Ex 8.1 NCERT Solutions - Quadrilaterals | Mathematics (Maths) Class 9

Q1. The angles of a quadrilateral are in the ratio 3: 5: 9: 13. Find all angles of the quadrilateral.

Answer
 Let x be the common ratio between the angles.
 Sum of the interior angles of the quadrilateral = 360°
 Now,
 3x + 5x + 9x + 13x = 360°
 ⇒ 30x = 360°
 ⇒ x = 12°
 Angles of the quadrilateral are:
 3x = 3×12° = 36°
 5x = 5×12° = 60°
 9x = 9×12° = 108°
 13x = 13×12° = 156°

Q2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer

Given,
 AC = BD
 To show,
 To show ABCD is a rectangle we have to prove that one of
 its interior angle is right-angled.
 Proof,
 In ΔABC and ΔBCD,
 BC = BC (Common)
 AB = DC (Opposite sides of a parallelogram are equal)
 AC = BD (Given)
 Therefore, ΔABC ≅ ΔBCD by SSS congruence condition.
 ∠B = ∠C (by CPCT)
 also,
 ∠A + ∠B = 180° (Sum of the angles on the same side of the
 transversal)
 ⇒ 2∠A = 180°
 ⇒ ∠A = 90° = ∠B
 Thus ABCD is a rectangle.

Q3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Answer -

Let ABCD be a quadrilateral whose diagonals bisect each
 other at right angles.
 Given,
 OA = OC, OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA =
 90°
 To show,
 ABCD is parallelogram and AB = BC = CD = AD
 Proof,
 In ΔAOB and ΔCOB,
 OA = OC (Given)
 ∠AOB = ∠COB (Opposite sides of a parallelogram are
 equal)
 OB = OB (Common)
 Therefore, ΔAOB ≅ ΔCOB by SAS congruence condition.
 Thus, AB = BC (by CPCT)
 Similarly we can prove,
 AB = BC = CD = AD
 Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.
 Thus, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.

Q4. Show that the diagonals of a square are equal and bisect each other at right angles.

Answer - 

Let ABCD be a square and its diagonals AC and BD
 intersect each other at O.
 To show,
 AC = BD, AO = OC and ∠AOB = 90°
 Proof,
 In ΔABC and ΔBAD,
 BC = BA (Common)
 ∠ABC = ∠BAD = 90°
 AC = AD (Given)
 Therefore, ΔABC ≅ ΔBAD by SAS congruence condition.
 Thus, AC = BD by CPCT. Therefore, diagonals are equal.
 Now,
 In ΔAOB and ΔCOD,
 ∠BAO = ∠DCO (Alternate interior angles)
 ∠AOB = ∠COD (Vertically opposite)
 AB = CD (Given)
 Therefore, ΔAOB ≅ ΔCOD by AAS congruence condition.
 Thus, AO = CO by CPCT. (Diagonal bisect each other.)
 Now,
 In ΔAOB and ΔCOB,
 OB = OB (Given)
 AO = CO (diagonals are bisected)
 AB = CB (Sides of the square)
 Therefore, ΔAOB ≅ ΔCOB by SSS congruence condition.
 also, ∠AOB = ∠COB
 ∠AOB + ∠COB = 180° (Linear pair)
 Thus, ∠AOB = ∠COB = 90° (Diagonals bisect each other at

Q5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Answer - 

Q8. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
 (i) ABCD is a square
 (ii) diagonal BD bisects ∠B as well as ∠D.

Answer - 

(i)∠DAC = ∠DCA (AC bisects ∠A as well as ∠C)
 ⇒ AD = CD (Sides opposite to equal angles of a triangle are
 equal)
 also, CD = AB (Opposite sides of a rectangle)
 Therefore, AB = BC = CD = AD
 Thus, ABCD is a square.
 

(ii) In ΔBCD,
 BC = CD
 ⇒ ∠CDB = ∠CBD (Angles opposite to equal sides are equal)
 also, ∠CDB = ∠ABD (Alternate interior angles)
 ⇒ ∠CBD = ∠ABD
 Thus, BD bisects ∠B
 Now,
 ∠CBD = ∠ADB
 ⇒ ∠CDB = ∠ADB
 Thus, BD bisects ∠D

Q9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig.). Show that:
 (i) ΔAPD ≅ΔCQB
 (ii) AP = CQ
 (iii) ΔAQB ≅ ΔCPD
 (iv) AQ = CP
 (v) APCQ is a parallelogram

Answer - 

(i) In ΔAPD and ΔCQB,
 DP = BQ (Given)
 ∠ADP = ∠CBQ (Alternate interior angles)
 AD = BC (Opposite sides of a ||gm)
 Thus, ΔAPD ≅ ΔCQB by SAS congruence condition.

(ii) AP = CQ by CPCT as ΔAPD ≅ ΔCQB.

(iii) In ΔAQB and ΔCPD,
 BQ = DP (Given)
 ∠ABQ = ∠CDP (Alternate interior angles)
 AB = BCCD (Opposite sides of a ||gm)
 Thus, ΔAQB ≅ ΔCPD by SAS congruence condition.

(iv) AQ = CP by CPCT as ΔAQB ≅ ΔCPD.

(v) From (ii) and (iv), it is clear that APCQ has equal
 opposite sides also it has equal opposite angles. Thus,
 APCQ is a ||gm.

Q10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig.). Show that
(i) ΔAPB ≅ ΔCQD
(ii) AP = CQ

 Answer

(i) In ΔAPB and ΔCQD,
 ∠ABP = ∠CDQ (Alternate interior angles)
 ∠APB = ∠CQD (equal to right angles as AP and CQ are
 perpendiculars)
 AB = CD (ABCD is a parallelogram)
 Thus, ΔAPB ≅ ΔCQD by AAS congruence condition.

(ii) AP = CQ by CPCT as ΔAPB ≅ ΔCQD.

Q11. In ΔABC and ΔDEF,AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig). Show that
 (i) quadrilateral ABED is a parallelogram
 (ii) quadrilateral BEFC is a parallelogram
 (iii) AD || CF and AD = CF
 (iv) quadrilateral ACFD is a parallelogram
 (v) AC = DF
 (vi) ΔABC ≅ ΔDEF

Answer:

(i) AB = DE and AB || DE (Given)
 Thus, quadrilateral ABED is a parallelogram because two
 opposite sides of a quadrilateral are equal and parallel to
 each other.

(ii) Again BC = EF and BC || EF.
 Thus, quadrilateral BEFC is a parallelogram.

(iii) Since ABED and BEFC are parallelograms.
 ⇒ AD = BE and BE = CF (Opposite sides of a parallelogram
 are equal)
 Thus, AD = CF.
 Also, AD || BE and BE || CF (Opposite sides of a
 parallelogram are parallel)
 Thus, AD || CF

(iv) AD and CF are opposite sides of quadrilateral ACFD
 which are equal and parallel to each other. Thus, it is a
 parallelogram.

(v) AC || DF and AC = DF because ACFD is a parallelogram.

(vi) In ΔABC and ΔDEF,
 AB = DE (Given)
 BC = EF (Given)
 AC = DF (Opposite sides of a parallelogram)
 Thus, ΔABC ≅ ΔDEF by SSS congruence condition.

Q12. ABCD is a trapezium in which AB || CD and AD = BC (see Fig). Show that
 (i) ∠A = ∠B
 (ii) ∠C = ∠D
 (iii) ΔABC ≅ ΔBAD
 (iv) diagonal AC =diagonal BD
 [Hint : Extend AB anddraw a line through C parallel to DA intersecting AB produced at E.]

Answer - 

Construction: Draw a line through C parallel to DA intersecting AB produced at E.

(i) CE = AD (Opposite sides of a parallelogram)
 AD = BC (Given)
 Therefor, BC = CE
 ⇒ ∠CBE = ∠CEB
 also,
 ∠A + ∠CBE = 180° (Angles on the same side of transversal
 and ∠CBE = ∠CEB)
 ∠B + ∠CBE = 180° (Linear pair)
 ⇒ ∠A = ∠B

(ii) ∠A + ∠D = ∠B + ∠C = 180° (Angles on the same side of
 transversal)
 ⇒ ∠A + ∠D = ∠A + ∠C (∠A = ∠B)
 ⇒ ∠D = ∠C

(iii) In ΔABC and ΔBAD,
 AB = AB (Common)
 ∠DBA = ∠CBA
 AD = BC (Given)
 Thus, ΔABC ≅ ΔBAD by SAS congruence condition.

(iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA