SEBA Class 9 Mathematics Chapter 8 Quadrilaterals

SEBA Class 9 Mathematics Chapter 8 Quadrilaterals Solutions, SEBA Class 9 Maths Textbook Notes in English Medium, SEBA Class 9 Mathematics Chapter 8 Quadrilaterals Solutions in English to each chapter is provided in the list so that you can easily browse throughout different chapter Assam Board SEBA Class 9 Mathematics Chapter 8 Quadrilaterals Notes and select needs one.

SEBA Class 9 Mathematics Chapter 8 Quadrilaterals

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Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 9 Mathematics Chapter 8 Quadrilaterals Question Answer. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 9 Mathematics Chapter 8 Quadrilaterals Solutions for All Subject, You can practice these here.

Quadrilaterals

Chapter – 8

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

Ans: Let ABCD be a quadrilateral in which 

∠A: ∠B: ∠C: ∠D = 3:5:9:13 

Sum of the ratios = 3 + 5 + 9 + 13 = 30 

Also, ∠A+∠B+∠C+∠D = 360⁰  | Sum of all the angles of a quadrilateral is 360⁰

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

Ans: Given: In parallelogram ABCD, AC = BD

To Prove: || gm ABCD is a rectangle.

Proof: In ∆ ACB and ∆ BDA 

AC = BD | Given

AB = BA | Common

BC = AD | Opposite sides of || gm ABCD

∴ ACB ≅ ∆BDA | ISSS Rule

∴ ∠ ABC = ∠ BAD …(1) c.p.c.t.

Again, ∵ AD || BC | Opp. sides of || gm ABCD and transversal AB intersects them

∴ ∠BAD+∠ABC = 180⁰…(2) Sum of consecutive interior angles on the same side of the transversal is 180⁰

From (1) and (2), 

∠BAD = ∠ABC = 90⁰ ∴ ∠ A = 90⁰ 

∴ || gm ABCD is a rectangle.

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

Ans: Given: ABCD is a quadrilateral where diagonals AC and BD intersect each other at right angles at O.

To Prove: Quadrilateral ABCD is a rhombus. 

Proof: In ∆AOB and ∆AOD

AO = AO | Common

OB = OD | Given

∠AOB =∠AOD | Each = 90⁰

∴ ∆AOB ≅ ∆AOD | SAS Rule

∴ AB = AD …(1) I c.p.c.t.

Similarly, we can prove that 

AB = BC 

BC = CD 

CD = AD

In view of (1), (2), (3) and (4), we obtain 

AB = BC = CD = DA

∴ Quadrilateral ABCD is a rhombus. 

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

Ans: Given: ABCD is a square

To Prove: (i) AC = BD

(ii) AC and BD bisect each other at right angles. 

Proof: (i) In ∆ABC and ∆BAD,

AB = BA | Common 

BC = AD | Opp. sides of square ABCD 

∠ ABC = ∠ BAD | Each = 90⁰ (∵ ABCD is a square) 

∴ ∆ ABC ≅ ∆BAD | SAS Rule 

AC = BD I c.p.c.t. 

(ii) In ∆ OAD and ∆OCB

AD = CB | Opp. sides of square ABCD

∠ OAD = ∠ OCB | ∵ AD || BC and transversal AC intersects them

∠ ODA = ∠ OBC | ∵ AD || BC and transversal BD intersects them

∴ ∆ OAD ≅ ∆OCB | ASA Rule 

∴ OA = OC …(1)

Similarly, we can prove that 

OB = OD …(2)

In view of (1) and (2) 

AC and BD bisect each other. 

Again, in ∆OBA and ∆ODA 

OB = OD | From (2) above

BA = DA | Opp. sides of square ABCD

OA = OA | Common

∴ ∆ OBA ≅ ∆ ODA | SSS Rule

∴ ∠AOB = ∠AOD | c.p.c.t. 

But ∠AOB + ∠AOD = 180⁰  | Linear Pair Axiom

∴ ∠AOB = ∠AOD = 90⁰ 

∴ AC and BD bisect each other at right angles.

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

Ans: Given: The diagonals AC and BD of a quadrilateral ABCD and equal and bisect each other at right angles.

To Prove: Quadrilateral ABCD is a square.

Proof: In ∆OAD and ∆OCB, 

OA = OC | Given

OD = OB | Given

∠ AOD = ∠ COB  | Vertically Opposite angles 

∴ ∆ OAD ≅ ∆OCB | SAS Rule 

∴ AD = CB | c.p.c.t.

∠ODA = ∠ OBC | c.p.c.t.

∴ ∠ODA =∠OBC 

∴ AD || BC

Now, AD = CB and AD || CB

∴ Quadrilateral ABCD is a || gm. 

In ∆AOB and ∆AOD

AO = AO  | Common

OB = OD | Given

∠AOB = ∠ AOD  | Each = 90⁰ (Given) 

∴ ∆ AOB ≅ ∆AOD  |SAS Rule

∴ AB = AD 

Now, ∵ ABCD is a parallelogram and AB = AD

∴ ABCD is a rhombus.

Again, in ∆ ABC and ∆BAD,

AC = BD | Given

BC = AD  | ∵ ABCD is a rhombus

AB = BA | Common

∴ ∆ABC ≅ ∆BAD | SSS Rule

∴ ∠ABC ≅ ∠ BAD | c.p.c.t.

∵ AD || BC | Opp. sides of || gm ABCD and transversal AB intersects them.

∴ ∠ABC+∠BAD = 180⁰ | Sum of consecutive interior angles on the same side of the transversal is 180⁰

∴ ∠ABC = ∠BAD = 90⁰ 

Similarly, ∠BCD = ∠ADC = 90⁰ 

∴ ABCD is a square.

6. Diagnosal AC of a parallelogram ABCD bisects ∠A. Show that:

(i) it bisects ∠C also

(ii) ABCD is a rhombus

Ans: Given: Diagonal AC of a parallelogram ABCD bisects ∠A.

To Prove: (i) it bisects ∠C also.

(ii) ABCD is a rhombus.

Proof: (i) In ∆ADC and ∆ CВА,

AD = BC | Opp. sides of || gm ABCD

CA = CA | Opp. sides of ||gm ABCD

∴ ∆ADCA ≅ ∆СВА | SSS Rule

∴ ∠ACD = ∠CAB | c.p.c.t.

and ∠DAC = ∠BCA  | c.p.c.t.

But ∠CAB = ∠DAC | Given 

∴ ∠ACD = ∠ BCA

∴ AC bisects ∠C also

(ii) From above,

∠ACD = ∠CAD

∴ AD = CD | Opposite sides of equal angles of a triangle are equal 

∴ AB = BC = CD = DA | ∵ ABCD is a || gm

∴ ABCD is a rhombus.

7. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Ans: Given: ABCD is a rhombus. 

To Prove: (i) Diagonal AC bisects ∠A as well as ∠C

(ii) Diagonal BD bisects ∠B as well as ∠D

Proof: ∵ ABCD is a rhombus 

∴ AD = CD 

∴ ∠DAC = ∠DCA …(1) | Angles opposite to equal sides of a triangle are equal 

Also, CD || AB and transversal AC intersects them 

∴ ∠DCA = ∠BCA ..(2) | Alt. Int. ∠S 

Frin (1) and (2) 

∠DCA = ∠BCA 

⇒ AC bisects ∠C 

Similarly AC bisects ∠A 

(ii) Proceeding similarly as in (i) above, we can prove that BD bi-sects ∠B as well as ∠D.

8. ABCD is a rectangle in which diagonal AC bisects ∠A. Show that 

(i) ABCD is a square.

(ii) diagonal BD bisects ∠B as well as ∠D.

Ans: Given: ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C

To Prove: (i) ABCD is a square.

(ii) diagonal BD bisects ∠B as well as ∠D.

Proof: (i) ∵ AB || DC and transversal AC inresects them

∴ ∠ACD = ∠CAB || Alt. Int. ∠S 

But ∠CAB = ∠CAD 

∴ ∠ACD = ∠CAD 

∴ AD = CD  | Sides opposite to equal angles of a triangle are equal

∴ ABCD is a square 

(ii) In ∆BDA and ∆DBC 

BD = DB | Common

DA = BC | Sides of a square ABCD

AB = DC | Sides of a square ABCD

∴ ∆BDA ≅ ∆DBC | SSS Rule

∴ ∠ABD = ∠ CDB | c.p.c.t.

But ∠CDB = ∠CBD | ∵ CB = CD (Sides of a square ABCD)

∴ ∠ ABD = ∠ CBD

∴ BD bisects ∠ B 

Now, ∠ ABD = ∠ CBD

∠ABD = ∠ ADB | ∵ AB = AD

∠CBD =∠CDB  | ∵ CB = CD

∴ ∠ ADB = ∠ CDB

∴ BD bisects ∠ D

9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ. Show that:

(i) ∆ APD ≅ ∆CQB

(ii) AP = CQ

(iii) ∆ AQB ≅ ∆CPD

(iv) AQ = CP

(v) APQC is a parallelogram.

Ans: Given: In parallelogram ABCD, two points P and Q are on di-agonal BD such that DP = BQ

Construction: Join AC to intersect BD at O.

Proof: (i) In ∆APD and ∆CQB | From (ii)

PD = QB | Given

AD = BC | Opp. sides of || gm ABCD    

∴ ∆ APD ≅ ∆ CQB | SSS Rule

(ii) ∵ APCQ is a || gm   | Proved in (i) above

AP = CO   | Op. sides of a || gm are equal 

(iii) In ∆AQB and ∆ CPD, 

AQ = CP | From (iii)

QB = PD | Given

AB = CD | Opp. sides of || gm ABCD

∴ ∆ AQB ≅ ∆CPD | SSS Rule

(iv) APQC is a || gm   | Proved in (i) above

∴ AQ = CP   | Opp. sides of a || gm are equal

(v) ∵ The diagonals of a parallelogram bisect each other. 

∴ OB = OD 

∴ OB-BQ = OD-DP   | ∵ BQ = DP (given)

∴ OQ = OP …(1)

Also, OA = OC…(2)  | ∵ Diagonals of a || gm bisect each other

In view of (1) and (2), APQC is a parallelogram.

10. ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD respectively. Show that:

(i) ΔΑΡΒ ≅ ∆ CQD

(ii) AP = CQ

Ans: Given: ABCD is a parallelo-gram and AP and CQ are perpendiculars from verti-ces A and Con diagonal BD respec-tively.

To Prove: (i) ∆ APB ≅ ∆ CQD

(ii) AP = CQ

Proof: (i) In ∆APB and ∆ CQD,

AB = CD  | Opp. sides || gm ABCD

∠ ABP = ∠ CDQ | ∵ AB || DC and transversal BD intersects them

∠APB = ∠CQD | Each = 90⁰

∴ ΔΑΡΒ  ≅ Δ CQD | AAS Rule

(ii) ∵ ∆ APB ≅ ∆CQD | Proved above in (i)

∴ AP = CQ | c.p.c.t.

11. In ∆ABC and ∆ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respec-tively. 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) ∆ ABC ≅ DEF

Ans: Given: In ∆ ABC and∆ DEF,

AB = DE, AB || DE, BC || EF and BC || EF.

Vertices A, B and C are joined to vertices D, E and F respectively.

To Prove: (i) quadrilateral ABED is a parallelogram

(ii) quadrilateral BECF is a parallelogram

(iii) AD || CF and AD = CF

(iv) quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) ∆ ABC ≅ ∆ DEF

Proof: (i) In quadrilateral ABED,

AB = DE and AB || DE   | Given

∴ quadrilateral ABED is a parallelogram. 

|∵ A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length

(ii) In quadrilateral BECF, 

BC = EF and BC || EF | Given

∴ quadrilateral BECF is a parallelogram. 

| ∵ A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length

(iii) ∵ ABED is a parallelogram  | Proved in (i)

∴ AD || BE and AD = BE …(1) | ∵ Opposite sides of a || gm are parallel and equal

∵ BEFC is a parallelogram | Proved in (ii)

∴ BE|| CF and BE = CF …(2) | ∵ Opposite sides of a || gm are parallel and equal 

From (1) and (2), we obtains AD || CF and AD = CF

(iv) In quadrilateral ACFD 

AD || CF and AD = CF | From (iii)

∴ quadrilateral ACFD is a parallelogram. 

∵ A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length

(v) ∵ ACFD is a parallelogram | Proved in (iv)

∴ AC || DF and AC = DF | In a parallelogram opposite sides and parallel and of equal length

(vi) In ∆ ABC and ∆ DEF,

AB = DE | ∵ ABED is a parallelogram

BC = EF | ∵ BEFC is a parallelogram

AC = DF | Proved in (v)

∴ Δ ABC ≅ ∆DEF | SSS Rule

12. ABCD is a trapezium in which AB || CD and AD = BC. Show that:

(i) ∠ A = ∠ B

(ii) ∠ C = ∠D

(iii) ∆ ABC ≅ ∆ BAD 

(iv) diagonal AC = diagonal BD

[Hing. Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] 

Ans: Given: ABCD is a trapezium in which AB || CD and AD = BC

To Prove: (i) ∠ A = ∠ B

(ii) ∠ C = ∠D

(iii) ∆ ABC ≅ ∆ BAD 

(iv) diagonal AC = diagonal BD

Construction: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.

Proof: (i) AB || CD | Given

and AD || EC | By construction

∴ AECD is a parallelogram | A quadrillelogram if a pair of opposite sides is parallel and is of equal length

∴ AD = EF | Opp. sides of a || gm are equal

But AD = BC | Given

∴ EC = BC

∴ ∠CBE = ∠CEB …(1) | Angles of opposite to equal sides of a triangle are equal 

∠B+∠CBE = 180⁰ …(2) | Linear Pair Axiom

∵ AD || FC | By construction and transversal AE intersects them

∴ ∠A+∠CEB = 180⁰ …(3) | The sum of consecutive interior angles on the same side of the transversal is 180⁰

From (2) and (3), 

∠B+∠CBE = ∠A+ ∠ CEB 

But ∠ CBE = ∠CEB | From (1)

∴ ∠B = ∠ A or ∠ A = ∠ B

(ii) ∵ AB || BD

∴ ∠ A + ∠ D = 180⁰ | The sum of consecutive interior angles on the same side of the transversal is 180⁰

and ∠ B+ ∠ C = 180⁰

∴ ∠A +∠D = ∠ B+ ∠ C

But ∠ A = ∠ B | Proved in (i)

∴ ∠ D = ∠C or∠C = ∠D

(iii) In ∆ABC and ∆BAD,

AB = BA | Common

BC = AD | Given

∠ ABC = ∠ BAD | From (i)

∴ ∆ ABC ≅ ∆BAD | SAS Rule

(iv) ∵ ∆ ABC ≅ ∆ BAD | From (iii) above

∴ AC = BD | c.p.c.t.

1. ABCD is a quadrilateral in which P, Q R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that:

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Ans: Given: ABCD is a quadrilat-eral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA, AC is a diagonal.

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Proof: In ∆ DAC

∵ Is the mid-point of DA and R is the midpoint of DC 

| Mid-point theorem

(ii) In ∆BAC

∵ P is the midpoint of AB and Q is the midpoint of BC 

| Mid-point theorem

∴ PQ = SR

(iii) PQ || AC  | From (i)

SR || AC  | From (ii)

∴ PQ|| SR  | Two lines parallel to the same line are parallel to each other 

Also, PQ = SR | From (ii)

∴ PQRS is a parallelogram  | A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length

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. 

Ans: PQRS is a rectangle.

Construction: Join AC and BD

Proof: In triangles RDS and PBQ

DS = QB | Half of opposite sides of || gm ABCD which are equal

DR = PB | Half of opposite sides of || gm ABCD which are equal 

∠SDR = ∠QBP | Opposite ∠S of || gm ABCD which are equal

∴ ∆ARDS ≅ ∆ PBQ | SAS Axiom

∴ SR = PQ | c.p.c.t.

In triangles RCQ and PAS,

RC = AP | Half of opposite sides of || gm ABCD which are equal 

CQ = AS | Half of opposite sides of || gm ABCD which are equal 

∠ CRQ = ∠ PAS | Opposite ∠S of || gm ABCD which are equal 

∴ ∆ CRCQ ≅ ΔΡΑS | SAS Axiom

∴ RQ = SP | c.p.c.t. 

∴ In PQRS, 

SR = PQ and RQ = SP 

∴ PQRS is a parallelogram,

In ∆ CDB,

∵ R and Q are the mid-points of DC and CB respectively. 

∴ RQ || DB ⇒ RF|| EO 

Similarly, RE || FO ∴ OFRE is a || gm ∴ ∠ R = ∠ EOF = 90⁰

| ∵ Opposite ∠ S of a || gm are equal and diagonals of a rhombus intersect at 90⁰

Thus PQRS is a rectangle

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.

Ans: Given: ABCD is a rectangle. P, Q, R and S are the mid-points of AB, BC, CD and DA respectively, PQ, QR, RS and SP are joined.

To prove: Quadrilateral PQRS is a rhombus.

Construction: Joined AC

Proof: In A ABC,

∴ P and Q are the mid-points of AB 

and BC respectively.

In ∆ ADC,

S and R are the mid-points of AD and DC respectively.

…(2)

From (1) and (2), PQ || SR and PQ = SR 

∴ Quadrilateral PQRS is a parallelogram …(3) 

In rectangle ABCD 

AD = BC  | Opposite sides

| Halves of equals are equal

⇒ AS = BQ 

In ∆ APS and ∆ BPQ, 

AP = BP  | ∵ P is the midpoint of AB 

AS = BQ | Proved above

∠ PAS = ∠PBO | Each = 90⁰

∴ ∆APS = ∆BPQ | SAS Axiom

∴ PS = PQ …(4) | c.p.c.t.

In view of (3) and (4), PQRS is a rhombus.

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the midpoint of AD. A line is drawn through E parallel to AB intersect-ing BC at F. Show that F is the midpoint of BC.

Ans: Given: ABCD is a trapezium in which AB || DC, BD is a diago-nal and E is the midpoint of AD. A line is drawn through E parallel to AB intersecting BC at F.

To Prove: F is the midpoint of BC. 

Proof: Let BD intersect EF at G.

In ∆ DAB,

∵ E is the mid-point of DA and EG || AB

∴ G is the midpoint of DB  | By converse of mid-point theorem Again, in ∆ BDC,

∵ G is the mid-point of BD and GF || AB || DC

∴ F is the midpoint of BC. | By converse of mid-point theorem.

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. Show that the segments AF and EC trisect the diagonal BD. 

Ans: Given: In a parallelogram ABCD, E and the mid-points of sides AB and CD respectively.

To Prove: Line segments AF and EC trisect the diagonal BD.

Proof: ∵ AB || DC | Opposite sides of || gm ABCD

∴ AE || FC …(1)

∵ AB = DC | Opposite sides of || gm ABCD

  | Half of equals are equal 

⇒ AE = CF …(2)

In view of (1) and (2),

AECF is a parallelogram  |A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length

∴ EC || AF …(3) | Opposite sides of || gm AECF

In ∆ DBC,

∵ F is the midpoint of DC 

and FP || CQ  | EC ||  AF

∴ P is the mid-point of DQ | By converse of mid-point theorem

⇒ DP = PQ …(4)

Similarly, in ∆BAP,

BQ = PQ …(5) 

From (4) and (5), we obtain 

DP = PQ = BQ 

⇒ Line segments AF and EC trisect the diagonal BD.

6. Show that the line segments joining the midpoints of the oppo-site side of a quadrilateral bisect each other.

Ans: Given: ABCD is a quadrilateral. P, Q, R and S are the mid-points of the sides DC, CB, BA and AD respectively.

To Prove: PR and QS bisect each other.

Construction: Join PQ, QR, RS, SP, AC and BD 

Proof: In ∆ABC,

∵ R and Q are the mid-points of AB and BC respectively.

Similarly, we can that 

∴ RQ || PS and RQ = PS

Thus a pair of opposite sides of a quadrilateral PQRS are parallel and equal.

∴ PQRS is a parallelogram.

Since the diagonals of a parallelogram bisect each other.

∴  PR and QS bisect each other.

7. ABC is a triangle right angled at C, A line through the midpoint M of hypotenuse AB and parallel to BC intersects AC at D. Show that:

(i) D is the midpoint of AC

Ans: Given: ABC is a triangle right angled at C.

A line through the midpoint M of hypotenuse 

AB and parallel to BC intersects AC at D.

To Prove: (i) D is the midpoint of AC.

(ii) MD ⊥ AC

Proof: (i) In ∆ ACB

∵ M is the mid-point of AB and MD || BC

∴ D is the midpoint of AC. | By converse of mid-point theorem

(ii) ∵ MD || BC and AC intersects them

∴ ∠ ADM = ∠ACB | Corresponding angles

But ∠ACB = 90⁰ | Given

∴ ∠ADM = 90⁰ ⇒ MD⊥AC

(iii) Now, ∠ ADM + ∠ CDM = 180⁰ | Linear Pair Axiom

∴ ∠ ADM = ∠ CDM = 90⁰

In ∆ ADM and ∆ CDM, 

AD = CD | ∵ D is the midpoint of AC

∠ADM = ∠ CDM | Each = 90⁰

DM = DM | Common

∴ ∆ ADM ≅ ∆ CDM | SAS Rule

∴ MA = MC | c.p.c.t.

But M is the midpoint of AB

Parimol

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