Class 9 Maths Ganita Manjari Chapter 9 Solutions — Propositions and their Converses

In triangle ABC assume $\angle B=\angle C$. Draw AD perpendicular to BC. Since B and C are equal, both are acute and D lies on segment BC. Right triangles ABD and ACD have equal angles at B and C and the common side AD. They are congruent by AAS, so $\mathrm{AB}=\mathrm{AC}$. This proves the converse independently.

The rain example assumes an exposed road and enough rain to wet it. Even then, a wet road does not imply rain: a tanker may spill water. For divisibility, every multiple of 6 is a multiple of 3 because $6k=3(2k)$. The converse is false: 3 is not divisible by 6.

For a positive integer n, pair each positive divisor d with $\frac{n}{d}$. If n is not a square, the divisors form distinct pairs, giving an even count. If $n=k^2$, exactly one divisor k pairs with itself; all other divisors form distinct pairs, so the count is odd. Thus both directions of the perfect-square statement hold. The textbook’s first factor-pair argument proves “odd number of divisors implies square”; the reverse argument must also mention that there is exactly one repeated pair.

This also verifies the counterexample to the proposed Fermat-prime rule.

Equal area does not imply congruence. Right triangles with perpendicular sides 3, 4 and 2, 6 both have area 6 square units, but different side lengths. Congruent triangles do have equal area because a rigid motion preserves all distances, including a corresponding base and height.

Sample example with both directions false: “If an integer is divisible by 2, it is divisible by 3” fails for 2. Its converse fails for 3.

Let a, b, c be positive triangle side lengths and suppose $a^2+b^2=c^2$. Construct a right triangle XYZ with $\mathrm{YZ}=a$ and $\mathrm{XZ}=b$. Its hypotenuse satisfies:

The constructed triangle and the given triangle have all three corresponding sides equal, so they are congruent by SSS. Hence the given triangle is right-angled, with c opposite its right angle. Notation clarification: in the forward theorem, c must denote the hypotenuse; arbitrary side labels do not guarantee this equation.

Sample responses: If today is Monday, today is a weekday; the converse fails on Tuesday. If a quadrilateral is a square, it is a rectangle; a 2-by-3 rectangle disproves the converse. If a positive integer ends in 0, it is divisible by 5; 15 disproves the converse. State the context and give one counterexample that satisfies the converse’s hypothesis but not its conclusion.

Original: true. Parallel lines cut by a transversal have equal corresponding angles. Converse: If a transversal makes a pair of corresponding angles equal, the two lines are parallel. True. This is the corresponding-angle criterion for parallelism in plane geometry. The lines and transversal must be distinct and meet in the usual configuration.

Original: true: each angle of a square is 90°. Converse: If all four angles of a quadrilateral are equal, it is a square. False. Their sum is 360°, so each is 90° and the figure is a rectangle; its sides need not all be equal. A rectangle of sides 2 cm and 3 cm is a counterexample.

Original: true. If $\mathrm{AB}=\mathrm{AC}$, triangle ABC is symmetric about its apex-angle bisector AI. This reflection interchanges B and C, the corresponding angle bisectors BE and CF, and their intersection points E and F. It fixes I, so $\mathrm{IE}=\mathrm{IF}$.

Converse: If $\mathrm{IE}=\mathrm{IF}$, then $\mathrm{AB}=\mathrm{AC}$. False. A useful counterexample is a triangle with angles $A=60^\circ$, $B=90^\circ$, $C=30^\circ$. Its sides AB and AC are unequal (AC is the hypotenuse). Here is a proof that its IE and IF are nevertheless equal.

The angle bisectors give $\angle \mathrm{ABE}=45^\circ$ and $\angle \mathrm{ACF}=15^\circ$. Therefore $\angle \mathrm{AEI}=75^\circ$ and $\angle \mathrm{AFI}=105^\circ$. Drop perpendiculars ID to AC and IG to AB. The incentre is equally distant from the two sides, so $\mathrm{ID}=\mathrm{IG}$. In this configuration D lies between A and E, and G lies between F and B; hence $\angle \mathrm{IED}=75^\circ$ and $\angle \mathrm{IFG}=180^\circ -105^\circ =75^\circ$. Right triangles IED and IFG are congruent by AAS, proving $\mathrm{IE}=\mathrm{IF}$.

For a fully specified construction, take $\mathrm{AB}=1$ unit, BC = $\sqrt3$ units and angle $B=90^\circ$; then $\mathrm{AC}=2$ units. The diagram shows the actual counterexample, not an isosceles triangle.

Original and converse are true. Equality is preserved when the same number a is added to both sides. Conversely, subtracting a from both sides of $a+x=a+y$ gives $x=y$. No division by an unknown quantity is involved.

Original: true. Write $a=u^2$ and $b=v^2$ for integers u, v. Then $ab=(uv)^2$. Converse: If ab is a perfect square, then a and b are perfect squares. False: $a=2$, $b=8$ give $ab=16$, although neither factor is a square.

Original: true. Squaring equal real numbers gives equal results. Converse: If $x^2=y^2$, then $x=y$. False: $x=2$ and $y=-2$. Factorisation gives $(x-y)(x+y)=0$, so the conclusion is $x=y$ or $x=-y$.

Both directions are true. The forward direction follows by cubing. Conversely:

The second factor is positive unless $x=y=0$. Thus either both are zero, or $x-y=0$. In both cases $x=y$.

Original: true: $24k=4\left(6k\right)=6\left(4k\right)$. Converse: A positive integer divisible by both 4 and 6 is divisible by 24. False: 12 is divisible by 4 and 6, but not by 24. Their least common multiple is 12, not 24.

Both directions are true. A multiple of 60 is a multiple of 5 and 12. Conversely, 5 and 12 are coprime, so their least common multiple is 60. Equivalently, write $n=12k$; since 5 divides 12k and not 12, it divides k, giving $n=60t$.

Both directions are true. For prime p, the factors of $p^2$ are exactly $1,p,p^2$. Conversely, if n has exactly three positive factors, write them 1, d, n. Factor pairing forces $d^2=n$. If d were composite, a proper prime divisor of d would be another divisor of n. Therefore d is prime and n is its square.

Original: true. For unequal primes p and q, pq has divisors 1, p, q, pq. Converse: A positive integer with four divisors is a product of two unequal primes. False. 8 has divisors 1, 2, 4, 8 but $8=2^3$. Numbers with four divisors can also be cubes of primes.

Interpret “no factors in common” as “no common factor other than 1”, since all positive integers have factor 1. Both directions are then true. Any common divisor divides their difference 3, so their greatest common divisor is 1 or 3. If n were divisible by 3, both numbers would share 3. Conversely, when n is not divisible by 3, their only common positive divisor is 1.

Each example is enough to refute its universal claim.

(i) Choose the prime exponent $n=11$:

(ii) Choose the even exponent $n=6$:

The exponents satisfy the stated hypotheses, but the results are composite.

(i) If $n=8k$, then $n=2\left(4k\right)=4\left(2k\right)$, so the statement is true. (ii) No. The converse fails: 12 is divisible by 2 and 4 but not by 8. Divisibility by 4 already includes divisibility by 2; the extra check adds no information. The usual decimal shortcut for 8 tests the number formed by the last three digits.

Forward: If a positive integer is divisible by 3, its digit sum is divisible by 3. Converse: If the digit sum is divisible by 3, the integer is divisible by 3. Both are true.

For a three-digit illustration, write $N=100a+10b+c$. Then:

Thus N and its digit sum differ by a multiple of 3. The same argument works for any number of digits because every power of 10 leaves remainder 1 on division by 3.

(i) Given: type Q implies equal diagonals. (a) Yes, equal stick lengths are necessary for constructing a type-Q quadrilateral. (b) The condition alone does not tell us which arrangements work. Equal diagonals may be insufficient: if Q denotes squares, the sticks must also bisect each other at right angles. We cannot infer the converse from the given implication.

(ii) Given: equal diagonals imply type Q. (a) Equal sticks are a sufficient way to guarantee type Q. They are not proved necessary by this implication; Q may also contain quadrilaterals with unequal diagonals. (b) No further restriction follows from the stated property, provided the equal sticks are positioned as the diagonals of a valid, non-degenerate quadrilateral. For the usual convex construction, their interiors intersect and the endpoints are joined in cyclic order. Avoid collinear sticks or overlapping endpoints.

Key distinction: “must be equal” in (i) expresses necessity; “choose equal to guarantee Q” in (ii) expresses sufficiency.