RS Aggarwal Solutions: Areas of Parallelograms and Triangles | Mathematics (Maths) Class 9 PDF Download

 Page 1


Question:1
Which of the following figures lie on the same base and between the same parallels. In such a case, write the comon base and the two parallels.
Solution:
i
No, it doesnt lie on the same base and between the same parallels.
ii
No, it doesnt lie on the same base and between the same parallels.
iii
 Yes, it lies on the same base and between the same parallels. The same base is AB and the parallels are AB and DE.
iv
 No, it doesnt lie on the same base and between the same parallels.
v
 Yes, it lies on the same base and between the same parallels. The same base is BC and the parallels are BC and AD.
vi
 Yes, it lies on the same base and between the same parallels. The same base is CD and the parallels are CD and BP.
Question:2
In the adjoining figure, show that ABCD is a parallelogram.
Calculate the area of || gm ABCD.
Solution:
Given: A quadrilateral ABCD and BD is a diagonal.
To prove: ABCD is a parallelogram.
Construction: Draw AM ? DC and CL ? AB   extendDCandAB
. Join AC, the other diagonal of ABCD.
Proof: ar(quad. ABCD) = ar(?ABD) + ar( ??DCB)
                                      = 2 ar( ??ABD)                    [ ? ar ?(?ABD) = ar( ??DCB)]
? ar( ??ABD) = 
1
2
ar(quad. ABCD)                 ...i
Again, ar(quad. ABCD) = ar(?ABC) + ar( ??CDA)
                                    = 2 ar( ?? ABC)                    [ ? ar ?(?ABC) = ar( ??CDA)]
? ar( ??ABC) = 
1
2
ar(quad. ABCD)                ...ii
From i
and ii
, we have:
Page 2


Question:1
Which of the following figures lie on the same base and between the same parallels. In such a case, write the comon base and the two parallels.
Solution:
i
No, it doesnt lie on the same base and between the same parallels.
ii
No, it doesnt lie on the same base and between the same parallels.
iii
 Yes, it lies on the same base and between the same parallels. The same base is AB and the parallels are AB and DE.
iv
 No, it doesnt lie on the same base and between the same parallels.
v
 Yes, it lies on the same base and between the same parallels. The same base is BC and the parallels are BC and AD.
vi
 Yes, it lies on the same base and between the same parallels. The same base is CD and the parallels are CD and BP.
Question:2
In the adjoining figure, show that ABCD is a parallelogram.
Calculate the area of || gm ABCD.
Solution:
Given: A quadrilateral ABCD and BD is a diagonal.
To prove: ABCD is a parallelogram.
Construction: Draw AM ? DC and CL ? AB   extendDCandAB
. Join AC, the other diagonal of ABCD.
Proof: ar(quad. ABCD) = ar(?ABD) + ar( ??DCB)
                                      = 2 ar( ??ABD)                    [ ? ar ?(?ABD) = ar( ??DCB)]
? ar( ??ABD) = 
1
2
ar(quad. ABCD)                 ...i
Again, ar(quad. ABCD) = ar(?ABC) + ar( ??CDA)
                                    = 2 ar( ?? ABC)                    [ ? ar ?(?ABC) = ar( ??CDA)]
? ar( ??ABC) = 
1
2
ar(quad. ABCD)                ...ii
From i
and ii
, we have:
 ar( ??ABD) = ar( ??ABC) = 
1
2
AB ? BD = 
1
2
AB ? CL
 ? CL = BD
 ? DC | ??| AB
Similarly, AD | ??| BC.
Hence, ABCD is a paralleogram.
? ar( ?| ?| gm ABCD) = base ?? height = 5 ?? 7 = 35 cm
2
Question:3
In a parallelogram ABCD, it is being given that AB = 10 cm and the altitudes corresponding to the sides AB and AD are DL = 6 cm and BM = 8 cm,
respectively. Find AD.
Solution:
ar(parallelogram ABCD) = base ?? height
? AB ??DL = AD ?? BM
? 10 ??? 6 = AD ?? BM
? AD ?? 8 = 60 cm
2
? AD =  7.5 cm
? AD = 7.5 cm
Question:4
Find the area of a figure formed by joining the midpoints of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm.
Solution:
Let ABCD be a rhombus and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively. 
Join the diagonals, AC and BD.
In ? ABC, we have:
PQ | | AC and PQ = 
1
2
AC                    
Bymidpointtheorem
PQ =
1
2
×16 = 8 cm
Again, in ?DAC, the points S and R are the midpoints of AD and DC, respectively.
? SR | | AC and SR = 
1
2
AC                    
Bymidpointtheorem
SR =
1
2
×12 = 6 cm
Area of PQRS = length ×breadth = 6 ×8 = 48 cm
2
Question:5
Find the area of a trapezium whose parallel sides are 9 cm and 6 cm respectively and the distance between these sides is 8 cm.
Solution:
artrapezium
 = 
1
2
? sumofparallelsides
? distancebetweenthem
 = 
1
2
 ? 9 +6
? 8
Page 3


Question:1
Which of the following figures lie on the same base and between the same parallels. In such a case, write the comon base and the two parallels.
Solution:
i
No, it doesnt lie on the same base and between the same parallels.
ii
No, it doesnt lie on the same base and between the same parallels.
iii
 Yes, it lies on the same base and between the same parallels. The same base is AB and the parallels are AB and DE.
iv
 No, it doesnt lie on the same base and between the same parallels.
v
 Yes, it lies on the same base and between the same parallels. The same base is BC and the parallels are BC and AD.
vi
 Yes, it lies on the same base and between the same parallels. The same base is CD and the parallels are CD and BP.
Question:2
In the adjoining figure, show that ABCD is a parallelogram.
Calculate the area of || gm ABCD.
Solution:
Given: A quadrilateral ABCD and BD is a diagonal.
To prove: ABCD is a parallelogram.
Construction: Draw AM ? DC and CL ? AB   extendDCandAB
. Join AC, the other diagonal of ABCD.
Proof: ar(quad. ABCD) = ar(?ABD) + ar( ??DCB)
                                      = 2 ar( ??ABD)                    [ ? ar ?(?ABD) = ar( ??DCB)]
? ar( ??ABD) = 
1
2
ar(quad. ABCD)                 ...i
Again, ar(quad. ABCD) = ar(?ABC) + ar( ??CDA)
                                    = 2 ar( ?? ABC)                    [ ? ar ?(?ABC) = ar( ??CDA)]
? ar( ??ABC) = 
1
2
ar(quad. ABCD)                ...ii
From i
and ii
, we have:
 ar( ??ABD) = ar( ??ABC) = 
1
2
AB ? BD = 
1
2
AB ? CL
 ? CL = BD
 ? DC | ??| AB
Similarly, AD | ??| BC.
Hence, ABCD is a paralleogram.
? ar( ?| ?| gm ABCD) = base ?? height = 5 ?? 7 = 35 cm
2
Question:3
In a parallelogram ABCD, it is being given that AB = 10 cm and the altitudes corresponding to the sides AB and AD are DL = 6 cm and BM = 8 cm,
respectively. Find AD.
Solution:
ar(parallelogram ABCD) = base ?? height
? AB ??DL = AD ?? BM
? 10 ??? 6 = AD ?? BM
? AD ?? 8 = 60 cm
2
? AD =  7.5 cm
? AD = 7.5 cm
Question:4
Find the area of a figure formed by joining the midpoints of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm.
Solution:
Let ABCD be a rhombus and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively. 
Join the diagonals, AC and BD.
In ? ABC, we have:
PQ | | AC and PQ = 
1
2
AC                    
Bymidpointtheorem
PQ =
1
2
×16 = 8 cm
Again, in ?DAC, the points S and R are the midpoints of AD and DC, respectively.
? SR | | AC and SR = 
1
2
AC                    
Bymidpointtheorem
SR =
1
2
×12 = 6 cm
Area of PQRS = length ×breadth = 6 ×8 = 48 cm
2
Question:5
Find the area of a trapezium whose parallel sides are 9 cm and 6 cm respectively and the distance between these sides is 8 cm.
Solution:
artrapezium
 = 
1
2
? sumofparallelsides
? distancebetweenthem
 = 
1
2
 ? 9 +6
? 8
= 60 cm
2
Hence, the area of the trapezium is 60 cm
2
.
Question:6
i
Calculate the area of quad. ABCD, given in Fig. i
.
ii
Calculate the area of trap. PQRS, given in Fig. ii
.
Solution:
i
In ?
BCD, 
DB
2
+BC
2
= DC
2
? DB
2
= 17
2
-8
2
= 225 ? DB = 15 cm
Ar( ?
BCD) = 
1
2
×b ×h =
1
2
×8 ×15 = 60 cm
2
In ?
BAD,
DA
2
+AB
2
= DB
2
? AB
2
= 15
2
-9
2
= 144 ? AB = 12 cm
Ar( ?
DAB) = 
1
2
×b ×h =
1
2
×9 ×12 = 54 cm
2
Area of quad. ABCD = Ar( ?
DAB) + Ar( ?
BCD) = 54 + 60 = 114 cm
2 
.
ii
Area of trapPQRS
= 
1
2
(8 +16)×8 = 96 cm
2
Question:7
In the adjoining figure, ABCD is a trapezium in which AB || DC; AB = 7 cm; AD = BC = 5 cm and the distance between AB and DC is 4 cm. Find the
length of DC and hence, find the area of trap. ABCD.
Solution:
?ADL is a right angle triangle.
So, DL = 
v
5
2
 - 4
2
  = 
v
9 = 3 cm
Similarly, in ?BMC, we have:
MC = 
v
5
2
 - 4
2
  = 
v
9 = 3 cm
? DC =  DL + LM + MC =  3 + 7 + 3 = 13 cm
Now, ar(trapezium. ABCD) = 
1
2
? sumofparallelsides
? distancebetweenthem
=
1
2
? 7 +13
? 4
= 40 cm
2
( )
( )
Page 4


Question:1
Which of the following figures lie on the same base and between the same parallels. In such a case, write the comon base and the two parallels.
Solution:
i
No, it doesnt lie on the same base and between the same parallels.
ii
No, it doesnt lie on the same base and between the same parallels.
iii
 Yes, it lies on the same base and between the same parallels. The same base is AB and the parallels are AB and DE.
iv
 No, it doesnt lie on the same base and between the same parallels.
v
 Yes, it lies on the same base and between the same parallels. The same base is BC and the parallels are BC and AD.
vi
 Yes, it lies on the same base and between the same parallels. The same base is CD and the parallels are CD and BP.
Question:2
In the adjoining figure, show that ABCD is a parallelogram.
Calculate the area of || gm ABCD.
Solution:
Given: A quadrilateral ABCD and BD is a diagonal.
To prove: ABCD is a parallelogram.
Construction: Draw AM ? DC and CL ? AB   extendDCandAB
. Join AC, the other diagonal of ABCD.
Proof: ar(quad. ABCD) = ar(?ABD) + ar( ??DCB)
                                      = 2 ar( ??ABD)                    [ ? ar ?(?ABD) = ar( ??DCB)]
? ar( ??ABD) = 
1
2
ar(quad. ABCD)                 ...i
Again, ar(quad. ABCD) = ar(?ABC) + ar( ??CDA)
                                    = 2 ar( ?? ABC)                    [ ? ar ?(?ABC) = ar( ??CDA)]
? ar( ??ABC) = 
1
2
ar(quad. ABCD)                ...ii
From i
and ii
, we have:
 ar( ??ABD) = ar( ??ABC) = 
1
2
AB ? BD = 
1
2
AB ? CL
 ? CL = BD
 ? DC | ??| AB
Similarly, AD | ??| BC.
Hence, ABCD is a paralleogram.
? ar( ?| ?| gm ABCD) = base ?? height = 5 ?? 7 = 35 cm
2
Question:3
In a parallelogram ABCD, it is being given that AB = 10 cm and the altitudes corresponding to the sides AB and AD are DL = 6 cm and BM = 8 cm,
respectively. Find AD.
Solution:
ar(parallelogram ABCD) = base ?? height
? AB ??DL = AD ?? BM
? 10 ??? 6 = AD ?? BM
? AD ?? 8 = 60 cm
2
? AD =  7.5 cm
? AD = 7.5 cm
Question:4
Find the area of a figure formed by joining the midpoints of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm.
Solution:
Let ABCD be a rhombus and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively. 
Join the diagonals, AC and BD.
In ? ABC, we have:
PQ | | AC and PQ = 
1
2
AC                    
Bymidpointtheorem
PQ =
1
2
×16 = 8 cm
Again, in ?DAC, the points S and R are the midpoints of AD and DC, respectively.
? SR | | AC and SR = 
1
2
AC                    
Bymidpointtheorem
SR =
1
2
×12 = 6 cm
Area of PQRS = length ×breadth = 6 ×8 = 48 cm
2
Question:5
Find the area of a trapezium whose parallel sides are 9 cm and 6 cm respectively and the distance between these sides is 8 cm.
Solution:
artrapezium
 = 
1
2
? sumofparallelsides
? distancebetweenthem
 = 
1
2
 ? 9 +6
? 8
= 60 cm
2
Hence, the area of the trapezium is 60 cm
2
.
Question:6
i
Calculate the area of quad. ABCD, given in Fig. i
.
ii
Calculate the area of trap. PQRS, given in Fig. ii
.
Solution:
i
In ?
BCD, 
DB
2
+BC
2
= DC
2
? DB
2
= 17
2
-8
2
= 225 ? DB = 15 cm
Ar( ?
BCD) = 
1
2
×b ×h =
1
2
×8 ×15 = 60 cm
2
In ?
BAD,
DA
2
+AB
2
= DB
2
? AB
2
= 15
2
-9
2
= 144 ? AB = 12 cm
Ar( ?
DAB) = 
1
2
×b ×h =
1
2
×9 ×12 = 54 cm
2
Area of quad. ABCD = Ar( ?
DAB) + Ar( ?
BCD) = 54 + 60 = 114 cm
2 
.
ii
Area of trapPQRS
= 
1
2
(8 +16)×8 = 96 cm
2
Question:7
In the adjoining figure, ABCD is a trapezium in which AB || DC; AB = 7 cm; AD = BC = 5 cm and the distance between AB and DC is 4 cm. Find the
length of DC and hence, find the area of trap. ABCD.
Solution:
?ADL is a right angle triangle.
So, DL = 
v
5
2
 - 4
2
  = 
v
9 = 3 cm
Similarly, in ?BMC, we have:
MC = 
v
5
2
 - 4
2
  = 
v
9 = 3 cm
? DC =  DL + LM + MC =  3 + 7 + 3 = 13 cm
Now, ar(trapezium. ABCD) = 
1
2
? sumofparallelsides
? distancebetweenthem
=
1
2
? 7 +13
? 4
= 40 cm
2
( )
( )
?Hence, DC = 13 cm and area of trapezium = 40 cm
2
Question:8
BD is one of the diagonals of a quad. ABCD. If AL ? BD and CM ? BD, show that 
ar quad. ABCD =
1
2
×BD × AL + CM .
Solution:
ar(quad. ABCD) = ar(? ?ABD) + ar (?DBC)
ar(?ABD) = 
1
2
? base ? height = 
1
2
? BD ? AL             ...i
 
ar(?DBC) = 
1
2
? BD ? CL                  ...ii
From i
and ii
, we get:
?ar(quad ABCD) = 
1
2
 ?BD ? ? AL + 
1
2
 ? BD ? ? CL
?
?ar(quad ABCD) = 
1
2
 ? BD ? ?(AL + CL)
Hence, proved.
Question:9
M is the midpoint of the side AB of a parallelogram ABCD. If ar(AMCD) = 24 cm
2
, find ar(?ABC).
Solution:
Join AC. 
AC divides parallelogram ABCD into two congruent triangles of equal areas. 
ar( ? ABC) = ar( ? ACD) =
1
2
ar(ABCD)
M is the midpoint of AB. So, CM is the median. 
CM divides ?
ABC in two triangles with equal area. 
ar( ? AMC) = ar( ? BMC) =
1
2
ar( ? ABC)
arAMCD
= ar( ?
ACD) + ar( ?
AMC) = ar( ?
ABC) + ar( ?
AMC) = ?ar( ?
ABC) + 
1
2
?ar( ?
ABC)
? 24 =
3
2
ar( ? ABC) ? ar( ? ABC) = 16 cm
2
Question:10
( ) ( )
Page 5


Question:1
Which of the following figures lie on the same base and between the same parallels. In such a case, write the comon base and the two parallels.
Solution:
i
No, it doesnt lie on the same base and between the same parallels.
ii
No, it doesnt lie on the same base and between the same parallels.
iii
 Yes, it lies on the same base and between the same parallels. The same base is AB and the parallels are AB and DE.
iv
 No, it doesnt lie on the same base and between the same parallels.
v
 Yes, it lies on the same base and between the same parallels. The same base is BC and the parallels are BC and AD.
vi
 Yes, it lies on the same base and between the same parallels. The same base is CD and the parallels are CD and BP.
Question:2
In the adjoining figure, show that ABCD is a parallelogram.
Calculate the area of || gm ABCD.
Solution:
Given: A quadrilateral ABCD and BD is a diagonal.
To prove: ABCD is a parallelogram.
Construction: Draw AM ? DC and CL ? AB   extendDCandAB
. Join AC, the other diagonal of ABCD.
Proof: ar(quad. ABCD) = ar(?ABD) + ar( ??DCB)
                                      = 2 ar( ??ABD)                    [ ? ar ?(?ABD) = ar( ??DCB)]
? ar( ??ABD) = 
1
2
ar(quad. ABCD)                 ...i
Again, ar(quad. ABCD) = ar(?ABC) + ar( ??CDA)
                                    = 2 ar( ?? ABC)                    [ ? ar ?(?ABC) = ar( ??CDA)]
? ar( ??ABC) = 
1
2
ar(quad. ABCD)                ...ii
From i
and ii
, we have:
 ar( ??ABD) = ar( ??ABC) = 
1
2
AB ? BD = 
1
2
AB ? CL
 ? CL = BD
 ? DC | ??| AB
Similarly, AD | ??| BC.
Hence, ABCD is a paralleogram.
? ar( ?| ?| gm ABCD) = base ?? height = 5 ?? 7 = 35 cm
2
Question:3
In a parallelogram ABCD, it is being given that AB = 10 cm and the altitudes corresponding to the sides AB and AD are DL = 6 cm and BM = 8 cm,
respectively. Find AD.
Solution:
ar(parallelogram ABCD) = base ?? height
? AB ??DL = AD ?? BM
? 10 ??? 6 = AD ?? BM
? AD ?? 8 = 60 cm
2
? AD =  7.5 cm
? AD = 7.5 cm
Question:4
Find the area of a figure formed by joining the midpoints of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm.
Solution:
Let ABCD be a rhombus and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively. 
Join the diagonals, AC and BD.
In ? ABC, we have:
PQ | | AC and PQ = 
1
2
AC                    
Bymidpointtheorem
PQ =
1
2
×16 = 8 cm
Again, in ?DAC, the points S and R are the midpoints of AD and DC, respectively.
? SR | | AC and SR = 
1
2
AC                    
Bymidpointtheorem
SR =
1
2
×12 = 6 cm
Area of PQRS = length ×breadth = 6 ×8 = 48 cm
2
Question:5
Find the area of a trapezium whose parallel sides are 9 cm and 6 cm respectively and the distance between these sides is 8 cm.
Solution:
artrapezium
 = 
1
2
? sumofparallelsides
? distancebetweenthem
 = 
1
2
 ? 9 +6
? 8
= 60 cm
2
Hence, the area of the trapezium is 60 cm
2
.
Question:6
i
Calculate the area of quad. ABCD, given in Fig. i
.
ii
Calculate the area of trap. PQRS, given in Fig. ii
.
Solution:
i
In ?
BCD, 
DB
2
+BC
2
= DC
2
? DB
2
= 17
2
-8
2
= 225 ? DB = 15 cm
Ar( ?
BCD) = 
1
2
×b ×h =
1
2
×8 ×15 = 60 cm
2
In ?
BAD,
DA
2
+AB
2
= DB
2
? AB
2
= 15
2
-9
2
= 144 ? AB = 12 cm
Ar( ?
DAB) = 
1
2
×b ×h =
1
2
×9 ×12 = 54 cm
2
Area of quad. ABCD = Ar( ?
DAB) + Ar( ?
BCD) = 54 + 60 = 114 cm
2 
.
ii
Area of trapPQRS
= 
1
2
(8 +16)×8 = 96 cm
2
Question:7
In the adjoining figure, ABCD is a trapezium in which AB || DC; AB = 7 cm; AD = BC = 5 cm and the distance between AB and DC is 4 cm. Find the
length of DC and hence, find the area of trap. ABCD.
Solution:
?ADL is a right angle triangle.
So, DL = 
v
5
2
 - 4
2
  = 
v
9 = 3 cm
Similarly, in ?BMC, we have:
MC = 
v
5
2
 - 4
2
  = 
v
9 = 3 cm
? DC =  DL + LM + MC =  3 + 7 + 3 = 13 cm
Now, ar(trapezium. ABCD) = 
1
2
? sumofparallelsides
? distancebetweenthem
=
1
2
? 7 +13
? 4
= 40 cm
2
( )
( )
?Hence, DC = 13 cm and area of trapezium = 40 cm
2
Question:8
BD is one of the diagonals of a quad. ABCD. If AL ? BD and CM ? BD, show that 
ar quad. ABCD =
1
2
×BD × AL + CM .
Solution:
ar(quad. ABCD) = ar(? ?ABD) + ar (?DBC)
ar(?ABD) = 
1
2
? base ? height = 
1
2
? BD ? AL             ...i
 
ar(?DBC) = 
1
2
? BD ? CL                  ...ii
From i
and ii
, we get:
?ar(quad ABCD) = 
1
2
 ?BD ? ? AL + 
1
2
 ? BD ? ? CL
?
?ar(quad ABCD) = 
1
2
 ? BD ? ?(AL + CL)
Hence, proved.
Question:9
M is the midpoint of the side AB of a parallelogram ABCD. If ar(AMCD) = 24 cm
2
, find ar(?ABC).
Solution:
Join AC. 
AC divides parallelogram ABCD into two congruent triangles of equal areas. 
ar( ? ABC) = ar( ? ACD) =
1
2
ar(ABCD)
M is the midpoint of AB. So, CM is the median. 
CM divides ?
ABC in two triangles with equal area. 
ar( ? AMC) = ar( ? BMC) =
1
2
ar( ? ABC)
arAMCD
= ar( ?
ACD) + ar( ?
AMC) = ar( ?
ABC) + ar( ?
AMC) = ?ar( ?
ABC) + 
1
2
?ar( ?
ABC)
? 24 =
3
2
ar( ? ABC) ? ar( ? ABC) = 16 cm
2
Question:10
( ) ( )
In the adjoining figure, ABCD is a quadrilateral in which diag. BD = 14 cm. If AL ? BD and CM ? BD such that AL = 8 cm and CM = 6 cm, find the area
of quad. ABCD.
Solution:
?ar(quad ABCD) = ar( ?
ABD) + ar( ?
BDC)
= 
1
2
 ?BD ? ? AL  +
1
2
 ?BD ? ? CM
= 
1
2
 ?BD ? ? ( AL + CM)
By substituting the values, we have;
ar(quad ABCD) = ? 1
2
 ? 14 ? 8 +6
= 7 ??14
= 98 cm
2
Question:11
If P and Q are any two points lying respectively on the sides DC and AD of a parallelogram ABCD then show that ar(?APB) = ar(?BQC).
Solution:
We know
ar? APB
= 
1
2
ar(ABCD)
           .....1
                   
Ifatriangleandaparallelogramareonthesamebaseandbetweenthesameparallelsthentheareaofthetriangleisequaltohalftheareaoftheparallelogram
Similarly, 
ar? BQC
= 
1
2
ar(ABCD)
           .....2
From 1
and 2
ar? APB
= ar? BQC
Hence Proved
Question:12
In the adjoining figure, MNPQ and ABPQ are parallelograms and T is any point on the side BP. Prove that
i
ar(MNPQ) = are(ABPQ)
ii
ar(?ATQ) = 
1
2
ar(MNPQ).