Exercise 8.1 (14 Multiple Choice Questions and
Question: 1
Three angles of a quadrilateral are 75º, 90º and 75º.
The fourth angle is
(a) 90º
(b) 95º
(c) 105º
(d) 120º
Solution:
d
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
x
x
x
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
(c) 40º
(d) 25º
Solution:
b
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,
in
,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
is:
(a) 40º
(b) 45º
(c) 50º
(d) 60º
Solution:
c
We are given that ABCD is a rhombus such that
40º.ACB
We need to find
.
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
(c) Diagonals of PQRS are perpendicular
(d) Diagonals of PQRS are equal
Solution:
c
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
(c) Diagonals of PQRS are perpendicular
(d) Diagonals of PQRS are equal
Solution:
d
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,
1
2
EF PR
(By the midpoint theorem) ….
(i)
Similarly, in ∆SQR,
F and G are the midpoints of RQ and RS respectively.
So,
1
2
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
(c) Trapezium
(d) Kite
Solution:
c
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
2
18
2
0 36
6
0
0
3 0
a a a a
a
a
The angles of the quadrilateral are
3 18 54
7 18 126
6 18 108
A
B
C
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,
(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
(c) Parallelogram
(d) Quadrilateral whose opposite angles are
supplementary
Solution:
d
Given: In quadrilateral ABCD, angle bisectors form a
Since the sum of all angles in quadrilateral is 360
o
,
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
( )
2
2
A B C D
A B C D
i
(
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
(c) A rectangle
(d) A trapezium
Solution:
c
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
90
9
( )
(
0
)
RPQ QPS
RPQ QPS
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
(c) A square
(d) A kite
Solution:
b
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
1
2
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
1
2
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
(c) A rhombus
(d) A parallelogram
Solution:
d
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
1
2
DE BC
( D & E are midpoint of AB and AC respectively, by
the midpoint theorem.)
Then,
1 1
2 2
2 2
DE BP PO OQ QC PO OQ
and
.
[ ]BP PQ OQ QC
DE PO OQ
DE PQ ii
Now, in ∆ABO,
and
1
2
DP AO DP AO iii
(
D and P are the midpoint of AB and BO
respectively.)
Similarly, in ∆ACO,
and
1
2
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
(c) Diagonals of ABCD are equal and perpendicular
(d) Diagonals of ABCD are perpendicular.
Solution:
c
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,
and
HF CD
Thus, all the sides of quadrilateral ABCD are equal.
Now, in ∆ABC,
since E and F are midpoints of AB and BC
respectively,
1
2
EF AC
…….(i) (By midpoint
theorem)
since E and H are midpoints of AB and AD
respectively,
1
2
EH DB
(By midpoint theorem)
1
.
2
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.
If
32ºDAC
and
70ºAOB
, then DBC is equal
to:
(a) 24º
(b) 86º
(c) 38º
(d) 32º
Solution:
c
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
(c) Opposite angles are bisected by the diagonals
(d) Diagonals bisect each other
Solution:
c
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)
Solution:
c
In ∆ABC, since, D and E are the midpoints of AB and
AC respectively, therefore,
DE BC
and
1
.
2
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)
and
AB CF
(
D is the midpoint of
AB)
and
(
AD is a part of line
AB).
Hence, option(c)
DE EF
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.
If
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
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
Question: 4
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
4 360º
90º
x
x
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
Solution:
No; Diagonals of a rectangle are not perpendicular to
each other.
Question: 7
Can all the four angles of a quadrilateral be obtuse
Solution:
No; because sum of the angles of a quadrilateral is
360°. A quadrilateral can have a maximum of three
obtuse angles.
Question: 8
In
, 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:
In
, 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,
DEAC and DE =
1
2
AC
3
2
5
7
.
1
cm
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?
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?
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
as
and
AB
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)
,
.
4
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
360
108 3 360
3 360 108
3 252
252
3
84
A B C D
A x x x
x
x
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
.
In
,AXCY
using the angle sum property of a quadrilateral,
360
60 90 90 360
240 360
360 240
120
120
XAY AYC C CXA
C
C
C
C
A C
(Opposite angle of parallelogram)
Now,
180 C D
(Adjacent angles of a
parallelogram)
120 180
180 120 60
D
D
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,
Then,
Thus,
ABD
is an equilateral triangle.
60
60
A
C A
Opposite angles of rhombus are equal
Now,
Sum of the adjacent angles of a rhombus is
supplementary
60 180
180 60
120
120
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
,
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
1
.
2
BC QR
Solution:
Given:
,AB QP BC RQ
and
.CA PR
We need to show that
1
.
2
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
2
1
2
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,
1
....
2
FE BC i
(By midpoint theorem)
Similarly,
1
...
2
DE AB ii
(
&D E
are midpoint of
& BC CA
respectively.)
and
1
....
2
DF AC iii
(
&D F
are midpoint of
& BC AB
respectively.)
But,
AB BC CA
(Sides of an equilateral
Δ
)
.....( )
1 1 1
2 2 2
AB BC CA iv
(Dividing by 2)
Using
, ,i ii iii
in
( )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
QP
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
Solution:
Since
(Opposite sides of
parallelogram
ABCD
) and
AP
is transversal,
2 3 
Alternate interior angles
But,
1 2
(Given)
1 3
In
, 1 3ABP
BP AB
(Sides opposite to equal angles are equal)
1
2
BC AB
(
P
is the midpoint of side
BC
)
1
2
(
, opposite sides of a
parallelogram)
1
2
(
,AB CD
opposite sides of a
parallelogram)
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
.
The bisector of
A
meets
DC
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
of
A
)
In ,
AFB FAB
ABF
10BF AB cm
( Sides opposite to equal angle
are equal)
But,
(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
1
..
2
PQ AC i
(By the midpoint
theorem)
S and R are the midpoints of AD and DC respectively.
||SR AC
and
1
..
2
SR AC ii
(By the midpoint
theorem)
Similarly, in ΔDBC,
||RQ BD
and
1
..
2
RQ BD iii
(By the midpoint
theorem)
And, in ΔDBA,
||SP BD
and
1
..
2
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 ACBD. 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 ACBD.
In ΔABC,
P and Q are the midpoints of AB and BC respectively.
||PQ AC
and
1
..
2
PQ AC i
(By the midpoint
theorem)
S and R are the midpoints of AD and DC respectively.
||SR AC
and
1
..
2
SR AC ii
(By the midpoint
theorem)
From (i) and (ii), we get
||PQ SR
and
.PQ SR
Similarly, in ΔDBC,
||RQ BD
and
1
..
2
RQ BD iii
(By the midpoint
theorem)
In ΔDBA,
||SP BD
and
1
..
2
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
1
..
2
PQ AC i
(By the midpoint
theorem)
S and R are the midpoints of AD and DC respectively.
||SR AC
and
1
..
2
SR AC ii
(By the midpoint
theorem)
From (i) and (ii), we get
||PQ SR
and
PQ SR
Similarly, in ΔDBC,
||RQ BD
and
1
..
2
RQ BD iii
(By the midpoint
theorem)
and, in ΔDBA,
||SP BD
and
1
..
2
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
AC
is transversal,
......BAC ACD ii
(Pair of alternate angles)
From (i) and (ii) we have
In ,
DAC ACD
ACD
DAC ACD
(Sides opposite to equal angles are
equal)
Since
ABCD
is a parallelogram,
AB CD
and
(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
P
and
Q
are the midpoints of
AB
and
CD
,
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
Prove that
A B
and
.C D
Solution:
Let us construct
and DP AB CQ AB
Now, in
APD
and
,BQC
90APD BQC
(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
And
(Opposite sides of a
gram)
Now,
AB DE
and
AB DE
ABED
is a
gram
And
......( )AD BE iv
(Opposite sides of a
gram)
From (i), (ii), (iii) and (iv), we get
CF BE
and
||CF BE
BCFE
is
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
1
3
.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,
.
3
AF
AC
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
1
||
2
PQ AC PQ AC
(By the midpoint
theorem) …. (i)
SR AC
and
1
2
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
1
2
||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
1
2
.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
as
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 EGQ
as above proved)
F is a midpoint of BC (Given)
By the midpoint theorem
and .
1
||
2
EF AB EF CG
But,
CG CD DG CD AB
(As proved above,
DG AB
)
Hence
||EF AB
and
)( .
1
2
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
bisects
,ABC CG
bisect
BCD
and DH
bisect
.
We need to prove that LKJI is a rectangle.
Since ABCD is a parallelogram,
(Adjacent angles of a
parallelogram are supplementary).
1
1 1
2
180
2 2
But,
1
2
And
1
2
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,
(Opposite angles of a parallelogram
are equal)
(Opposite sides of a parallelogram are
equal)
DAI BKC
(Opposite angles of a parallelogram
are equal)
(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
(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 
In
ΔBDA
and
ΔBDC
1 2
(Given)
BD BD
(Common)
3 4 
(As proved above)
 BDA DBC
(By ASA congruency rule)
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
1
2
DF BC DF BC BE EC
(Using the midpoint theorem)
Similarly, DE joins midpoints of AB and BC
respectively.
and
1
2
DE AC DE AC AF FC
Also, EF joins the midpoints of BC and AC
respectively.
1
2
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,
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
(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.