NCERT Solutions · Class 9 Mathematics Quadrilaterals

(i) True. Let \(\displaystyle \angle A = 90^\circ\) in parallelogram \(\displaystyle ABCD\). \[\angle A + \angle B = 180^\circ \quad (AD \parallel BC) \] \[\angle B = 90^\circ \] \[\angle C = \angle A = 90^\circ, \quad \angle D = \angle B = 90^\circ \quad \text{(opposite angles)} \] All four angles are right angles, so \(\displaystyle ABCD\) is a rectangle.(ii) False. Every rhombus already has perpendicular diagonals. Counterexample: rhombus \(\displaystyle PQRS\) with diagonals \(\displaystyle PR = 8\), \(\displaystyle QS = 6\) (figure, left); the half-diagonals are $\displaystyle 4$ and 3. \[PQ = \sqrt{4^2 + 3^2} = 5 = QR = RS = SP \] \[QS^2 = 36 \neq 50 = PQ^2 + PS^2 \;\Rightarrow\; \angle P \neq 90^\circ \] So it is not a square.(iii) True. Let \(\displaystyle ABCD\) be a parallelogram with \(\displaystyle AC = BD\) (figure, right). Compare \(\displaystyle \triangle ABC\) and \(\displaystyle \triangle DCB\). \[AB = DC \quad \text{(opposite sides)} \] \[BC = CB \quad \text{(common)} \] \[AC = DB \quad \text{(given)} \] \[\triangle ABC \cong \triangle DCB \quad \text{(SSS)} \;\Rightarrow\; \angle ABC = \angle DCB \] \[\angle ABC + \angle DCB = 180^\circ \quad (AB \parallel DC) \] \[\angle ABC = \angle DCB = 90^\circ \] A parallelogram with a right angle is a rectangle, by (i).Answer: (i) True; (ii) False; (iii) True.

\[\angle BAC = \angle DAC \quad \text{(given)} \] \[AB \parallel DC \;\Rightarrow\; \angle DCA = \angle BAC \quad \text{(alternate angles)} \] \[AD \parallel BC \;\Rightarrow\; \angle BCA = \angle DAC \quad \text{(alternate angles)} \] \[\angle DCA = \angle BAC = \angle DAC = \angle BCA \] So \(\displaystyle AC\) bisects \(\displaystyle \angle C\).In \(\displaystyle \triangle ABC\): \[\angle BAC = \angle BCA \;\Rightarrow\; AB = BC \quad \text{(sides opposite equal angles)} \] \[AB = CD, \quad BC = AD \quad \text{(opposite sides of a parallelogram)} \] \[AB = BC = CD = DA \] All four sides are equal, so \(\displaystyle ABCD\) is a rhombus.Answer: \(\displaystyle AC\) bisects \(\displaystyle \angle C\), and \(\displaystyle AB = BC = CD = DA\), so \(\displaystyle ABCD\) is a rhombus.

(i) Converse: if the diagonals bisect the angles, the quadrilateral is a rhombus. Yes. Use \(\displaystyle AC\), which bisects \(\displaystyle \angle A\) and \(\displaystyle \angle C\), and \(\displaystyle BD\), which bisects \(\displaystyle \angle B\) and \(\displaystyle \angle D\). \[\angle BAC = \angle DAC, \quad AC = AC, \quad \angle BCA = \angle DCA \;\Rightarrow\; \triangle ABC \cong \triangle ADC \quad \text{(ASA)} \] \[AB = AD, \quad CB = CD \] \[\angle ABD = \angle CBD, \quad BD = BD, \quad \angle ADB = \angle CDB \;\Rightarrow\; \triangle BAD \cong \triangle BCD \quad \text{(ASA)} \] \[BA = BC, \quad DA = DC \] \[AB = BC = CD = DA \] All four sides are equal: a rhombus.(ii) Converse: if the diagonals bisect each other at right angles, the quadrilateral is a rhombus. Yes. Let the diagonals meet at \(\displaystyle O\), so \(\displaystyle OA = OC\), \(\displaystyle OB = OD\) and every angle at \(\displaystyle O\) is \(\displaystyle 90^\circ\). \[AB^2 = OA^2 + OB^2 \] \[BC^2 = OB^2 + OC^2 = OB^2 + OA^2 \] \[CD^2 = OC^2 + OD^2 = OA^2 + OB^2 \] \[DA^2 = OD^2 + OA^2 = OB^2 + OA^2 \] \[AB = BC = CD = DA \] A rhombus.(iii) Converse: equal diagonals make a rectangle. No. Counterexample: isosceles trapezium \(\displaystyle PQRS\) (figure, above) with \[P(0,0), \quad Q(4,0), \quad R(3,3), \quad S(1,3) \] \[PR = QS = \sqrt{3^2 + 3^2} = 3\sqrt{2} \] \[PQ = 4 \neq 2 = SR \] Opposite sides unequal, so it is not a parallelogram, hence not a rectangle. Extra condition: \(\displaystyle ABCD\) is a parallelogram (Q1 (iii)).Answer: (i) Yes; (ii) Yes; (iii) No, but adding that \(\displaystyle ABCD\) is a parallelogram makes it Yes.

Let \(\displaystyle \angle A = \angle C = 2\alpha\) and \(\displaystyle \angle B = \angle D = 2\beta\), with \(\displaystyle P, Q, R, S\) as in the given figure. \[\angle A + \angle B = 180^\circ \quad (AD \parallel BC) \] \[2\alpha + 2\beta = 180^\circ \;\Rightarrow\; \alpha + \beta = 90^\circ \] In \(\displaystyle \triangle ABP\), \(\displaystyle \angle PAB = \alpha\) and \(\displaystyle \angle PBA = \beta\): \[\angle APB = 180^\circ - \alpha - \beta = 90^\circ \] The same argument in \(\displaystyle \triangle BCQ\), \(\displaystyle \triangle CDR\), \(\displaystyle \triangle DAS\) gives \[\angle BQC = \angle CRD = \angle DSA = 90^\circ \] \(\displaystyle \angle P = \angle APB\) and \(\displaystyle \angle R = \angle CRD\); \(\displaystyle \angle Q\) and \(\displaystyle \angle S\) are vertically opposite \(\displaystyle \angle BQC\) and \(\displaystyle \angle DSA\): \[\angle P = \angle Q = \angle R = \angle S = 90^\circ \] So \(\displaystyle PQRS\) is a rectangle.Why \(\displaystyle AB \neq BC\). If \(\displaystyle AB = BC\), then \(\displaystyle ABCD\) is a rhombus, and \[\angle BAC = \angle BCA = \angle DAC \quad (AD \parallel BC) \] so \(\displaystyle AC\) bisects \(\displaystyle \angle A\); likewise the diagonals bisect all its angles. The bisectors of \(\displaystyle A\) and \(\displaystyle C\) are both \(\displaystyle AC\); those of \(\displaystyle B\) and \(\displaystyle D\) are both \(\displaystyle BD\). The four points merge into the centre, leaving no rectangle.Answer: \(\displaystyle PQRS\) is a rectangle; for \(\displaystyle AB = BC\) it shrinks to a point.