Lesson: Quadrilaterals

Exercise 8.1 (14 Multiple Choice Questions and

Answers)

Question: 1

Three angles of a quadrilateral are 75º, 90º and 75º.

The fourth angle is

(a) 90º

(b) 95º

(d) 120º

Solution:

Let the fourth angle be ‘x’.

Since the sum of all four angles of a quadrilateral is

360º, so

75 0 5 360

240 360

º 360º 240º 120º

   

  

   

Hence, the fourth angle is 120º.

Question: 2

A diagonal of a rectangle is inclined to one side of the

rectangle at 25º. The acute angle between the

diagonals is:

(a) 55°

(b) 50º

(d) 25º

Solution:

Let us consider a rectangle PQRS and diagonal SQ is

inclined to PQ at 25º.

We need to find the acute angle between the

diagonals.

Let both the diagonals intersect at point ‘O’.

Since the diagonals of a rectangle bisect each other,

 ,OPQ

OP OQ

     25ºOQP OPQ

Since

ROQ

is an exterior angle to

 ,OPQ

25º 25º 50ºROQ OQP OPQ       

Also

180ºROQ QOP   

(∵ PR is a line segment),

180º 50º 130ºQOP    

Hence, the acute angle between the diagonals is 50º.

Question: 3

ABCD is a rhombus such that

40º.ACB 

Then

ADB

is:

(a) 40º

(b) 45º

(d) 60º

Solution:

We are given that ABCD is a rhombus such that

40º.ACB 

We need to find

ADB

Since the diagonals of a rhombus are perpendicular

bisector of each other,

in ∆BOC,

90ºBOC 

Now,

180ºBOC OCB OBC     

(Sum of all angles in

a triangle)

 

90º 40º 180º

40º 40º

180º 90º 40º

180º 130º 50º

(

. I

)

OBC

ACB OCB

OBC

   

    

    

   



Since

AB CD

BDA DBC   

(Alternate interior angle)

50º

BDA OBC

BDA

   

  

Question: 4

The quadrilateral formed by joining the midpoints of

the sides of a quadrilateral

PQRS, taken in order, is a rectangle, if:

(a) PQRS is a rectangle

(b) PQRS is a parallelogram

(d) Diagonals of PQRS are equal

Solution:

Let KLMN be the quadrilateral formed by joining the

midpoints of quadrilateral PQRS.

Now, if PQRS is a rectangle, then

diagonal PR = diagonal QS.

Then KLMN will be a square having all 4 sides equal.

If PQRS is a parallelogram, then the opposite sides

will be parallel to each other and equal, but the angles

may not be right angle.

If the diagonals of PQRS are equal, then all 4 sides of

the quadrilateral KLMN will be equal.

So, option (a), (b) and (d) can’t be the answer.

Now, if the diagonals of PQRS are perpendicular, then

by the midpoint theorem, the opposite sides are equal

and parallel to each other.

Also, each angle of quadrilateral KLMN will be right

angle. So, KLMN will be a rectangle.

Hence, option(c) is the answer.

Question: 5

The quadrilateral formed by joining the midpoints of

the sides of a quadrilateral

PQRS, taken in order, is a rhombus, if

(a) PQRS is a rhombus

(b) PQRS is a parallelogram

(d) Diagonals of PQRS are equal

Solution:

Let EFGH be the rhombus formed by joining the

midpoints of the sides of a quadrilateral PQRS.

So,

EF FG GH HE  

Now, in ∆PQR,

E and F are the midpoints of PQ and QR respectively.

So,

EF PR

(By the midpoint theorem) ….

(i)

Similarly, in ∆SQR,

F and G are the midpoints of RQ and RS respectively.

So,

GF SQ

(By the midpoint theorem) ….

(ii)

As,

EF GF

(Since EFGH is a rhombus),

1 1

2 2

PR SQ 

(Using (i) & (ii))

PR SQ 

Hence, the diagonals of PQRS are equal.

Question: 6

If angles A, B, C and D of the quadrilateral ABCD,

taken in order, are in the ratio

3:7:6:4, then ABCD is a

(a) Rhombus

(b) Parallelogram

(d) Kite

Solution:

Given:

Ratio of the angles of quadrilateral ABCD is

3: 7: 6: 4.

Let the angles of the quadrilateral ABCD be 3a, 7a, 6a

and 4a respectively.

Since the sum of all angles of a quadrilateral is 360°,

Therefore,

3 7 6 4 360

0 36

3 0

a a a a

    









 





The angles of the quadrilateral are

3 18 54

7 18 126

6 18 108

     

and

4 18 72D     

Now, let’s Produce AB to point M.

Then,

180ABC CBM    

(Linear pair axiom)

126 180

180 126 54

CBM

CBM A

     

         

Hence,

AD BC

(Since, the corresponding angles are

equal)

Now, the sum of co-interior angles,

126 54 180A B        

and

108 72 180C D        

Hence, ABCD is a trapezium.

Question: 7

If bisectors of ∠A and ∠B of a quadrilateral ABCD

intersect each other at P, bisectors of

∠B and ∠C at Q, bisectors of ∠C and ∠D at R and

bisectors of ∠D and ∠A at S, then PQRS is a:

(a) Rectangle

(b) Rhombus

(d) Quadrilateral whose opposite angles are

supplementary

Solution:

Given: In quadrilateral ABCD, angle bisectors form a

quadrilateral PQRS.

Since the sum of all angles in quadrilateral is 360

therefore,

360 .A B C D        

On dividing both sides by 2,

 

360

1 1 1 1

180

2 2 2 2

7 8 6 5 180

1 1

( )

A B C D

        

         

          



(



AP, BP, CR and DR are angle bisectors of ∠A, ∠B,

∠C & ∠D respectively.)

Now, let’s consider ∆CDR,

6 5 2 180      

(Sum of all angles in a triangle)

 

6 5 180 2 ii        

Similarly, on considering ∆ABP,

7 8 1 180      

(Sum of all angles in a triangle)

 

7 8 180 1 iii        

Adding (ii) & (iii),

7 8 6 5 180 1 180 2               

180 180 1 180 2          

(Using (i))

1 2 180     

But,

1 4  

and

2 3  

(Vertically opposite

angles)

Hence,

3 4 180 .    

Hence, PQRS is a quadrilateral whose opposite angles

are supplementary.

Question: 8

If APB and CQD are two parallel lines, then the

bisectors of the angles APQ,

BPQ, CQP and PQD form

(a) A square

(b) A rhombus

(d) A trapezium

Solution:

Given: APB and CQD are two parallel lines.

Let the bisectors of angles APQ and CQP meet at a

point R and the bisectors of angles DQP and BPQ

meet at a point S.

Since APB is a line,

180APR RPQ QPS SPB        

(Linear pair

axiom)

180RPQ RPQ QPS QPS         

(As PR and PS are angle bisectors for ∠APQ and

∠QPB respectively.)

2 2 180

2 180

( )

(

)

RPQ QPS

RPS

     

   

Now,

APQ PQD  

(



APB CQD

and PQ is a traversal

line)

1 1

2 2

APQ PQD

RPQ PQS

   

(PR and QS are angle bisectors for

APQ

and

PQD

respectively.)

PR QS 

(Alternate interior angle)

Similarly,

BPQ PQC  

(Since

APB CQD

and PQ

is a traversal line)

PS RQ 

Hence



PRQS is a rectangle.

Question: 9

The figure obtained by joining the midpoints of the

sides of a rhombus, taken in order, is:

(a) A rhombus

(b) A rectangle

(d) A kite

Solution:

Let KLMN be the quadrilateral which is formed by

joining the mid points of the rhombus SRQP.

Let’s join SQ, LN and KM.

Let us consider ∆PSQ.

Since N and M are mid points of sides PS and PQ

respectively,

NM SQ 

and

 

NM SQ i 

(By the midpoint theorem)

Similarly, on considering ∆RSQ,

Since K and L are the midpoints of the sides SR and

RQ respectively,

KL SQ 

And

 

KL SQ ii 

(By the midpoint

theorem)

Using (i) and (ii), we get

 

and ..KL NM KL NM iii 

Hence, KLMN is a parallelogram.

Also, NLRS is a parallelogram.

 

.NL SR iv  

(Opposite sides of a

parallelogram)

Similarly, MKRQ is a parallelogram.

MK QR 

(Opposite sides of a parallelogram)

MK SR 

(∵

SR QR

, sides of a rhombus)

⇒ MK=NL (Using (iv))

So, diagonals of a parallelogram are equal.

Hence, KLMN is a rectangle.

Question: 10

D and E are the mid-points of the sides AB and AC of

∆ABC and O is any point on side BC.

O is joined to A. If P and Q are the midpoints of OB

and OC respectively, then DEQP is:

(a) A square

(b) A trapezium

(d) A parallelogram

Solution:

We are given that D and E are the mid-points of the

sides AB and AC of the ∆ABC.

P and Q are the midpoints of OB and OC respectively.

Let’s Join DE, DP and EQ.

In ∆ABC,

 

DE BC i

and

DE BC

(∵ D & E are midpoint of AB and AC respectively, by

the midpoint theorem.)

Then,

   

1 1

2 2

DE BP PO OQ QC PO OQ     

 

and

[ ]BP PQ OQ QC

DE PO OQ

DE PQ ii

 

  

  



Now, in ∆ABO,

 

and

DP AO DP AO iii 

(



D and P are the midpoint of AB and BO

respectively.)

Similarly, in ∆ACO,

 

and

EQ AO EQ AO iv 

(



E and Q are midpoint of AC and OC respectively.)

Using eqn. (iii) & (iv), we get

DP EQ

and

.DP EQ

Using eqn. (i) & (ii), we get,

DE BC

⇒

DE PQ

and

.DE PQ

Hence, DEQP is a parallelogram.

Question: 11

The figure formed by joining the mid-points of the

sides of a quadrilateral ABCD, taken in order, is a

square only if,

(a) ABCD is a rhombus

(b) Diagonals of ABCD are equal

(d) Diagonals of ABCD are perpendicular.

Solution:

Let us consider EFGH be the quadrilateral formed by

joining the midpoints of the sides of a quadrilateral

ABCD. Then EFGH is a square.

EF FG GH HE   

and

HF EG

But,

EG AD

and

HF CD

AD CD 

Thus, all the sides of quadrilateral ABCD are equal.

Now, in ∆ABC,

since E and F are midpoints of AB and BC

respectively,

EF AC

…….(i) (By midpoint

theorem)

Similarly, in ∆ADB,

since E and H are midpoints of AB and AD

respectively,

EH DB

(By midpoint theorem)

 

EF DB ii  

( )EH EF

From Eq. (i) and (ii),

AC DB

Thus, all the sides of quadrilateral ABCD are equal

and diagonals of ABCD are equal. Therefore, the

quadrilateral ABCD is a square.

So, the diagonals of the quadrilateral are also

perpendicular.

Question: 12

The diagonals AC and BD of a parallelogram ABCD

intersect each other at the point O.

32ºDAC 

and

70ºAOB 

, then ∠DBC is equal

to:

(a) 24º

(b) 86º

(d) 32º

Solution:

We are given that ABCD is a parallelogram,

32DAC  

and

70AOB  

Since

32DAC ACB    

(Alternate interior angle),

also,

ACB OCB  

32OCB   

Now, let’s consider ∆BOC.

∠AOB is an exterior angle to ∆BOC.

OBC OCB AOB     

(Exterior angle property)

70 32 38OBC AOB OCB           

Also,

DBC OBC  

Thus,

38DBC  

Question: 13

Which of the following is not true for a parallelogram?

(a) Opposite sides are equal

(b) Opposite angles are equal

(d) Diagonals bisect each other

Solution:

In a parallelogram, the opposite sides are equal, the

opposite angles are equal, the diagonals bisect each

other but the opposite angles are not bisected by the

diagonals.

Question: 14

D and E are the midpoints of the sides AB and AC

respectively of triangle ABC. DE is produced to F. To

prove that CF is equal and parallel to DA, we need

additional information which is:

(a)

DAE EFC  

(b)

AE EF

(c)

DE EF

(d)

ADE ECF  

Solution:

In ∆ABC, since, D and E are the midpoints of AB and

AC respectively, therefore,

DE BC

and

DE BC

DF BC 

and

 

2 .......DE BC i

Suppose,

DE EF

then

2DF DE

Then, using Eq. (i) we get,

DF BC

and

DF BC

Hence, the quadrilateral DBCF is a parallelogram.

DB CF 

and

DB CF

(Property of a

parallelogram)

AD CF 

and

AB CF

(



D is the midpoint of

AB)

AD CF 

and

AD CF

(



AD is a part of line

AB).

Hence, option(c)

DE EF

is the additional

information that we need to prove the desired result.

Exercise 8.2

Question: 1

Diagonals AC and BD of a parallelogram ABCD

intersect each other at O.

3OA cm

and

2OD cm

, determine the lengths of

AC and BD.

Solution:

We are given that

ABCD

is a parallelogram,

3OA cm

and

2 .OD cm

We need to find the length of AC and BD.

Since

ABCD

is a parallelogram and the diagonals of

a parallelogram bisects each other,

2 2 3 6AC OA cm cm    

and,

2 2 2 4 .BD OD cm cm   

Hence it is proved.

Question: 2

Diagonals of a parallelogram are perpendicular to each

other. Is this statement true? Give reason for your

answer.

Solution:

No; if the parallelogram is a rhombus then the

diagonals are perpendicular to each other. Otherwise,

in all parallelograms, diagonals bisect each other.

Question: 3

Can the angles 110º, 80º, 70º and 95º be the angles of

a quadrilateral? Why or why not?

Solution:

We know that the sum of all angles of a quadrilateral

is 360°.

Here,

110 80 70 95 355 360     

So, 110 , 80 , 70 , 95   

cannot be the angles of a

quadrilateral.

Question: 4

In quadrilateral ABCD,

180A D   

. What special

name can be given to this quadrilateral?

Solution:

Since

180A D   

AB DC 

(Sum of co-interior angles is

180°

)

ABCD 

can be called a trapezium.

Question: 5

All the angles of a quadrilateral are equal. What

special name is given to this quadrilateral?

Solution:

Given, all the angles of a quadrilateral are equal.

Let each angle of the quadrilateral be ‘x’.

Then,

360º x x x x   

(Sum of all angles of a

quadrilateral)

4 360º

90º

 

Since all the angles of the quadrilateral are 90º, the

given quadrilateral is a rectangle.

Question: 6

Diagonals of a rectangle are equal and perpendicular

to each other. Is this statement true? Give reason for

your answer.

Solution:

No; Diagonals of a rectangle are not perpendicular to

each other.

Question: 7

Can all the four angles of a quadrilateral be obtuse

angles? Give reason for your answer.

Solution:

No; because sum of the angles of a quadrilateral is

360°. A quadrilateral can have a maximum of three

obtuse angles.

Question: 8

, 5 , 8ABC AB cm BC cm  

and

7 .CA cm

If D

and E are the midpoints of AB and BC, determine the

length of DE.

Solution:

Given:

, 5 , 8ABC AB cm BC cm  

and

7 .CA cm

D and E are respectively the midpoints of AB and BC.

Let’s join DE.

By the midpoint theorem,

DE∥AC and DE =

 



Question: 9

In the following figure, it is given that BDEF and

FDCE are parallelograms.

Can you say that

BD CD

? Why or why not?

Solution:

It is given that

BDEF

is a

,gram

 

.....BD FE i 

Opposite sides of

gram

Also,

FDCE

is a

,gram

 

...DC FE ii 

From eq. (i) and (ii)

BD CD

Question: 10

In the given figure, ABCD and AEFG are two

parallelograms. If ∠C = 55º, determine ∠F.

Solution:

Since ABCD is a

,gram

 

55 ..A C i     

(Opposite angles of a

parallelogram are equal)

Again, AEFG is a

,gram

55 F A    

(Using Eq. (i)).

Question: 11

Can all the angles of a quadrilateral be acute angles?

Give reason for your answer.

Solution:

No; the angle sum of a quadrilateral is 360°. Therefore,

a quadrilateral should have at least one obtuse angle.

Question: 12

Can all the angles of a quadrilateral be right angles?

Give reason for your answer.

Solution:

Yes, because the angle sum will be 360°, which is a

required property of a quadrilateral.

Question: 13

Diagonals of a quadrilateral ABCD bisect each other.

If ∠A = 35º, determine ∠B.

Solution:

Since diagonals of the given quadrilateral bisect each

other,

∴ ABCD is a

gram

Now,

and DAB ABC 

are consecutive interior angles

AD BC

and

is a transversal and we know that

some of the consecutive interior angles

180

180

180 180 35

145

DAB ABC

ABC DAB

ABC

   

      

   

Question: 14

Opposite angles of a quadrilateral ABCD are equal. If

4AB cm

, determine CD.

Solution:

Given

is a gram

(Opposite sides of a parallelogram)

A C B D

ABCD

CD AB cm

   



  

 

Exercise 8.3

Question: 1

One angle of a quadrilateral is of 108º and the

remaining three angles are equal.

Find each of the three equal angles.

Solution:

Let ABCD be a quadrilateral such that

108A  

and

B C D    

Let

B C D x     

Now by the angle sum property we have,

360

108 3 360

3 360 108

3 252

252

A B C D

A x x x

        

      

    

    

  



 

 

Hence, the measure of each of the equal angle is 84°.

Question: 2

ABCD is a trapezium in which

||AB DC

and

45º.A B   

Find angles C and D of the trapezium.

Solution:

We are given that ABCD is a trapezium and

45ºA B   

And

DC AB

So,

A

and

D

are co-interior angles.

180A D   

(Sum of co-interior angles)

45 180D   

180 45 135D    

Similarly,

B

and

C

are co-interior angles.

180B C   

(Sum of co-interior angles)

45 180C   

180 45 135C    

Hence, the angles C and D are 135° each.

Question: 3

The angle between two altitudes of a parallelogram

through the vertex of an obtuse angle of the

parallelogram is 60º. Find the angles of the

parallelogram.

Solution:

Let

ABCD

is a parallelogram and

,AX BC AY CD 

and

60XAY  

,AXCY

using the angle sum property of a quadrilateral,

360

60 90 90 360

240 360

360 240

120

XAY AYC C CXA

A C

     

     

   

    

   

    

(Opposite angle of parallelogram)

Now,

180    C D

(Adjacent angles of a

parallelogram)

120 180

180 120 60

   

    

60B D    

(Opposite angles of parallelogram are

equal).

Question: 4

ABCD is a rhombus in which altitude from D to side

AB bisects AB. Find the angles of the rhombus.

Solution:

Given: ABCD is a rhombus and DE is the altitude on

AB such that

.AE EB

In ∆AED and ∆BED,

DE DE

(Common side)

 

90DEA DEB   

AE EB

(Given)

AED BED  

(By SAS congruence)

( )AD BD C.P.C.T 

Also,

AD AB

Then,

AD AB BD  

Thus,

ABD

is an equilateral triangle.

C A

  

     

Opposite angles of rhombus are equal

Now,

180ADC DAB    

Sum of the adjacent angles of a rhombus is

supplementary

60 180

180 60

120

ADC

ABC ADC

    

     

  

    

Opposite angles of a rhombus are equal

Thus, the angles of rhombus are 60°, 120°, 60° and

120°.

Question: 5

E and F are points on diagonal AC of a parallelogram

ABCD such that

AE CF

Show that BFDE is a parallelogram.

Solution:

Since ABCD is a

 gram

the diagonals of a

parallelogram bisect each other.

 

OD OB .... i 

 

OA OC .... ii

 

....AE CF iii

(Given)

   

 

....

iii ii

OA AE OC CF

OE OF iv



   

 

Hence, in

BFDE

diagonals bisect each other.

Therefore,

BFDE

is parallelogram.

Question: 6

E is the midpoint of the side AD of the trapezium

ABCD with

| .|AB DC

A line through E drawn

parallel to AB intersect BC at F. Show that F is the

midpoint of BC. [Hint: Join AC]

Solution:

Given: ABCD is a trapezium with

| .|AB DC

E is midpoint of AD and

| .|EF AB

Since AB || DC and EF || AB,

||EF DC

Now, in ∆ADC, E is midpoint of AD and

EF DC

||EO DC

O

is midpoint of AC (Converse of the midpoint

theorem)

In ∆CAB, O is the midpoint of AC (As proved above)

And

OF AB

(

EF AB 

and OF lies on EF.)

F

is the midpoint of BC. (Converse of the

midpoint theorem)

Hence, we proved that

.CF FB

Question: 7

Through A, B and C, lines RQ, PR and QP have been

drawn, respectively parallel to sides BC, CA and AB of

a ΔABC as shown in the figure. Show that

BC QR

Solution:

Given:

,AB QP BC RQ 

and

.CA PR

We need to show that

BC QR

Let us consider

.RBCA

Since

RA BC

and

BR CA

(Given),

RBCA 

is a parallelogram,

so,

 

....RA BC i

Opposite sides of

parallelogram

Similarly, on considering

,BCQA

Since,

AQ BC

and

AB CQ

BCQA 

is parallelogram

so,

 

....AQ BC ii

Opposite side of a

parallelogram

Adding (i) and (ii), we get

QR BC

BC QR

 

Hence, it is proved.

Question: 8

D, E and F are the midpoints of the sides BC, CA and

AB, respectively of an equilateral triangle ABC. Show

that DEF is also an equilateral triangle.

Solution:

Since FE is a line segment joining the midpoints of

sides AB and AC respectively,

 

....

FE BC i 

(By midpoint theorem)

Similarly,

 

...

DE AB ii

(

&D E

are midpoint of

& BC CA

respectively.)

and

 

....

DF AC iii

(

&D F

are midpoint of

& BC AB

respectively.)

But,

AB BC CA 

(Sides of an equilateral

)

.....( )

1 1 1

2 2 2

AB BC CA iv  

(Dividing by 2)

Using

     

, ,i ii iii

( )iv

DE EF FD 

DEF

is an equilateral triangle.

Question: 9

Points P and Q have been taken on opposite sides are

on the opposite sides AB and CD, respectively of a

parallelogram ABCD such that

AP CQ

as shown

below. Show that AC and PQ bisect each other.

Solution:

Let us consider ∆AOP and ∆QOC

1 2 

(

DC AB 

and alternate interior angles are

equal)

AP CQ

(Given)

3 4 

(

AB CD 

and

is a transversal)

APO CQO  

by ASA congruency rule

 

  . . .OA OC C PCT

and

OP OQ

Hence we have proved that AC and PQ bisect each

other.

Question 10

In the given below figure, P is the midpoint of side BC

of a parallelogram ABCD such that

.BAP DAP  

Prove that

2 .AD CD

Solution:

Since

AD BC

(Opposite sides of

parallelogram

ABCD

) and

is transversal,

2 3 

Alternate interior angles

But,

1 2 

(Given)

1 3 

   , 1 3ABP

BP AB 

(Sides opposite to equal angles are equal)

BC AB 

(



is the midpoint of side

)

AD AB 

(

BC AD

, opposite sides of a

parallelogram)

AD CD 

(

,AB CD

opposite sides of a

parallelogram)

2AD CD 

Hence, it is proved.

Exercise 8.4

Question: 1

A square is inscribed in an isosceles right triangle so

that the square and the triangle have one angle

common. Show that the vertex of the square opposite

the vertex of the common angle bisects the

hypotenuse.

Solution:

Given: An isosceles right ∆ABC such that

90ACB  

and

.BC AC

A square CMPN is inscribed in it.

We need to prove that P bisects the hypotenuse AB

that is

AP PB

Since CMPN is a square,

CM MP PN CN   

(All sides are equal)

Also, ∆ABC is isosceles with

AC BC

(

)

AN NC CM MB

AN MB CN CM

   

  

Let’s consider ∆ANP and ∆BMP.

AN = MB (As proved above)

90ANP PMB    

PN PM

(∵CMPN is a square)

ANP BMP  

(By SAS congruency)

AP PB 

(By CPCT).

Hence, it is proved.

Question: 2

In a parallelogram

, 10ABCD AB cm

and

6AD cm

The bisector of

A

meets

at E. AE

and BC produced to meet at F. Find the length of CF.

Solution:

Given: ABCD is a parallelogram and

cm cm10 ; 6 .AB AD 

AE is the angle bisector of angle A.

Now,

DAF AFB 

(Alternate interior angle)

But,

DAF FAB 

(Since AE is the angle bisector

A

)

In ,

AFB FAB

ABF

 

 

10BF AB cm  

(∵ Sides opposite to equal angle

are equal)

But,

6BC AD cm 

(Opposite sides of a

parallelogram are equal)

Now,

CF BF BC 

10 6

4cm

 



Hence, the length of CF=4 cm.

Question: 3

P, Q, R and S are respectively the midpoints of the

sides AB, BC, CD and DA of a quadrilateral ABCD in

which

.AC BD

Prove that PQRS is a rhombus.

Solution:

Given: ABCD is a quadrilateral in which P, Q, R and S

are the midpoints of AB, BC, CD and DA respectively.

Let’s join PQ, QR, RS and SP.

Also,

.AC BD

Now, in ΔABC,

P and Q are the midpoints of AB and BC respectively.

||PQ AC

and

 

PQ AC i 

(By the midpoint

theorem)

Similarly, in ΔADC,

S and R are the midpoints of AD and DC respectively.

||SR AC

and

 

SR AC ii 

(By the midpoint

theorem)

Similarly, in ΔDBC,

||RQ BD

and

 

RQ BD iii 

(By the midpoint

theorem)

And, in ΔDBA,

||SP BD

and

 

SP BD iv 

(By the midpoint

theorem)

Now,

AC BD

(Given)

Therefore, from eqn. (i), (ii), (iii) & (iv), we get

PQ SR RQ SP  

and

.PQ SR RQ SP  

Therefore, each side of the quadrilateral PQRS are

equal and parallel.

Hence,

PQRS is a rhombus.

Question: 4

P, Q, R and S are respectively the mid-points of the

sides AB, BC, CD and DA of a quadrilateral ABCD

such that AC⊥BD. Prove that PQRS is a rectangle.

Solution:

Given: P, Q, R and S are respectively the midpoints of

the sides AB, BC, CD and DA of a quadrilateral ABCD

such that AC⊥BD.

In ΔABC,

P and Q are the midpoints of AB and BC respectively.

||PQ AC

and

 

PQ AC i 

(By the midpoint

theorem)

Similarly, in ΔADC,

S and R are the midpoints of AD and DC respectively.

||SR AC

and

 

SR AC ii 

(By the midpoint

theorem)

From (i) and (ii), we get

||PQ SR

and

.PQ SR

Similarly, in ΔDBC,

||RQ BD

and

 

RQ BD iii 

(By the midpoint

theorem)

In ΔDBA,

||SP BD

and

 

SP BD iv 

(By the midpoint

theorem)

From (iii) and (iv), we get,

||RQ SP

and

RQ SP

Hence

PQRS

is a parallelogram.

Now,

AC BD

(Given)

And,

||PQ AC

(From eqn. (i))

PQ BD 

But,

||RQ BD

(From eqn. (iii))

PQ RQ 

Similarly, we can show that

,RQ SR

SR PS

And

PS PQ

Hence, it is proved that

PQRS

is a rectangle.

Question 5

P, Q, R and S are respectively the midpoints of sides

AB, BC, CD and DA of quadrilateral ABCD in which

AC BD

and AC ⊥BD. Prove that PQRS is a square.

Solution:

In ΔABC,

P and Q are the midpoints of AB and BC respectively.

||PQ AC

and

 

PQ AC i 

(By the midpoint

theorem)

Similarly, in ΔADC,

S and R are the midpoints of AD and DC respectively.

||SR AC

and

 

SR AC ii 

(By the midpoint

theorem)

From (i) and (ii), we get

||PQ SR

and

PQ SR

Similarly, in ΔDBC,

||RQ BD

and

 

RQ BD iii 

(By the midpoint

theorem)

and, in ΔDBA,

||SP BD

and

 

SP BD iv 

(By midpoint

theorem)

From (iii) and (iv), we get,

||RQ SP

and

RQ SP

Hence,

PQRS

is a parallelogram.

Now,

AC BD

(Given)

PQ QR RS SP   

(Using (i), (ii), (iii) & (iv))

Now,

AC BD

(Given)

And, PQ || AC (From eqn. (i))

PQ BD 

But,

||RQ BD

(From eqn. (iii))

PQ RQ 

Similarly, we can show that

,RQ SR

,SR PS

and

PS PQ

Hence, it is proved that

PQRS

is a square.

Question 6

A diagonal of a parallelogram bisects one of its angles.

Show that it is a rhombus.

Solution:

Let ABCD be the parallelogram. AC bisects

A

We need to prove that ABCD is a rhombus.

Since AC bisect ∠A,

 

.....DAC BAC i 

Since

AB CD

and

is transversal,

 

......BAC ACD ii 

(Pair of alternate angles)

From (i) and (ii) we have

In ,

DAC ACD

ACD

DAC ACD

 



 

 

.....CD AD iii 

(Sides opposite to equal angles are

equal)

Since

ABCD

is a parallelogram,

AB CD

and

 

.....BC AD iv

(Opposite sides of

parallelogram are equal)

From (iii) and (iv) we have

AB BC CD DA  



Parallelogram

ABCD

is a rhombus.

Question 7

P and Q are the midpoints of the opposite sides AB

and CD of a parallelogram ABCD. AQ intersects DP at

S and BQ intersects CP at R. Show that PRQS is a

parallelogram.

Solution:

It is given that ABCD is a parallelogram,

and AB DC AB DC 

AB DC 

and

1 1

2 2

AB DC

Since

and

are the midpoints of

and

AP CQ 

and

AP CQ

APCQ

is a parallelogram.

Then

 

.....AQ PC i

(Opposite sides of a

parallelogram are parallel).

Similarly,

DPBQ

is a parallelogram.

 

.....DP QB ii 

From (i) and (ii), we get

AQ PC

and

DP QB

SQ PR 

and

SP QR

Hence, it is proved that

PRQSY

is a parallelogram.

Question 8

ABCD is a quadrilateral in which

||AB DC

and

.AD BC

Prove that

A B  

and

.C D  

Solution:

Let us construct

  and DP AB CQ AB

Now, in

APD

and

,BQC

   90APD BQC

AD BC

(Given)

DP CQ

(Distance between



lines)

APD BQC  

(By RHS property)

So,

A B 

 

. . .C PCT

We are given that

DC AB

So, A C B D

C D A B

 

 

   

   

Hence, it is proved.

Question 9

In Fig. 8.11, AB || DE, AB = DE, AC || DF and AC =

DF. Prove that BC || EF and BC = EF.

Solution:

It is given that

AC DF

and

AC DF

ACFD 

is a



gram

 

.....AD CF i 

And

 

....AD CF ii

(Opposite sides of a



gram)

Now,

AB DE

and

AB DE

ABED 

is a

gram

 

....AD BE iii 

And

......( )AD BE iv

(Opposite sides of a



gram)

From (i), (ii), (iii) and (iv), we get

CF BE

and

||CF BE

BCFE 

gram

Hence, we can say that

BC EF

and

BC EF

(Opposite sides of a



gram).

Hence, it is proved.

Question 10

E is the midpoint of a median AD of ABC and BE is

produced to meet AC at F.

Show that

.AF AC

Solution:

Given: AD is the median of ΔABC and E is the

midpoint of AD, also BE meets AC at F.

Let’s Draw

||DG BF

In ΔADG, E is the midpoint of AD and

| .|EF DG

By converse of the midpoint theorem, we have

F is the midpoint of AG .



 

..... iAF FG

Similarly, in ΔBCF, D is the midpoint of BC and

||DG BF

Therefore, G is the midpoint of CF.

Hence,

 

..... iiFG GC

From equations (i) and (ii), we get

 

..... iiiAF FG GC 

From the figure we have,

AF FG GC AC  

AF AF AF AC   

(From (iii))

3AF AC 

Hence,



Question: 11

Show that the quadrilateral formed by joining the

midpoints of the consecutive sides of a square is also a

square.

Solution:

Let ABCD be the square such that

.AB BC CD DA  

Also,

AC BD

and let P, Q, R and S are the mid

points of the sides AB, BC, CD and DA respectively.

In ∆ABC,

P and Q are the mid-points of AB and BC respectively.

  and

PQ AC PQ AC

(By the midpoint

theorem) …. (i)

Similarly in ∆ADC,

SR AC

and

SR AC

(By the mid-point

theorem) ... (ii)

Clearly,

PQ SR

and

PQ SR

Since, in the 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 a

parallelogram are equal and parallel) …. (iii)

In ∆BCD, Q and R are the mid-points of side BC and

CD respectively.

  and

||QR BD QR BD

(By the midpoint

theorem) ... (iv)

However, the diagonals of a square are equal,

... ( )AC BD v

By using equation (i), (ii), (iii), (iv), and (v), we obtain

PQ QR SR PS  

We know that diagonals of a square are perpendicular

bisector of each other.

90AOB COD BOC DOA        

Now, in the quadrilateral EOHS, we have

SE OH

Therefore,

    180AOD AES

(Corresponding

angles)

180 90 90AES       

Again,

180AES SEO   

(Linear pair)

180 90 90SEO      

Similarly

SH EO

Therefore,

180AOD DHS   

(Corresponding

angles)

180 90 90DHS      

Again,

180DHS SHO   

(Linear pair)

180 90 90SHO      

Again, in the quadrilateral EOHS, we

have

90SEO SHO EOH     

Therefore, by the angle sum property of quadrilateral

EHOS, we get

360SEO SHO EOH ESH      

90 90 90 360ESH        

90ESH   

In the same manner, in quadrilaterals EPFO, FQGO

and GOHR, we get

90HRG FQG EPF     

Therefore, in the quadrilateral PQRS, we have

PQ QR SR PS  

and

ESH HRG

90FQG EPF   

Hence, PQRS is a square.

Question: 12

E and F are respectively the midpoints of the

non-parallel sides AD and BC of a trapezium ABCD.

Prove that

||EF AB

and

 

.EF AB CD 

(Hint: Join CE and produce it to meet AB produced at

G.)

Solution:

Let’s join BE, which on producing, meet CD produced

at G.

In ΔBEA and ΔEDG,

BEA GED  

(Vertically opposite angles)

EA ED

(∵E is midpoint of AD)

EAB EDG  

(Alternate interior angles are equal

DC B CA AB G

and AD is a transversal line)

EAB EDG  

(By ASA congruency rule)

AB DG 

and

BE EG

(By CPCT)

In ΔCGB,

E is midpoint of BG (

;BE EGQ

as above proved)

F is a midpoint of BC (Given)

∴By the midpoint theorem

and .

EF AB EF CG

But,

CG CD DG CD AB   

(As proved above,

DG AB

)

Hence

||EF AB

and

)( .

EF AB CD 

Hence, it is proved.

Question: 13

Prove that the quadrilateral formed by the bisectors of

the angles of a parallelogram is a rectangle.

Solution:

Let ABCD is a parallelogram. AE bisects

,BAD BF

bisects

,ABC CG

bisect

BCD

and DH

bisect

ADC

We need to prove that LKJI is a rectangle.

Since ABCD is a parallelogram,

180BAD ABC    

(Adjacent angles of a

parallelogram are supplementary).

     

1 1

180

2 2

BAD ABC

But,

BAJ BAD  

And

ABJ ABC  

(Given)

90BAJ ABJ     

ΔABJ is a right triangle since its acute interior angles

are complementary.

90 90AJB IJK       

Similarly in ΔCDL,

90CDL DCL    

90DLC  

In ΔADI and ΔCBK,

  ADI CBK

(Opposite angles of a parallelogram

are equal)

AD BC

(Opposite sides of a parallelogram are

equal)

  DAI BKC

(Opposite angles of a parallelogram

are equal)

ADI CBK  

(By ASA congruency rule)

So,

  AID BKC

(By CPCT)

Now,

AID JIL  

and

BKC LKJ  

(Vertical

opposite angles)

LKJ JIL   

90JIL LKJ     

Since all four angles of quadrilateral LKJI are right

angle,

therefore, LKJI is a rectangle.

Hence, it is proved.

Question: 14

P and Q are points on the opposite sides AD and BC of

a parallelogram ABCD such that PQ passes through

the point of intersection O of its diagonals AC and BD.

Show that

.OP OQ

Solution:

It is given that

AD BC

(Opposite sides of a

parallelogram are parallel),

So,

PD BQ

1 2 and 3 4     

(Alternate interior angles are

equal)

Now, In

DOP

and

BOQ

3= 4 

1 2 

(As proved above)

OD OB

(Diagonals of a parallelogram bisect each

other).

DOP BOQ  

(By AAS congruency rule)

OP OQ 

(By CPCT)

Question: 15

ABCD is a rectangle in which diagonal BD bisects

B

Show that ABCD is a square.

Solution:

Let’s join AC.

We have

DC AB

(Opposite sides of a rectangle),

So,

 

...4 1 i  

Similarly,

 

....3 2 ii  

 

..... given1 2 iii  

From

     

, we geti , ii , iii

3 4 

ΔBDA

and

ΔBDC

1 2 

(Given)

BD BD

(Common)

3 4 

(As proved above)

  BDA DBC

(By ASA congruency rule)

AD CD

AB CB

(CPCT)



Rectangle ABCD is a square.

Question: 16

D, E and F are respectively the midpoints of the sides

AB, BC and CA of a triangle ABC. Prove that by

joining these midpoints D, E and F, the triangle ABC

is divided into four congruent triangles.

Solution:

In ∆ABC,

DF joins the midpoints of the sides AB and AC

respectively.

Then,

and

DF BC DF BC BE EC  

(Using the midpoint theorem)

Similarly, DE joins midpoints of AB and BC

respectively.

and

DE AC DE AC AF FC   

Also, EF joins the midpoints of BC and AC

respectively.

EF AB and EF AB AD DB   

Now, in ∆BDE and ∆DEF

BD EF

(As proved above)

DE DE

(Common)

BE DF

(Given)

So, by SSS rule,

.BDE FED  

Similarly,

EFC FED 

And,

EFD ADF  

So, all 4 triangles are congruent.

Question: 17

Prove that the line joining the mid-points of the

diagonals of a trapezium is parallel to the parallel

sides of the trapezium.

Solution:

Let PQRS be a trapezium in which

PQ RS

and let X

and Y be the midpoints of the diagonals PR and SQ,

respectively.

We need to prove that

XY SR PQ 

Now, let us join SX and produce it to intersect PQ

produced at T.

In ∆SXR and ∆PXT, we have

SRX TPX  

(Alternate interior angles) ,

PX XR

(Given)

and

SXR PXT  

(Vertically opposite angles)

SXR TXP   

(By ASA congruence)

SR PT 

and

SX XT 

(by CPCT)

Thus, in ∆STQ, the points X and Y are the midpoints

ST and SQ respectively.

XY TQ  

(By the converse of mid-point theorem)

XY PQ SR   

Hence, it is proved.

Question: 18

P is the midpoint of the side CD of a parallelogram

ABCD. A line through C parallel to PA intersects AB

at Q and DA produced at R. Prove that

DA AR

and

.CQ QR

Solution:

In ∆CDR, P is midpoint DR and

AP CR

By the converse of the midpoint theorem, we have

DA AR

But, AD BC

(Opposite sides of a parallelogram are

equal)

.......(i)AR BC 

Again, in

ΔCDR,

A is midpoint of DR and

AQ CD

ABCD

is a parallelogram and its opposite sides

are parallel to each other

Q

Midpoint of CR (By the midpoint theorem)

CQ QR 

Hence, it is proved.