NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 12 Quadrilaterals Exercise 12.2

Exercise Set 12.2

Question 1. True or false?
(i) A parallelogram with a right angle is a rectangle.
(ii) A rhombus with perpendicular diagonals is a square.
(iii) If the diagonals of a parallelogram are equal, then it is a rectangle.
Answer:
(i) True. In a parallelogram, adjacent angles add to 180° and opposite angles are equal. If ∠A = 90°, then ∠B = 180° − 90° = 90°, ∠C = ∠A = 90° and ∠D = ∠B = 90°. All angles are right angles, so it is a rectangle.
(ii) False. Every rhombus already has perpendicular diagonals. A rhombus with angles 60° and 120° has perpendicular diagonals but is not a square. (A rhombus is a square only when it has a right angle, or equivalently equal diagonals.)
(iii) True. Let ABCD be a parallelogram with AC = BD. In △ABC and △BAD: AB = BA (common), BC = AD (opposite sides), AC = BD (given).
So △ABC ≅ △BAD (SSS), which gives ∠ABC = ∠BAD. These are adjacent angles of a parallelogram, so they add to 180°; hence each is 90° and ABCD is a rectangle.

Teacher's Note:
1. In (ii) the extra condition is always true for a rhombus, so it cannot make it a square.
2. In (iii) the parallelogram property BC = AD is what makes the SSS step work.

Question 2. The diagonal AC of a parallelogram ABCD bisects ∠A. Show that it also bisects ∠C and that ABCD is a rhombus.

Answer:
Given: ABCD is a parallelogram and ∠BAC = ∠DAC (angles 1 and 2).
To prove: ∠BCA = ∠DCA, and AB = BC = CD = DA.
1. AB ∥ DC with transversal AC: ∠BAC = ∠DCA (alternate angles) ... (1)
2. AD ∥ BC with transversal AC: ∠DAC = ∠BCA (alternate angles) ... (2)
3. Since ∠BAC = ∠DAC, (1) and (2) give ∠DCA = ∠BAC = ∠DAC = ∠BCA. So ∠DCA = ∠BCA, i.e. AC bisects ∠C.
4. Also ∠BAC = ∠BCA, so in △ABC the sides opposite these equal angles are equal: BC = AB.
5. Opposite sides of a parallelogram are equal: AB = DC and BC = AD. Hence AB = BC = CD = DA, and ABCD is a rhombus.

Teacher's Note:
1. Alternate angles carry the bisection from A across to C.
2. The key move is spotting the isosceles triangle ABC.

Question 3. The following questions examine converses of true properties. Answer them with Yes or No. If your answer is No, what extra condition can you add so that the answer becomes Yes?
(i) If the diagonals of a quadrilateral ABCD bisect its angles, must ABCD be a rhombus?
(ii) If the diagonals of a quadrilateral ABCD bisect each other at right angles, must ABCD be a rhombus?
(iii) If the diagonals of a quadrilateral ABCD are of equal length, must ABCD be a rectangle? (In the ancient Indian study of quadrilaterals, equality of diagonals was considered significant. Quadrilaterals were first classified according to whether their diagonals were equal or not, before considering equality of sides.)
Answer:
(i) Yes. Suppose AC bisects ∠A and ∠C, and BD bisects ∠B and ∠D.
In △ABC and △ADC: ∠BAC = ∠DAC, AC is common, ∠BCA = ∠DCA. So △ABC ≅ △ADC (ASA), giving AB = AD and CB = CD.
Using BD in the same way, △BAD ≅ △BCD (ASA), giving BA = BC and DA = DC.
So AB = BC = CD = DA. Opposite sides are equal, so ABCD is a parallelogram (Theorem 2), and with all sides equal it is a rhombus.

(ii) Yes. Let the diagonals meet at O with OA = OC, OB = OD and ∠AOB = 90°.
Since the diagonals bisect each other, ABCD is a parallelogram (Theorem 4).
In △AOB and △AOD: OB = OD, ∠AOB = ∠AOD = 90°, OA common. So △AOB ≅ △AOD (SAS) and AB = AD.
A parallelogram with two adjacent sides equal has all four sides equal, so ABCD is a rhombus.
(iii) No. An isosceles trapezium (AB ∥ DC, AD = BC) has equal diagonals AC and BD, yet its angles are not right angles. (A kite with equal diagonals is another counterexample.)
Extra condition: the diagonals should also bisect each other. Then ABCD is a parallelogram with equal diagonals, which is a rectangle by Q1(iii).
So: if the diagonals of a quadrilateral are equal and bisect each other, it is a rectangle.

Teacher's Note:
1. A single counterexample is enough to answer 'No'.
2. In (iii) the fix is to add 'bisect each other', which brings in the parallelogram.

Question 4. Let ABCD be a parallelogram with AB ≠ BC. Show that the pairwise intersection points of the four angle bisectors form the vertices of a rectangle (see Fig. 12.11). Why did we assume AB ≠ BC?

Answer:
Let the bisectors of ∠A and ∠B meet at R, of ∠B and ∠C at Q, of ∠C and ∠D at P, and of ∠D and ∠A at S.
Angle at S. Adjacent angles of a parallelogram are supplementary: ∠A + ∠D = 180°.
In △ASD the angles at A and D are half of ∠A and half of ∠D, so ∠SAD + ∠SDA = \( \frac{1}{2} \)(∠A + ∠D) = \( \frac{1}{2} \) × 180° = 90°.
Hence ∠ASD = 180° − 90° = 90°. The angle of PQRS at S is formed by the same two bisector lines, so it is also 90°.
Other angles. In the same way:
1. In △ARB: \( \frac{1}{2} \)(∠A + ∠B) = 90°, so ∠ARB = 90° and the angle at R is 90°.
2. In △BQC: \( \frac{1}{2} \)(∠B + ∠C) = 90°, so the angle at Q is 90°.
3. In △CPD: \( \frac{1}{2} \)(∠C + ∠D) = 90°, so the angle at P is 90°.
All four angles of PQRS are right angles. Its opposite angles are equal, so it is a parallelogram (Theorem 3), and a parallelogram with a right angle is a rectangle. So PQRS is a rectangle.
Why AB ≠ BC? If AB = BC, ABCD is a rhombus. Then each diagonal bisects the angles at its ends (Q2), so the bisectors of ∠A and ∠C both lie on AC, and those of ∠B and ∠D on BD. All four meeting points coincide with the centre O, and we get a single point, not a rectangle.

Teacher's Note:
1. The whole proof rests on 'half of 180° is 90°'.
2. Discuss the rhombus case: the rectangle shrinks to a single point.