**Chapter 8 Quadrilaterals Ex 8.1**

**Ex 8.1 Class 9 Maths 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°

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°.**

**Ex 8.1 Class 9 Maths 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. [

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

From (1) and (2), we have

**∠**

**ABC =**

**∠**

**DCB = 90°**

i.e., ABCD is a parallelogram having an angle equal to 90°.

i.e., ABCD is a parallelogram having an angle equal to 90°.

**∴**

**ABCD is a rectangle.**

**Ex 8.1 Class 9 Maths 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)

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.

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.

**Ex 8.1 Class 9 Maths 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)

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,

[Alternate interior angles are equal]

Similarly,

**∠**

**2 =**

**∠**

**4**

Now, in ∆OAD and ∆OCB, we have

AD = CB [Sides of a square ABCD]

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)

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.

From (1), (2) and (4), we get AC and BD are equal and bisect each other at right angles.

**Ex 8.1 Class 9 Maths 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

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]

In ∆AOD and ∆COB, we have

AO = CO [Given]

OD = OB [Given]

**∠**

**AOD =**

**∠**

**COB [Vertically opposite angles]**

So, ∆AOD

So, ∆AOD

**≅**

**∆COB [By SAS congruency]**

**∴∠**

**1 =**

**∠**

**2 [By C.P.C.T.]**

But, they form a pair of alternate interior angles.

But, they form a pair of alternate interior angles.

**∴**

**AD || BC**

Similarly, AB || DC

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]

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.

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.

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

Thus, ABCD is a square.

**Ex 8.1 Class 9 Maths Question
6.**

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

**∠**

**C also,**

(ii) ABCD is a rhombus.

(ii) ABCD is a rhombus.

Solution:

We have a parallelogram ABCD in which diagonal AC bisects

Solution:

We have a parallelogram ABCD in which diagonal AC bisects

**∠**

**A**

**⇒**

**∠**

**DAC =**

**∠**

**BAC**

(i) Since, ABCD is a parallelogram.

(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.

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

**∴**

**∠**

**2 =**

**∠**

**4 …(2)**

[ v Alternate interior angles are equal]

Also,

[ v Alternate interior angles are equal]

Also,

**∠**

**1 =**

**∠**

**2 …(3)**

[

[

**∵**

**AC bisects**

**∠**

**A]**

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

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]

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.

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

AB = BC = CD = DA

Thus, ABCD is a rhombus.

**Ex 8.1 Class 9 Maths 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
**

**⇒**

**∠**

**1 =**

**∠**

**2 …….(1)**

[

[

**∵**

**Angles opposite to equal sides of a triangle are equal]**

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

[

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

From (1) and (2), we have

**∠**

**2 =**

**∠**

**3 …(3)**

Since, AB || DC and AC is transversal.

Since, AB || DC and AC is transversal.

**∴**

**∠**

**2 =**

**∠**

**4 …(4)**

[

[

**∵**

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

we have

we have

**∠**

**1 =**

**∠**

**4**

**∴**

**AC bisects**

**∠**

**C as well as**

**∠**

**A.**

Similarly, we can prove that BD bisects

Similarly, we can prove that BD bisects

**∠**

**B as well as**

**∠**

**D.**

**Ex 8.1 Class 9 Maths 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

Solution:

We have a rectangle ABCD such that AC bisects

**∠**

**A as well as**

**∠**

**C.**

i.e.,

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

From (1) and (2), we have

**∠**

**3 =**

**∠**

**4**

In ∆ABC,

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.

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.**

**Ex 8.1 Class 9 Maths 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]

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]

Now, in ∆AQB and ∆CPD, we have

QB = PD [Given]

**∠**

**ABQ =**

**∠**

**CDP [Proved]**

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

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.**

**Ex 8.1 Class 9 Maths 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

Solution:

(i) In ∆APB and ∆CQD, we have

**∠**

**APB =**

**∠**

**CQD [Each 90°]**

AB = CD [

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.]**

**Ex 8.1 Class 9 Maths 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

(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.

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)

[

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

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]**

**Ex 8.1 Class 9 Maths Question
12.**

**ABCD is a trapezium in which AB || CD and
AD = BC (see figure). Show that
(i )**

**∠**

**A=**

**∠**

**B**

(ii )

(ii )

**∠**

**C=**

**∠**

**D**

(iii) ∆ABC

(iii) ∆ABC

**≅**

**∆BAD**

(iv) diagonal AC = diagonal

(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.

[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

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,

Also,

**∠**

**ABC +**

**∠**

**CBE = 180° … (4)**

[Linear pair]

and

[Linear pair]

and

**∠**

**A +**

**∠**

**CEB = 180° …(5)**

[Co-interior angles of a parallelogram ADCE]

From (4) and (5), we get

[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,

Similarly,

**∠**

**B +**

**∠**

**C = 180° … (8)**

From (7) and (8), we get

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.]**

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