NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

NCERT
SOLUTIONS
CLASS - 9th

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Book : Mathematics Ncert Solutions | Chapter-8 Maths

Class : 9th
Subject : Maths
Chapter : 8
Chapter Name : QUADRILATERALS

Exercise 8.1

Q1 The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Answer. Let the common ratio between the angles be x. Therefore, the angles will be 3x, 5x, 9x,
and 13x respectively.
As the sum Of all interior angles Of a quadrilateral is 360 ,
∘

Therefore, 3x + 5x + 9x + 13x = 360
∘

30x = 360
∘

x = 12
∘

Hence, the angles are
3x = 3 × 12 = 360
∘
5x = 5 × 12 = 60
∘
9x = 9 × 12 = 108
∘
13x = 13 × 12 = 156

Page : 146 , Block Name : Exercise 8.1

Q2 If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer.

Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of its
interior angles is 90 .
∘

In △ABC and △DCB,
AB = DC (Opposite sides Of a parallelogram are equal)
BC = BC (Common)
AC = DB (Given)
∴ △ABC ≅ △DCB (By SSS Congruence rule)

⇒ ∠ABC = ∠DCB

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Book : Mathematics Ncert Solutions | Chapter-8 Maths
It is known that the sum of the measures of angles on the same side of transversal is 180 .
∘

∘
∠ABC + ∠DCB = 180 (AB∥CD)

∘
⇒ ∠ABC + ∠ABC = 180
∘
⇒ 2∠ABC = 180
∘
⇒ ∠ABC = 90

Since ABCD is a parallelogram and one of its interior angles is 90 , ABCD is a rectangle.
∘

Page : 146 , Block Name : Exercise 8.1

Q3 Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a
rhombus.

Answer.

Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each Other at right
angle i.e., OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90 . To prove ABCD
∘

a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD are equal.
In △AOD and ΔCOD ,
OA = OC (Diagonals bisect each other )
∠AOD = ∠COD (Given)

OD = OD (Common)
∴ △AOD = ΔCOD(By SAS congruence rule)

∴ AD = CD(1)

Similarly, it can be proved that
AD = AB and CD = BC (2)
From equations (1) and (2) ,
AB = BC = CD = AD
Since opposite sides of quadrilateral ABCD are equal, it can be said that ABCD is a parallelogram.
Since all sides of a parallelogram ABCD are equal, it can be said that ABCD is a rhombus.

Page : 146 , Block Name : Exercise 8.1

Q4 Show that the diagonals of a square are equal and bisect each other at right angles.

Answer.

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Book : Mathematics Ncert Solutions | Chapter-8 Maths

Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O. To prove that
the diagonals of a square are equal and bisect each other at right angles, we have to prove AC = BD,
OA = OC, OB = OD, and ∠AOB = 90 .
∘

In △ABC and ΔDCB,
AB = DC (Sides Of a square are equal to each Other)
∠ABC = ∠DCB (All interior angles are of 90 )
∘

BC = CB (Common side)
∴ △ABC = ΔDCB(By SAS congruency )

∴ AC = DB(ByCPCT)

Hence, the diagonals of a square are equal in length.
In △AOB and ΔCOD ,
∠AOB = ∠COD (Vertically Opposite Angle )

□ABO = □CDO (Alternate interior angles)

AB = CD (Sides of a square are always equal )
□△AOB□ΔCOD(By AAS congruence rule)

□AO = CO and OB = OD(ByCP CT )

Hence, the diagonals of a square bisect each other.
In △AOB and ΔCOB ,
As we had proved that diagonals bisect each other , therefore ,
AO = CO
AB = CB (Sides of a square are equal)
BO = BO (Common)
□△AOB□ΔCOB(By SSS congruence rule)

□□AOB = □COB(ByCP CT )

However, □AOB + □COB = 180 (Linear pair )
∘

∘
2□AOB = 180
∘
□AOB = 90

Hence, the diagonals of a square bisect each other at right angles.

Page : 146 , Block Name : Exercise 8.1

Q5 Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles,
then it is a square.

Answer.

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Book : Mathematics Ncert Solutions | Chapter-8 Maths

Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each
other at O. It is given that the diagonals of ABCD are equal and bisect each other at right angles.
Therefore, AC = BD, OA = OC, 0B = OD, and ∠AOB = ∠BOC = ∠COD= ∠AOD = 90 . To ∘

prove ABCD is a square, we have to prove that ABCD is a parallelogram, AB = BC = CD = AD, and
one of its interior angles is 90 .
∘

In △AOB and ΔCOD,
AO = CO (Diagonals bisect each other)
OB = OD (Diagonals bisect each other)
∠AOB = ∠COD( Vertically opposite angles )

□△AOB = □ΔCOD( SAS congruence rule)

□AB = CD(ByCP CT ) … (1)

And, □OAB = □OCD(ByCP CT )

However, these are alternate interior angles for line AB and CD and alternate interior
angles are equal to each other only when the two lines are parallel.
□AB∥CD … (2)

From equations (1) and (2), we obtain
ABCD is a parallelogram.
In △AOD and △COD ,
AO = CO (diagonals bisect each other)
∘
□AOD = □COD ( Given that each is 90 )

OD = OD (Common)
□△AOD□ΔCOD(SAS congruence rule)

□AD = DC … (3)

However, AD = BC and AB = CD (Opposite sides of parallelogram ABCD)
□AB = BC = CD = DA

Therefore, all the sides of quadrilateral ABCD are equal to each other.
In △ADC and ΔBCD, ,
AD = BC (Already proved)
AC = BD (Given)
DC =CD (Common)
□△ADC□ΔBCD(SSS Congruence rule)

□□ADC = □BCD(ByCPCT)

However, □ADC + □BCD = 180 ( Co-interior angles )
∘

∘
□□ADC + □ADC = 180
∘
□2□ADC = 180
∘
□□ADC = 90

One of the interior angles of quadrilateral ABCD is a right angle.
Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = AD and one of its interior

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Book : Mathematics Ncert Solutions | Chapter-8 Maths
angles is 90 .Therefore, ABCD is a square.
∘

Page : 146 , Block Name : Exercise 8.1

Q6 Diagonal AC of a parallelogram ABCD bisects ∠ A

Show that
(i) it bisects ∠ C also,
(ii) ABCD is a rhombus.

Answer. (i) ABCD is a parallelogram.
□□DAC = □BCA (Alternate interior angles ) …..(1)

And, □BAC = □DCA (Alternate interior angles ) …..(2)
However, it is given that AC bisects □A .
□□DAC = □BAC … (3)

From equations (1), (2), and (3), we obtain
□DAC = □BCA = □BAC = □DCA … (4)

□□DCA = □BCA

Hence, AC bisects □C .

(ii) From equation (4) , we obtain
□DAC = □DCA

□DA = DC( Side opposite to equal angles are equal)

However, DA = BC and AB = CD (Opposite sides of a parallelogram)
□AB = BC = CD = DA

Hence, ABCD is a rhombus.

Page : 146 , Block Name : Exercise 8.1

Q7 ABCD is a rhombus. Show that diagonal AC bisects ∠ A as well as ∠ C and diagonal BD bisects
∠ B as well as ∠ D.

Answer.

Let us join AC.

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Book : Mathematics Ncert Solutions | Chapter-8 Maths
In △ABC ,
BC = AB (Sides of a rhombus are equal to each other )
∠1=∠2 (Angles opposite to equal sides of a triangle are equal)
However, ∠1=∠3 (Alternate interior angles for parallel lines AB and CD)
∠2 = ∠3
Therefore, AC bisects ∠C.
Also, □2 = □4 (Alternate interior angles for ∥ lines BC and DA

□□1 = □4

Therefore, AC bisects □A .

Similarly, it can be proved that BD bisects \anglB and □D as well.

Page : 146 , Block Name : Exercise 8.1

Q8 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.

Answer.

(i) It is given that ABCD is a rectangle.
∠A = ∠C

1 1
⇒ ∠A = ∠C (AC bisects ∠A and ∠C)
2 2

⇒ ∠DAC = ∠DCA

CD = DA (Sides opposite to equal angles are also equal)
However, DA = BC and AB = CD (Opposite sides of a rectangle are equal)
□AB = BC = CD = DA

ABCD is a rectangle and all of its sides are equal.
Hence, ABCD is a square.

(ii) Let us join BD.
In ΔBCD ,
BC = CD (Sides of a square are equal to each other.)
∠CDB = ∠CBD (Angles opposite to equal sides are equal)

However, ∠CDB = ∠ABD (Alternate interior angles for AB∥CD)
△CBD = △ABD

BD bisects ∠B

Also, △CBD = △ADB (Alternate interior angles for BC II AD )

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Book : Mathematics Ncert Solutions | Chapter-8 Maths
∠CDB = ∠ABD

BD bisects ∠D

Page : 146 , Block Name : Exercise 8.1

Q9 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) APCQ is a parallelogram.

Answer. (i) In △AP D and ΔCQB, ,
□ADP = □CBQ (Alternate interior angles for BC II AD )

AD = CB (Opposite sides of a parallelogram ABCD)
DP = BQ (Given )
□△AP D□ΔCQB (Using SAS Congruence rule)

(ii) As we have observed that △APD□ΔCQB,
□AP = CQ(CP CT )

(iii) In △AQB and ΔCPD
□ABQ = □CDP ( Alternate interior angles for AB∥CD)

AB = CD (Opposite sides of a parallelogram ABCD )
BQ = DP (Given)
□△AQB□ΔCP D(U sing SAS congruence rule)

(iv) As we had observed that △AQB□ΔCPD,
□AQ = CP (CP CT )

(v) From the result obtained in (ii) and (iv) ,
AQ = CP and AP = CQ
Since opposite sides in quadrilateral APCQ are equal to each other, APCQ is a parallelogram.

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Book : Mathematics Ncert Solutions | Chapter-8 Maths
Page : 147 , Block Name : Exercise 8.1

Q10 ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal
BD

.
Show that
(i) ∆ APB ≅ ∆ CQD
(ii) AP = CQ

Answer. (i) ∆ APB and ∆ CQD
∘
□AP B = □CQD (Each90 )

AB = CD (Opposite sides of a parallelogram ABCD )
□ABP = □CDQ (Alternate interior angles for AB II CD )

□△APB□ΔCQD(By AAS congruency )

(ii) By using the above result
△AP B□ΔCQD , we obtain

AP = CQ (By CPCT)

Page : 147 , Block Name : Exercise 8.1

Q11 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

.
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.

Answer. It is given that AB = DE and AB II DE.
If two opposite sides of a quadrilateral are equal and parallel to each other , then it will be a
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parallelogram.
Therefore, quadrilateral ABED is a parallelogram.

(ii) Again, BC = EF and BC II EF
Therefore, quadrilateral BCEF is a parallelogram.

(iii) As we had observed that ABED and BEFC are parallelograms, therefore
AD = BE and AD II BE
(Opposite sides of a parallelogram are equal and parallel)
And, BE = CF and BE II CF
(Opposite sides of a parallelogram are equal and parallel)
□AD = CF and AD∥CF

(iv) As we had observed that one pair of opposite sides (AD and CF ) of a quadrilateral ACFD are
equal and parallel to each other, therefore , it is a parallelogram.

(v) As ACFD is a parallelogram, therefore, the pair of opposite sides will be equal and parallel to
each other.
□AC∥DF and AC = DF

(vi) △ABC and ΔDEF ,
AB = DE ( Given )
BC = EF (Given)
AC = DF (ACFD is parallelogram)
□△ABC□ΔDEF ( By SSS Congruence Rule)

Page : 147 , Block Name : Exercise 8.1

Q12 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
[Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Answer. Let us extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at
point E. It is clear that AECD is a parallelogram.
(i) AD = CE (Opposite sides of a parallelogram AECD)
However, AD = BC (Given)
Therefore, BC = CE
□CEB = □CBE ( Angle opposite to equal sides are also equal)

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Book : Mathematics Ncert Solutions | Chapter-8 Maths
Consider parallel lines AD and CE. AE is the transversal line for them.
□A + □CEB = 180 (Angles on the same side of transversal )
∘

∘
□A + □CBE = 180 ( Using the relation □CEB = □CBE) … (1)

However, □B + □CBE = 180 ( Linear pair angles ) … . (2)
∘

From equations (1) and (2) , we obtain
□A = □B

(ii) AB II CD
□A + □D = 180 (Angles on the same side are transversal)
∘

Also, □C + □B = 180 (Angles on the same side are transversal)
∘

□□A + □D = □C + □B

However, □A = □B [ Using the result obtained in (i) ]
□□C = □D

(iii) In △ABC and ΔBAD, ,
AB = BA (Common side )
BC = AD (Given)
□B = □A( Proved before )

□△ABC□ΔBAD (SAS congruence rule)

(iv) We had observed that,
△ABC□ΔBAD

□AC = BD(ByCPCT)

Page : 147 , Block Name : Exercise 8.1

Exercise 8.2

Q1 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 :
(i) SR || AC and SR = $ $ AC
1

2

(ii) PQ = SR
(iii) PQRS is a parallelogram.

Answer. (i) In △ADC, S and R are the mid-points of sides AD and CD respectively.
In a triangle, the line segment joining the mid-points of any two sides of the triangle is parallel to
the 3rd side and is half of it.
1
□ SR ∥
∥AC and SR = AC … (1)
2

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(ii) In △ABC . P and Q are the mid-points of sides AB and BC respectively. Therefore, by using
mid-point theoram,
1
PQ ∥
∥AC and PQ = AC … (2)
2

Using equations (1) and (2) , we obtain
PQ II SR and PQ = SR ….. (3)
□PQ = SR

(iii) From equation (3), we obtained
PQ II SR and PQ = SR
Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal.
Hence, PQRS is a parallelogram.

Page : 150 , Block Name : Exercise 8.2

Q2 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.

Answer.

In △ABC, P and Q are the mid-points of sides AB and BC respectively.
1
□PQ ∥
∥AC and PQ = AC( Using mid-point theorem ) … (1)
2

In △ADC,
R and S are the mid-points of sides CD and AD respectively.
1
□RS ∥
∥AC and RS = AC( Using mid-point theorem ) … (2)
2

From equations (1) and (2) , we obtain
PQ II RS and PQ = RS
Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a
parallelogram.
Let the diagonals of rhombus ABCD intersect each other at point O.
In quadrilateral OMPQ,
MQ II ON ( (∵ PQ∥AC)
QN II OM ( (∵ QR∥BD)
Therefore, OMQN is a parallelogram.
□□M QN = □N OM

□□P QR = □N OM

However, □NOM = 90 (Diagonals of a rhombus are perpendicular to each other).
∘

∘
□□P QR = 90

Clearly, PQRS is a parallelogram having one of its interior angles as 90 .
∘

Hence, PQRS is a rectangle.

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Book : Mathematics Ncert Solutions | Chapter-8 Maths

Page : 150 , Block Name : Exercise 8.2

Q3 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.

Answer.

Let us join AC and BD.
In △ABC ,
P and Q are the mid-points of AB and BC respectively.
1
□PQ ∥
∥AC and PQ = AC( Mid-point theorem) … (1)
2

Similarly in △ADC,
1
SR ∥
∥AC and SR = AC( Mid-point theorem ) … (2)
2

Clearly, PQ II SR and PQ = SR
Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a
parallelogram.
□ PS ∥ QR and PS = QR (Opposite sides of parallelogram)... (3)

In △BCD , Q and R are the mid-points of side BC and CD respectively.
BD (Mid-point theoram) …… (4)
1
□QR ∥∥BD and QR =
2

However,the diagonals of a rectangle are equal.
□AC = BD … (5)

By using equation (1), (2), (3), (4), and (5), we obtain
PQ = QR = SR = PS
Therefore, PQRS is a rhombus.

Page : 150 , Block Name : Exercise 8.2

Q4 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

Show that F is the mid-point of BC.

Answer. Let EF intersect DB at G.

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By converse of mid-point theoram, we know that a line drawn through the mid-point of any side of
a triangle and parallel to another side, bisects the third side.
In △ABD,
EF II AB and E is the mid-point of AD.
Therefore, G will be the mid-point of DB.
As EF II AB and AB II CD ,
□EF∥CD (Two lines parallel to the same line are parallel to each other.)

In △BCD , GF II CD and G is the mid-point of line BD. Therefore, by using converse of mid-point
theoram, F is the mid-point of BC.

Page : 150 , Block Name : Exercise 8.2

Q5 In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively

.
Show that the line segments AF and EC trisect the diagonal BD.

Answer. ABCD is a parallelogram.
□AB∥CD

And hence, AE II FC
Again, AB = CD (Opposite sides of a parallelogram ABCD)
1 1
AB = CD
2 2

AE = FC ( E and F are mid-points of side AB and CD)
In quadrilateral AECF, one pair of opposite sides (AE and CF ) is parallel and equal to each other.
Therefore, AECF is a parallelogram.
□ AF ∥EC (Opposite sides of a parallelogram)

In ΔDQC, F is the mid-point of side DC and FP II CQ (as AF II EC). Therefore, by using converse
of mid-point theoram , it can be said that P is the mid-point of DQ.
DP=PQ
Similarly, in △APB, E is the mid-point of side AB and EQ II AP (as AF II EC ).
Therefore, by using converse of mid-point theoram, it can be said that Q is the mid-point of PB.
PQ=QB
From equations (1) and (2),

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Book : Mathematics Ncert Solutions | Chapter-8 Maths
DP = PQ = BQ
Hence, the line segments AF and EC trisect the diagonal BD.

Page : 151 , Block Name : Exercise 8.2

Q6 Show that the line segments joining the mid-points of the opposite sides of a quadrilateral
bisect each other.

Answer.

Let ABCD is a quadrilateral in which P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA
respectively. JOin PQ, QR, RS, SP, and BD.
In △ABD, S and P are the mid-points Of AD and AB respectively. Therefore, by using mid-point
theorem, it can be said that
1
SP ∥
∥BD and SP = BD … (1)
2

Similarly in △BCD,
1
QR ∥
∥BD and QR = BD … (1)
2

From equations (1) and (2), we obtain

SP ∥ QR and SP = QR

In quadrilateral SPQR, one pair of opposite sides is equal and parallel to each other. Therefore,
SPQR is a parallelogram.
We know that diagonals Of a parallelogram bisect each other.
Hence, PR and QS bisect each other.

Page : 151 , Block Name : Exercise 8.2

Q7 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
1

2

Answer.

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Book : Mathematics Ncert Solutions | Chapter-8 Maths
(i) In △ABC,
It is given that M is the mid-point of AB and MD II BC.
Therefore, D is the mid-point of AC. (Converse of mid-point theorem)
(ii) As DM II CB and AC is a transversal line for them, therefore,
∘
∠M DC + ∠DCB = 180 (Co − interior angles )

∘ ∘
□M DC + 90 = 180
∘
□M DC = 90

□M D□AC

(iii) Join MC.

In △AMD and ΔCMD,
AD = CD (D is the mid-point Of side AC)
∘
∠ADM = ∠CDM (Each90 )

DM = DM (Common)
□△AM D = □ΔCM D (By SAS Congruence Rule)
Therefore, AM = CM (By CPCT)
However, AM = (M is the mid-point of AB)
1

2 AB

Therefore, it can be said that
1
CM = AM = AB
2

Page : 151 , Block Name : Exercise 8.2

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