NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1.

## NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1

Question 1.

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

Solution:

Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.

∴ 3x + 5x + 9x + 13x = 360°

[Angle sum property of a quadrilateral]

⇒ 30x = 360°

⇒ x = = 12°

∴ 3x = 3 x 12° = 36°

5x = 5 x 12° = 60°

9x = 9 x 12° = 108°

13a = 13 x 12° = 156°

⇒ The required angles of the quadrilateral are 36°, 60°, 108° and 156°.

Question 2.

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

Solution:

Let ABCD is a parallelogram such that AC = BD.

In ∆ABC and ∆DCB,

AC = DB [Given]

AB = DC [Opposite sides of a parallelogram]

BC = CB [Common]

∴ ∆ABC ≅ ∆DCB [By SSS congruency]

⇒ ∠ABC = ∠DCB [By C.P.C.T.] …(1)

Now, AB || DC and BC is a transversal. [ ∵ ABCD is a parallelogram]

∴ ∠ABC + ∠DCB = 180° … (2) [Co-interior angles]

From (1) and (2), we have

∠ABC = ∠DCB = 90°

i.e., ABCD is a parallelogram having an angle equal to 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:

Let ABCD be a quadrilateral such that the diagonals AC and BD bisect each other at right angles at O.

∴ In ∆AOB and ∆AOD, we have

AO = AO [Common]

OB = OD [O is the mid-point of BD]

∠AOB = ∠AOD [Each 90]

∴ ∆AQB ≅ ∆AOD [By,SAS congruency

∴ AB = AD [By C.P.C.T.] ……..(1)

Similarly, AB = BC .. .(2)

BC = CD …..(3)

CD = DA ……(4)

∴ From (1), (2), (3) and (4), we have

AB = BC = CD = DA

Thus, the quadrilateral ABCD is a rhombus.

Alternatively : ABCD can be proved first a parallelogram then proving one pair of adjacent sides equal will result in rhombus.

Question 4.

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

Solution:

Let ABCD be a square such that its diagonals AC and BD intersect at O.

(i) To prove that the diagonals are equal, we need to prove AC = BD.

In ∆ABC and ∆BAD, we have

AB = BA [Common]

BC = AD [Sides of a square ABCD]

∠ABC = ∠BAD [Each angle is 90°]

∴ ∆ABC ≅ ∆BAD [By SAS congruency]

AC = BD [By C.P.C.T.] …(1)

(ii) AD || BC and AC is a transversal. [∵ A square is a parallelogram]

∴ ∠1 = ∠3

[Alternate interior angles are equal]

Similarly, ∠2 = ∠4

Now, in ∆OAD and ∆OCB, we have

AD = CB [Sides of a square ABCD]

∠1 = ∠3 [Proved]

∠2 = ∠4 [Proved]

∴ ∆OAD ≅ ∆OCB [By ASA congruency]

⇒ OA = OC and OD = OB [By C.P.C.T.]

i.e., the diagonals AC and BD bisect each other at O. …….(2)

(iii) In ∆OBA and ∆ODA, we have

OB = OD [Proved]

BA = DA [Sides of a square ABCD]

OA = OA [Common]

∴ ∆OBA ≅ ∆ODA [By SSS congruency]

⇒ ∠AOB = ∠AOD [By C.P.C.T.] …(3)

∵ ∠AOB and ∠AOD form a linear pair.

∴∠AOB + ∠AOD = 180°

∴∠AOB = ∠AOD = 90° [By(3)]

⇒ AC ⊥ BD …(4)

From (1), (2) and (4), we get AC and BD are equal and 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:

Let ABCD be a quadrilateral such that diagonals AC and BD are equal and bisect each other at right angles.

Now, in ∆AOD and ∆AOB, We have

∠AOD = ∠AOB [Each 90°]

AO = AO [Common]

OD = OB [ ∵ O is the midpoint of BD]

∴ ∆AOD ≅ ∆AOB [By SAS congruency]

⇒ AD = AB [By C.P.C.T.] …(1)

Similarly, we have

AB = BC … (2)

BC = CD …(3)

CD = DA …(4)

From (1), (2), (3) and (4), we have

AB = BC = CD = DA

∴ Quadrilateral ABCD have all sides equal.

In ∆AOD and ∆COB, we have

AO = CO [Given]

OD = OB [Given]

∠AOD = ∠COB [Vertically opposite angles]

So, ∆AOD ≅ ∆COB [By SAS congruency]

∴∠1 = ∠2 [By C.P.C.T.]

But, they form a pair of alternate interior angles.

∴ AD || BC

Similarly, AB || DC

∴ ABCD is a parallelogram.

∴ Parallelogram having all its sides equal is a rhombus.

∴ ABCD is a rhombus.

Now, in ∆ABC and ∆BAD, we have

AC = BD [Given]

BC = AD [Proved]

AB = BA [Common]

∴ ∆ABC ≅ ∆BAD [By SSS congruency]

∴ ∠ABC = ∠BAD [By C.P.C.T.] ……(5)

Since, AD || BC and AB is a transversal.

∴∠ABC + ∠BAD = 180° .. .(6) [ Co – interior angles]

⇒ ∠ABC = ∠BAD = 90° [By(5) & (6)]

So, rhombus ABCD is having one angle equal to 90°.

Thus, ABCD is a square.

Question 6.

Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that

(i) it bisects ∠C also,

(ii) ABCD is a rhombus.

Solution:

We have a parallelogram ABCD in which diagonal AC bisects ∠A

⇒ ∠DAC = ∠BAC

(i) Since, ABCD is a parallelogram.

∴ AB || DC and AC is a transversal.

∴ ∠1 = ∠3 …(1)

[ ∵ Alternate interior angles are equal]

Also, BC || AD and AC is a transversal.

∴ ∠2 = ∠4 …(2)

[ v Alternate interior angles are equal]

Also, ∠1 = ∠2 …(3)

[ ∵ AC bisects ∠A]

From (1), (2) and (3), we have

∠3 = ∠4

⇒ AC bisects ∠C.

(ii) In ∆ABC, we have

∠1 = ∠4 [From (2) and (3)]

⇒ BC = AB …(4)

[ ∵ Sides opposite to equal angles of a ∆ are equal]

Similarly, AD = DC ……..(5)

But, ABCD is a parallelogram. [Given]

∴ AB = DC …(6)

From (4), (5) and (6), we have

AB = BC = CD = DA

Thus, ABCD is a rhombus.

Question 7.

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

Solution:

Since, ABCD is a rhombus.

⇒ AB = BC = CD = DA

Also, AB || CD and AD || BC

Now, CD = AD ⇒ ∠1 = ∠2 …….(1)

[ ∵ Angles opposite to equal sides of a triangle are equal]

Also, AD || BC and AC is the transversal.

[ ∵ Every rhombus is a parallelogram]

⇒ ∠1 = ∠3 …(2)

[ ∵ Alternate interior angles are equal]

From (1) and (2), we have

∠2 = ∠3 …(3)

Since, AB || DC and AC is transversal.

∴ ∠2 = ∠4 …(4)

[ ∵ Alternate interior angles are equal] From (1) and (4),

we have ∠1 = ∠4

∴ AC bisects ∠C as well as ∠A.

Similarly, we can prove that BD bisects ∠B as well as ∠D.

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.

Solution:

We have a rectangle ABCD such that AC bisects ∠A as well as ∠C.

i.e., ∠1 = ∠4 and ∠2 = ∠3 ……..(1)

(i) Since, every rectangle is a parallelogram.

∴ ABCD is a parallelogram.

⇒ AB || CD and AC is a transversal.

∴∠2 = ∠4 …(2)

[ ∵ Alternate interior angles are equal]

From (1) and (2), we have

∠3 = ∠4

In ∆ABC, ∠3 = ∠4

⇒ AB = BC

[ ∵ Sides opposite to equal angles of a A are equal]

Similarly, CD = DA

So, ABCD is a rectangle having adjacent sides equal.

⇒ ABCD is a square.

(ii) Since, ABCD is a square and diagonals of a square bisect the opposite angles.

So, BD bisects ∠B as well as ∠D.

Question 9.

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that

Solution:

We have a parallelogram ABCD, BD is the diagonal and points P and Q are such that PD = QB

(i) Since, AD || BC and BD is a transversal.

∴ ∠ADB = ∠CBD [ ∵ Alternate interior angles are equal]

⇒ ∠ADP = ∠CBQ

Now, in ∆APD and ∆CQB, we have

AD = CB [Opposite sides of a parallelogram ABCD are equal]

PD = QB [Given]

∠ADP = ∠CBQ [Proved]

∴ ∆APD ≅ ∆CQB [By SAS congruency]

(ii) Since, ∆APD ≅ ∆CQB [Proved]

⇒ AP = CQ [By C.P.C.T.]

(iii) Since, AB || CD and BD is a transversal.

∴ ∠ABD = ∠CDB

⇒ ∠ABQ = ∠CDP

Now, in ∆AQB and ∆CPD, we have

QB = PD [Given]

∠ABQ = ∠CDP [Proved]

AB = CD [ Y Opposite sides of a parallelogram ABCD are equal]

∴ ∆AQB = ∆CPD [By SAS congruency]

(iv) Since, ∆AQB = ∆CPD [Proved]

⇒ AQ = CP [By C.P.C.T.]

(v) In a quadrilateral ∆PCQ,

Opposite sides are equal. [Proved]

∴ ∆PCQ is a parallelogram.

Question 10.

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that

Solution:

(i) In ∆APB and ∆CQD, we have

∠APB = ∠CQD [Each 90°]

AB = CD [ ∵ Opposite sides of a parallelogram ABCD are equal]

∠ABP = ∠CDQ

[ ∵ Alternate angles are equal as AB || CD and BD is a transversal]

∴ ∆APB = ∆CQD [By AAS congruency]

(ii) Since, ∆APB ≅ ∆CQD [Proved]

⇒ AP = CQ [By C.P.C.T.]

Question 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, respectively (see figure).

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

Solution:

(i) We have AB = DE [Given]

and AB || DE [Given]

i. e., ABED is a quadrilateral in which a pair of opposite sides (AB and DE) are parallel and of equal length.

∴ ABED is a parallelogram.

(ii) BC = EF [Given]

and BC || EF [Given]

i.e. BEFC is a quadrilateral in which a pair of opposite sides (BC and EF) are parallel and of equal length.

∴ BEFC is a parallelogram.

(iii) ABED is a parallelogram [Proved]

∴ AD || BE and AD = BE …(1)

[ ∵ Opposite sides of a parallelogram are equal and parallel] Also, BEFC is a parallelogram. [Proved]

BE || CF and BE = CF …(2)

[ ∵ Opposite sides of a parallelogram are equal and parallel]

From (1) and (2), we have

AD || CF and AD = CF

(iv) Since, AD || CF and AD = CF [Proved]

i.e., In quadrilateral ACFD, one pair of opposite sides (AD and CF) are parallel and of equal length.

∴Quadrilateral ACFD is a parallelogram.

(v) Since, ACFD is a parallelogram. [Proved]

So, AC =DF [∵ Opposite sides of a parallelogram are equal]

(vi) In ∆ABC and ∆DFF, we have

AB = DE [Given]

BC = EF [Given]

AC = DE [Proved in (v) part]

∆ABC ≅ ∆DFF [By SSS congruency]

Question 12.

ABCD is a trapezium in which AB || CD and AD = BC (see 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].

Solution:

We have given a trapezium ABCD in which AB || CD and AD = BC.

(i) Produce AB to E and draw CF || AD.. .(1)

∵ AB || DC

⇒ AE || DC Also AD || CF

∴ AECD is a parallelogram.

⇒ AD = CE …(1)

[ ∵ Opposite sides of the parallelogram are equal]

But AD = BC …(2) [Given]

By (1) and (2), BC = CF

Now, in ∆BCF, we have BC = CF

⇒ ∠CEB = ∠CBE …(3)

[∵ Angles opposite to equal sides of a triangle are equal]

Also, ∠ABC + ∠CBE = 180° … (4)

[Linear pair]

and ∠A + ∠CEB = 180° …(5)

[Co-interior angles of a parallelogram ADCE]

From (4) and (5), we get

∠ABC + ∠CBE = ∠A + ∠CEB

⇒ ∠ABC = ∠A [From (3)]

⇒ ∠B = ∠A …(6)

(ii) AB || CD and AD is a transversal.

∴ ∠A + ∠D = 180° …(7) [Co-interior angles]

Similarly, ∠B + ∠C = 180° … (8)

From (7) and (8), we get

∠A + ∠D = ∠B + ∠C

⇒ ∠C = ∠D [From (6)]

(iii) In ∆ABC and ∆BAD, we have

AB = BA [Common]

BC = AD [Given]

∠ABC = ∠BAD [Proved]

∴ ∆ABC = ∆BAD [By SAS congruency]

(iv) Since, ∆ABC = ∆BAD [Proved]

⇒ AC = BD [By C.P.C.T.]

### NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2

Question 1.

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

(i) SR || AC and SR = AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Solution:

(i) In ∆ACD, We have

∴ S is the mid-point of AD and R is the mid-point of CD.

SR = AC and SR || AC …(1)

[By mid-point theorem]

(ii) In ∆ABC, P is the mid-point of AB and Q is the mid-point of BC.

PQ = AC and PQ || AC …(2)

[By mid-point theorem]

From (1) and (2), we get

PQ = AC = SR and PQ || AC || SR

⇒ PQ = SR and PQ || SR

(iii) In a quadrilateral PQRS,

PQ = SR and PQ || SR [Proved]

∴ 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.

Solution:

We have a rhombus ABCD and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Join AC.

In ∆ABC, P and Q are the mid-points of AB and BC respectively.

∴ PQ = AC and PQ || AC …(1)

[By mid-point theorem]

In ∆ADC, R and S are the mid-points of CD and DA respectively.

∴ SR = AC and SR || AC …(2)

[By mid-point theorem]

From (1) and (2), we get

PQ = AC = SR and PQ || AC || SR

⇒ PQ = SR and PQ || SR

∴ PQRS is a parallelogram. …….(3)

Now, in ∆ERC and ∆EQC,

∠1 = ∠2

[ ∵ The diagonals of a rhombus bisect the opposite angles]

CR = CQ [ ∵ = ]

CE = CE [Common]

∴ ∆ERC ≅ ∆EQC [By SAS congruency]

⇒ ∠3 = ∠4 ...(4) [By C.P.C.T.]

But ∠3 + ∠4 = 180° ......(5) [Linear pair]

From (4) and (5), we get

⇒ ∠3 = ∠4 = 90°

Now, ∠RQP = 180° - ∠b [ Y Co-interior angles for PQ || AC and EQ is transversal]

But ∠5 = ∠3

[ ∵ Vertically opposite angles are equal]

∴ ∠5 = 90°

So, ∠RQP = 180° - ∠5 = 90°

∴ One angle of parallelogram PQRS is 90°.

Thus, PQRS is a rectangle.

Question 3.

ABCD is a rectangle and P, Q, R ans S are mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rhombus.

Solution:

We have,

Now, in ∆ABC, we have

PQ = AC and PQ || AC ...(1)

[By mid-point theorem]

Similarly, in ∆ADC, we have

SR = AC and SR || AC ...(2)

From (1) and (2), we get

PQ = SR and PQ || SR

∴ PQRS is a parallelogram.

Now, in ∆PAS and ∆PBQ, we have

∠A = ∠B [Each 90°]

AP = BP [ ∵ P is the mid-point of AB]

AS = BQ [∵ AD = BC]

∴ ∆PAS ≅ ∆PBQ [By SAS congruency]

⇒ PS = PQ [By C.P.C.T.]

Also, PS = QR and PQ = SR [∵opposite sides of a parallelogram are equal]

So, PQ = QR = RS = SP i.e., PQRS is a parallelogram having all of its sides equal.

Hence, 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 figure). Show that F is the mid-point of BC.

Solution:

We have,

In ∆DAB, we know that E is the mid-point of

AD and EG || AB [∵ EF || AB]

Using the converse of mid-point theorem, we get, G is the mid-point of BD.

Again in ABDC, we have G is the midpoint of BD and GF || DC.

[∵ AB || DC and EF || AB and GF is a part of EF]

Using the converse of the mid-point theorem, we get, F is the mid-point of BC.

Question 5.

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

Solution:

Since, the opposite sides of a parallelogram are parallel and equal.

∴ AB || DC

⇒ AE || FC ...(1)

and AB = DC

⇒ AB = DC

⇒ AE = FC ...(2)

From (1) and (2), we have

AE || PC and AE = PC

∴ ∆ECF is a parallelogram.

Now, in ∆DQC, we have F is the mid-point of DC and FP || CQ

[∵ AF || CE]

⇒ DP = PQ ...(3)

[By converse of mid-point theorem] Similarly, in A BAP, E is the mid-point of AB and EQ || AP [∵AF || CE]

⇒ BQ = PQ ...(4)

[By converse of mid-point theorem]

∴ From (3) and (4), we have

DP = PQ = BQ

So, 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.

Solution:

Let ABCD be a quadrilateral, where P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

Join PQ, QR, RS and SP.

Let us also join PR, SQ and AC.

Now, in ∆ABC, we have P and Q are the mid-points of its sides AB and BC respectively.

∴ PQ || AC and PQ = AC ...(1)

[By mid-point theorem]

Similarly, RS || AC and RS = AC ...(2)

∴ By (1) and (2), we get

PQ || RS, PQ = RS

∴ PQRS is a parallelogram.

And the diagonals of a parallelogram bisect each other, i.e., PR and SQ bisect each other. Thus, the line segments joining the midpoints of opposite sides of a quadrilateral ABCD 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

(ii) MD ⊥ AC

(iii) CM = MA = AB

Solution:

we have

(i) In ∆ACB, We have

M is the mid-point of AB. [Given]

MD || BC , [Given]

∴ Using the converse of mid-point theorem,

D is the mid-point of AC.

(ii) Since, MD || BC and AC is a transversal.

∠MDA = ∠BCA

[ ∵ Corresponding angles are equal] As

∠BCA = 90° [Given]

∠MDA = 90°

⇒ MD ⊥AC.

(iii) In ∆ADM and ∆CDM, we have

∠ADM = ∠CDM [Each equal to 90°]

MD = MD [Common]

AD = CD [∵ D is the mid-point of AC]

∴ ∆ADM ≅ ∆CDM [By SAS congruency]

⇒ MA = MC [By C.P.C.T.] .. .(1)

∵ M is the mid-point of AB [Given]

MA = AB ...(2)

From (1) and (2), we have

CM = MA = AB

## NCERT Solutions for Class 9 Maths

- Chapter 1 Number systems
- Chapter 2 Polynomials
- Chapter 3 Coordinate Geometry
- Chapter 4 Linear Equations in Two Variables
- Chapter 5 Introduction to Euclid Geometry
- Chapter 6 Lines and Angles
- Chapter 7 Triangles
- Chapter 8 Quadrilaterals
- Chapter 9 Areas of Parallelograms and Triangles
- Chapter 10 Circles
- Chapter 11 Constructions
- Chapter 12 Heron’s Formula
- Chapter 13 Surface Areas and Volumes
- Chapter 14 Statistics
- Chapter 15 Probability
- Class 9 Maths (Download PDF)

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