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

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

Edited By Ravindra Pindel | Updated on Aug 31, 2022 12:52 PM IST

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.

This Story also Contains
  1. NCERT Exemplar Class 9 Maths Solutions Chapter 9-Exercise 9.1
  2. NCERT Exemplar Class 9 Maths Solutions Chapter 9-Exercise 9.2
  3. NCERT Exemplar Class 9 Maths Solutions Chapter 9-Exercise 9.3
  4. NCERT Exemplar Class 9 Maths Solutions Chapter 9-Exercise 9.4
  5. Important Topics for NCERT Exemplar Solutions Class 9 Maths Chapter 9:
  6. NCERT Class 9 Exemplar Solutions for Other Subjects:
  7. NCERT Class 9 Maths Exemplar Solutions for Other Chapters:
  8. Features of NCERT Exemplar Class 9 Maths Solutions Chapter 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 \triangle ABC
`
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 = \frac{1}{2} (Base) (Height)
Area of \triangle ABN= \frac{1}{2} (BN)(AM) …(ii)
(AM is the perpendicular distance between point A and BN. Hence AM is the height of \triangle ABN )
Area of \triangle ACN = \frac{1}{2}(CN)(AM)
(AM is the perpendicular distance between point A and CN. Hence AM is the height of \triangle ACN )
From (i), we have BN = NC
Area of \triangle ACN = \frac{1}{2}(BN)(AM)…(iii)
Comparing equations (ii) and (iii), we get
Area of \triangle ABN = Area of \triangle ACN
Hence the median of a triangle divides it into two triangles of equal area.
Hence, option (A) is the correct answer.

Question:2

In which of the following figures (Fig. 9.3), you find two polygons on the same base and between the same parallels?

Answer: [d]
Solution.
(a) We have a triangle and a parallelogram but we can easily see that they are not between the same parallel lines. Hence it is incorrect.
(b) We have a triangle and a parallelogram on the same base BC, but we can easily see that they are not between the same parallel lines. Hence it is also incorrect.
(c) Same as option a, this is incorrect.
(d) Here, the parallelograms, PQRA and BQRS are on the same base QR and between the same parallels QR and PS.
Hence, option (d) 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 \triangle HAE, using Pythagoras theorem
EH^{2}= AE^{2}+AH^{2}
EH^{2}=3^{2}+4^{2}=9+16=25
So, EH = 5 cm.
Similarly we can find out HG = GF = FE = 5 cm
Hence EFGH is a rhombus.
Now, area of rhombus = \frac{1}{2}d_{1}d_{2}
Where d_{1},d_{2} are the diagonals of the rhombus
area of rhombus = \frac{1}{2}(EG)(HF)
H ere, EG = AD = 8 cm
and HF = AB = 6 cm
Area of rhombus EFGH = \frac{8\times 6}{2}=4\times 6=24cm^{2}
Therefore option (D) is correct.

Question:4

In Fig. 9.4, the area of parallelogram ABCD is :


(A) AB × BM
(B) BC × BN
(C) DC × DL
(D) AD × DL

Answer: [C]
Solution.
We know that
Area of parallelogram = Base × corresponding altitude
Area of parallelogram ABCD = AB × DL … (1)
Now, AB = DC (opposite sides of a parallelogram are equal)
Substituting in (1), we get
Area of parallelogram ABCD = DC × DL
Hence, option (C) is the correct answer.
We can check for other options as well.
In option (A), AB × BM; BM is not the corresponding altitude to the side AB
In option (B), BC × BN; BC is not the corresponding altitude to the side BN
In option (D), AD × DL; AD is not the corresponding altitude to the side DL
Therefore option (C) 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
\Rightarrow AB + BC + CD + AD > AB + CD + BE + AM (rearranging LHS)
\Rightarrow AB + BC + CD + AD > AB + BE + EM + AM[ Q CD = EM (from eq. 3)]
\Rightarrow 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) \frac{1}{2}ar(ABC)
(B) \frac{1}{3}ar(ABC)
(C) \frac{1}{4}ar(ABC)
(D) ar(ABC)

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=\frac{1}{2}BC
DF = BE = EC …(i)
DE=\frac{1}{2}AC
DE = AF = FC …(ii)
FE=\frac{1}{2}AB
FE = AD = DB …(iii)
In \triangle ADF and \triangle FEC
AD = FE (from iii)
DF = EC (from i)
AF = FC (given)
\triangle ADF\cong \triangle FEC(SSS congruence)
Similarly we can prove that,
\triangle ADF \cong \triangle FEC \cong \triangle DBE \cong \triangle EFD
So the bigger triangle is divided into 4 smaller congruent triangles.
So area of \triangle ABC = 4 (area of one smaller triangle) …(iv)
If we consider the parallelograms formed then we can see that one \parallel gmconsists of two such triangles.
\parallel BDFE = \triangle DBE + \triangle EFD
\parallel DFCE = \triangle FEC + \triangle EFD
\parallel ADEF = \triangle ADF + \triangle EFD
So area of every parallelogram = 2 (area of one smaller triangle) …(v)
Using (iv) and (v),
area of every \parallel gm = \frac{1}{2}ar(ABC)
Therefore option (A) is correct.

Question:7

Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is
(A) 1 : 2
(B) 1 : 1
(C) 2 : 1
(D) 3 : 1

Answer: [B]
Solution.
Area of a parallelogram is given as the product of its base and corresponding altitude.
Altitude is the perpendicular distance between the two parallel sides.
Now, parallelogram on the same or equal bases and between the same parallels will have the same area. This is because the base is the same, and also as they are between the same bases so the altitude will be the same for them.
So, the ratio of their area is 1 : 1.
Hence, (B) is the correct answer.

Question:8

ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD
(A) is a rectangle
(B) is always a rhombus
(C) is a parallelogram
(D) need not be any of (A), (B) or (C)

Answer: [D]
Solution.
Since diagonal of a parallelogram divides it into two triangles of equal area and rectangle and a rhombus are also parallelogram. It may be a kite as diagonals of a kite divides it two triangles of equal areas. Then ABCD need not be any of (a), (b) or (c).
Hence, (d) is the correct option.

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: \frac{1}{2} (Base) (Height)
Area of \triangle BEC=\frac{1}{2}(BC)(EM) …(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)
=\frac{1}{2}(BC)(EM)/(BC)(EM)
=\frac{1}{2}
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 AB \parallel DC,
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, EF=\frac{1}{2}(a+b)
Now, ABFE and EFCD are also trapeziums.
ar(ABFE)=\frac{1}{2}(EF+AB)h
={\frac{1}{2}}\left [ \frac{1}{2}(a+b)+a \right ]\times h
=\frac{h}{4}(3a+b)
ar(EFCD)=\frac{1}{2}(CD+EF)h
=\frac{1}{2}\left [ b+\frac{1}{2}(a+b) \right ]\times h
=\frac{h}{4}(a+3b)
Required ratio
\therefore \frac{ar(ABEF)}{ar(EFCD)}=\frac{\frac{h}{4}(3a+b)}{\frac{h}{4}(a+3b)}=\frac{(3a+b)}{(a+3b)}
So, the required ratio is (3a+b):(a+3b)
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, ar(ABCD) = ar(AXCD) + ar(\triangle XBC) …(1)
Draw XG perpendicular to CD
Area of \parallel gm ABCD = (base) (corresponding altitude)

= (AB) (GX) …(2)

Area of trapezium AXCD = \frac{1}{2} (Sum of parallel sides) (Distance between them)

= \frac{1}{2} (DC + AX) (GX)

Now, AX = \frac{1}{2}AB and AB = CD (Given)

Area of trapezium AXCD = \frac{1}{2} (AB + \frac{1}{2} AB) (GX)…(3)

Area of \triangle ABC = \frac{1}{2} (base) (height)

= \frac{1}{2} (AB) (GX)

= \frac{1}{2} (AB) (GX) …(4)

Now,
Area of trapezium AXCD = \frac{1}{2} (AB + \frac{1}{2}AB) (GX) = 24 cm^{2}
 (given)

\frac{3}{4}(AB) (GX) = 24 cm^{2}
\frac{1}{4}(AB)(GX)=8 cm^{2}
\frac{1}{2}(AB)(GX)=16 cm^{2}

From equation (4),
\frac{1}{2}(AB)(GX)= Area of \triangle ABC=16cm^{2}

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,
PS^{2}+SR^{2}=PR^{2}
5^{2}+SR^{2}=13^{2}
SR^{2}=169-25=144
SR = 12 cm = PQ (opposite sides of a rectangle are equal)
Area of a triangle is given as \frac{1}{2}\times base\times height
Now, ar(\triangle PQR)=\frac{1}{2}\times PQ\times QR =\frac{1}{2}\times 12\times 5=30cm^{2}
As, PA< PQ

ar(\triangle PAS)<ar(\triangle PQR)
or ar(\triangle PAS)<30cm^{2}
But it is given that ar( PAS)=30cm^{2}

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 180cm^{2}

We know that diagonal of a parallelogram divides it into two triangles of equals area, so
\therefore ar(\triangle QRS)=\frac{1}{2}ar(\parallel gmPQRS)
=\frac{1}{2}\times 180=90cm^{2}

Given that A is any point on SQ
So, ar (\triangle ASR) < ar(\triangle QRS)
Hence,ar (\triangle ASR) < 90 cm^{2}
But given that area of \triangle ASR = 90 cm^{2}

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) = \frac{1}{4} ar ( ABC).

Answer:

Answer: [True]
Solution.
Given \triangle ABC and \triangle BDE are two equilateral triangles.
D is the mid-point of BC, so BD = CD = \frac{1}{2} BC


Let each sides of \triangle ABC be x.
So each side of \triangle BDE is \frac{x}{2}.
Area of an equilateral triangle is given as \frac{\sqrt{3}}{4}(side)^{2}
Now, \frac{ar(\triangle BDE)}{ar(\triangle ABC)}=\frac{\frac{\sqrt{3}}{4}\left ( \frac{x}{2} \right )^{2}}{\frac{\sqrt{3}}{4}x^{2}}=\frac{1x^{2}}{4x^{2}}=\frac{1}{4}

Hence, ar(\triangle BDE)=\frac{1}{4}ar(\triangle ABC)

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) = \frac{1}{2}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 (\parallel gm) = (base) (corresponding altitude)
ar (\parallel EFGD) = (GD) (h)
ar (\parallel ABCD) = (CD) (h)
CD = 2 GD (Given that G is mid-point of CD)
So, ar (\parallel ABCD) = 2 ar (\parallel EFGD)
\ar (\parallel EFGD) = \frac{1}{2} ar (\parallel ABCD) ....(i)
Now, DPDC and parallelogram ABCD are on the same base BC and between the same parallels AB and CD
So, ar (\triangle DPC) = \frac{1}{2} ar (\parallel ABCD) …(ii)
From (i) and (ii)
\Rightarrow ar (\triangle DPC) = ar (\parallel EFGD)
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 PA \parallel QB \parallel RC. 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 PA\parallel QB\parallel RC
To prove: ar(PQE) = ar(CFD)
Proof:
PQ = QR = RS
PS = PQ + QR + RS = 3 PQ
Also, PA \parallel QB \parallel RC and PS \parallel AD
So PABQ, QBCR, RCDS – all are parallelograms.
Hence, AB=BC=CD=\frac{1}{3}AD
We know that opposite sides of a parallelogram are equal, so in \parallel gm APSD:
PS = AD
\Rightarrow \frac{1}{3}PS=\frac{1}{3}AD
\Rightarrow PQ=CD …(1)
Consider \triangle PEQ and \triangle CFD
PS\parallel AD and PD is transversal
\angle QPE = \angle CDF [Alternate interior angles] …(2)
Now, QB \parallel RC and AD is a transversal
\angle QBD = \angle RCD [Corresponding angles] …(3)
PS \parallel AD and PD is transversal
\therefore \angle PQB = \angle QBC[Alternate interior angles] …(4)
From (3) and (4)
\angle PQE=\angle FCD\angle PQE = \angle FCD???????\angle PQE = \angle FCD…(5)
From (1), (2), (5)
\triangle PEQ\cong \triangle CFD [ASA congruence rule]
Hence, ar (\triangle PQE) = ar(\triangle CED)[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: \triangle LMN with X and Y points on the side LN
LX = XY = YN and XZ \parallel LM
To prove: ar(\triangle LZY) = ar(MZYX)
Proof:
Since XZ \parallel LM and \triangle XMZ and \triangle XLZ are on the same base XZ and between the same parallel lines LM and XZ. Then,
ar(\triangle XMZ)=ar (\triangle XLZ) …(1)
Adding ar(\triangle XYZ) both sides of eq. (1), we get
ar(\triangle XMZ) + ar(\triangle XYZ) = ar(\triangle XLZ) + ar(\triangle XYZ)
\Rightarrow ar(XMZY) = ar(LZY)
This can be rearranged as:
ar (LZY) = ar (MZYX)
Hence proved.

Question:3

The area of the parallelogram ABCD is 90 cm^{2} (see Fig.9.13). Find

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

Answer:

Answer:
(i) 90cm^{2}
(ii) 45cm^{2}
(iii) 45cm^{2}
Solution.
(i) Here, ABCD is a parallelogram with area 90cm^{2}
\therefore AB \parallel CD
So, AB \parallel CF [Extended part of CD is CF]
\therefore AB\parallel EF
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.
\therefore ar(ABCD) = ar(ABEF)
\therefore ar(ABEF) = 90 cm^{2}
(ii)\triangle ABD and parallelogram ABCD lies between same parallel lines and have same base AB. So we know that
\therefore ar(ABD)=\frac{1}{2}\times ar(ABCD)
Now, ar(ABCD)=90cm^{2}
\Rightarrow ar (ABD)=\frac{1}{2}\times 90cm^{2}
\therefore ar(ABD)=45cm^{2}
(iii) Here, \triangle BEF and parallelogram ABEF are on the same base EF and between the same parallels AB and EF. So we know that,
ar(\triangle BEF)=\frac{1}{2}ar(ABEF)=90cm^{2}
ar(\triangle BEF)=\frac{1}{2}\times 90=45cm^{2}

Question:4

In \triangle ABC, D is the mid-point of AB and P is any point on BC. If CQ\parallel PD meets AB in Q (Fig.), then prove that ar (BPQ)=\frac{1}{2}ar(ABC).

Answer:

Solution.
In \triangle ABC, D is the mid-point of AB and P is any point on BC.
CQ\parallel PD
Join D and C

Consider \triangle ABC,
Since D is the mid-point of AB
So CD is the median of \triangle ABC
Using mid-point theorem,
ar(\triangle BCD)=\frac{1}{2}ar(\triangle ABC)…(i)
Now, CQ \parallel PD
Since \triangle PDQ and \triangle PDC are on the same base PD and between the same parallel lines PD & QC. So we have:
\therefore ar(\triangle PDQ)=ar(\triangle PDC) …(ii)
From (i) & (ii)
ar (\triangle BCD) = \frac{1}{2}ar(\triangle ABC)
ar(\triangle BPD)+ar(\triangle PDC)=\frac{1}{2}ar(\triangle ABC)
Now, area of \triangle PDC = area of \triangle PDQ
ar(\triangle BPD)+ar(\triangle PDQ)=\frac{1}{2}ar(\triangle ABC)
\Rightarrow ar(\triangle BPQ)=\frac{1}{2}ar(\triangle ABC)
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 \triangle ABE and \triangle ADF
We know that BC = DC (All sides of square are equal)
\Rightarrow \frac{BC}{2}=\frac{DC}{2}
\Rightarrow EB=DF(Since E and F are mid points of BC and CD)
\angle ABE=\angle ADF=90^{\circ}(All angles in a square are right angle)
\Rightarrow AB=AD(All sides in a square are equal)
\therefore \triangle ABE\cong \triangle ADF(By SAS criterion)
Hence, AE = AF (by C.P.C.T)
Now, ER = RF ( R is the midpoint of EF)
AR = AR (Common)
\therefore \triangle AER\cong \triangle AFR(By SSS criterion)
Hence, AR divides the triangle into two triangles of equal area.
\Rightarrow ar(\triangle AER)=ar(\triangle AFR)
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 \triangle PQS, PT is the median of side QS
\Rightarrow ar(\triangle PTS)=ar(\triangle PTQ) …(i)
In \triangle SOQ, OT is the median of side QS
\Rightarrow ar(\triangle STO)=(\triangle QTO) …(ii)
Adding (i) and (ii), we have
ar(\triangle PTS)+ar(\triangle STO)=ar (\triangle PTQ)+(\triangle QTO)
\Rightarrow ar(\triangle PSO)=ar(\triangle PQO)
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) = 3cm^{2}, find the area of the parallelogram ABCD.

Answer:

Answer: \left [ ar(ABCD)=12cm^{2} \right ]
Solution.
Given: ABCD is a parallelogram.
CE = BC
The lines BC and AF when produced, meet at E.
ar\triangle DFB=3cm^{2}
Now,
BC = EC & AB \parallel FC (given)
\therefore FC=\frac{1}{2}AB (By mid-point theorem)
Also, AB = DC (opposite sides of a parallelogram are equal)
\therefore FC=\frac{1}{2}DC
\Rightarrow FC=DF=\frac{1}{2}DC
Now, let perpendicular distance between parallel lines AB and CD be h
ar(\triangle DFB)=\frac{1}{2}(Base)(Height)
=\frac{1}{2}(DF)(h)
=\frac{1}{2}\left ( \frac{1}{2}CD \right )(h)
=\frac{1}{4}(CD)(h) …(i)
Area of \parallel ABCD = (Base) (corresponding altitude)
= (CD) (h) …(ii)
From (i) and (ii)
Area of \parallel ABCD = 4 (ar (\triangle DFB))
Area of \parallel ABCD = 4 (3 cm^{2}) = 12 cm^{2}

Question:8

In trapezium ABCD, AB\parallel DC and L is the mid-point of BC. Through L, a line PQ\parallel AD 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 AB \parallel CD and BL = CL
PQ \parallel AD
To prove: ar(ABCD) = ar(APQD)
Proof: DC is produced at Q and AB \parallel DC
So, DQ \parallel AP
Also, PQ \parallel AD(given)
Then APQD is a parallelogram
In \triangle CLQ and \triangle BLP
\ CL = BL [L is mid-point of BC]
\angle LCQ=\angle LBP [alternate interior angles]
\angle CQL=\angle LPB [alternate interior angle as PQ is a transverse]
\therefore \triangle CLQ\cong \triangle BLP[AAS congruence rule]
ar(\triangle CLQ)=ar(\triangle BLP) [Congruent triangles have equal area] …(i)


Now,
ar(ABCD) = ar(APLCD) + ar(\triangle BLP)…(ii)
ar(APQD) = ar(APLCD) + ar(\triangle CLQ)…(iii)
From eq (i), (ii) and (iii)
\Rightarrow 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) =\frac{1}{2} 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 \triangle BDA into two triangles of equal area.
\therefore ar(\triangle BRA) =\frac{1}{2} ar(\triangle BDA)…(i)
Similarly, median RS divides \triangle BRA into two triangles of equal area.
\therefore ar(\triangle ASR)=\frac{1}{2}ar(\triangle BRA)…(ii)
From eq. (i) and (ii)
\therefore ar(\triangle ASR)=\frac{1}{4}ar(\triangle BDA) …(iii)
Similarly,
ar(\triangle CFP)=\frac{1}{4}ar(\triangle BCD) ….(iv)
On adding equations (iii) and (iv), we get
ar(\triangle ASR)+ar(\triangle CFP)=\frac{1}{4}ar(\triangle BDA+\triangle BCD)
\Rightarrow ar(\triangle ASR)+ar(\triangle CFP)=\frac{1}{4}ar (Quadrilateral BCDA) …(v)
Similarly,
ar(\triangle DRF)+ar(\triangle BSP)=\frac{1}{4}ar(BCDA)….(vi)
On adding (v) and (vi), we get

ar(\triangle ASR)+ar(\triangle CFP)+ar (\triangle BSP)+ar(\triangle BSP)
=\frac{1}{2}ar (quadrilateral BCDA) …(vii)
But
ar(\triangle ASR) + ar(\triangle CFP) + ar(\triangle DRF) + ar(\triangle BSP) = ar(\parallel gm PFRS)
=\frac{1}{2}ar (quadrilateral BCDA) …(viii)
from subtracting eq. (vii) from eq. (viii) we get
ar(parallelogram PFRS) =\frac{1}{2}ar(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,
\triangle ABC and \triangle ABF are on same base AB and between parallels, AB and DF
Area (\triangle ABC) = area (\triangle ABF) …(1)
We also know that,
Diagonal of a parallelogram divides it into two triangles of equal area.
So, Area (\triangle ABC) = area (\triangle ACD) …(2)
Now,
Area (\triangle ADF) = area (\triangle ACD) + area (\triangle ACF)
\ Area (\triangle ADF) = area (\triangle ABC) + area(\triangle ACF) [ Q From equation (2)]
\RightarrowArea (\triangle ADF) = area (\triangle ABF) + area (\triangle ACF) [ Q From equation (1)]
\RightarrowArea (\triangle ADF) = 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 \triangle AOB and \triangle COD
OA = OC and OB = OD [From eq. (i)]
and
ÐAOB = Ð COD [Vertically opposite angles]
\therefore \triangle AOB\cong \triangle COD [SAS congruence rule]
ar(\triangle AOB)=ar(\triangle COD) …(ii) [Congruent figures have equal area]
In \triangle AOP and \triangle COQ
Ð PAO = Ð OCQ [alternate interior angles]
OA = OC [From eq. (i)]
ÐAOP = Ð COQ [Vertically opposite angles]
\triangle AOP\cong \triangle COQ [ASA congruence rule]
ar(\triangle AOP)=ar(\triangle COQ) …(iii) [Congruent figures have equal area]
In \triangle POD and \triangle BOQ
Ð PDO = Ð OBQ [alternate interior angles]
OD = OB [From eq. (i)]
Ð DOP = Ð BOQ [Vertically opposite angles]
\triangle DOP\cong \triangle BOQ [ASA congruence rule]
ar(CDPQ)=ar(\triangle COQ)+ar(\triangle COD)+ar(\triangle POD) …(iv) [Congruent figures have equal area]
=ar(\triangle AOP)+ar(\triangle AOB)+ar(\triangle BOQ) [From ii, iii, iv]
= ar(ABQP)
\Rightarrow 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 \triangle GBC = 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, ar(\triangle BEC)=\frac{1}{2}ar(\triangle ABC) …(1)
And, ar(\triangle ACF)=\frac{1}{2}ar(\triangle ABC) ….(2)
from (1) and (2)
ar(\triangle ACF)=ar(\triangle BEC)
\therefore ar(\triangle BGC)=ar(GFAE)
This can be written as:
Area of \triangle GBC = 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: CD\parallel AE and CY \parallel BA
To prove: ar (\triangle CBX) = ar(\triangle AXY)
Proof: We know that, triangles on the same base and between the same parallels have equal area.
Consider CY\parallel BA
\triangle ABYand \triangle ABC both lie on the same base AB and between the same parallels CY and BA.
ar(\triangle ABY) = ar(\triangle ABC)…(1)
Now, ar(\triangle ABY) = ar(ABX) + ar(AXY)
and, ar(\triangle ABC) = ar(ABX) + ar(CBX)
Putting the above values in eq. (1)
\Rightarrow ar(ABX) + ar(AXY) = ar(ABX) + ar(CBX)
\Rightarrow ar(AXY) = ar(CBX)
Hence proved

Question:6

In \triangle ABC, if L and M are the points on AB and AC, respectively such that LM \parallel BC. Prove that ar (LOB) = ar (MOC)

Answer:

Solution.
Given: \triangle ABC with L and M points on AB and AC respectively such that LM \parallel BC.
To prove: ar (\triangle LOB) = ar(\triangle MOC)
Proof:
We know that, triangles on the same base and between the same parallels are equal in area.
Hence, \triangle LBC and \triangle MBC lie on the same base BC and between the same parallels BC and LM.
So, ar(\triangle LBC) = ar(\triangle MBC) …(1)
ar(\triangle LBC) = ar(\triangle LOB) + ar(\triangle BOC)
ar(\triangle MBC)=ar(\triangle MOC)+ar(\triangle BOC)
Using these values in equation 1, we get:
\Rightarrow ar(\triangle LOB) + ar(\triangle BOC) = ar(\triangle MOC) + ar(\triangle BOC)
On eliminating ar(\triangle BOC) form both sides, we get
ar (\triangle LOB) = ar(\triangle MOC)
Hence proved.

Question:5

ABCD is a trapezium in which AB\parallel DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively the mid-points of AD and BC, prove that ar (DCYX) = \frac{7}{9 } ar (XYBA)

Answer:

Solution.
Given: Trapezium ABCD with AB\parallel DC,
DC = 30 cm and AB = 50 cm.
X and Y are the mid-points of AD and BC, respectively.
To prove: ar(DCYX)=\frac{7}{9}ar(XYBA)

Construction, Join DY and extend it to meet AB produced at P.
Proof:
In \triangle DCY and \triangle PBY
CY = BY [Given, Y is the mid-point of BC]
\angle DCY = \angle PBY [alternate interior angles]
\angle 2 = \angle 3[Vertically opposite angles]
\angle DCY \cong \angle PBY[ASA congruence rule]
t hen, DC = BP [CPCT]
DC = 30 cm (given)
Q DC = BP = 30 cm
Now,
AP = AB + BP = 50 + 30 = 80 cm
In \triangle ADP; XY \parallel AP, X is the mid-point of AD and Y is the mid-point of DP.
XY=\frac{1}{2}AP=\frac{1}{2}\times 80=40cm [mid-point theorem ]
Let distance between AB, XY and XY, DC be h cm
Area of trapezium is given as \frac{1}{2} [Sum of parallel side × distance between them]
Now, area of trapezium DCYX = \frac{1}{2}h(30+40)

= \frac{1}{2}h(70)=35hcm^{2}

Area of trapezium XYBA = \frac{1}{2}h(40+50)=\frac{1}{2}h\times 90=45hcm^{2}
\frac{ar(DCYX)}{ar(XYBA)}=\frac{35h}{45h}=\frac{7}{9}
ar(DCYX)=\frac{7}{9}ar(XYBA)
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, BP \parallel AC and EQ\parallel AD.
To prove:ar(ABCDE) = ar(\triangle APQ)
Proof: We know that triangles on the same base and between the same parallel lines are equal in area.
Now, \triangle ADQ and \triangle ADE lie on the same base AD and lie between the same parallels AD and EQ. So,
ar(\triangle ADE) = ar(\triangle ADQ)…(i)
Similarly,
\triangle CAB and \triangle CAP lie on the same base AC and lie between the same parallels AC and PB
ar(\triangle CAB)=ar(\triangle CAP) …(ii)
On adding equations (i) and (ii), we get
ar (\triangle ADE) + ar(\triangle CAB) = ar(\triangle ADQ) + ar (\triangle CAP)
Adding ar (DCDA) to both sides, we get

ar(\triangle ADE)+ar(\triangle CAB)+ar(\triangle CDA)=ar(\triangle ADQ)+ar(\triangle CAP)+ar(\triangle CDA)
\Rightarrowar(ABCDE) = ar(\triangle APQ)
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: ar(\triangle PEA) = ar(\triangle QFD)
Proof:
PQDA is a parallelogram, so AP \parallel DQ, PQ \parallel AD
Now, parallelograms PQDA and AEFD are on the same base AD and lie between the same parallel lines AD and EQ. Hence,
ar(\parallel gmPQDA)=ar(\parallel gmAEFD)
ar(PFDA)+ar(\triangle QFD)=ar(\triangle PEA)+ar(PFDA)
So, we get:
ar(\triangle QFD) = ar(\triangle PEA)
Hence proved

Question:9

In Fig., X and Y are the mid-points of AC and AB respectively, QP \parallel BC 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.
QP \parallel BC
CYQ, BXP are straight lines,
To prove: ar(\triangle ABP) = ar(\triangle ACQ)
Proof:
Since, X and Y are the mid-point of AC and AB.
So, XY \parallel BC
We know that triangles on the same base and between the same parallels are equal in area
Consider parallel lines BC and XY;
\triangle BYC and \triangle BXC lie on same base BC and between the same parallels BC and XY.
So, ar(\triangle BYC) = ar(\triangle BXC)
ar(\triangle BOC)+ar(\triangle BOY)=ar(\triangle BOC)+ar(\triangle XOC)
\Rightarrow ar(\triangle BOY)=ar(\triangle COX)
Adding ar(XOY) to LHS and RHS
\Rightarrow ar(BOY) + ar(XOY) = ar(\triangle COX) + ar(\triangle XOY)
\Rightarrow ar(\triangle BYX) = ar(\triangle CXY)…(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.
\because ar(XYAP)=ar(YXAQ) …(ii)
on adding eq. (i) and (ii), we get
ar (\triangle BYX) + ar(XYAP) = ar(\triangle CXY) + ar(YXAQ)
\Rightarrow ar(\triangle ABP) = ar(\triangle ACQ)
Hence Proved.

Question:8

If the medians of a \triangle ABC intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = \frac{1}{3} ar (ABC)

Answer:

Solution.
Given: \triangle ABC with medians AM, BN & CL
ar (AGB) = ar (AGC) = ar (BGC) =\frac{1}{3} ar (ABC)


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
ar (\triangle AMC) = ar (\triangle AMB)
ar(1) + ar(2) + ar(6) = ar(3) + ar(4) + ar(5) …(i)
CL is the median
ar(\triangle CAL)=ar(\triangle CBL)
ar(1)+ar(6)+ar(5)=ar(2)+ar(3)+ar(4)…(ii)
BN is the median
ar(\triangle BNA)=ar(\triangle BNC)
ar(4)+ar(5)+ar(6)=ar(1)+ar(2)+ar(3)
(i) – (ii)
ar(1)+ar(2)+ar(6)-ar(1)-ar(6)-ar(5)=ar(3)+ar(4)+ar(5)-ar(2)-ar(3)-ar(4)

So, ar(2)=ar(5)
(ii) – (iii)
ar(1)+ar(6)+ar(5)-ar(4)-ar(5)-ar(6)=ar(2)+ar(3)+ar(4)-ar(1)-ar(2)-ar(3)
ar(1)-ar(4)=ar(4)-ar(1)
So, ar(1)=ar(4)
Similarly we can prove that area of ar(1)=ar(2)=ar(3)=ar(4)=ar(5)=ar(6)
Hence the triangle is divided into 6 triangles of equal area.
\triangle AGB, \triangle AGCand \triangle BGC consists of 2 triangles each
So they have equal area which is equal to twice the area of the smaller triangles.
ar(\triangle AGB)=ar(\triangle AGC)=ar(\triangle BGC)=2 (area of 1 small triangle)
Hence, ar(\triangle AGB)=ar(\triangle AGC)=ar(\triangle BGC)=\frac{1}{3}ar(ABC)
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.

NCERT Class 9 Exemplar Solutions for Other Subjects:

NCERT Class 9 Maths Exemplar Solutions for Other Chapters:

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.

To view/download these solutions in an offline environment, students can use the NCERT exemplar Class 9 Maths solutions chapter 9 pdf download feature, enabling them to have a pdf version of the chapter handy while attempting the NCERT exemplar Class 9 Maths chapter 9.

Check the Solutions of Questions Given in the Book

Also, Read NCERT Solution Subject Wise

Check NCERT Notes Subject Wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (FAQs)

1. Two similar triangles have sides ratio 1:2. What is the ratio of their area?

 If sides are in the ratio 1:2 that means base length will be in the ratio 1:2 as well as height will be in the ratio 1:2. Therefore, we can say that area will be in the ratio 1:4.

2. Why is the area of the rectangle equal to the product of length and breadth?

By various experiences with comparison of rectangular lands, the formula of area of rectangle is devised. It is the simplest formula we assume for the area of the rectangle.

Even if we had assumed that the area of the rectangle is twice the product of length and breadth, it does not harm our understanding of space and areas.

3. Two triangles with the same base have the same area; are they necessarily congruent?

No, these two triangles must have the same height but their shapes may be different.

4. What is the weightage of the chapter Area of Parallelograms and Triangles in the final examination?

Generally, this chapter accounts for 8-10 % weightage of the final paper. These NCERT exemplar Class 9 Maths solutions chapter 9 are sufficient to attempt and excel in the chapter of Area of Parallelograms and Triangles.

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A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
Environmental Engineer

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems. 

2 Jobs Available
Safety Manager

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

2 Jobs Available
Conservation Architect

A Conservation Architect is a professional responsible for conserving and restoring buildings or monuments having a historic value. He or she applies techniques to document and stabilise the object’s state without any further damage. A Conservation Architect restores the monuments and heritage buildings to bring them back to their original state.

2 Jobs Available
Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software. 

2 Jobs Available
Highway Engineer

Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

2 Jobs Available
Field Surveyor

Are you searching for a Field Surveyor Job Description? A Field Surveyor is a professional responsible for conducting field surveys for various places or geographical conditions. He or she collects the required data and information as per the instructions given by senior officials. 

2 Jobs Available
Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
Veterinary Doctor
5 Jobs Available
Speech Therapist
4 Jobs Available
Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

4 Jobs Available
Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available
Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Anatomist

Are you searching for an ‘Anatomist job description’? An Anatomist is a research professional who applies the laws of biological science to determine the ability of bodies of various living organisms including animals and humans to regenerate the damaged or destroyed organs. If you want to know what does an anatomist do, then read the entire article, where we will answer all your questions.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
Producer

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story. 

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.  

2 Jobs Available
Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Production Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software. 

2 Jobs Available
Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available
QA Manager
4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
ITSM Manager
3 Jobs Available
Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process. 

2 Jobs Available

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