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**SEBA Class 9 Mathematics Chapter 8 Quadrilaterals**

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**Solutions****SEBA Class 9 Mathematics Chapter 8 Quadrilaterals****Quadrilaterals**

**Quadrilaterals****Chapter – 8**

Exercise 8.1 |

**Q.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

**Q.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.

**Q.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.

**Q.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 Ic.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.

**Q.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.

**Q.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.

**Q.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

Friv (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.

**Q.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 intersects 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

**Q.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.

**Q.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.

**Q.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

**Q.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

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

| Aquadrillelogram 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.

Exercise 8.2 |

**Q.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

**Q.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: 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

**Q.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.

**Q.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- in 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.

**Q.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.

**Q.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.

**Q.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