NCERT Exemplar Class 9 Maths Solutions Chapter 9 Areas of Parallelograms and Triangles

NCERT exemplar Class 9 Maths solutions chapter 9 provides the fundamentals of the area of parallelogram and area of a triangle to develop the formulas of area for different structures. These fundamentals are essential and useful for higher mathematics. These NCERT exemplar Class 9 Maths chapter 9 solutions are prepared by experts with 6+ years of subject matter expertise. These NCERT exemplar Class 9 Maths chapter 9 solutions are designed to provide the students with detailed steps while studying NCERT Book for Class 9 Maths. They develop a better understanding of the concept of the area of parallelogram as they are pretty expressive. The NCERT exemplar Class 9 Maths solutions chapter 9 follows the recommended CBSE Syllabus for Class 9.

NCERT Exemplar Class 9 Maths Solutions Chapter 9-Exercise 9.1

Question:1

The median of a triangle divides it into two
(A) triangles of equal area
(B) congruent triangles
(C) right triangles
(D) isosceles triangles

Answer: [A]
Solution:
Let AN be the median of
`
We have drawn AM perpendicular to BC
We know that median bisects a line into 2 equal parts.
Hence BN = NC …(i)
Now, area of a triangle is given as (Base) (Height)
Area of …(ii)
(AM is the perpendicular distance between point A and BN. Hence AM is the height of )
Area of
(AM is the perpendicular distance between point A and CN. Hence AM is the height of )
From (i), we have BN = NC
Area of …(iii)
Comparing equations (ii) and (iii), we get
Area of Area of
Hence the median of a triangle divides it into two triangles of equal area.
Hence, option (A) is the correct answer.

Question:3

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is :
(A) a rectangle of area 24 cm ²
(B) a square of area 25 cm ²
(C) a trapezium of area 24 cm ²
(D) a rhombus of area 24 cm ²

Answer: [D]
Solution.
Let ABCD be a rectangle.
Given sides 8 cm and 6 cm.

Let length of rectangle = 8 cm
Breadth of rectangle = 6 cm
Let E, F, G and H are the mid-points of the sides of rectangle
AE = EB = DG = GC = 3 cm
AH = HD = BF = FC = 4 cm
Then we have to find out about the figure EFGH.
In , using Pythagoras theorem


So, EH = 5 cm.
Similarly we can find out HG = GF = FE = 5 cm
Hence EFGH is a rhombus.
Now, area of rhombus =
Where are the diagonals of the rhombus
area of rhombus =
H ere, EG = AD = 8 cm
and HF = AB = 6 cm
Area of rhombus EFGH =
Therefore option (D) is correct.

Question:5

In Fig. 9.5, if parallelogram ABCD and rectangle ABEM are of equal area, then :

(A) Perimeter of ABCD = Perimeter of ABEM
(B) Perimeter of ABCD < Perimeter of ABEM
(C) Perimeter of ABCD > Perimeter of ABEM
(D) Perimeter of ABCD = (Perimeter of ABEM)

Answer:

Answer: [C]
Solution.

In a rectangle opposite sides are equal.
So in ABEM,
AB = EM … (1)
In a parallelogram also the opposite sides are equal.
So in ABCD,
CD = AB …(2)
Adding eq. (1) and (2)
We get AB + CD = EM + AB
So, CD = EM …(3)
We know that, the shortest distance between any two parallel lines is the perpendicular distance between them. So, perpendicular distance between two parallel sides of a parallelogram is always less than the length of the other parallel sides.
BE < BC and AM < AD
On adding these inequalities, we get
BE + AM < BC + AD
or
BC + AD > BE + AM
On adding AB + CD on both sides, we get
AB + CD + BC + AD > AB + CD + BE + AM
(rearranging LHS)
[ Q CD = EM (from eq. 3)]
Perimeter of ABCD Perimeter of ABEM
Therefore option (C) is correct.

Question:6

The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to
(A)
(B)
(C)
(D)

Answer: [A]
Solution.
Let the given triangle ABC be shown as below with medians AE, BF, CD.

D is the mid-point of AB, E is the mid-point of BC, F is the mid-point of AC
Mid-point theorem states that the line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side
So,
DF = BE = EC …(i)

DE = AF = FC …(ii)

FE = AD = DB …(iii)
In and
AD = FE (from iii)
DF = EC (from i)
AF = FC (given)
(SSS congruence)
Similarly we can prove that,

So the bigger triangle is divided into 4 smaller congruent triangles.
So area of (area of one smaller triangle) …(iv)
If we consider the parallelograms formed then we can see that one consists of two such triangles.



So area of every (area of one smaller triangle) …(v)
Using (iv) and (v),
area of every
Therefore option (A) is correct.

Question:9

If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is
(A) 1 : 3
(B) 1 : 2
(C) 3 : 1
(D) 1 : 4

Answer: [B]
Solution.
We know that a triangles and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram. Let us prove the same. We have a parallelogram ABCD and triangle BEC on the same base BC.


Construct EM perpendicular to BC
Area of a triangle is given as: (Base) (Height)
Area of …(i)

Area of a parallelogram is given as: (Base) (corresponding altitude)
Area of a parallelogram ABCD= (BC) (EM) …(ii)
Ratio of the area of the triangle to the area of parallelogram = (i) : (ii)


Hence the ratio of the area of the triangle to the area of parallelogram is 1 : 2.
Hence, (B) is the correct answer.

Question:10

ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (Fig.). E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is


(A) a : b
(B) (3a + b) : (a + 3b)
(C) (a + 3b) : (3a + b)
(D) (2a + b) : (3a + b)

Answer:B

Given: ABCD is a trapezium in which ,
E and F are the mid-points of AD and BC, respectively

(Mid-segment Theorem on Trapezium: the line segment joining the midpoints of the nonparallel sides of a trapezium is half the sum of the lengths of the parallel sides and is also parallel to them.)
So,
Now, ABFE and EFCD are also trapeziums.






Required ratio

So, the required ratio is
Hence, (B) is the correct answer.

NCERT Exemplar Class 9 Maths Solutions Chapter 9-Exercise 9.2

Question:1

Write true or false and justify your answer.
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm², then ar (ABC) = 24 cm².

Answer: [False]
Solution.
We have parallelogram ABCD and X is a mid point of AB.

Now, …(1)
Draw XG perpendicular to CD
Area of = (base) (corresponding altitude)

= (AB) (GX) …(2)

Area of trapezium AXCD = (Sum of parallel sides) (Distance between them)

=

Now, and (Given)

Area of trapezium …(3)

Area of (base) (height)

…(4)

Now,
Area of trapezium
(given)

From equation (4),
Area of

Hence, the given statement is false.

Question:2

Write true or false and justify your answer.
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ.
If PS = 5 cm,then ar (PAS) = 30 cm².

Answer: [False]
Solution.
It is given that PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm.
A is any point on PQ, therefore, PA < PQ.
It is given that A is any point on PQ, therefore PA < PQ.

PS = 5 cm, PR = 13 cm
In D PSR, using Pythagoras theorem,



SR = 12 cm = PQ (opposite sides of a rectangle are equal)
Area of a triangle is given as
Now,
As,

or
But it is given that

Hence, the given statement is false.

Question:3

Write true or false and justify your answer.
PQRS is a parallelogram whose area is 180 cm² and A is any point on the diagonal QS. The area of DASR = 90 cm².

Answer:

Answer: [False]
Solution.
Given, PQRS is a parallelogram whose area is

We know that diagonal of a parallelogram divides it into two triangles of equals area, so

Given that A is any point on SQ
So,
Hence,
But given that area of

Hence, the given statement is false.

Question:4

Write true or false and justify your answer.
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then ar (BDE) = ar ( ABC).

Answer:

Answer: [True]
Solution.
Given and are two equilateral triangles.
D is the mid-point of BC, so BD = CD = BC


Let each sides of be x.
So each side of is .
Area of an equilateral triangle is given as
Now,

Hence,

Therefore the given statement is true.

Question:5

Write true or false and justify your answer.
In Fig., ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = ar ( EFGD).

Answer:[False]

Solution.
Given that ABCD and EFGD are two parallelogram and G is mid-point of CD.
Let perpendicular distance between AB and CD be h
ar = (base) (corresponding altitude)
ar = (GD) (h)
ar = (CD) (h)
CD = 2 GD (Given that G is mid-point of CD)
So,
\ ....(i)
Now, DPDC and parallelogram ABCD are on the same base BC and between the same parallels AB and CD
So, …(ii)
From (i) and (ii)

Therefore the given statement is false.

NCERT Exemplar Class 9 Maths Solutions Chapter 9-Exercise 9.3

Question:1

In Fig., PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR = RS and . Prove that ar (PQE) = ar (CFD).

Answer:

Solution.
Given: a parallelogram PSDA with points Q and R taken on PS such that PQ = RS = QR
and
To prove: ar(PQE) = ar(CFD)
Proof:
PQ = QR = RS
PS = PQ + QR + RS = 3 PQ
Also, and
So PABQ, QBCR, RCDS – all are parallelograms.
Hence,
We know that opposite sides of a parallelogram are equal, so in APSD:
PS = AD

…(1)
Consider and
and PD is transversal
[Alternate interior angles] …(2)
Now, and AD is a transversal
[Corresponding angles] …(3)
and PD is transversal
[Alternate interior angles] …(4)
From (3) and (4)
…(5)
From (1), (2), (5)
[ASA congruence rule]
Hence, [Congruent figures have equal area]
Hence proved

Question:2

X and Y are points on the side LN of the triangle LMN such that LX=XY=YN. Through X, a line is drawn parallel to LM to meet MN at Z (see Fig.). Prove that ar (LZY) = ar (MZYX)

Answer:

Solution.
Given: with X and Y points on the side LN
LX = XY = YN and
To prove:
Proof:
Since and and are on the same base XZ and between the same parallel lines LM and XZ. Then,
…(1)
Adding both sides of eq. (1), we get

ar(XMZY) = ar(LZY)
This can be rearranged as:
ar (LZY) = ar (MZYX)
Hence proved.

Question:3

The area of the parallelogram ABCD is (see Fig.9.13). Find

(i) ar (ABEF)
(ii) ar (ABD)
(iii) ar (BEF)

Answer:

Answer:
(i)
(ii)
(iii)
Solution.
(i) Here, ABCD is a parallelogram with area

So, [Extended part of CD is CF]

Hence, ABEF is a parallelogram.
We know that parallelogram that lies between same parallel lines and having same base have equal areas.
Now, parallelogram ABCD and ABEF lies between same parallel lines AB and CF have same base AB. So these have equal area.


(ii) and parallelogram ABCD lies between same parallel lines and have same base AB. So we know that

Now,


(iii) Here, and parallelogram ABEF are on the same base EF and between the same parallels AB and EF. So we know that,

Question:4

In , D is the mid-point of AB and P is any point on BC. If meets AB in Q (Fig.), then prove that .

Answer:

Solution.
In , D is the mid-point of AB and P is any point on BC.

Join D and C

Consider ,
Since D is the mid-point of AB
So CD is the median of
Using mid-point theorem,
…(i)
Now,
Since and are on the same base PD and between the same parallel lines PD & QC. So we have:
…(ii)
From (i) & (ii)
ar

Now, area of = area of


Hence proved.

Question:5

ABCD is a square. E and F are respectively the midpoints of BC and CD. If R is the mid-point of EF (Fig.), prove that ar (AER) = ar(AFR)

Answer:

Solution.
Given that ABCD is a square
E and F are the midpoints of BC and CD
R is the mid-point of EF
Proof: Consider and
We know that BC = DC (All sides of square are equal)

(Since E and F are mid points of BC and CD)
(All angles in a square are right angle)
(All sides in a square are equal)
(By SAS criterion)
Hence, AE = AF (by C.P.C.T)
Now, ER = RF ( R is the midpoint of EF)
AR = AR (Common)
(By SSS criterion)
Hence, AR divides the triangle into two triangles of equal area.

Hence proved.

Question:6

O is any point on the diagonal PR of a parallelogram PQRS (Fig.). Prove that ar (PSO) = ar (PQO).

Answer:

Solution.
Given: O is any point on the diagonal PR of a parallelogram PQRS
Construction: Join QS
Let diagonals PR and QS intersect each other at T.

We know that diagonals of a parallelogram bisect each other.
\ T is the mid-point of QS and PR
Since a median of a triangle divides it into two triangle of equal area.
\ In , PT is the median of side QS
…(i)
In , OT is the median of side QS
…(ii)
Adding (i) and (ii), we have


Hence proved

Question:7

ABCD is a parallelogram in which BC is produced to E such that CE = BC (Fig.). AE intersects CD at F.
If ar (DFB) = , find the area of the parallelogram ABCD.

Answer:

Answer:
Solution.
Given: ABCD is a parallelogram.
CE = BC
The lines BC and AF when produced, meet at E.

Now,
BC = EC & (given)
(By mid-point theorem)
Also, AB = DC (opposite sides of a parallelogram are equal)


Now, let perpendicular distance between parallel lines AB and CD be h



…(i)
Area of = (Base) (corresponding altitude)
= (CD) (h) …(ii)
From (i) and (ii)
Area of =
Area of =

Question:8

In trapezium ABCD, and L is the mid-point of BC. Through L, a line has been drawn which meets AB in P and DC produced in Q (Fig.). Prove that ar (ABCD) = ar (APQD)

Answer:

Solution.
Given: Trapezium ABCD with and BL = CL

To prove: ar(ABCD) = ar(APQD)
Proof: DC is produced at Q and
So,
Also, (given)
Then APQD is a parallelogram
In and
\ CL = BL [L is mid-point of BC]
[alternate interior angles]
[alternate interior angle as PQ is a transverse]
[AAS congruence rule]
[Congruent triangles have equal area] …(i)


Now,
…(ii)
…(iii)
From eq (i), (ii) and (iii)
ar(ABCD) = ar(APQD)

Question:9

If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral (Fig.).

Answer:

Solution.
Given that if the mid-points of the sides of a quadrilateral are joined in order, a parallelogram is formed. We have to find the area of this parallelogram.
To prove: ar (parallelogram PFRS) ar(quadrilateral ABCD)

Let ABCD is a quadrilateral and P, F, R and S are the mid-points of the sides BC, CD, AD and AB respectively and PFRS is a parallelogram.
Construction:- Join BD and BR.
We know that median of a triangle divides it into two triangles of equal area.
So, BR divides into two triangles of equal area.
…(i)
Similarly, median RS divides into two triangles of equal area.
…(ii)
From eq. (i) and (ii)
…(iii)
Similarly,
….(iv)
On adding equations (iii) and (iv), we get

(Quadrilateral BCDA) …(v)
Similarly,
….(vi)
On adding (v) and (vi), we get


(quadrilateral BCDA) …(vii)
But

(quadrilateral BCDA) …(viii)
from subtracting eq. (vii) from eq. (viii) we get
ar(parallelogram PFRS) (quadrilateral BCDA)
Hence proved

NCERT Exemplar Class 9 Maths Solutions Chapter 9-Exercise 9.4

Question:1

A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ADF) = ar (ABFC)

Answer:

Solution.
According to the question, ABCD is a parallelogram.
E is a point on BC.
AE and DC are produced to meet at F.
Using the above information, we can make the figure as follows:

We know that, triangles on the same base and between the same parallels are equal in area.
Here, we have,
and are on same base AB and between parallels, AB and DF
Area = area …(1)
We also know that,
Diagonal of a parallelogram divides it into two triangles of equal area.
So, Area = area …(2)
Now,
Area = area + area
\ Area = area + area [ Q From equation (2)]
Area = area + area [ Q From equation (1)]
Area = area (ABFC)

Hence proved.

Question:2

The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area.

Answer:

Solution.
Given in parallelogram ABCD, diagonals intersect at O and draw a line PQ, which intersects AD at P and BC at Q.
To prove: PQ divides the parallelogram ABCD into two part of equals area.
ar (ABQP) = ar(CDPQ)

We know that, diagonals of a parallelogram bisect each other.
\ OA = OC and OB = OD …(i)
In and
OA = OC and OB = OD [From eq. (i)]
and
ÐAOB = Ð COD [Vertically opposite angles]
[SAS congruence rule]
…(ii) [Congruent figures have equal area]
In and
Ð PAO = Ð OCQ [alternate interior angles]
OA = OC [From eq. (i)]
ÐAOP = Ð COQ [Vertically opposite angles]
[ASA congruence rule]
…(iii) [Congruent figures have equal area]
In and
Ð PDO = Ð OBQ [alternate interior angles]
OD = OB [From eq. (i)]
Ð DOP = Ð BOQ [Vertically opposite angles]
[ASA congruence rule]
…(iv) [Congruent figures have equal area]
[From ii, iii, iv]
= ar(ABQP)
ar(ABQP) = ar(CDPQ)
Hence Proved.

Question:3

The medians BE and CF of a triangle ABC intersect at G. Prove that the area of = area of the quadrilateral AFGE.

Answer:

Solution.
Given that the medians BE and CF of a triangle ABC intersect at G
We know that the median divides a triangle into two triangles of same area.


So, …(1)
And, ….(2)
from (1) and (2)


This can be written as:
Area of = area of the quadrilateral AFGE.
Hence proved

Question:4

In Fig., CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY)

Answer:

Solution.
Given: and
To prove:
Proof: We know that, triangles on the same base and between the same parallels have equal area.
Consider
and both lie on the same base AB and between the same parallels CY and BA.
…(1)
Now,
and,
Putting the above values in eq. (1)
ar(ABX) + ar(AXY) = ar(ABX) + ar(CBX)
ar(AXY) = ar(CBX)
Hence proved

Question:6

In , if L and M are the points on AB and AC, respectively such that . Prove that ar (LOB) = ar (MOC)

Answer:

Solution.
Given: with L and M points on AB and AC respectively such that

To prove:
Proof:
We know that, triangles on the same base and between the same parallels are equal in area.
Hence, and lie on the same base BC and between the same parallels BC and LM.
So, …(1)


Using these values in equation 1, we get:

On eliminating form both sides, we get

Hence proved.

Question:5

ABCD is a trapezium in which , DC = 30 cm and AB = 50 cm. If X and Y are, respectively the mid-points of AD and BC, prove that ar (DCYX) = ar (XYBA)

Answer:

Solution.
Given: Trapezium ABCD with
DC = 30 cm and AB = 50 cm.
X and Y are the mid-points of AD and BC, respectively.
To prove:

Construction, Join DY and extend it to meet AB produced at P.
Proof:
In and
CY = BY [Given, Y is the mid-point of BC]
[alternate interior angles]
[Vertically opposite angles]
[ASA congruence rule]
t hen, DC = BP [CPCT]
DC = 30 cm (given)
Q DC = BP = 30 cm
Now,

In , X is the mid-point of AD and Y is the mid-point of DP.
[mid-point theorem ]
Let distance between AB, XY and XY, DC be h cm
Area of trapezium is given as [Sum of parallel side × distance between them]
Now, area of trapezium

=

Area of trapezium


Hence proved.

Question:7

In Fig., ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (APQ)

Answer:

Solution.
Given: a pentagon ABCDE, and
To prove:
Proof: We know that triangles on the same base and between the same parallel lines are equal in area.
Now, and lie on the same base AD and lie between the same parallels AD and EQ. So,
…(i)
Similarly,
and lie on the same base AC and lie between the same parallels AC and PB
…(ii)
On adding equations (i) and (ii), we get

Adding ar (DCDA) to both sides, we get



Hence proved.

Question:10

In Fig., ABCD and AEFD are two parallelograms. Prove that ar (PEA) = ar (QFD)

Answer:

Solution.
Given: Two parallelograms ABCD and AEFD.
To prove:
Proof:
PQDA is a parallelogram, so
Now, parallelograms PQDA and AEFD are on the same base AD and lie between the same parallel lines AD and EQ. Hence,


So, we get:

Hence proved

Question:9

In Fig., X and Y are the mid-points of AC and AB respectively, and CYQ and BXP are straight lines. Prove that ar (ABP) = ar (ACQ).

Answer:

Solution.
Given: X and Y are the mid points of AC and AB respectively.

CYQ, BXP are straight lines,
To prove:
Proof:
Since, X and Y are the mid-point of AC and AB.
So,
We know that triangles on the same base and between the same parallels are equal in area
Consider parallel lines BC and XY;
and lie on same base BC and between the same parallels BC and XY.
So,


Adding ar(XOY) to LHS and RHS

…(i)
We observe that the quadrilateral XYAP and XYAQ have the same base XY and these are between the same parallel lines XY and PQ.
…(ii)
on adding eq. (i) and (ii), we get


Hence Proved.

Question:8

If the medians of a intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = ar (ABC)

Answer:

Solution.
Given: with medians AM, BN & CL



Proof: We know that a median divides the triangle into two triangles of same area.
Let the area of small triangles be denoted as 1, 2, 3, 4, 5, 6 as shown in the figure
AM is the median

…(i)
CL is the median

…(ii)
BN is the median


(i) – (ii)


So,
(ii) – (iii)


So,
Similarly we can prove that area of
Hence the triangle is divided into 6 triangles of equal area.
and consists of 2 triangles each
So they have equal area which is equal to twice the area of the smaller triangles.
(area of 1 small triangle)
Hence,
Hence proved.

Important Topics for NCERT Exemplar Solutions Class 9 Maths Chapter 9:

Highlights of the key topics covered in NCERT exemplar Class 9 Maths solutions chapter 9 are as follows:

• Area of parallelogram between two parallel lines and a given base will always be the same for all the parallelograms.
• It discusses that any triangle with a given base will have the same area if the triangle’s height is the same.
• The basic formula for area of a rectangle is used in formulation of the area of parallelogram and triangles.
• NCERT exemplar Class 9 Maths solutions chapter 9 deals with the role of median and how it creates two triangles of the same area.

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Features of NCERT Exemplar Class 9 Maths Solutions Chapter 9:

These class 9 maths NCERT exemplar chapter 9 solutions provide fundamental knowledge of parallelograms and triangle’s area, which will be very useful for higher mathematics for competitive exams like JEE Mains and JEE Advanced. The Class 9 Maths NCERT exemplar solutions chapter 9 Area of parallelograms is a good starting point for practicing the problems based on this chapter and widening the knowledge base. Once the students are thorough with the problems and solutions of exemplar, it will be easier to solve other books such as NCERT Class 9 Maths, RD Sharma Class 9 Maths, RS Aggarwal Class 9 Maths, et cetera.

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