# NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

**
NCERT solutions for class 9 maths chapter 9 Areas of Parallelograms and Triangles:
**
A four side shape having equal opposite sides and whose diagonals bisect each other is called a parallelogram. Triangle is a 3 side shape having 3 sides and 3 angles. In this particular chapter, you will learn about the areas of different triangles and parallelograms. Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles will provide you the solutions to all types of triangles and parallelograms covered in the chapter. Many structures have the shape of the triangle and parallelograms. Understanding the area is mandatory in the field of calculating the area of land. If a land has an irregular shape, the land will be divided into triangles and the area will be calculated.

CBSE NCERT solutions for class 9 maths chapter 9 Areas Of Parallelograms And Triangles is also covering the solutions to the application based questions as well. There are several interesting problems in the chapter. Solving these problems will help you in improving the concepts of the chapter and also in exams like the olympiads. In total there are 4 practice exercises having 51 questions. NCERT solutions for class 9 maths chapter 9 Areas of Parallelograms and Triangles have covered all the exercises including the optional ones in a detailed manner. NCERT solutions are available for other classes and chapters as well. Four exercises of this chapter are explained below.

##
**
NCERT solutions for class 9 maths chapter 9 Areas Of Parallelograms And Triangles Excercise: 9.1
**

###
**
Q
**
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.

###
**
Answer:
**

In figure (i), (iii) and (v) we can see that. they lie on the same base and between the same parallel lines.

In figure (i) figure (iii) figure (v)

Common base DC QR AD

Two parallels DC and AB QR and PS AD and BQ

##
**
NCERT solutions for class 9 maths chapter 9 Areas Of Parallelograms And Triangles Excercise: 9.2
**

**
Q1
**
In Fig.
, ABCD is a parallelogram,
and
. If
,
and
, find AD.

###
**
Answer:
**

We have,

AE
DC and CF
AD

AB = 16 cm, AE = 8 cm and CF = 10 cm

Since ABCD is a parallelogram,

therefore, AB = DC = 16 cm

We know that, area of parallelogram (ABCD) = base . height

= CD
AE = (16
8 )

SInce, CF
AD

therefore area of parallelogram = AD
CF = 128

= AD = 128/10

=
**
AD = 12.8 cm
**

Thus the required length of AD is 12.8 cm.

**
Q2
**
If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that

###
**
Answer:
**

Join GE and HE,

Since F, F, G, H are the mid-points of the parallelogram ABCD. Therefore, GE || BC ||AD and HF || DC || AB.

It is known that if a triangle and the parallelogram are on the same base and between the same parallel lines. then the area of the triangle is equal to the half of the area of the parallelogram.

Now,
EFG and ||gm BEGC are on the same base and between the same parallels EG and BC.

Therefore, ar (
EFG) =
ar (||gm BEGC)...............(i)

Similarly, ar (
EHG) = 1/2 . ar(||gm AEGD)..................(ii)

By adding eq (i) and eq (ii), we get

ar (
EFG) + ar (
EHG) = 1/2 (ar (||gm BEGC) + ar(||gm AEGD))

ar (EFGH) = 1/2 ar(ABCD)

Hence proved

###
**
Answer:
**

We have,

ABCD is a parallelogram, therefore AB || CD and BC || AD.

Now,
APB and ||gm ABCD are on the same base AB and between two parallels AB and DC.

Therefore, ar (
APB) = 1/2 . ar(||gm ABCD)...........(i)

Also,
BQC and ||gm ABCD are on the same base BC and between two parallels BC and AD.

Therefore, ar(
BQC) = 1/2 . ar(||gmABCD)...........(ii)

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

ar (
APB) = ar(
BQC)

Hence proved.

**
Q4 (i)
**
In Fig.
, P is a point in the interior of a parallelogram ABCD. Show that

[
**
Hint
**
: Through P, draw a line parallel to AB.]

###
**
Answer:
**

We have a ||gm ABCD and AB || CD, AD || BC. Through P, draw a line parallel to AB

Now,
APB and ||gm ABEFare on the same base AB and between the same parallels EF and AB.

Therefore, ar (
APB) = 1/2 . ar(ABEF)...............(i)

Similarly, ar (
PCD ) = 1/2 . ar (EFDC) ..............(ii)

Now, by adding both equations, we get

Hence proved.

**
Q4 (ii)
**
In Fig.
, P is a point in the interior of a parallelogram ABCD. Show that

[
**
Hint
**
: Through P, draw a line parallel to AB.]

###
**
Answer:
**

We have a ||gm ABCD and AB || CD, AD || BC. Through P, draw a line parallel to AB

Now,
APD and ||gm ADGHare on the same base AD and between the same parallels GH and AD.

Therefore, ar (
APD) = 1/2 . ar(||gm ADGH).............(i)

Similarily, ar ( PBC) = 1/2 . ar(||gm BCGH)............(ii)

By adding the equation i and eq (ii), we get

Hence proved.

**
Q5
**
In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that

###
**
Answer:
**

(i) Parallelogram PQRS and ABRS are on the same base RS and between the same parallels RS and PB.

Therefore,
............(i)

Hence proved

(ii)
AXS and ||gm ABRS are on the same base AS and between same parallels AS and RB.

Therefore, ar (
AXS) = 1/2 . ar(||gm ABRS)............(ii)

Now, from equation (i) and equation (ii), we get

Hence proved.

###
**
Answer:
**

We have a field in the form of parallelogram PQRS and a point A is on the side RS. Join AP and AQ. The field is divided into three parts i.e,
APS,
QAR and
PAQ.

Since
APQ and parallelogram, PQRS is on the same base PQ and between same parallels RS and PQ.

Therefore,
............(i)

We can write above equation as,

ar (||gm PQRS) - [ar (
APS) + ar(
QAR)] = 1/2 .ar(PQRS)

from equation (i),

Hence, she can sow wheat in APQ and pulses in [ APS + QAR] or wheat in [ APS + QAR] and pulses in APQ.

##
**
NCERT solutions for class 9 maths chapter 9 Areas Of Parallelograms And Triangles Excercise: 9.3
**

**
Q1
**
In Fig.
, E is any point on median AD of a
. Show that.

###
**
Answer:
**

We have
ABC such that AD is a median. And we know that median divides the triangle into two triangles of equal areas.

Therefore, ar(
ABD) = ar(
ACD)............(i)

Similarly, In triangle
BEC,

ar(
BED) = ar (
DEC)................(ii)

On subtracting eq(ii) from eq(i), we get

ar(
ABD) - ar(
BED) =

Hence proved.

**
Q2
**
In a triangle ABC, E is the mid-point of median AD. Show that
.

###
**
Answer:
**

We have a triangle ABC and AD is a median. Join B and E.

Since the median divides the triangle into two triangles of equal area.

ar(
ABD) = ar (
ACD) = 1/2 ar(
ABC)..............(i)

Now, in triangle
ABD,

BE is the median [since E is the midpoint of AD]

ar (
BED) = 1/2 ar(
ABD)........(ii)

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

ar (
BED) = 1/2 . (1/2 ar(ar (
ABC))

**
ar (
BED) = 1/4 .ar(
ABC)
**

Hence proved.

**
Q3
**
Show that the diagonals of a parallelogram divide it into four triangles of equal area.

###
**
Answer:
**

Let ABCD is a parallelogram. So, AB || CD and AD || BC and we know that Diagonals bisects each other. Therefore, AO = OC and BO = OD

Since OD = BO

Therefore, ar (
BOC) = ar (
DOC)...........(a) ( since OC is the median of triangle CBD)

Similarly, ar( AOD) = ar( DOC) ............(b) ( since OD is the median of triangle ACD)

and, ar ( AOB) = ar( BOC)..............(c) ( since OB is the median of triangle ABC)

From eq (a), (b) and eq (c), we get

ar ( BOC) = ar ( DOC)= ar( AOD) = ( AOB)

Thus, the diagonals of ||gm divide it into four equal triangles of equal area.

###
**
Answer:
**

We have,
ABC and
ABD on the same base AB. CD is bisected by AB at point O.

OC = OD

Now, in
ACD, AO is median

ar (
AOC) = ar (
AOD)..........(i)

Similarly, in
BCD, BO is the median

ar (
BOC) = ar (
BOD)............(ii)

Adding equation (i) and eq (ii), we get

ar ( AOC) + ar ( BOC) = ar ( AOD) + ar ( BOD)

Hence proved.

###
**
Answer:
**

We have a triangle
ABC such that D, E and F are the midpoints of the sides BC, CA and AB respectively.

Now, in
ABC,

F and E are the midpoints of the side AB and AC.

Therefore according to mid-point theorem, the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and half of the third side.

EF || BC or EF || BD

also, EF = 1/2 (BC)

[ D is the midpoint of BC]

Similarly, ED || BF and ED = FB

Hence BDEF is a parallelogram.

**
Q5 (ii)
**
D, E and F are respectively the mid-points of the sides BC, CA and AB of a
. Show that

###
**
Answer:
**

We already proved that BDEF is a ||gm.

Similarly, DCEF and DEAF are also parallelograms.

Now, ||gm BDEF and ||gm DCEF is on the same base EF and between same parallels BC and EF

Ar (BDEF) = Ar (DCEF)

Ar(
BDF) = Ar (
DEF) .............(i)

It is known that diagonals of ||gm divides it into two triangles of equal area.

Ar(DCE) = Ar (DEF).......(ii)

and, Ar( AEF) = Ar ( DEF)...........(iii)

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

Ar( BDF) = Ar(DCE) = Ar( AEF) = Ar ( DEF)

Thus, Ar (
ABC) = Ar(
BDF) + Ar(DCE) + Ar(
AEF) + Ar (
DEF)

Ar (
ABC) = 4 . Ar(
DEF)

Hence proved.

**
Q5 (iii)
**
D, E and F are respectively the mid-points of the sides BC, CA and AB of a
. Show that

###
**
Answer:
**

Since we already proved that,

ar(
DEF) = ar (
BDF).........(i)

So, ar(||gm BDEF) = ar(
BDF) + ar (
DEF)

= 2 . ar(
DEF) [from equation (i)]

Hence proved.

**
Q6 (i)
**
In Fig.
, diagonals AC and BD of quadrilateral ABCD intersect at O such that. If
, then show that:

[
**
Hint:
**
From D and B, draw perpendiculars to AC.]

###
**
Answer:
**

We have ABCD is quadrilateral whose diagonals AC and BD intersect at O. And OB = OD, AB = CD

Draw DE
AC and FB
AC

In
DEO and
BFO

DOE =
BOF [vertically opposite angle]

OED =
BFO [each
]

OB = OD [given]

Therefore, by AAS congruency

DEO
BFO

DE = FB [by CPCT]

and ar( DEO) = ar( BFO) ............(i)

Now, In
DEC and
ABF

DEC =
BFA [ each
]

DE = FB

DC = BA [given]

So, by RHS congruency

DEC
BFA

1 =
2 [by CPCT]

and, ar(
DEC) = ar(
BFA).....(ii)

By adding equation(i) and (ii), we get

Hence proved.

**
Q6 (ii)
**
In Fig.
, diagonals AC and BD of quadrilateral ABCD intersect at O such that
. If
, then show that:

[
**
Hint:
**
From D and B, draw perpendiculars to AC.]

###
**
Answer:
**

We already proved that,

Now, add
on both sides we get

Hence proved.

**
Q6 (iii)
**
In Fig.
, diagonals AC and BD of quadrilateral ABCD intersect at O such that
. If
, then show that:

[
**
Hint
**
: From D and B, draw perpendiculars to AC.]

###
**
Answer:
**

Since
DCB and
ACB both lie on the same base BC and having equal areas.

Therefore, They lie between the same parallels BC and AD

CB || AD

also
1 =
2 [ already proved]

So, AB || CD

Hence ABCD is a || gm

**
Q7
**
D and E are points on sides AB and AC respectively of
such that
. Prove that
.

###
**
Answer:
**

We have
ABC and points D and E are on the sides AB and AC such that ar(
DBC ) = ar (
EBC)

Since
DBC and
EBC are on the same base BC and having the same area.

They must lie between the same parallels DE and BC

Hence DE || BC.

###
**
Answer:
**

We have a
ABC such that BE || AC and CF || AB

Since XY || BC and BE || CY

Therefore, BCYE is a ||gm

Now, The ||gm BCEY and
ABE are on the same base BE and between the same parallels AC and BE.

ar(
AEB) = 1/2 .ar(||gm BEYC)..........(i)

Similarly, ar(
ACF) = 1/2 . ar(||gm BCFX)..................(ii)

Also, ||gm BEYC and ||gmBCFX are on the same base BC and between the same parallels BC and EF.

ar (BEYC) = ar (BCFX).........(iii)

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

ar(
ABE) = ar(
ACF)

Hence proved.

###
**
Answer:
**

Join the AC and PQ.

It is given that ABCD is a ||gm and AC is a diagonal of ||gm

Therefore, ar(
ABC) = ar(
ADC) = 1/2 ar(||gm ABCD).............(i)

Also, ar( PQR) = ar( BPQ) = 1/2 ar(||gm PBQR).............(ii)

Since
AQC and
APQ are on the same base AQ and between same parallels AQ and CP.

ar(
AQC) = ar (
APQ)

Now, subtracting ABQ from both sides we get,

ar(
AQC) - ar (
ABQ) = ar (
APQ) - ar (
ABQ)

ar(
ABC) = ar (
BPQ)............(iii)

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

Hence proved.

**
Q10
**
Diagonals AC and BD of a trapezium ABCD with
intersect each other at O. Prove that
.

###
**
Answer:
**

We have a trapezium ABCD such that AB || CD and it's diagonals AC and BD intersect each other at O

ABD and
ABC are on the same base AB and between same parallels AB and CD

ar(
ABD) = ar (
ABC)

Now, subtracting AOB from both sides we get

ar ( AOD) = ar ( BOC)

Hence proved.

**
Q11
**
In Fig.
, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

###
**
Answer:
**

We have a pentagon ABCDE in which BF || AC and CD is produced to F.

(i) Since
ACB and
ACF are on the same base AC and between same parallels AC and FB.

ar(
ACB) = ar (
ACF)..................(i)

(ii) Adding the ar (AEDC) on both sides in equation (i), we get

ar(
ACB) + ar(AEDC) = ar (
ACF) + ar(AEDC)

Hence proved.

###
**
Answer:
**

We have a quadrilateral shaped plot ABCD. Draw DF || AC and AF || CF.

Now,
DAF and
DCF are on the same base DF and between same parallels AC and DF.

ar (
DAF) = ar(
DCF)

On subtracting
DEF from both sides, we get

ar(
ADE) = ar(
CEF)...............(i)

The portion of
ADE can be taken by the gram panchayat and on adding the land
CEF to his (Itwaari) land so as to form a triangular plot.(
ABF)

**
We need to prove that ar(
ABF) = ar (quad. ABCD)
**

Now, adding ar(quad. ABCE) on both sides in eq (i), we get

ar (
ADE) + ar(quad. ABCE) = ar(
CEF) + ar(quad. ABCE)

ar (ABCD) = ar(
ABF)

###
**
Answer:
**

We have a trapezium ABCD, AB || CD

XY ||AC meets AB at X and BC at Y. Join XC

Since
ADX and
ACX lie on the same base CD and between same parallels AX and CD

Therefore, ar(
ADX) = ar(
ACX)..........(i)

Similarly ar(
ACX) = ar(
ACY).............(ii) [common base AC and AC || XY]

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

ar( ADX) = ar ( ACY)

Hence proved.

###
**
Answer:
**

We have, AP || BQ || CR

BCQ and
BQR lie on the same base (BQ) and between same parallels (BQ and CR)

Therefore, ar (
BCQ) = ar (
BQR)........(i)

Similarly, ar ( ABQ) = ar ( PBQ) [common base BQ and BQ || AP]............(ii)

Add the eq(i) and (ii), we get

ar ( AQC) = ar ( PBR)

Hence proved.

###
**
Answer:
**

We have,

ABCD is a quadrilateral and diagonals AC and BD intersect at O such that ar(
AOD) = ar (
BOC) ...........(i)

Now, add ar ( BOA) on both sides, we get

ar(
AOD) + ar (
BOA) = ar (
BOA) + ar (
BOC)

ar (
ABD) = ar (
ABC)

Since the
ABC and
ABD lie on the same base AB and have an equal area.

Therefore, AB || CD

Hence ABCD is a trapezium.

**
Q16
**
In Fig.
,
and
. Show that both the quadrilaterals ABCD and DCPR are trapeziums.

###
**
Answer:
**

Given,

ar(
DPC) = ar(
DRC) ..........(i)

and ar(
BDP) = ar(ARC)............(ii)

from equation (i),

Since
DRC and
DPC lie on the same base DC and between same parallels.

CD || RP (opposites sides are parallel)

Hence quadrilateral DCPR is a trapezium

Now, by subtracting eq(ii) - eq(i) we get

ar(
BDP) - ar(
DPC) = ar(
ARC) - ar(
DRC)

ar(
BDC) = ar(
ADC) (Since theya are on the same base DC)

AB || DC

Hence ABCD is a trapezium.

##
**
NCERT solutions for class 9 maths chapter 9 Areas Of Parallelograms And Triangles Excercise: 9.4
**

###
**
Answer:
**

We have ||gm ABCD and a rectangle ABEF both lie on the same base AB such that,
**
ar(||gm ABCD) = ar(ABEF)
**

for rectangle, AB = EF

and for ||gm AB = CD

CD = EF

AB + CD = AB + EF ...........(i)

SInce
BEC and
AFD are right angled triangle

Therefore, AD > AF and BC > BE

(BC + AD ) > (AF + BE)...........(ii)

Adding equation (i) and (ii), we get

(AB + CD)+(BC + AD) > (AB + BE) + (AF + BE)

(AB + BC + CD + DA) > (AB + BE + EF + FA)

Hence proved, perimeter of ||gm ABCD is greater than perimeter of rectangle ABEF.

**
Q2
**
In Fig.
, D and E are two points on BC such that
. Show that
.

###
**
Answer:
**

In
ABC, D and E are two points on BC such that BD = DE = EC

AD is the median of ABE, therefore,

area of ABD= area of AED...................(1)

AE is the median of ACD, therefore,

area of AEC= area of AED...................(2)

From (1) and (2)

area of ABD=area of AED= area of AEC

Hence proved.

**
Q3
**
In Fig.
, ABCD, DCFE and ABFE are parallelograms. Show that
.

###
**
Answer:
**

Given,

ABCD is a parallelogram. Therefore, AB = CD & AD = BC & AB || CD .......(i)

Now,
ADE and
BCF are on the same base AD = BC and between same parallels AB and EF.

Therefore, ar (
ADE) = ar(
BCF)

Hence proved.

###
**
Answer:
**

Given,

ABCD is a ||gm and AD = CQ. Join AC.

Since
DQC and
ACD lie on the same base QC and between same parallels AD and QC.

Therefore, ar(
DQC) = ar(
ACD).......(i)

Subtracting ar(
PQC) from both sides in eq (i), we get

ar(
DPQ) = ar(
PAC).............(i)

Since
PAC and
PBC are on the same base PC and between same parallel PC and AB.

Therefore, ar(
PAC) = ar(
PBC)..............(iii)

From equation (ii) and eq (ii), we get

Hence proved.

[
**
Hint:
**
Join EC and AD. Show that
and
, etc.]

###
**
Answer:
**

Let join the CE and AD and draw
. It is given that
ABC and
BDE is an equilateral triangle.

So, AB =BC = CA =
and D i sthe midpoint of BC

therefore,

(i) Area of
ABC =
and

Area of
BDE =

Hence,

[
**
Hint:
**
Join EC and AD. Show that
and
, etc.]

###
**
Answer:
**

Since
ABC and
BDE are equilateral triangles.

Therefore,
ACB =
DBE =

BE || AC

BAE and
BEC are on the same base BE and between same parallels BE and AC.

Therefore, ar (
BAE) = ar(
BEC)

ar(
BAE) = 2 ar(
BED) [since D is the meian of
BEC ]

Hence proved.

[
**
Hint
**
: Join EC and AD. Show that
and
, etc.]

###
**
Answer:
**

We already proved that,

ar(
ABC) = 4.ar(
BDE) (in part 1)

and, ar(
BEC) = 2. ar(
BDE) (in part ii )

ar( ABC) = 2. ar( BEC)

Hence proved.

###
**
Answer:
**

Since
ABC and
BDE are equilateral triangles.

Therefore,
ACB =
DBE =

BE || AC

BDE and
AED are on the same base ED and between same parallels AB and DE.

Therefore, ar(
BED) = ar(
AED)

On subtracting EFD from both sides we get

Hence proved.

[
**
Hint
**
: Join EC and AD. Show that
and
, etc.]

###
**
Answer:
**

In right angled triangle
ABD, we get

So, in
PED,

So,

Therefore, the Area of ..........(i)

And, Area of triangle ...........(ii)

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

ar(
AFD) = 2. ar(
EFD)

Since ar(
AFD) = ar(
BEF)

[
**
Hint:
**
Join EC and AD. Show that
and
, etc.]

###
**
Answer:
**

.....(from part (v) ar(
BFE) = 2. ar(
FED) ]

Hence proved.

**
Q6
**
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that

[
*
Hint:
*
From A and C, draw perpendiculars to BD.]

###
**
Answer:
**

Given that,

A quadrilateral ABCD such that it's diagonal AC and BD intersect at P. Draw
and

Now, ar(
APB) =
and,

Therefore, ar(
APB)
ar(
CDP)=

....................(i)

Similarly, ar(
APD)
ar (
BPC) =

................(ii)

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

Hence proved.

###
**
Answer:
**

We have
ABC such that P, Q and R are the midpoints of the side AB, BC and AP respectively. Join PQ, QR, AQ, PC and RC as shown in the figure.

Now, in
APC,

Since R is the midpoint. So, RC is the median of the
APC

Therefore, ar(
ARC) = 1/2 . ar (
APC)............(i)

Also, in
ABC, P is the midpoint. Thus CP is the median.

Therefore, ar(
APC) = 1/2. ar (
ABC)............(ii)

Also, AQ is the median of
ABC

Therefore, 1/2. ar (
ABC) = ar (ABQ)............(iii)

In
APQ, RQ is the median.

Therefore, ar (
PRQ) = 1/2 .ar (
APQ).............(iv)

In
ABQ, PQ is the median

Therefore, ar(
APQ) = 1/2. ar(
ABQ).........(v)

From eq (i),

...........(vi)

Now, put the value of ar(
APC) from eq (ii), we get

Taking RHS;

(from equation (iii))

(from equation (v))

(from equation (iv))

Hence proved.

###
**
Answer:
**

In
RBC, RQ is the median

Therefore ar(
RQC) = ar(
RBQ)

= ar (PRQ) + ar (BPQ)

= 1/8 (ar
ABC) + ar(
BPQ) [from eq (vi) & eq (A) in part (i)]

= 1/8 (ar
ABC) + 1/2 (ar
PBC) [ since PQ is the median of
BPC]

= 1/8 (ar
ABC) + (1/2).(1/2)(ar
ABC) [CP is the medain of
ABC]

= 3/8 (ar
ABC)

Hence proved.

###
**
Answer:
**

QP is the median of
ABQ

Therefore, ar(
PBQ) = 1/2. (ar
ABQ)

= (1/2). (1/2) (ar
ABC) [since AQ is the median of
ABC

= 1/4 (ar
ABC)

= ar (
ARC) [from eq (A) of part (i)]

Hence proved.