 Lesson: Areas of Parallelograms and Triangles
Exercise 9.1
Question: 1
Which of the following figures lie on the same base
and between the same parallels?
In such a case, write the common base and the two
parallels. Solution:
(i) Yes, Δ
PDC
and Trapezium
ABCD
lie on
the same base
CD
and between the same
parallel lines
AB
and
DC
.
(ii) No. Parallelogram
PQRS
and trapezium
MNRS
lie on the same base
SR
but not
between the same parallel lines.
(iii) Yes, as
TRQ
and parallelogram
PQRS
lie
between same base
RQ
and between the
same parallel lines
PS
and
QR
.
(iv)No, because parallelogram
ABCD
and Δ
PQR
do
not lie on the same base.
(v) Yes, because parallelograms
ABCD
and
APQD
lie on the same base
and
between the same parallel lines
and
BQ
.
(vi)No, because every possible parallelograms are
on different bases. Exercise 9.2
Question: 1
In Fig. 9.15,
ABCD
is a parallelogram,
AE
DC
and
CF
. If
cm cm16 , 8AB AE
and
10 cmCF
, find
.
Fig. 9.15
Solution:
It is given that
ABCD
is a parallelogram,
AE
DC
and
CF
Also,
16 cmAB CD
(Opposite sides of a parallelogram)
cm10CF
and
cm8AE
(Given)
Now,
Area of a parallelogram = Base Altitude 16 8 10
128
10
12.8
Question: 2
If
E
,
F
,
G
and
H
are respectively the midpoints of the
sides of a parallelogram
ABCD
, show that
( )aar EFG r AH BCD
1
2
Solution:
Let us join
HF
.
It is given that
ABCD
is a parallelogram,
and
(Opposite sides of a
parallelogram)
and
1 1 1 1
2 2 2 2 So,
AH BF
and
DH CF
(
H
and
F
are the mid
points of
and
BC
respectively).
Thus,
ABFH
and
HFCD
are parallelograms.
Now,
EFH
and parallelogram
ABFH
both lie on the
same base
HF
and between the same parallel lines
AB
and
HF
.
Area Area( gm
1
( ) ) ...
2
EFH ABFH i
Also,
Area Area
1
( ) ( ) ...
2
GHF HFCD ii
Adding (i) and (ii), we get
=
1 1
( ) ( )) ( )(
2 2
ar EFH ar GHF HABF HFCD
1
( ) ( )
2
Area EFGH Area ABCD
ar EFGH ar ABCD
1
2
( )
Hence, it is proved. Question: 3
P
and
Q
are any two points lying on the sides
DC
and
respectively of a parallelogram
ABCD.
Show
that
.ar APB ar BQC
Solution:
We have,
APB
and parallelogram
ABCD
are on the same
base
AB
and between the same parallel lines
AB
and
DC
.
So,
ar APB ar gm ABCD i
1
.........| .....
2
|
Similarly,
BQC
and parallelogram
ABCD
are on
the same base
BC
and between the same parallel lines
and
BC
.
ar BQC ar gm ABCD ii
1
...........| ...
2
|
Comparing (i) and (ii),
we have
ar APB ar BQC
Question: 4
In Fig. 9.16,
P
is a point in the interior of a
parallelogram
ABCD
. Show that
(i) ( ) ( ) ( )
(ii) ( ) ( ) ( ) ( )
ar APB ar PCD ar ABCD
ar APD ar PBC ar APB ar PCD
1
2
[
Hint:
Through
P
, draw a line parallel to
AB
.]
Solution: (i)
Let us draw a line through
P
and parallel to
AB
such
that it intersects
at
G
and
BC
at
H
.
ABHG,
||AB GH
(by construction) …………. (i)
and
||AG BH
…………….. (ii)
From Eq. (i) and (ii), we have
ABHG
is a parallelogram.
Similarly, we can show that
GHCD
is also a
parallelogram.
Now, Δ
APB
and parallelogram
ABHG
lie on the
same base
AB
and between the same parallel lines
AB
and
GH
.
ar APB ar gmABHG iii
1
................
2
Also,
Δ
PCD
and parallelogram
CDGH
lie on the same base CD
and between the same parallel lines
CD
and
GH
.
ar PCD ar gmCDGH iv
1
...............
2
ar APB ar PCD ar ABHG ar CDGH
1
2
ar APB ar PCD ar ABCD
1
2
(ii)Let us draw a line through
P
and parallel to
such that it intersects
AB
at
E
and
CD
at
F
.
AEFD,
(by construction)………… (i)
And
(AE DF AB CD)
…………………. (ii)
From eq. (i) and (ii), we get
AEFD
is a parallelogram.
Similarly, we can show that
EBCF
is also a
parallelogram.
Now,
Δ
APD
and parallelogram
AEFD
lie on the same base
and between the same parallel lines
and
EF.
ar APD ar AEFD iii
1
.................
2
Also, Δ
PBC
and parallelogram
BCFE
lie on the same base
BC
and between the same parallel lines
BC
and
EF
.
ar PBC ar BCFE iv
1
..................
2
Adding Eq. (iii) and (iv), we get
1
2
1
( ( ))
2
ar APD ar PBC ar AEFD ar BCFE
ar APD ar PBC ar ABCD
Using part (i), we have
ar APD ar PBC ar APB ar PCD
Hence, it is proved. Question: 5
In Fig. 9.17,
PQRS
and
ABRS
are parallelograms and
X
is any point on side
BR.
Show that
i
ii
ar PQRS ar ABRS
ar AXS ar PQRS
1
2
Solution:
(i)
Parallelograms
PQRS
and
ABRS
lie on the same base
SR
and between the same parallel lines
SR
and
PB
.
ar gmPQRS ar gmABRS i ......
(ii)
AXS
and parallelogram
ABRS
lie on the same
base AS and between the same parallel lines
AS
and
BR. ar AXS ar ABRS ii
1
......
2
From (i) and (ii), we have
1
2
ar AXS ar gmPQRS
Question: 6
A farmer was having a field in the form of a
parallelogram
PQRS
. She took any point
A
on
RS
and
joined it to points
P
and
Q
. In how many parts the
fields is divided? What are the shapes of these parts?
The farmer wants to sow wheat and pulses in equal
portions of the field separately. How should she do
it?
Solution:
The field is divided into three parts. The three parts
are in the shape of triangles. The three triangles are Δ
PSA
, Δ
PAQ
and Δ
QAR
( ) ( ) ( ) ( . ) ..ar PSA ar PAQ ar QAR ar PQRS i
Since
PAQ
and parallelogram
PQRS
lie on same
base
PQ
and between the same parallel lines
PQ
and
SR
,
1
( ) ( ) ...
2
ar PAQ ar gmPQRS ii
So, from (i) and (ii),
( gm
1
( ) ( ) )...
2
ar PSA ar QAR ar PQRS iii
Clearly, from (ii) and (iii),
we can say that the farmer should sow wheat in
PAQ
and pulses in both
PSA
and
QAR
or vice
versa. Exercise 9.3
Question: 1
In Fig. 9.23,
E
is any point on median
of a Δ
ABC
.
Show that
( .( ) )ABE ar A Ear C
Solution:
It is given that
is the median of Δ
ABC
.
It will divide Δ
ABC
into two triangles of equal
area.
...ar ABD ar ACD i Also,
E
is a point on
.
ED
is also the median of Δ
EBC
.
... ar EBD ar ECD ii
Subtracting Eq. (ii) from Eq. (i),
ar ABD ar EBD ar ACD ar ECD
ar ABE ar ACE
Hence, it is proved.
Question: 2
In a triangle
ABC
,
E
is the mid-point of median
.
Show that
ar BED ar ABC
1
4
Solution: Since,
is the median of
ABC
; it divides
ABC
into two triangles of equal area.
...
1
( ) ( ) ( ) ( )
2
ar ABD ar ACD ar ABC i
Now, in
ABD
, E
is the midpoint of
.
So,
BE
is the median and hence, it divides
ABD
into
two triangles of equal area.
Thus,
1
( ( ) .) ( ) .. (ii)
2
ar BED ar ABE ar ABD
Using (i) in (ii), we get
1 1 1
( ) ( )
2 2 4
( )ar BED ar ABC ar ABC
Hence, it is proved.
Question: 3 Show that the diagonals of a parallelogram divide it
into four triangles of equal area.
Solution:
Let
ABCD
be a parallelogram. Diagonals
AC
and
BD
intersect each other at
O
.
Then,
O
is the midpoint of
AC
and
BD
. (Diagonals of
a parallelogram bisect each other)
In Δ
ABC
,
BO
is the median.
...ar AOB ar BOC i
Also,
In Δ
BCD
,
CO
is the median.
...ar BOC ar COD ii
In
ΔACD, OD
is the median.
...ar AOD ar COD iii Using (i), (ii) and (iii), we get
ar AOB ar BOC ar COD ar AOD
So, the diagonals of a parallelogram divide it into
four triangles of equal area.
Question: 4 In Fig. 9.24,
ABC
and
ABD
are two triangles on the
same base
AB
. If line segment
CD
is bisected by
AB
at
O
, show that
ar ABC ar ABD
.
Solution:
In
ΔACD,
AO
is the median (
CD
is bisected by
AB
at
O
)
...ar AOC ar AOD i
Also, In Δ
BCD
,
BO
is the median (
CD
is bisected by
AB
at
O
)
...ar BOC ar BOD ii
Adding (i) and (ii), we get,
ar AOC ar BOC ar AOD ar BOD
ar ABC ar ABD
Question: 5
D, E
and
F
are respectively the mid-points of the
sides
BC
,
CA
and
AB
of a Δ
ABC
.
Show that
(i)
BDEF
is a parallelogram
(ii)
1
4
ar DEF ar ABC
(iii)
1
2
ar BDEF ar ABC
Solution: (i) In Δ
ABC
,
E
and
F
are the midpoints of
AC
and
AB
respectively.
||EF BC
and
1
2
EF BC
(by midpoint
theorem)
Also,
1
2
BD BC
(
D
is the midpoint of
BC
) So,
BD EF
and
BD EF
( || )EF BC
BDEF
is a parallelogram.
(ii)
BDEF
is a parallelogram [From part (i)]
ar BFD ar DEF i ( ) ( )...............
(As
DF
is
the diagonal)
Similarly,
AFDE
is a parallelogram,
ar AFE ar DEF ii ..................
and
CDFE
is a parallelogram,
ar CDE ar DEF iii ............... Now,
ar BFD ar AFE ar CDE ar DEF
ar ABC
From (i), (ii) and (iii), we have
ar DEF ar DEF ar DEF ar DEF
ar ABC
( )4 ( )Area DEF Area ABC
1
4
ar DEF ar ABC
(iii)
( ) ( ) ( )ar gmBDEF ar DEF ar BDF
2
1
2
4
1
2
ar gram BDEF ar DEF ar DEF
ar gram BDEF ar DEF
ar gram BDEF ar ABC
ar gram BDEF ar ABC Question: 6
In Fig. 9.25, diagonals
AC
and
BD
ABCD
intersect at
O
such that
OB OD
.
If
AB CD
, then show that:
(i)
(ar DOC ar AOB )
(ii)
ar DCB ar ACB
(iii)
is a .||DA CB or ABCD gram
[Hint: From
D
and
B
, draw perpendiculars to
AC
.] Solution:
Let us draw
and DE AC BF AC
.
(i) In
DOE
and
BOF
,
DEO BFO
(Both are 90°)
DOE BOF
(Vertically opposite angles)
OD OB
(Given)
Therefore,
DOE BOF
by
AAS
congruence
condition.
Thus,
DE BF
(By CPCT) (i)
Also,
( ) ( )ar DOE ar BOF
(ii) (Congruent
triangles have equal area)
Now, in
DEC
and
BFA
,
DEC BFA
(Both are 90°)
CD AB
(Given)
DE BF
(From (i))
Therefore,
DEC BFA
by RHS congruence condition.
Thus,
( ) ( )ar DEC ar BFA
(iii) (Congruent triangles
have equal area.)
Adding (ii) and (iii), we get
( ) ( ) ( ) ( )ar DOE ar DEC ar BOF ar BFA
( ) ( )ar DOC ar AOB
(ii) From part (i), we have
( ) ( )ar DOC ar AOB
( )ar OCB
on both the sides,
( ) ( )
( ) ( )ar DOC ar OCB ar AOB ar OCB
( ) ( ) ar DCB ar ACB
(iii) In triangle
BCD
,
OC
is the median
Therefore
( ) ( )ar COD ar BOC
(iv)
Similarly, in triangle
is the median
Therefore
( ) ( )ar AOD ar AOB
(v)
Now, adding (iv) and (v), we get
( ) ( ) ( ) ( )ar COD ar AOD ar BOC ar AOB
Area ( ) ( )ar ACD ABC vii
Using the result of part (ii) in the above equation, we
get
( ) ( )ar ACD ar BCD
If two triangles have same base and equal area, then
they will lie between the same parallel lines.
Therefore,
||DC AB
Also,
AB CD
(Given)
ABCD
, one pair of opposite sides is
equal and parallel.
ABCD
is a parallelogram. Question: 7
D
and
E
are points on sides
AB
and
AC
respectively
of Δ
ABC
such that
.( ) ( )ar DBC ar EBC
Prove that
||DE BC
.
Solution: We know that area of a triangle is
1
2
base height
Since both triangle Δ
DBC
and Δ
EBC
are on the same
base
BC
and
( ) ( )ar DBC ar EBC
Therefore corresponding height is same for both
triangles.
i.e., if perpendiculars are drawn from points
D
and
E
on the base
BC
then the length of both
perpendiculars will be same.
Thus distance between
DE
and
BC
is same
throughout.
Hence,
||DE BC
. Question: 8
XY
is a line parallel to side
BC
of a triangle
ABC
.
If
||BE AC
and
||CF AB
meet
XY
at
E
and
F
respectively, show that
( ) ( )ar ABE ar ACF
Solution:
We have
EY
||
BC
(as
XY
||
BC
is given) (i)
and,
BE
CY
(as
BE
||
AC
is given) (ii)
From (i) and (ii),
EBCY
is a parallelogram.
(Both the pairs of opposite sides are parallel.)
Similarly, we can prove that BXFC
is a parallelogram.
Parallelograms
BXFC
and
EBCY
both lie on the same
base
BC
and between the same parallels
EF
and
BC
.
( ) ( ) ar BEYC ar BXFC iii
Also,
ΔAEB and parallelogram BEYC lies on the same base
BE and between the same parallels BE and AC.
i v
1
...
2
( )ar AEB ar BEYC
Similarly,
ΔACF and parallelogram BXFC lies on the same base
CF and between the same parallels CF and AB.
v
1
...
2
ar ACF ar BXFC
From (iii), (iv) and (v),
( ) ( )ar AEB ar ACF Question: 9
The side
AB
of a parallelogram
ABCD
is produced to
any point
P
. A line through
A
and parallel to
CP
meets
CB
produced at
Q
and then parallelogram
PBQR
is completed (see Fig. 9.26). Show that
( ) ( )ar ABCD ar PBQR
[Hint: Join
AC
and
PQ
. Now compare ar (
ACQ
) and
ar (
APQ
).]
Fig. 9.26
Solution: Let us join
AC
and
QP
.
( ) ( )ar ACQ ar APQ
(Both triangles lie on the
same base
AQ
and between the same parallel lines
AQ
and
CP.
)
( ) ( ) ( ) ( )ar ACQ ar ABQ ar APQ ar ABQ
( ) ) (ar ABC ar QBP i Now,
AC
is a diagonal of the parallelogram
ABCD
.
Thus,
ar ABC ar ABCD ii
1
..........
2
And
QP
is a diagonal of parallelogram
PBQR
.
ar QBP ar PBQR iii
1
.............
2
From (i), (ii) and (iii),
1 1
2 2
ar ABCD ar PBQR
ar ABCD ar PBQR
Hence, it is proved.
Question: 10
Diagonals
AC
and
BD
of a trapezium
ABCD
with
AB
||
DC
intersects each other at
O
.
Prove that
( ) ( )ar AOD ar BOC Solution:
Since ΔDAC and ΔDBC lie on the same base DC and
between the same parallels AB and CD,
ar DAC ar DBC
ar DAC ar DOC ar DBC ar DOC
ar AOD ar BOC
Hence, it is proved.
Question: 11
In Fig. 9.27,
ABCDE
is a pentagon. A line through
B
parallel to
AC
meets
DC
produced at
F
.
Show that
(i)
( ) ( )ar ACB ar ACF
(ii)
( ) ( )ar AEDF ar ABCDE Solution:
Let us join
AC
and
AF
.
(i) Since Δ
ACB
and Δ
ACF
lie on the same base
AC
and between the same parallels
AC
and
BF
.
( ) ( )ar ACB ar ACF
(ii) From part (i), we have
( ) ( )ar ACB ar ACF
ar ACB ar ACDE ar ACF ar ACDE
ar ABCDE ar AEDF
Hence, it is proved.
Question: 12 A villager Itwaari has a plot of land of the shape of a
quadrilateral. The Gram Panchayat of the village
decided to take over some portion of his plot from
one of the corners to construct a Health Centre.
Itwaari agrees to the above proposal with the
condition that he should be given equal amount of
land in lieu of his land adjoining his plot so as to
form a triangular plot. Explain how this proposal will
be implemented.
Solution:
Let
ABCD
be the plot of land in the shape of a Let us join
BD
.
Let
AE
be drawn parallel to
BD
. Let us join
BE
which
intersected
at
O
.
Let Δ
AOB
be the area for constructing the health
centre. The triangle BEC is the shape of the field he
is expecting.
We need to prove that
.( ) ( )ar DEO ar AOB
We know that Δ
DEB
and Δ
DAB
lie on the same base
BD
and between the same parallel lines
BD
and
AE
.
Therefore,
( ) ( )ar DEB ar DAB
( ) ( ) ( ) ) (ar DEB ar DOB ar DAB ar DOB
( ) ( )ar DEO ar AOB
Hence, it is proved.
Question: 13
ABCD
is a trapezium with
AB
||
DC
. A line parallel to
AC
intersects
AB
at
X
and
BC
at
Y
. Prove that ( ) ( )ar ADX ar ACY
. [Hint: Join
CX
.]
Solution:
Since Δ
and Δ
AXC
lie on same base
AX
and
between the same parallels
AB
and CD.
... ( ) ( )ar ADX ar AXC i
Also, Δ
AXC
and Δ
ACY
lie on same base
AC
and
between the same parallels
XY
and
AC.
( ) ( )ar AXC ar ACY ii
From (i) and (ii),
( ) ( )ar ADX ar ACY
Hence, it is proved.
Question: 14
In Fig. 9.28,
AP
||
BQ
||
CR
. Prove that
( ) ( )ar AQC ar PBR Solution:
Since Δ
AQB
and Δ
PBQ
lie on same base
BQ
and
between same parallel lines
AP
and
BQ,
( ) ( )ar AQB ar PBQ i
Also, Δ
BQC
and Δ
BQR
lie on the same base
BQ
and
between same parallels
CR
and
BQ,
( ) ( )ar BQC ar BQR ii
( ) ( ) ( ) ( )ar AQB ar BQC ar PBQ ar BQR
) ( ( )ar AQC ar PBR
Hence, it is proved.
Question: 15 Diagonals
AC
and
BD
ABCD
intersect at
O
in such a way that
( ) ( )ar AOD ar BOC
. Prove that
ABCD
is a
trapezium.
Solution:
It is given that
( ) ( )ar AOD ar BOC
( ) ( ) ( ) ( )ar AOD ar AOB ar BOC ar AOB
( ) ( ) ar ADB ar ACB
Areas of Δ
and Δ
ACB
are equal and both lie on
the same base
AB
. Therefore, they must lie between
the same parallel lines.
Thus,
AB CD
Therefore,
ABCD
is a trapezium.
Question: 16 In Fig.9.29,
and ) (( ) ( )) (ar DRC ar DPC ar BDP ar ARC
ABCD
and
DCPR
are trapeziums.
Solution:
It is given that
( ) ( )ar BDP ar ARC i
and
( ) ( )ar DPC ar DRC ii
Subtracting Eq. (ii) from Eq. (i), we get
( ) ( ) ( ) ( )ar BDP ar DPC ar ARC ar DRC
( ) ( )ar BDC ar ADC
Now, area of Δ
BDC
and Δ
are equal and lie on
the same base
DC
. Therefore, they must lie between the same parallel
lines.
Thus,
AB CD
Therefore,
ABCD
is a trapezium.
Also,
( ) ( )ar DRC ar DPC
and both lie on the same base
DC
.
Therefore, they must lie between the same parallel
lines.
Thus,
DC PR
Therefore,
DCPR
is a trapezium.
Exercise 9.4
Question: 1
Parallelogram
ABCD
and rectangle
ABEF
are on the
same base
AB
and have equal areas. Show that the
perimeter of the parallelogram is greater than that of
the rectangle.
Solution: Since parallelogram
ABCD
and rectangle
ABEF
are
on the same base
AB
and have equal areas.
So,
The height of the parallelogram
ABCD
is equal to
that of the rectangle
ABEF
.
But,
Height of the rectangle
ABEF
=
BE
Now,
BC
> Height of the parallelogram
ABCD
= Height of the rectangle
ABEF
=
BE
Thus,
BC
>
BE
As,
Perimeter of the rectangle
ABEF
= 2(
AB
BE
)
Further, Perimeter of the parallelogram
ABCD
= 2(
AB
BC
)
> 2(
AB
BE
) (
BC
>
BE
)
= Perimeter of the rectangle
ABEF
Therefore,
The perimeter of the parallelogram is greater than
that of the rectangle.
Alternate solution:
We need to show that perimeter of parallelogram
ABCD
> perimeter of rectangle
ABEF.
We know that perimeter of parallelogram
ABCD
AB BC CD DA
AB BC AB BC
(Opposite sides of
parallelogram are equal)
2( )...........( )AB BC i
Perimeter of rectangle
ABEF
( )AB BE EF AF
( )AB BE AB BE
(Opposite sides of rectangle
are equal)
2( )...........( )AB BE ii
Now, in
BEC
,
BEC 90
(Since
ABEF
is a rectangle)
So
BEC
is a right angled triangle.
Thus,
...........( )BC BE iii
(Since length of the
hypotenuse is greater than other sides)
From equation (i), we have
Perimeter of parallelogram
ABCD
=
2( ) 2( )AB BC AB BE
(From (iii)) Parallelogram
ABCD
> perimeter of rectangle
ABEF
(Using (ii))
Hence, it is proved.
Question: 2
In Fig. 9.30,
D
and
E
are two points on
BC
such
that
BD DE EC
. Show that
ar ABD ar ADE ar AEC .
Can you now answer the question that you have left
in the ‘Introduction’ of this chapter, whether the field of Budhia has been actually divided into three
parts of equal area?
[Remark: Note that by taking
BD DE EC
, the
triangle
ABC
is divided into three triangles
ABD
,
and
AEC
of equal areas. In the same way, by
dividing
BC
into
n
equal parts and joining the points
of division so obtained to the opposite vertex of
BC
, you can divide
ABC
into
n
triangles of equal
areas.]
Solution:
We are given that
BD DE EC
We need to proof that
ar ABD ar ADE ar AEC .
Since
BD DE
,
D
is the midpoint of
BE
in
ABE
.
So,
is the median of
ABE
.
Thus,
... ( )ar ABD ar ADE i
(Since median of
a triangle divides it into two triangles of equal area)
Similarly,
AE
is the median of
(Since
DE EC
is given)
Thus,
... ( )ar ADE ar AEC ii
(Since median of
a triangle divides it into two triangles of equal area)
From (i) and (ii), we have ar ABD ar ADE ar AEC
Now, we can say that the field of Budhia has been
actually divided into three parts of equal area as she
divided one side of the triangle into three equal parts
and then join these two points of division to the
opposite vertex.
Hence, it is proved.
Question: 3
In Fig. 9.31,
ABCD, DCFE
and
ABFE
are
parallelograms. Show that
.
Solution:
It is given that
ABCD, DCFE
and
ABFE
are
parallelograms.
We need to show that
In
and
BCF
,
(Since
ABCD
is a parallelogram and
opposite sides of a parallelogram are equal)
DE CF
(Since
DCFE
is a parallelogram and
opposite sides of a parallelogram are equal)
EA FB
(Since
ABFE
is a parallelogram and
opposite sides of a parallelogram are equal)
So,
by using SSS congruence rule.
(Congruent triangles have
equal areas.)
Hence, it is proved. Question: 4
In Fig. 9.32,
ABCD
is a parallelogram and
BC
is
produced to a point
Q
such that
. If
AQ
intersect
DC
at
P
, show that
( ) ( )ar BPC ar DPQ
.
[Hint: Join
AC
.] Solution:
It is given that
Since
ABCD
is a parallelogram,
(Opposite sides of a parallelogram
are equal.)
From (i) and (ii), we have
BC CQ
Then, in
BPQ
,
CP
is the median. ... ( ) ( ) ( )ar CPQ ar BPC iii
(Median of a
triangle divides it into two triangles of equal
areas.)
Now, in
ABQ
,
C
is the midpoint of
BQ
(
BC CQ
as above
proved) and
CP BA
(
CD BA
being opposite
sides of the parallelogram
ABCD
and
CP
lies on
CD
.)
Then, by using midpoint theorem, we have
P
is the
midpoint of
AQ
and
1
2
CP AB
But
AB CD
(Opposite sides of a parallelogram
ABCD
)
1
CP CD
2
So,
P
is the midpoint of
CD
.
Thus,
PQ
is the median of
CDQ
.
... ( ) ( ) ( )ar DPQ ar CPQ iv
From (iii) and (iv), we have
( ) ( )ar BPC ar DPQ
Hence, it is proved.
Question: 5 In Fig.9.33,
ABC
and
BDE
are two equilateral
triangles such that
D
is the mid-point of
BC
. If
AE
intersects
BC
at
F
, show that
(i)
ar BDE ar ABC
1
( ) ( )
4
(ii)
ar BDE ar BAE
1
( ) ( )
2
(iii)
ar ABC ar BEC2
(iv)
ar BFE ar AFD
(v)
ar BFE ar FED2
(vi)
ar FED ar AFC
1
( )
8 [Hint: Join
EC
and
. Show that
BE AC||
and
DE AB||
, etc.]
Solution:
Given,
ABC
is an equilateral triangle.
So,
AB BC CA a say( )
Also,
BDE
is an equilateral triangle where
D
is the
mid-point of
BC
. (Given)
Then,
2 2
BC a
BE DE BD
Let us join
EC
and
.
(i) We know that area of an equilateral triangle
side
2
3
( )
4
So, 2
2
2
3
( )
4 2
3
4 4
1 3 1
( )
4 4 4
1
( ) ( )
4
a
ar BDE
a
a ar ABC
ar BDE ar ABC
(ii) We know that
CBE 60
(
BDE
is an
equilateral triangle) and
BCA 60
(
ABC
is an equilateral
triangle)
CBE BCA
But these are alternative angles with respect
to the line segments
BE
and
AC
and
transversal
BC
. Thus,
BE AC
... ( ) ( ) ( )ar ABE ar BEC i
(Both triangles
lie on same base
BE
and between same
parallel lines
BE
and
AC
)
Now,
D
is the midpoint of
BC
(Given).
So,
DE
is the median of
BEC
.
(Using(i))ar BDE ar BEC ar ABE
ar BDE ar ABE
1 1
( ) ( ) ( )
2 2
1
( ) ( )
2
Hence, it is proved.
(iii) Since
D
is the midpoint of
BC
(Given),
DE
is the median of
BEC
.
...
1
( ) ( ) ( )
2
ar BDE ar BEC i
From part (i), we have
...
1
( ) ( ) ( )
4
ar BDE ar ABC ii
From (i) and (ii), we have
ar ABC ar BEC
ar ABC ar BEC
1 1
( ) ( )
4 2
( ) 2 ( )
Hence, it is proved.
(iv) We know that
BDE 60
(
BDE
is an
equilateral triangle) and
ABD 60
(
ABC
is an equilateral
triangle)
BDE ABD
But these are alternative angles with respect
to the line segments
AB
and
DE
and
transversal
BD
.
Thus,
AB DE
ar BED ar AED( ) ( )
(Both lie on same base DE and between
same parallel lines AB and DE)
ar BFE ar DFE ar AFD ar DFE
ar BFE ar AFD
( ) ( ) ( ) ( )
( ) ( )
Hence, it is proved.
(v) Let us draw
EM BD
. Then,
...
1
( ) ( )
2
ar FED EM FD i
In
BED
,
1 1
( ) ......( )
2 2 2 4
a a
ar BED EM BD EM EM ii
But we know that
...
2
2
3 3
( ) ( )
4 2 4 4
a a
ar BED iii
On comparing (ii) and (iii), we have
2
3
4 4 4
a a
EM
3
4
EM a Now putting
3
4
EM a
in equation (i), we
get
...
1 3
( )
2 4
3
( ) ( )
8
ar FED a FD
ar FED a FD iv
Now,
ABC
is an equilateral triangle and
is
the median.
So,
is the perpendicular bisector of
BC
.
1 1
( )
2 2
But we know that
ar ABC a
2
3
( )
4
2
3 1
4 2
3
.......( )
2
Now,
(Using( ))
1 1 3
( )
2 2 2
3
( ) ............( )
4
ar AFD AD FD a FD v
ar AFD a FD vi
From part (iv) we know that
( ) ( )ar BFE ar AFD Using
3
( ) ( ( ))
4
ar BFE a FD vi
On dividing both sides by 2, we get
Using
( ) 3
2 8
( )
( )( ( ))
2
( ) 2 ( )
ar BFE
a FD
ar BFE
ar FED iv
ar BFE ar FED
Hence, it is proved.
(vi) From part (i), we have
ar BDE ar ABC
1
( ) ( )
4
1
( ) ( ) (2 ( ))
4
ar BFE ar FED ar ADC
(Median of a triangle divides it into two
triangles of equal areas)
1
2 ( ) ( ) ( ( ))
2
ar FED ar FED ar ADC
(Using part (v))
1
3 ( ) ( ( ) ( ))
2
ar FED ar AFC ar AFD
1
3 ( ) ( ( ) ( ))
2
ar FED ar AFC ar AFD
1 1
3 ( ) ( ) ( )
2 2
ar FED ar AFD ar AFC Using part
1 1
3 ( ) ( ) ( )( ( ))
2 2
ar FED ar BFE ar AFC iv
Using part
1 1
3 ( ) 2 ( ) ( )( ( ))
2 2
ar FED ar FED ar AFC v
1
4 ( ) ( )
2
ar FED ar AFC
1
( ) ( )
8
ar FED ar AFC
Hence, it is proved.
Question: 6
Diagonals
AC
and
BD
ABCD
intersect each other at
P
. Show that
ar APB ar CPD ar APD ar BPC
[Hint: From
A
and
C
, draw perpendiculars to
BD
.]
Solution:
It is given that
ABCD
diagonals
AC
and
BD
intersect each other at
P
. Let us draw
AX BD
and
CY BD
as shown in
figure.
Now,
1
..........( )
2
ar APB AX PB i
1
..........( )
2
ar APD AX PD ii
1
..........( )
2
ar CPD CY PD iii
1
..........( )
2
ar BCP CY PB iv
On multiplying (i) and (iii), we get 1 1
2 2
1
..........( )
4
ar APB ar CPD AX PB CY PD
ar APB ar CPD AX PB CY PD v
On multiplying (ii) and (iv), we get
1 1
2 2
1
..........( )
4
ar APD ar BCP AX PD CY PB
ar APD ar BCP AX PD CY PB vi
On comparing (v) and (vi), we have
ar APB ar CPD ar APD ar BPC
Hence it is proved.
Question: 7
P
and
Q
are respectively the mid-points of sides
AB
and
BC
of a triangle
ABC
and
R
is the mid-point of
AP
, show that
(i)
)
2
(
1
ar PRQ ar ARC
(ii)
3
8
ar RQC ar ABC
(iii)
ar PBQ ar ARC Solution:
In
ABC
,
P
and
Q
are the mid-points of sides
AB
and
BC
respectively and
R
is the mid-point of
AP
.
Let us join
AQ
and
PC
.
(i) Since
P
is the midpoint of
AB
,
CP
is the
median of
ABC
.
1
( ) ( )......( )
2
ar APC ar ABC i
Now,
R
is the midpoint of
AP
,
CR
is the median
of
APC
.
1
( ) ( )
2
ar ARC ar APC
Using
1 1
( ) ( ) ( ( ))
2 2
ar ARC ar ABC i 1
( ) ( )..........( )
4
ar ARC ar ABC ii
Similarly,
Since
Q
is the midpoint of
BC
,
AQ
is the
median of
ABC
.
1
( ) ( )......( )
2
ar ABQ ar ABC iii
Since
P
is the midpoint of
AB
,
PQ
is the
median of
ABQ
.
Using
...
1
( )
2
1 1
( ) ( ( ))
2 2
1
( ) ( )
4
ar APQ ar ABQ
ar APQ ar ABC iii
ar APQ ar ABQ iv
Now, R is the midpoint of AP, QR is the median
of
APQ
.
... Using
Using
1
( ) ( )
2
1 1
( ) ( ) ( )( ( ))
2 4
1
( ) ( ( ))
2
ar PRQ ar APQ
ar PRQ ar ABC v iv
ar PRQ ar ARC ii
Hence, it is proved.
(ii) In
BRC
,
Q
is the midpoint of
BC
, so,
RQ
is the
median. 1
( ) ( )
2
1
( ) ( ) ( )
2
ar RQC ar BRC
ar BPQ ar PRQ ar RQC
1 1
( ) ( ) ( ) ( )
2 2
ar RQC ar RQC ar BPQ ar PRQ
...
1 1
( ) ( ) ( ) ( )
2 2
ar RQC ar BPQ ar PRQ vi
Since
PQ
is the midpoint of
ABQ
,
1 1 1
( ) ( ) ( )
2 2 2
ar BPQ ar ABQ ar ABC
(Since
AQ
is the median, so it divides
( )ar ABC
into two
equal parts)
1
( ) ( ).....( )
4
ar BPQ ar ABC vii
From equation (v), (vi) and (vii), we get
1 1 1 1
( ) ( ) ( )
2 2 4 8
1 1 1 1
( ) ( )
2 2 4 8
2 1
( ) ( )
8
3
( ) ( )
8
ar RQC ar ABC ar ABC
ar RQC ar ABC
ar RQC ar ABC
ar RQC ar ABC
(iii) From (vii) we have 1
( ) ( )
4
ar BPQ ar ABC
( ) ( )ar BPQ ar ARC
(Using (ii))
Question: 8
In Fig. 9.34,
ABC
is a right triangle right angled at
A
.
BCED
,
ACFG
and
ABMN
are squares on the sides
BC
,
CA
and
AB
respectively. Line segment
AX DE
meets
BC
at
Y
. Show that: (i)
MBC ABD  
(ii)
2ar BYXD ar MBC
(iii)
ar BYXD ar A( BMN)
(iv)
FCB ACE
(v)
2ar CYXE ar FCB
(vi)
( )ar CYXE ar ACFG
(vii)
ar BCED ar ABMN ar F( ) AC G Note: Result (vii) is the famous Theorem of
Pythagoras. You shall learn a simpler proof of this
theorem in Class X.
Solution:
(i) In
MBC
,
MBC MBA ABC
90MBC ABC
(i) (Since
MBAN
is
a square)
In
ABD
,
ABD CBD ABC
90ABD ABC
(ii) (Since
BCED
is
a square)
From (i) and (ii), we have
... ( )MBC ABD iii
In
MBC
and
ABD
,
MB AB
(Sides of the square
MBAN
)
MBC ABD
(Using (iii))
BC BD
(Sides of the square
BCED
)
MBC ABD  
by SAS congruence rule.
(ii) Since
AX DE
(Given) and
BD DE
BD YX
Also,
BY DX
(Opposite sides of
square
BCED
)
Thus
BYXD
is a parallelogram.
Also,
90YBD
(Since
BCED
is a
square)
Thus
BYXD
is a rectangle.
Now, rectangle BYXD and
ABD
both lie
on same base BD and between parallel
lines BD and AX.
So,
( ) 2 ( )ar BYXD ar ABD
( ) 2 ( )ar BYXD ar MBC
(Using part (i)
and the property of congruent triangles
having equal areas)
Hence, it is proved.
(iii) Square
ABMN
and
MBC
both lie on
same base
MB
and between the same
parallel lines
MB
and
NC
(Since
MB AN
being opposite sides of the square
ABMN
and
NC
is the extension of
NA
)
( ) 2 ( )ar ABMN ar MBC
( ) ( )ar ABMN ar BYXD
(Using part (ii))
(iv) In
FCB
,
FCB FCA ACB
90FCB ACB
... (i) (Since
ACFG
is a
square)
In
ACE
,
ACE ECB ACB
90ACE ACB
... (ii) (Since
BCED
is
a square)
From (i) and (ii), we have
... ( )FCB ACE iii
In
FCB
and
ACE
,
FC AC
(Sides of the square
ACFG
)
FCB ACE
(Using (iii))
BC CE
(Sides of the square
BCED
)
FCB ACE
by SAS congruence rule.
(v) Since
AX DE
(Given) and
CE DE
CE YX
Also,
CY EX
(Opposite sides of
square
BCED
)
Thus
CYXE
is a parallelogram.
Also,
90YCE
(Since,
BCED
is a
square)
Thus
CYXE
is a rectangle.
Now, rectangle
CYXE
and
ACE
both lie
on same base CE and between parallel lines
CE
and
AX
.
So,
( ) 2 ( )ar CYXE ar ACE
( ) 2 ( )ar CYXE ar FCB
(Using part (iv)
and the property of congruent triangles
having equal areas)
Hence, it is proved.
(vi) Square
ACFG
and
FCB
both lie on
same base
CF
and between the same
parallel lines
BG
and
CF
(Since
CF AG
being opposite sides of the square
ACFG
and
BG
is the extension of
GA
)
( ) 2 ( )ar ACFG ar FCB
( ) ( )ar ACFG ar CYXE
(Using part (v))
(vii) It is given that
BCED
,
ACFG
and
ABMN
are
squares on the sides
BC
,
CA
and
AB
respectively.
We know that area of a square
2
( )side
So,
2
( ) ( ) ........( )ar BCED BC i
2
( ) ( ) ........( )ar ACFG AC ii
2
( ) ( ) ........( )ar ABMN AB iii
Now, it is also given that
ABC
is a right
triangle right angled at A.
So, by Pythagoras theorem
2 2 2
( ) ( ) ( )BC AB AC
( ) ( ) ( )ar BCED ar ACFG ar ABMN
(Using (i), (ii) and (iii))
Hence, it is proved.