Question-1 :- The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Solution :-Let the common ratio between the angles be x. Therefore, the angles will be 3x, 5x, 9x, and 13x respectively. As the sum of all interior angles of a quadrilateral is 360°, 3x + 5x + 9x + 13x = 360° 30x = 360° x = 12° Hence, the angles are 3x = 3 × 12 = 36° 5x = 5 × 12 = 60° 9x = 9 × 12 = 108° 13x = 13 × 12 = 156°
Question-2 :- If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution :-Given that : Let ABCD be a parallelogram. Prove that : ABCD is a rectangle or one of its interior angles is 90°. Proof : In ∆ABC and ∆DCB, AB = DC (Opposite sides of a parallelogram are equal) BC = BC ( Common) AC = DB (Given) ∆ABC ≅ ∆DCB (By SSS Congruence rule) By CPCT Rule, ∠ABC = ∠DCB It is known that the sum of the measures of angles on the same side of transversal is 180°. ∠ABC + ∠DCB = 180° (AB || CD) ∠ABC + ∠ABC = 180° 2 ∠ABC = 180° ∠ABC = 90° Since ABCD is a parallelogram and one of its interior angles is 90°, ABCD is a rectangle.
Question-3 :- : Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Solution :-Given that : Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle i.e., OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°. Prove that : ABCD is a rhombus or ABCD is a parallelogram and all the sides of ABCD are equal. Proof : In ∆AOD and ∆COD, OA = OC (Diagonals bisect each other) ∠AOD = ∠COD (Given) OD = OD (Common) ∆AOD ≅ ∆COD (By SAS congruence rule) By CPCT Rule, AD = CD .....(1) Similarly, it can be proved that AD = AB and CD = BC .....(2) From equations (1) and (2), AB = BC = CD = AD Since opposite sides of quadrilateral ABCD are equal, It can be said that ABCD is a parallelogram. Since all sides of a parallelogram ABCD are equal, it can be said that ABCD is a rhombus.
Question-4 :- Show that the diagonals of a square are equal and bisect each other at right angles.
Solution :-Given that : Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O. Prove that : The diagonals of a square are equal and bisect each other at right angles or AC = BD, OA = OC, OB = OD, and ∠AOB = 90°. Proof : In ∆ABC and ∆DCB, AB = DC (Sides of a square are equal to each other) ∠ABC = ∠DCB (All interior angles are of 90 ) BC = CB (Common side) ∆ABC ≅ ∆DCB (By SAS congruency) AC = DB (By CPCT) Hence, the diagonals of a square are equal in length. In ∆AOB and ∆COD, ∠AOB = ∠COD (Vertically opposite angles) ∠ABO = ∠CDO (Alternate interior angles) AB = CD (Sides of a square are always equal) ∆AOB ≅ ∆COD (By AAS congruence rule) AO = CO and OB = OD (By CPCT) Hence, the diagonals of a square bisect each other. In ∆AOB and ∆COB, As we had proved that diagonals bisect each other, therefore, AO = CO AB = CB (Sides of a square are equal) BO = BO (Common) ∆AOB ≅ ∆COB (By SSS congruency) ∠AOB = ∠COB (By CPCT) However, AOB + COB = 180° (Linear pair) 2 ∠AOB = 180° ∠AOB = 90° Hence, the diagonals of a square bisect each other at right angles.
Question-5 :- Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Solution :-Given that : Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O. The diagonals of ABCD are equal and bisect each other at right angles. Therefore, AC = BD, OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = = 90°. Prove that : ABCD is a square or ABCD is a parallelogram, AB = BC = CD = AD, and one of its interior angles is 90°. Proof : In ∆AOB and ∆COD, AO = CO (Diagonals bisect each other) OB = OD (Diagonals bisect each other) ∠AOB = ∠COD (Vertically opposite angles) ∆AOB ≅ ∆COD (SAS congruence rule) By CPCT Rule, AB = CD ... (1) And, ∠OAB = ∠OCD However, these are alternate interior angles for line AB and CD and alternate interior angles are equal to each other only when the two lines are parallel. AB || CD ... (2) From equations (1) and (2), we obtain ABCD is a parallelogram. In ∆AOD and ∆COD, AO = CO (Diagonals bisect each other) ∠AOD = ∠COD (Given that each is 90°) OD = OD (Common) ∆AOD ≅ ∆COD (SAS congruence rule) By CPCT Rule, AD = DC ... (3) However, AD = BC and AB = CD (Opposite sides of parallelogram ABCD) AB = BC = CD = DA Therefore, all the sides of quadrilateral ABCD are equal to each other. In ∆ADC and ∆BCD, AD = BC (Already proved) AC = BD (Given) DC = CD (Common) ∆ADC ≅ ∆BCD (SSS Congruence rule) By CPCT Rule, ∠ADC = ∠BCD However, ∠ADC + ∠BCD = 180° (Co-interior angles) ∠ADC + ∠ADC = 180° 2 ∠ADC = 180° ∠ADC = 90° One of the interior angles of quadrilateral ABCD is a right angle. Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angles is 90°. Therefore, ABCD is a square.
Question-6 :- : Diagonal AC of a parallelogram ABCD bisects ∠ A. Show that
(i) it bisects ∠ C also,
(ii) ABCD is a rhombus.
Given that : ABCD is a parallelogram and AC bisects A. Prove that : (i) it bisects ∠ C also, (ii) ABCD is a rhombus. Proof : (i) ABCD is a parallelogram. ∠DAC = ∠BCA (Alternate interior angles) ... (1) And, ∠BAC = ∠DCA (Alternate interior angles) ... (2) However, it is given that AC bisects A. ∠DAC = ∠BAC ... (3) From equations (1), (2), and (3), we obtain ∠DAC = ∠BCA = ∠BAC = ∠DCA ... (4) ∠DCA = ∠BCA Hence, AC bisects C. (ii)From equation (4), we obtain ∠DAC = ∠DCA DA = DC (Side opposite to equal angles are equal) However, DA = BC and AB = CD (Opposite sides of a parallelogram) AB = BC = CD = DA Hence, ABCD is a rhombus.
Question-7 :- ABCD is a rhombus. Show that diagonal AC bisects ∠ A as well as ∠ C and diagonal BD bisects ∠ B as well as ∠ D.
Solution :-Given that : ABCD is a rhombus. Prove that : Diagonal AC bisects ∠ A as well as ∠ C and Diagonal BD bisects ∠ B as well as ∠ D. Proof : In ∆ABC, BC = AB (Sides of a rhombus are equal to each other) ∠1 = ∠2 (Angles opposite to equal sides of a triangle are equal) However, ∠1 = ∠3 (Alternate interior angles for parallel lines AB and CD) ∠2 = ∠3 Therefore, AC bisects C. Also, ∠2 = ∠4 (Alternate interior angles for || lines BC and DA) ∠1 = ∠4 Therefore, AC bisects A. Similarly, it can be proved that BD bisects B and D as well.
Question-8 :- 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.
Given that : ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠ C. Prove that : (i) ABCD is a square (ii) diagonal BD bisects ∠ B as well as ∠ D. Proof : (i) It is given that ABCD is a rectangle. ∠A = ∠C CD = DA (Sides opposite to equal angles are also equal) However, DA = BC and AB = CD (Opposite sides of a rectangle are equal) AB = BC = CD = DA ABCD is a rectangle and all of its sides are equal. Hence, ABCD is a square. (ii) Let us join BD. In ∆BCD, BC = CD (Sides of a square are equal to each other) ∠CDB = ∠CBD (Angles opposite to equal sides are equal) However, ∠CDB = ∠ABD (Alternate interior angles for AB || CD) ∠CBD = ∠ABD BD bisects B. Also, ∠CBD = ∠ADB (Alternate interior angles for BC || AD) ∠CDB = ∠ABD BD bisects D.
Question-9 :- : In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that:
(i) Δ APD ≅ Δ CQB
(ii) AP = CQ
(iii) Δ AQB ≅ Δ CPD
(iv) AQ = CP
(v) APCQ is a parallelogram
Given that : ABCD is a parallelogram and AC bisects A. Prove that : (i) Δ APD ≅ Δ CQB (ii) AP = CQ (iii) Δ AQB ≅ Δ CPD (iv) AQ = CP (v) APCQ is a parallelogram Proof : (i) In ∆APD and ∆CQB, ∠ADP = ∠CBQ (Alternate interior angles for BC || AD) AD = CB (Opposite sides of parallelogram ABCD) DP = BQ (Given) ∆APD ≅ ∆CQB (Using SAS congruence rule) (ii) As we had observed that ∆APD ≅ ∆CQB, AP = CQ (CPCT) (iii) In ∆AQB and ∆CPD, ∠ABQ = ∠CDP (Alternate interior angles for AB || CD) AB = CD (Opposite sides of parallelogram ABCD) BQ = DP (Given) ∆AQB ≅ ∆CPD (Using SAS congruence rule) (iv) As we had observed that ∆AQB ≅ ∆CPD, AQ = CP (CPCT) (v) From the result obtained in (ii) and (iv), AQ = CP and AP = CQ Since opposite sides in quadrilateral APCQ are equal to each other, APCQ is a parallelogram.
Question-10 :- ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that
(i) Δ APB ≅ Δ CQD
(ii) AP = CQ
Given that : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Prove that : (i) Δ APB ≅ Δ CQD (ii) AP = CQ Proof : (i) In ∆APB and ∆CQD, ∠APB = ∠CQD (Each 90°) AB = CD (Opposite sides of parallelogram ABCD) ∠ABP = ∠CDQ (Alternate interior angles for AB || CD) ∆APB ≅ ∆CQD (By AAS congruency) (ii) By using the above result ∆APB ≅ ∆CQD, we obtain AP = CQ (By CPCT)
Question-11 :- In Δ ABC and Δ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively . 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) Δ ABC ≅ Δ DEF.
Given that : In Δ ABC and Δ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Prove 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) Δ ABC ≅ Δ DEF. Proof : (i) It is given that AB = DE and AB || DE. If two opposite sides of a quadrilateral are equal and parallel to each other, then it will be a parallelogram. Therefore, quadrilateral ABED is a parallelogram. (ii) Again, BC = EF and BC || EF Therefore, quadrilateral BCEF is a parallelogram. (iii) As we had observed that ABED and BEFC are parallelograms, therefore AD = BE and AD || BE (Opposite sides of a parallelogram are equal and parallel) And, BE = CF and BE || CF (Opposite sides of a parallelogram are equal and parallel) AD = CF and AD || CF (iv) As we had observed that one pair of opposite sides (AD and CF) of quadrilateral ACFD are equal and parallel to each other, therefore, it is a parallelogram. (v) As ACFD is a parallelogram, therefore, the pair of opposite sides will be equal and parallel to each other. AC || DF and AC = DF (vi) ∆ABC and ∆DEF, AB = DE (Given) BC = EF (Given) AC = DF (ACFD is a parallelogram) ∆ABC ≅ ∆DEF (By SSS congruence rule)
Question-12 :- : ABCD is a trapezium in which AB || CD and AD = BC. Show that
(i) ∠ A = ∠ B
(ii) ∠ C = ∠ D
(iii) Δ ABC ≅ Δ BAD
(iv) diagonal AC = diagonal BD
Given that : ABCD is a trapezium in which AB || CD and AD = BC . Prove that : (i) ∠ A = ∠ B (ii) ∠ C = ∠ D (iii) Δ ABC ≅ Δ BAD (iv) diagonal AC = diagonal BD Construction : Let us extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at point E. Proof : (i) AD = CE (Opposite sides of parallelogram AECD) However, AD = BC (Given) Therefore, BC = CE ∠CEB = ∠CBE (Angle opposite to equal sides are also equal) Consider parallel lines AD and CE. AE is the transversal line for them. ∠A + ∠CEB = 180º (Angles on the same side of transversal) ∠A + ∠CBE = 180º (Using the relation ∠CEB = ∠CBE) ... (1) However, B + ∠CBE = 180º (Linear pair angles) ... (2) From equations (1) and (2), we obtain ∠A = ∠B (ii) AB || CD ∠A + ∠D = 180º (Angles on the same side of the transversal) Also, ∠C + ∠B = 180° (Angles on the same side of the transversal) ∠A + ∠D = ∠C + ∠B However, ∠A = ∠B [Using the result obtained in (i)] ∠C = ∠D (iii) In ∆ABC and ∆BAD, AB = BA (Common side) BC = AD (Given) ∠B = ∠A (Proved before) ∆ABC ≅ ∆BAD (SAS congruence rule) (iv) We had observed that, ∆ABC ≅ ∆BAD AC = BD (By CPCT)