# Class IX Chapter 8 Quadrilaterals Maths

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1 Class IX Chapter 8 Quadrilaterals Maths Exercise 8.1 Question 1: The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral. Answer: 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 360º, 3x + 5x + 9x + 13x = 360º 30x = 360º x = 12º Hence, the angles are 3x = 3 12 = 36º 5x = 5 12 = 60º 9x = 9 12 = 108º 13x =

2 13 12 = 156º Question 2: If the diagonals of a parallelogram are equal, then show that it is a rectangle. Answer: Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of its interior angles is 90º. 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) ABC = DCB It is known that the sum of the measures of angles on the same side of transversal is 180º. ABC + DCB = 180º (AB CD) ABC + ABC = 180º 2 ABC = 180º ABC = 90º

3 Since ABCD is a parallelogram and one of its interior angles is 90º, 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. Answer: 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 = 90º. To prove ABCD a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD are equal. In AOD and COD, OA = OC (Diagonals bisect each other) AOD = COD (Given) OD = OD (Common) AOD COD (By SAS congruence rule) AD = CD (1) Similarly, it can be proved that AD = AB and CD = BC (2) From equations (1) and (2),

4 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. Answer: Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O. To prove that the diagonals of a square are equal and bisect each other at right angles, we have to prove AC = BD, OA = OC, OB = OD, and AOB = 90º. 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,

5 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 = 180º (Linear pair) AOB = 180º 2 AOB = 90º 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. Answer:

6 Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O. It is given that 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 = = 90º. To prove ABCD is a square, we have to prove that ABCD is a parallelogram, AB = BC = CD = AD, and one of its interior angles is 90º. 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) AB = CD (By CPCT)... (1) And, OAB = OCD (By CPCT) 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,

8 (ii) ABCD is a rhombus. Answer: (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

9 BD bisects B as well as D. Answer: Let us join AC. 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.

10 Answer: ( 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)

11 1 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 the given figure). Show that:

12 i) APD CQB ( (ii) AP = CQ iii) AQB CPD ( (iv) AQ = CP (v) APCQ is a parallelogram Answer: (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

13 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 (See the given figure). Show that i) APB CQD ( (ii) AP = CQ Answer: (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

15 (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 (see the given figure). Show that i) A = B ( ii) C = D ( iii) ABC BAD ( (iv) diagonal AC = diagonal BD [Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

16 Answer: Let us extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at point E. It is clear that AECD is a parallelogram. (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)

17 (iv) We had observed that, ABC BAD AC = BD (By CPCT) Exercise 8.2 Question 1:

18 ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see the given figure). AC is a diagonal. Show that: ( i) SR AC and SR = AC ( ii) PQ = SR ( iii) PQRS is a parallelogram. Answer: (i) In ADC, S and R are the mid-points of sides AD and CD respectively. In a triangle, the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it. SR AC and SR = AC... (1) (ii) In ABC, P and Q are mid-points of sides AB and BC respectively. Therefore, by using mid-point theorem, PQ AC and PQ = AC... (2) Using equations (1) and (2), we obtain PQ SR and PQ = SR... (3) PQ = SR

19 (iii) From equation (3), we obtained PQ SR and PQ = SR Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal. Hence, PQRS is a parallelogram. Question 2: ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. Answer: In ABC, P and Q are the mid-points of sides AB and BC respectively. AC (Using mid-point theorem)... (1) R and S are the mid-points of CD and AD respectively. PQ AC and PQ = In ADC, (2) RS AC and RS = AC (Using mid-point theorem)... From equations (1) and (2), we obtain PQ RS and PQ = RS Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram. Let the diagonals of rhombus ABCD intersect each other at point O.

20 In quadrilateral OMQN, MQ ON ( PQ AC) QN OM ( QR BD) Therefore, OMQN is a parallelogram. MQN = NOM PQR = NOM However, NOM = 90 (Diagonals of a rhombus are perpendicular to each other) PQR = 90 Clearly, PQRS is a parallelogram having one of its interior angles as 90º. Hence, PQRS is a rectangle. Question 3: ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. Answer: Let us join AC and BD. In ABC, P and Q are the mid-points of AB and BC respectively.

21 PQ AC and PQ = AC (Mid-point theorem)... (1) Similarly ADC, in SR AC and SR = AC (Mid-point theorem)... (2) Clearly, PQ SR and PQ = SR Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram. PS QR and PS = QR (Opposite sides of parallelogram)... (3) In BCD, Q and R are the mid-points of side BC and CD respectively. QR BD and QR = BD (Mid-point theorem)... (4) However, the diagonals of a rectangle are equal. AC = BD (5) By using equation (1), (2), (3), (4), and (5), we obtain PQ = QR = SR = PS Therefore, PQRS is a rhombus. Question 4: ABCD is a trapezium in which AB DC, BD is a diagonal and E is the mid - point of AD. A line is drawn through E parallel to AB intersecting BC at F (see the given figure). Show that F is the mid-point of BC.

22 Answer: Let EF intersect DB at G. By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side, bisects the third side.

23 1 In ABD, EF AB and E is the mid-point of AD. Therefore, G will be the mid-point of DB. As EF AB and AB CD, EF CD (Two lines parallel to the same line are parallel to each other) In BCD, GF CD and G is the mid-point of line BD. Therefore, by using converse of mid-point theorem, F is the mid-point of BC. Question 5:

24 In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the given figure). Show that the line segments AF and EC trisect the diagonal BD. Answer: ABCD is a parallelogram. AB CD And hence, AE FC Again, AB = CD (Opposite sides of parallelogram ABCD) AB = CD AE = FC (E and F are mid-points of side AB and CD) In quadrilateral AECF, one pair of opposite sides (AE and CF) is parallel and equal to each other. Therefore, AECF is a parallelogram. AF EC (Opposite sides of a parallelogram) In DQC, F is the mid-point of side DC and FP CQ (as AF EC). Therefore, by using the converse of mid-point theorem, it can be said that P is the mid-point of DQ. DP = PQ... (1)

25 Similarly, in APB, E is the mid-point of side AB and EQ AP (as AF EC). Therefore, by using the converse of mid-point theorem, it can be said that Q is the mid-point of PB. PQ = QB... (2) From equations (1) and (2), DP = PQ = BQ Hence, the line segments AF and EC trisect the diagonal BD. Question 6: Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. Answer: Let ABCD is a quadrilateral in which P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively. Join PQ, QR, RS, SP, and BD. In ABD, S and P are the mid-points of AD and AB respectively. Therefore, by using mid-point theorem, it can be said that

26 SP BD and SP = BD... (1) Similarly BCD, in QR BD and QR = BD... (2) From equations (1) and (2), we obtain SP QR and SP = QR In quadrilateral SPQR, one pair of opposite sides is equal and parallel to each other. Therefore, SPQR is a parallelogram. We know that diagonals of a parallelogram bisect each other. Hence, PR and QS bisect each other. Question 7: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC

27 It is given that M is the mid-point of AB and MD BC. Therefore, D is the mid-point of AC. (Converse of mid-point theorem) (ii) As DM CB and AC is a transversal line for them, therefore, MDC + DCB = 180º (Co-interior angles) MDC + 90º = 180º MDC = 90º MD AC (iii) Join MC. In AMD and CMD,

28 AD = CD (D is the mid-point of side AC) ADM = CDM (Each 90º) DM = DM (Common) AMD CMD (By SAS congruence rule) Therefore, AM = CM (By CPCT) However, AM = AB (M is the mid-point of AB) Therefore, it can be said that CM = AM = AB

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