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

NCERT Solutions for Class
9 Maths Chapter 8 Quadrilaterals Ex 8.1 are the part
of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for
Chapter 8 Quadrilaterals Ex 8.1.

**NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1****NCERT Solutions for Class 9 Maths Chapter 8****Quadrilaterals**Ex 8.2

**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° [Using angle sum property of a quadrilateral]

⇒ 30x = 360°

⇒ x = 360°/30 = 12°

∴ 3x = 3 × 12° = 36°

5x = 5 × 12° = 60°

9x = 9 × 12° = 108°

13x = 13 × 12° = 156°

Thus,
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.

AC = DB [Given]

AB = DC [Opposite sides of a parallelogram]

BC = CB [Common]

∴ ∆ABC ≅ ∆DCB [By SSS congruency rule]

⇒ ∠ABC = ∠DCB [By CPCT] …(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 get

∠ABC = ∠DCB = 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.

AO = AO [Common]

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

∠AOB = ∠AOD [Each 90°]

∴ ∆AQB ≅ ∆AOD [By SAS congruency rule]

∴ AB = AD [By CPCT] ……..(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.

**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 of a square 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 rule]

AC = BD [By CPCT] …..…(1)

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

∴ ∠DAC = ∠ACB
[Alternate interior angles are equal]

Similarly, ∠ADB = ∠DBC

Now, in ∆OAD and ∆OCB, we have

AD = CB [Sides
of a square ABCD]

∠DAO = ∠OCB
[Proved above]

∠ADO = ∠OBC
[Proved above]

∴ ∆OAD ≅ ∆OCB [By
ASA congruency rule]

⇒ OA = OC and OD = OB [By CPCT]

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

(iii)
In ∆OBA and ∆ODA, we have

OB = OD [Proved above]

BA = DA [Sides of
a square ABCD]

OA = OA [Common]

∴ ∆OBA ≅ ∆ODA [By SSS congruency rule]

⇒ ∠AOB = ∠AOD [By CPCT] ….…(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 diagonals of a
square 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. So,
ABCD is a parallelogram.

BO = OD (Given)

AO = OA (Common)

∠AOB = ∠AOD (Given 90°)

∴ Î”ABO ≅ Î”ADO (By SAS congruency rule)

AB = AD (By CPCT)

Similarly, AB = BC, BC = CD and CD = AD

i.e.,
AB = BC = CD = DA ……..(i)

Again,
in Î”ABC and Î”BAD, we have

AB =
BA (Common)

AC
= BD
(Given)

BC =
AD
[From (i)]

∴ Î”ABC ≅ Î”BAD

∴ ∠ABC = ∠BAD ……..(ii)

But ∠ABC + ∠BAD = 180° (Sum of co-interior angles of a
parallelogram is 180°)

∴ ∠ABC = ∠BAD = 90° [From eq. (ii)]

Thus, AB = BC = CD = DA and ∠A = 90°

∴ 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.

**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) [ ∵ 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

Hence, ABCD is a rhombus.

**Ex 8.1 Class 9 Maths**** ****Question 7.
**ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD
bisects ∠B as well as ∠D.

**Solution:
**Since, ABCD is a rhombus.

⇒ AB = BC = CD = DA

Also, AB || DC 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 a transversal.

⇒ ∠1 = ∠3 ….…(2) [ ∵ Alternate interior
angles are equal]

From (1) and (2), we have

∠2 = ∠3 ….…(3)

Since, AB || DC and AC is a 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.

**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 ∠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 || DC 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.

**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

**(i)**
Î”APD ≅ Î”CQB

**(ii)**
AP = CQ

**(iii)**
Î”AQB ≅ Î”CPD

**(iv)** AQ = CP

**(v) **APCQ is a parallelogram.

**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
above]

∴ ∆APD ≅ ∆CQB [By
SAS congruency rule]

**(ii)**
Since ∆APD ≅ ∆CQB [Proved above]

⇒ AP = CQ [By CPCT]

**(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 above]

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

∴ ∆AQB ≅ ∆CPD [By
SAS congruency rule]

**(iv) **Since
∆AQB ≅ ∆CPD [Proved above]

⇒ AQ = CP
[By CPCT]

**(v)**
In a quadrilateral APCQ,

Opposite sides are equal. [Proved
above]

∴ APCQ 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

**(i)**
∆APB ≅ ∆CQD

**(ii)**
AP = CQ

**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 rule]

**(ii)**
Since ∆APB ≅ ∆CQD [Proved
above]

⇒ AP = CQ [By CPCT]

**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 ≅ ∆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 in (i)]

∴ AD || BE and AD = BE ….…(1)

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

Also, BEFC is a parallelogram. [Proved in (ii)]

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
above]

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
above]

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

**(vi)**
In ∆ABC and ∆DFF, we have

AB = DE [Given]

BC = EF [Given]

AC = DF [Proved
in (v)]

∆ABC ≅ ∆DFF [By
SSS congruency rule]

**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)**∠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 CE || AD

⇒ AE || DC Also AD || CE

∴ AECD is a parallelogram.

⇒ AD = CE …..…(1)

[ ∵ Opposite sides of the parallelogram are equal]

But AD = BC …..…(2) [Given]

From (1) and (2), we get BC = CE

Now, in ∆BCE, we have BC = CE

⇒ ∠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
above]

∴ ∆ABC ≅ ∆BAD [By
SAS congruency rule]

**(iv)**
Since ∆ABC ≅ ∆BAD [Proved above]

⇒ AC = BD [By CPCT]

**NCERT Solutions for Maths Class 10**