ = ∠ICB, the side opposite the larger angle is longer, so IC > IB.Furthermore, using the Sine Rule / angle comparison in ΔIFB and ΔIEC (where ∠FIB = ∠EIC = 90° − ½∠A), having AB < AC strictly forces IE < IF.Therefore, IE = IF is possible only when ∠B = ∠C, which implies AB = AC.
Exercise 9.1 — Question 4
Addition & Cancellation Law of Equality
Question 4:"If x = y, then a + x = a + y, where x, y and a are any three numbers."
Step 1: Frame the Converse Statement
Converse: "If a + x = a + y, then x = y (where x, y and a are any three numbers)."
Step 2: Truth Value & Justification of Original Proposition
Original Proposition: TRUE
Justification: By Euclid's Axiom (if equals are added to equals, the wholes are equal), adding the same number a to both sides of the true equality x = y preserves equality: a + x = a + y.
Step 3: Truth Value & Justification of Converse Statement
Converse Statement: TRUE
Justification: Given a + x = a + y, add the additive inverse (−a) to both sides (or subtract a from both sides):
(−a) + (a + x) = (−a) + (a + y) ⇒ 0 + x = 0 + y ⇒ x = y
Exercise 9.1 — Question 5
Product of Perfect Squares
Question 5:"If a and b are perfect squares, then ab is a perfect square."
Step 1: Frame the Converse Statement
Converse: "If ab is a perfect square, then a and b are perfect squares."
Step 2: Truth Value & Justification of Original Proposition
Original Proposition: TRUE
Justification: Since a and b are perfect squares, there exist integers m and n such that a = m2 and b = n2.Multiplying them gives:
ab = m2 · n2 = (mn)2
Since mn is an integer, ab = (mn)2 is a perfect square.
Step 3: Truth Value & Counterexample for Converse Statement
Converse Statement: FALSE
Counterexample: Take a = 2 and b = 8 (or a = 3, b = 12). Then:
ab = 2 × 8 = 16 = 42 (which is a perfect square)
However, neither a = 2 nor b = 8 is a perfect square!
Exercise Set 9.1 — Questions 6 to 12 (Page 6)
Exercise 9.1 — Question 6
Squaring Real Numbers
Question 6: (x and y are real numbers) "If x = y, then x2 = y2."
Step 1: Frame the Converse Statement
Converse: "If x2 = y2, then x = y (where x and y are real numbers)."
Step 2: Truth Value & Justification of Original Proposition
Original Proposition: TRUE
Justification: If x = y, multiplying both sides by x (and substituting x = y on the right) gives x · x = y · y ⇒ x2 = y2.
Step 3: Truth Value & Counterexample for Converse Statement
Converse Statement: FALSE
Counterexample: Take x = 3 and y = −3. Then:
x2 = 32 = 9 and y2 = (−3)2 = 9 ⇒ x2 = y2
However, x ≠ y because 3 ≠ −3.
Exercise 9.1 — Question 7
Cubing Real Numbers
Question 7: (x and y are real numbers) "If x = y, then x3 = y3."
Step 1: Frame the Converse Statement
Converse: "If x3 = y3, then x = y (where x and y are real numbers)."
Step 2: Truth Value & Justification of Original Proposition
Original Proposition: TRUE
Justification: If x = y, then cubing both sides yields x · x · x = y · y · y ⇒ x3 = y3.
Step 3: Truth Value & Algebraic Proof of Converse Statement
Converse Statement: TRUE
Justification: Suppose x3 = y3. Rearranging and factoring gives:
For real numbers x and y, the second factor (x + ½y)2 + ¾y2 is zero only when x = y = 0.In all cases, x − y = 0 ⇒ x = y (cubing preserves the sign of real numbers).
Exercise 9.1 — Question 8
Divisibility by 24 vs. 4 and 6
Question 8: (n is a positive integer) "If n is divisible by 24, then it is divisible by both 4 and 6."
Step 1: Frame the Converse Statement
Converse: "If n is divisible by both 4 and 6, then n is divisible by 24."
Step 2: Truth Value & Justification of Original Proposition
Original Proposition: TRUE
Justification: If n is divisible by 24, then n = 24k for some positive integer k. Since:
n = 24k = 4 × (6k) and n = 24k = 6 × (4k)
n is divisible by both 4 and 6.
Step 3: Truth Value & Counterexample for Converse Statement
Converse Statement: FALSE
Counterexample: Take n = 12 (or n = 36). Then 12 = 4 × 3 (divisible by 4) and 12 = 6 × 2 (divisible by 6), because LCM(4, 6) = 12.However, 12 is not divisible by 24.
Exercise 9.1 — Question 9
Divisibility by 60 vs. Coprime Factors 5 and 12
Question 9: (n is a positive integer) "If n is divisible by 60, then it is divisible by both 5 and 12."
Step 1: Frame the Converse Statement
Converse: "If n is divisible by both 5 and 12, then n is divisible by 60."
Step 2: Truth Value & Justification of Original Proposition
Original Proposition: TRUE
Justification: If n is divisible by 60, then n = 60k = 5(12k) = 12(5k) for some integer k. Thus, n is divisible by both 5 and 12.
Step 3: Truth Value & Justification of Converse Statement
Converse Statement: TRUE
Justification: Since n is a multiple of 12, we can write n = 12m. We are also given that 5 divides n = 12m.Since 5 and 12 are coprime (they share no common factor greater than 1, i.e., HCF(5, 12) = 1), prime 5 must divide m, so m = 5k. Substituting gives:
n = 12(5k) = 60k ⇒ n is divisible by 60
Exercise 9.1 — Question 10
Squares of Prime Numbers & 3 Factors
Question 10: (n is a positive integer) "If n is the square of a prime number, then it has exactly 3 factors."
Step 1: Frame the Converse Statement
Converse: "If a positive integer n has exactly 3 factors, then n is the square of a prime number."
Step 2: Truth Value & Justification of Original Proposition
Original Proposition: TRUE
Justification: Let n = p2 where p is a prime number.By the Fundamental Theorem of Arithmetic, the only positive divisors of p2 are 1, p, and p2.Since p > 1, these three divisors are distinct, so n has exactly 3 factors.
Step 3: Truth Value & Justification of Converse Statement
Converse Statement: TRUE
Justification: Since n has 3 factors (an odd number of factors), by Example 3 in the text, n must be a perfect square: n = m2 for some integer m > 1.Now, 1, m, and m2 are already 3 distinct factors of n.If m were composite, any proper factor d of m (\(1 < d < m\)) would also divide n, giving n more than 3 factors.Therefore, m must be prime, so n = m2 is the square of a prime number.
Exercise 9.1 — Question 11
Product of Two Primes & 4 Divisors
Question 11: (n is a positive integer) "If n is a product of two unequal prime numbers, then it has exactly 4 divisors."
Step 1: Frame the Converse Statement
Converse: "If a positive integer n has exactly 4 divisors, then n is a product of two unequal prime numbers."
Step 2: Truth Value & Justification of Original Proposition
Original Proposition: TRUE
Justification: Let n = p × q where p and q are distinct prime numbers (\(p \neq q\)).By unique prime factorisation, the only positive divisors of pq are 1, p, q, and pq.Since \(1 < p \neq q < pq\), all 4 divisors are distinct, so n has exactly 4 divisors.
Step 3: Truth Value & Counterexample for Converse Statement
Converse Statement: FALSE
Counterexample: Take n = 8 = 23 (or n = 27 = 33, the cube of any prime number).The divisors of 8 are:
However, 8 = 23 is the cube of a single prime, not a product of two unequal prime numbers!
Exercise 9.1 — Question 12
Common Factors of n and n + 3
Question 12: (n is a positive integer) "If n and n + 3 have no factors in common, then n is not a multiple of 3." (Note: In standard number theory terminology, "have no factors in common" means have no common factor greater than 1, i.e., HCF(n, n + 3) = 1.)
Step 1: Frame the Converse Statement
Converse: "If n is not a multiple of 3, then n and n + 3 have no factors in common (other than 1)."
Step 2: Truth Value & Justification of Original Proposition
Original Proposition: TRUE
Justification: Suppose on the contrary that n were a multiple of 3, say n = 3k.Then n + 3 = 3k + 3 = 3(k + 1) would also be a multiple of 3, so n and n + 3 would share the common factor 3.Therefore, if n and n + 3 have no common factor (> 1), n cannot be a multiple of 3.
Step 3: Truth Value & Justification of Converse Statement
Converse Statement: TRUE
Justification: Suppose d > 1 is any common factor of both n and n + 3.Since d divides both n and n + 3, d must also divide their difference:
(n + 3) − n = 3 ⇒ d divides 3
Since 3 is prime and d > 1, the only possible common factor is d = 3.However, we are given that n is not a multiple of 3, so d = 3 cannot divide n.Thus, n and n + 3 have no common factors (> 1).
Exercise Set 9.1 — Questions 13 to 17 (Pages 6–7)
Exercise 9.1 — Question 13
Disproving Prime-Generating Expressions
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.
Part (i): Counterexample to "All numbers of the form 4n2 + 1 are prime"
Substitute n = 3 (or n = 7) into the expression 4n2 + 1:
For n = 3: 4(3)2 + 1 = 4(9) + 1 = 36 + 1 = 37 (Prime)
Counterexample: For n = 10 (gives 121 = 112) or n = 11 (gives 143 = 11 × 13), the result is composite.
Part (iii): Counterexample to "All numbers of the form 4n + 3 are prime"
Test small positive integers n = 1, 2, 3:• For n = 1: 41 + 3 = 7 (Prime)• For n = 2: 42 + 3 = 19 (Prime)• For n = 3: 43 + 3 = 64 + 3 = 67 (Prime)• For n = 4: 44 + 3 = 256 + 3 = 259 = 7 × 37 (Composite!)
Counterexample: For n = 4, 44 + 3 = 259 = 7 × 37, which is composite (not prime).
Exercise 9.1 — Question 14
Counterexamples for Exponential Prime Claims
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.
Part (i): Counterexample to "If n is prime, then 2n − 1 is prime" (Mersenne Numbers)
For primes n = 2, 3, 5, 7, the values of 2n − 1 are 3, 7, 31, and 127 (all prime).However, for the next prime n = 11:
211 − 1 = 2048 − 1 = 2047 = 23 × 89
Counterexample: n = 11 is prime, but 211 − 1 = 2047 = 23 × 89 is composite.
Part (ii): Counterexample to "If n is even, then 2n + 1 is prime"
For n = 2, 22 + 1 = 5 (prime), and for n = 4, 24 + 1 = 17 (prime).However, for the next even number n = 6:
26 + 1 = 64 + 1 = 65 = 5 × 13
(Similarly, for n = 10: 210 + 1 = 1025 = 25 × 41.)
Counterexample: n = 6 is an even number, but 26 + 1 = 65 = 5 × 13 is composite.
Exercise 9.1 — Question 15
Divisibility by 8 vs. Divisibility by 2 and 4
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?
Part (i): Justification of the Given Statement
Suppose a number n is divisible by 8. Then n = 8k for some integer k.Since 8 can be factored as 2 × 4:
n = 8k = 2 × (4k) (so n is divisible by 2)
n = 8k = 4 × (2k) (so n is divisible by 4)
Hence, every multiple of 8 is automatically divisible by both 2 and 4.
Part (ii): Is Checking Divisibility by 2 and 4 Enough to Test Divisibility by 8?
Answer: NO, it is NOT enough!
Explanation: Testing whether a number is divisible by 2 and 4 and then concluding it is divisible by 8 relies on the converse statement: "If a number is divisible by both 2 and 4, then it is divisible by 8."This converse is FALSE because 2 and 4 are not coprime (every multiple of 4 is already a multiple of 2, so LCM(2, 4) = 4 ≠ 8).
Counterexamples: n = 4, 12, 20, 28 are divisible by both 2 and 4, but NOT divisible by 8!
Exercise 9.1 — Question 16
Divisibility by 3 as Proposition & Converse
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 'sum of the digits is a multiple of 3' using 'If-then' sentences.
Step 1: Statement 1 (Proposition P)
Proposition P: "If a number is divisible by 3, then the sum of its digits is a multiple of 3." (TRUE)
Step 2: Statement 2 (Converse Q — Used as the Divisibility Test)
Converse Q: "If the sum of the digits of a number is a multiple of 3, then the number is divisible by 3." (TRUE)
Because both the proposition and its converse are true, the two conditions imply each other (if and only if).
Exercise 9.1 — Question 17
Constructing Quadrilaterals of Type Q with Sticks
Question 17: Suppose we have identified a category of quadrilaterals called Q, and we have to construct a quadrilateral of this type using two thin sticks as diagonals (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? Why or why not?
(b) Will it matter how the two sticks are put together? (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?
Part (i): Given "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 MUST be of equal length.
Why: Because every quadrilateral of type Q is guaranteed to have equal-length diagonals. If we used sticks of unequal length, the resulting quadrilateral could never belong to category Q.(b) Will it matter how the two sticks are put together?
YES, it may matter how they are put together.
Why: The statement only says "If in Q ⇒ equal diagonals"; it does not say that the converse is true!For example, if Q is the category of rectangles (or squares), every rectangle has equal diagonals, so we need equal sticks — but we must also place the sticks so they bisect each other (and are perpendicular for a square), otherwise we won't get a rectangle.
Part (ii): Given "If a quadrilateral has equal diagonals, then it is of type Q"
(a) Should the two sticks be of equal length?
Choosing two sticks of equal length is SUFFICIENT (guarantees type Q), though not necessarily the only way unless the converse also holds.
Why: The given rule guarantees that any quadrilateral with equal diagonals automatically belongs to type Q. So picking two equal-length sticks guarantees success! (However, since the converse is not stated, type Q might also contain some quadrilaterals with unequal diagonals).(b) Will it matter how the two sticks are put together?
NO, it will NOT matter how the two equal sticks are put together!
Why: As long as the two equal-length sticks intersect to form the diagonals of a quadrilateral, the quadrilateral has equal diagonals — and by the given rule ("If equal diagonals ⇒ type Q"), it is automatically of type Q regardless of the angle or intersection point of the sticks.