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

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

Edited By Ramraj Saini | Updated on Sep 28, 2023 10:17 PM IST

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

Parallelograms and Triangles Class 9 Questions And Answers are discusses here. These NCERT solutions are developed by expert team at Careers360 team keeping in mind latest CBSE syllabus 2023-24. These solutions are simple, easy to understand and cover all the concepts step by step thus, ultimately help the students. In this particular NCERT book chapter, you will learn about the areas of different triangles and parallelograms.

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 class 9 maths NCERT syllabus . 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. Areas of parallelograms and triangles class 9 NCERT solutions have covered all the exercises including the optional ones in a detailed manner. Here you will get NCERT solutions for class 9 Maths also.

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Areas of Parallelograms and Triangles Class 9 Solutions - Important Formulae

Area of Parallelogram = Base × Height

Area of Triangle = (1/2) × Base × Height

Alternatively, it can be expressed as half the area of the parallelogram containing the triangle:

Area of Triangle = (1/2) × Area of Parallelogram

Area of Trapezium = (1/2) × (Sum of Parallel Sides) × Distance between Parallel Sides

Area of Rhombus = (1/2) × (Product of Diagonals)

Free download NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles for CBSE Exam.

Areas of Parallelograms and Triangles Class 9 NCERT Solutions (Intext Questions and Exercise)

Class 9 maths chapter 9 question answer - Exercise: 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.

1640236058747

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

Class 9 areas of parallelograms and triangles NCERT solutions - Exercise : 9.2

Q1 In Fig. \small 9.15 , ABCD is a parallelogram, \small AE\perp DC and \small CF\perp AD . If \small AB=16\hspace{1mm}cm , \small AE=8\hspace{1mm}cm and \small CF=10\hspace{1mm}cm , find AD.

1595868588337

1595868585372Answer:

We have,
AE \perp DC and CF \perp 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 \times AE = (16 \times 8 ) cm^2
SInce, CF \perp AD
therefore area of parallelogram = AD \times CF = 128 cm^2
= 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

\small ar(EFGH)=\frac{1}{2}ar(ABCD)

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.
15958686532791595868649957
Now, \Delta EFG and ||gm BEGC are on the same base and between the same parallels EG and BC.
Therefore, ar ( \Delta EFG) = 1/2 ar (||gm BEGC)...............(i)
Similarly, ar ( \Delta EHG) = 1/2 . ar(||gm AEGD)..................(ii)
By adding eq (i) and eq (ii), we get

ar ( \Delta EFG) + ar ( \Delta EHG) = 1/2 (ar (||gm BEGC) + ar(||gm AEGD))
ar (EFGH) = 1/2 ar(ABCD)
Hence proved

Q3 P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that \small ar(APB)=ar(BCQ) .

Answer:

15958687200951595868716112
We have,
ABCD is a parallelogram, therefore AB || CD and BC || AD.


Now, \Delta APB and ||gm ABCD are on the same base AB and between two parallels AB and DC.
Therefore, ar ( \Delta APB) = 1/2 . ar(||gm ABCD)...........(i)

Also, \Delta BQC and ||gm ABCD are on the same base BC and between two parallels BC and AD.
Therefore, ar( \Delta BQC) = 1/2 . ar(||gmABCD)...........(ii)

From eq(i) and eq (ii), we get,
ar ( \Delta APB) = ar( \Delta BQC)
Hence proved.

Q4 (i) In Fig. 9.16 , P is a point in the interior of a parallelogram ABCD. Show that
\small ar(APB)+ar(PCD) = \frac{1}{2}ar(ABCD)

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

1640236159415 Answer:

15958687797141595868777222
We have a ||gm ABCD and AB || CD, AD || BC. Through P, draw a line parallel to AB
Now, \Delta APB and ||gm ABEFare on the same base AB and between the same parallels EF and AB.
Therefore, ar ( \Delta APB) = 1/2 . ar(ABEF)...............(i)
Similarly, ar ( \Delta PCD ) = 1/2 . ar (EFDC) ..............(ii)
Now, by adding both equations, we get
\small ar(APB)+ar(PCD) = \frac{1}{2}ar(ABCD)

Hence proved.

Q4 (ii) In Fig. \small 9.16 , P is a point in the interior of a parallelogram ABCD. Show that

\small ar(APD)+ar(PBC)=ar(APB)+ar(PCD)

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


1640236187995

Answer:

15958688099861595868806399
We have a ||gm ABCD and AB || CD, AD || BC. Through P, draw a line parallel to AB
Now, \Delta APD and ||gm ADGHare on the same base AD and between the same parallels GH and AD.
Therefore, ar ( \Delta APD) = 1/2 . ar(||gm ADGH).............(i)

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

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

\small ar(APD)+ar(PBC)=ar(APB)+ar(PCD)

Hence proved.

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

(i) \small ar(PQRS)=ar(ABRS)
(ii) \small ar(AXS)=\frac{1}{2}ar(PQRS)

1640236228177 Answer:

15958688880441595868884843
(i) Parallelogram PQRS and ABRS are on the same base RS and between the same parallels RS and PB.
Therefore, \small ar(PQRS)=ar(ABRS) ............(i)
Hence proved

(ii) \Delta AXS and ||gm ABRS are on the same base AS and between same parallels AS and RB.
Therefore, ar ( \Delta AXS) = 1/2 . ar(||gm ABRS)............(ii)

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

\small ar(AXS)=\frac{1}{2}ar(PQRS)

Hence proved.

Q6 A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

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, \Delta APS, \Delta QAR and \Delta PAQ.
15958689508421595868947171
Since \Delta APQ and parallelogram, PQRS is on the same base PQ and between same parallels RS and PQ.
Therefore, ar(\Delta APQ) = \frac{1}{2}ar(PQRS) ............(i)
We can write above equation as,

ar (||gm PQRS) - [ar ( \Delta APS) + ar( \Delta QAR)] = 1/2 .ar(PQRS)
\Rightarrow ar(\Delta APS)+ar(\Delta QAR) = \frac{1}{2}ar(PQRS)
from equation (i),
\Rightarrow ar(\Delta APS)+ar(\Delta QAR) =ar(\Delta APQ)

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

Class 9 maths chapter 9 NCERT solutions - exercise : 9.3

Q1 In Fig. \small 9.23 , E is any point on median AD of a \small \Delta ABC . Show that. \small ar(ABE)=ar(ACE)

1595869014446

Answer: 1595869011521
We have \Delta ABC such that AD is a median. And we know that median divides the triangle into two triangles of equal areas.
Therefore, ar( \Delta ABD) = ar( \Delta ACD)............(i)

Similarly, In triangle \Delta BEC,
ar( \Delta BED) = ar ( \Delta DEC)................(ii)

On subtracting eq(ii) from eq(i), we get
ar( \Delta ABD) - ar( \Delta BED) =
\small ar(ABE)=ar(ACE)

Hence proved.

Q2 In a triangle ABC, E is the mid-point of median AD. Show that \small ar(BED)=\frac{1}{4}ar(ABC) .

Answer:

We have a triangle ABC and AD is a median. Join B and E.
15958691008381595869097063
Since the median divides the triangle into two triangles of equal area.
\therefore ar( \Delta ABD) = ar ( \Delta ACD) = 1/2 ar( \Delta ABC)..............(i)
Now, in triangle \Delta ABD,
BE is the median [since E is the midpoint of AD]
\therefore ar ( \Delta BED) = 1/2 ar( \Delta ABD)........(ii)

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

ar ( \Delta BED) = 1/2 . (1/2 ar(ar ( \Delta ABC))
ar ( \Delta BED) = 1/4 .ar( \Delta ABC)

Hence proved.

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

Answer:

15958691687331595869165497
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 ( \Delta BOC) = ar ( \Delta DOC)...........(a) ( since OC is the median of triangle CBD)

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

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

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

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

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

Q4 In Fig. \small 9.24 , ABC and ABD are two triangles on the same base AB. If line- segment CD is bisected by AB at O, show that \small ar(ABC)=ar(ABD) .

15958692325591595869229774
Answer:

We have, \Delta ABC and \Delta ABD on the same base AB. CD is bisected by AB at point O.
\therefore OC = OD
Now, in \Delta ACD, AO is median
\therefore ar ( \Delta AOC) = ar ( \Delta AOD)..........(i)

Similarly, in \Delta BCD, BO is the median
\therefore ar ( \Delta BOC) = ar ( \Delta BOD)............(ii)

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

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

\small ar(ABC)=ar(ABD)

Hence proved.

Q5 (i) D, E and F are respectively the mid-points of the sides BC, CA and AB of a \small \Delta ABC . Show that BDEF is a parallelogram.

Answer:

We have a triangle \Delta ABC such that D, E and F are the midpoints of the sides BC, CA and AB respectively.
15958693008621595869297243
Now, in \Delta 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.
\therefore EF || BC or EF || BD

also, EF = 1/2 (BC)
\Rightarrow EF = BD [ 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 \small \Delta ABC . Show that

\small ar(DEF)=\frac{1}{4}ar(ABC)

Answer:

15958693194201595869315903
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
\therefore Ar (BDEF) = Ar (DCEF)
\Rightarrow Ar( \Delta BDF) = Ar ( \Delta DEF) .............(i)

It is known that diagonals of ||gm divides it into two triangles of equal area.
\therefore Ar(DCE) = Ar (DEF).......(ii)

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

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

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

Thus, Ar ( \Delta ABC) = Ar( \Delta BDF) + Ar(DCE) + Ar( \Delta AEF) + Ar ( \Delta DEF)
Ar ( \Delta ABC) = 4 . Ar( \Delta DEF)
\Rightarrow ar(\Delta DEF) = \frac{1}{4}ar(\Delta ABC)

Hence proved.

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

\small ar(BDEF)=\frac{1}{2}ar(ABC)


Answer:

15958693444271595869343465
Since we already proved that,
ar( \Delta DEF) = ar ( \Delta BDF).........(i)

So, ar(||gm BDEF) = ar( \Delta BDF) + ar ( \Delta DEF)
= 2 . ar( \Delta DEF) [from equation (i)]
\\= 2[\frac{1}{4}ar(\Delta ABC)]\\ =\frac{1}{2} ar(\Delta ABC)

Hence proved.

Q6 (i) In Fig. \small 9.25 , diagonals AC and BD of quadrilateral ABCD intersect at O such that. If \small AB=CD , then show that:

\small ar(DOC)=ar(AOB)

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

1640236346940 Answer:
We have ABCD is quadrilateral whose diagonals AC and BD intersect at O. And OB = OD, AB = CD
Draw DE \perp AC and FB \perp AC
15958694082431595869404431
In \Delta DEO and \Delta BFO
\angle DOE = \angle BOF [vertically opposite angle]
\angle OED = \angle BFO [each 90^0 ]
OB = OD [given]

Therefore, by AAS congruency
\Delta DEO \cong \Delta BFO
\Rightarrow DE = FB [by CPCT]

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

Now, In \Delta DEC and \Delta ABF
\angle DEC = \angle BFA [ each 90^0 ]
DE = FB
DC = BA [given]
So, by RHS congruency
\Delta DEC \cong \Delta BFA
\angle 1 = \angle 2 [by CPCT]
and, ar( \Delta DEC) = ar( \Delta BFA).....(ii)

By adding equation(i) and (ii), we get
\small ar(DOC)=ar(AOB)
Hence proved.

Q6 (ii) In Fig. \small 9.25 , diagonals AC and BD of quadrilateral ABCD intersect at O such that \small OB=OD . If \small AB=CD , then show that: \small ar(DCB)=ar(ACB)

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


1640236384556

Answer:

15958694180161595869416364
We already proved that,
ar(\Delta DOC)=ar(\Delta AOB)
Now, add ar(\Delta BOC) on both sides we get

\\ar(\Delta DOC)+ar(\Delta BOC)=ar(\Delta AOB)+ar(\Delta BOC)\\ ar(\Delta DCB) = ar (\Delta ACB)
Hence proved.

Q6 (iii) In Fig. \small 9.25 , diagonals AC and BD of quadrilateral ABCD intersect at O such that \small OB=OD . If \small AB=CD , then show that:

\small DA\parallel CB or ABCD is a parallelogram.

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


1640236400803

Answer:

15958694319271595869429449
Since \Delta DCB and \Delta ACB both lie on the same base BC and having equal areas.
Therefore, They lie between the same parallels BC and AD
\Rightarrow CB || AD
also \angle 1 = \angle 2 [ already proved]
So, AB || CD
Hence ABCD is a || gm

Q7 D and E are points on sides AB and AC respectively of \small \Delta ABC such that \small ar(DBC)=ar(EBC) . Prove that \small DE\parallel BC .

Answer:

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

15958695226421595869519191
Since \Delta DBC and \Delta EBC are on the same base BC and having the same area.
\therefore They must lie between the same parallels DE and BC
Hence DE || BC.

Q8 XY is a line parallel to side BC of a triangle ABC. If \small BE\parallel AC and \small CF\parallel AB meet XY at E and F respectively, show that \small ar(ABE)=ar(ACF)

Answer:

We have a \Delta ABC such that BE || AC and CF || AB
Since XY || BC and BE || CY
Therefore, BCYE is a ||gm
1640236430685 Now, The ||gm BCEY and \Delta ABE are on the same base BE and between the same parallels AC and BE.
\therefore ar( \Delta AEB) = 1/2 .ar(||gm BEYC)..........(i)
Similarly, ar( \Delta 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.
\therefore ar (BEYC) = ar (BCFX).........(iii)

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

ar( \Delta ABE) = ar( \Delta ACF)
Hence proved.

Q9 The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Fig. \small 9.26 ). Show that \small ar(ABCD)=ar(PBQR) . [ Hint: Join AC and PQ. Now compare \small ar(ACQ) and \small ar(APQ) .]

1640236457256 Answer:

Join the AC and PQ.
15958697534921595869750167
It is given that ABCD is a ||gm and AC is a diagonal of ||gm
Therefore, ar( \Delta ABC) = ar( \Delta ADC) = 1/2 ar(||gm ABCD).............(i)

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

Since \Delta AQC and \Delta APQ are on the same base AQ and between same parallels AQ and CP.
\therefore ar( \Delta AQC) = ar ( \Delta APQ)

Now, subtracting \Delta ABQ from both sides we get,

ar( \Delta AQC) - ar ( \Delta ABQ) = ar ( \Delta APQ) - ar ( \Delta ABQ)
ar( \Delta ABC) = ar ( \Delta BPQ)............(iii)

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

\small ar(ABCD)=ar(PBQR)

Hence proved.

Q10 Diagonals AC and BD of a trapezium ABCD with \small AB\parallel DC intersect each other at O. Prove that \small ar(AOD)=ar(BOC) .

Answer:

15958698595061595869855980

We have a trapezium ABCD such that AB || CD and it's diagonals AC and BD intersect each other at O
\Delta ABD and \Delta ABC are on the same base AB and between same parallels AB and CD
\therefore ar( \Delta ABD) = ar ( \Delta ABC)

Now, subtracting \Delta AOB from both sides we get

ar ( \Delta AOD) = ar ( \Delta BOC)

Hence proved.

Q11 In Fig. \small 9.27 , ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

(i) \small ar(ACB)=ar(ACF)
(ii) \small ar(AEDF)=ar(ABCDE)

15958699297841595869926693

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

(i) Since \Delta ACB and \Delta ACF are on the same base AC and between same parallels AC and FB.
\therefore ar( \Delta ACB) = ar ( \Delta ACF)..................(i)

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

ar( \Delta ACB) + ar(AEDC) = ar ( \Delta ACF) + ar(AEDC)
\therefore \small ar(AEDF)=ar(ABCDE)

Hence proved.

Q12 A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given an equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.

Answer:

15958700115511595870008875
We have a quadrilateral shaped plot ABCD. Draw DF || AC and AF || CF.
Now, \Delta DAF and \Delta DCF are on the same base DF and between same parallels AC and DF.
\therefore ar ( \Delta DAF) = ar( \Delta DCF)
On subtracting \Delta DEF from both sides, we get

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

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

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

ar ( \Delta ADE) + ar(quad. ABCE) = ar( \Delta CEF) + ar(quad. ABCE)
ar (ABCD) = ar( \Delta ABF)

Q13 ABCD is a trapezium with \small AB\parallel DC . A line parallel to AC intersects AB at X and BC at Y. Prove that \small ar(ADX)=ar(ACY) . [ Hint: Join CX.]

Answer:

15958700795861595870076704
We have a trapezium ABCD, AB || CD
XY ||AC meets AB at X and BC at Y. Join XC

Since \Delta ADX and \Delta ACX lie on the same base CD and between same parallels AX and CD
Therefore, ar( \Delta ADX) = ar( \Delta ACX)..........(i)
Similarly ar( \Delta ACX) = ar( \Delta ACY).............(ii) [common base AC and AC || XY]
From eq (i) and eq (ii), we get

ar( \Delta ADX) = ar ( \Delta ACY)

Hence proved.

Q14 In Fig. \small 9.28 , \small AP\parallel BQ\parallel CR . Prove that \small ar(AQC)=ar(PBR) .

15958701927321595870189293
Answer:

We have, AP || BQ || CR
\Delta BCQ and \Delta BQR lie on the same base (BQ) and between same parallels (BQ and CR)
Therefore, ar ( \Delta BCQ) = ar ( \Delta BQR)........(i)

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

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

ar ( \Delta AQC) = ar ( \Delta PBR)

Hence proved.

Q15 Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that \small ar(AOD)=ar(BOC) . Prove that ABCD is a trapezium.

Answer:

15958702462921595870244150
We have,
ABCD is a quadrilateral and diagonals AC and BD intersect at O such that ar( \Delta AOD) = ar ( \Delta BOC) ...........(i)

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

ar( \Delta AOD) + ar ( \Delta BOA) = ar ( \Delta BOA) + ar ( \Delta BOC)
ar ( \Delta ABD) = ar ( \Delta ABC)
Since the \Delta ABC and \Delta ABD lie on the same base AB and have an equal area.
Therefore, AB || CD

Hence ABCD is a trapezium.

Q16 In Fig. \small 9.29 , \small ar(DRC)=ar(DPC) and \small ar(BDP)=ar(ARC) . Show that both the quadrilaterals ABCD and DCPR are trapeziums.

15958703519631595870349006
Answer:

Given,
ar( \Delta DPC) = ar( \Delta DRC) ..........(i)
and ar( \Delta BDP) = ar(ARC)............(ii)

from equation (i),
Since \Delta DRC and \Delta DPC lie on the same base DC and between same parallels.
\therefore CD || RP (opposites sides are parallel)

Hence quadrilateral DCPR is a trapezium

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

ar( \Delta BDP) - ar( \Delta DPC) = ar( \Delta ARC) - ar( \Delta DRC)
ar( \Delta BDC) = ar( \Delta ADC) (Since theya are on the same base DC)
\Rightarrow AB || DC

Hence ABCD is a trapezium.

Areas of parallelograms and triangles class 9 solutions - Exercise : 9.4

Q1 Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

Answer:

15958704265621595870423378

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

SInce \Delta BEC and \Delta AFD are right angled triangle
Therefore, AD > AF and BC > BE
\Rightarrow (BC + AD ) > (AF + BE)...........(ii)

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

(AB + CD)+(BC + AD) > (AB + BE) + (AF + BE)
\Rightarrow (AB + BC + CD + DA) > (AB + BE + EF + FA)
Hence proved, perimeter of ||gm ABCD is greater than perimeter of rectangle ABEF.

Q2 In Fig. \small 9.30 , D and E are two points on BC such that \small BD=DE=EC . Show that \small ar(ABD)=ar(ADE)=ar(AEC) .

1640236583768Answer:

In \Delta 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. \small 9.31 , ABCD, DCFE and ABFE are parallelograms. Show that \small ar(ADE)=ar(BCF) .

1640236638372 Answer:

Given,
15958705591711595870555496
ABCD is a parallelogram. Therefore, AB = CD & AD = BC & AB || CD .......(i)

Now, \Delta ADE and \Delta BCF are on the same base AD = BC and between same parallels AB and EF.
Therefore, ar ( \Delta ADE) = ar( \Delta BCF)

Hence proved.

Q4 In Fig. \small 9.32 , ABCD is a parallelogram and BC is produced to a point Q such that \small AD=CQ . If AQ intersects DC at P, show that \small ar (BPC) = ar (DPQ) . [ Hint : Join AC.]

1640236665039 Answer:

15958706193791595870615704
Given,
ABCD is a ||gm and AD = CQ. Join AC.
Since \Delta DQC and \Delta ACD lie on the same base QC and between same parallels AD and QC.
Therefore, ar( \Delta DQC) = ar( \Delta ACD).......(i)

Subtracting ar( \Delta PQC) from both sides in eq (i), we get
ar( \Delta DPQ) = ar( \Delta PAC).............(i)

Since \Delta PAC and \Delta PBC are on the same base PC and between same parallel PC and AB.
Therefore, ar( \Delta PAC) = ar( \Delta PBC)..............(iii)

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

\small ar (BPC) = ar (DPQ)

Hence proved.

Q5 (i) In Fig. \small 9.33 , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

\small ar(BDE)=\frac{1}{4}ar(ABC)

[ Hint: Join EC and AD. Show that \small BE\parallel AC and \small DE\parallel AB , etc.]

1640236707923 Answer:

15958706735671595870671414
Let join the CE and AD and draw EP \perp BC . It is given that \Delta ABC and \Delta BDE is an equilateral triangle.
So, AB =BC = CA = a and D i sthe midpoint of BC
therefore, BD = \frac{a}{2}= DE = BE
(i) Area of \Delta ABC = \frac{\sqrt{3}}{4}a^2 and
Area of \Delta BDE = \frac{\sqrt{3}}{4}(\frac{a}{2})^2= \frac{\sqrt{3}}{16}a^2

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


Q5 (ii) In Fig. \small 9.33 , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
\small ar(BDE)=\frac{1}{2}ar(BAE)

[ Hint: Join EC and AD. Show that \small BE\parallel AC and \small DE\parallel AB , etc.]


1640236728299

Answer:

15958707000691595870698541
Since \Delta ABC and \Delta BDE are equilateral triangles.
Therefore, \angle ACB = \angle DBE = 60^0
\Rightarrow BE || AC

\Delta BAE and \Delta BEC are on the same base BE and between same parallels BE and AC.
Therefore, ar ( \Delta BAE) = ar( \Delta BEC)
\Rightarrow ar( \Delta BAE) = 2 ar( \Delta BED) [since D is the meian of \Delta BEC ]
\Rightarrow \small ar(BDE)=\frac{1}{2}ar(BAE)
Hence proved.

Q5 (iii) In Fig. \small 9.33 , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
\small ar(ABC)=2ar(BEC)

[ Hint : Join EC and AD. Show that \small BE\parallel AC and \small DE\parallel AB , etc.]


1640236744718

Answer:

We already proved that,
ar( \Delta ABC) = 4.ar( \Delta BDE) (in part 1)
and, ar( \Delta BEC) = 2. ar( \Delta BDE) (in part ii )

\Rightarrow ar( \Delta ABC) = 2. ar( \Delta BEC)

Hence proved.

Q5 (iv) In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
\small ar(BFE)=ar(AFD)
[ Hint: Join EC and AD. Show that \small BE\parallel AC and \small DE\parallel AB , etc.]


1640236764354

Answer:

15958707074701595870706529
Since \Delta ABC and \Delta BDE are equilateral triangles.
Therefore, \angle ACB = \angle DBE = 60^0
\Rightarrow BE || AC

\Delta BDE and \Delta AED are on the same base ED and between same parallels AB and DE.
Therefore, ar( \Delta BED) = ar( \Delta AED)

On subtracting \Delta EFD from both sides we get

\small ar(BFE)=ar(AFD)

Hence proved.

Q5 (v) In Fig. \small 9.33 , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
\small ar(BFE)=2ar(FED)

[ Hint : Join EC and AD. Show that \small BE\parallel AC and \small DE\parallel AB , etc.]


1640236770366

Answer:

15958707161571595870714602
In right angled triangle \Delta ABD, we get

\\AB^2 = AD^2+BD^2\\ AD^2 = AB^2-BD^2
=a^2 - \frac{a^2}{4}=\frac{3a^2}{4}
AD=\frac{\sqrt{3}a}{2}

So, in \Delta PED,
PE^2 = DE^2-DP^2
\\=(\frac{a}{2})^2-(\frac{a}{4})^2\\ =\frac{3a^2}{16}
So, PE=\frac{\sqrt{3}a}{4}

Therefore, the Area of \Delta AFD =1/2 (FD)\frac{\sqrt{3}a}{2} ..........(i)

And, Area of triangle \Delta EFD =1/2 (FD)\frac{\sqrt{3}a}{4} ...........(ii)

From eq (i) and eq (ii), we get
ar( \Delta AFD) = 2. ar( \Delta EFD)
Since ar( \Delta AFD) = ar( \Delta BEF)

\Rightarrow \small ar(BFE)=2ar(FED)


Q5 (vi) In Fig. 9.33 , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
ar (FED) = \frac{1}{8} ar (AFC)

[ Hint: Join EC and AD. Show that BE\parallel AC and DE\parallel AB , etc.]

1640236806124 Answer:

ar(\Delta AFC) =
\\= ar(\Delta AFD) + ar(\Delta ADC)\\ = ar(\Delta BFE) + 1/2. ar(\Delta ABC)\\ = ar(\Delta BFE) + 1/2\times [4\times ar(\Delta BDE)]\\ = ar(\Delta BFE) + 2\times ar(\Delta BDE)
= 2\times ar (\Delta FED) + 2\times [ar(\Delta BFE) + ar(\Delta FED)] .....(from part (v) ar( \Delta BFE) = 2. ar( \Delta FED) ]
\\=2ar(\Delta FED) + 2[3\times ar(\Delta FED)]\\ =8\times ar(\Delta FED)

ar (FED) = \frac{1}{8} ar (AFC)
Hence proved.

Q6 Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) \times ar (CPD) = ar (APD) \times ar (BPC).

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

Answer:

15958707858231595870782539
Given that,
A quadrilateral ABCD such that it's diagonal AC and BD intersect at P. Draw AM \perp BD and CN \perp BD

Now, ar( \Delta APB) = \frac{1}{2}\times BP\times AM and,
ar(\Delta CDP)=\frac{1}{2}\times DP\times CN
Therefore, ar( \Delta APB) \times ar( \Delta CDP)=
\\=(\frac{1}{2}\times BP\times AM).(\frac{1}{2}\times DP \times CN)\\ =\frac{1}{4}\times BP \times DP\times AM\times CN ....................(i)

Similarly, ar( \Delta APD) \times ar ( \Delta BPC) =
=\frac{1}{4}\times BP \times DP\times AM\times CN ................(ii)

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

ar (APB) \times ar (CPD) = ar (APD) \times ar (BPC).
Hence proved.

Q7 (i) P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that

\small ar(PRQ)=\frac{1}{2}ar(ARC)


Answer:

15958708428571595870840637
We have \Delta 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 \Delta APC,
Since R is the midpoint. So, RC is the median of the \Delta APC
Therefore, ar( \Delta ARC) = 1/2 . ar ( \Delta APC)............(i)

Also, in \Delta ABC, P is the midpoint. Thus CP is the median.
Therefore, ar( \Delta APC) = 1/2. ar ( \Delta ABC)............(ii)

Also, AQ is the median of \Delta ABC
Therefore, 1/2. ar ( \Delta ABC) = ar (ABQ)............(iii)

In \Delta APQ, RQ is the median.
Therefore, ar ( \Delta PRQ) = 1/2 .ar ( \Delta APQ).............(iv)

In \Delta ABQ, PQ is the median
Therefore, ar( \Delta APQ) = 1/2. ar( \Delta ABQ).........(v)


From eq (i),
\frac{1}{2}ar(\Delta ARC)= \frac{1}{4}ar(\Delta APC) ...........(vi)
Now, put the value of ar( \Delta APC) from eq (ii), we get
Taking RHS;
= \frac{1}{8}[ar(\Delta ABC)]
= \frac{1}{4}[ar(\Delta ABQ)] (from equation (iii))

= \frac{1}{2}[ar(\Delta APQ)] (from equation (v))

=ar(\Delta PRQ) (from equation (iv))

\Rightarrow \small ar(PRQ)=\frac{1}{2}ar(ARC)

Hence proved.

Q7 (ii) P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that
\small ar(RQC)=\frac{3}{8}ar(ABC)

Answer:

15958708500221595870849049
In \Delta RBC, RQ is the median
Therefore ar( \Delta RQC) = ar( \Delta RBQ)
= ar (PRQ) + ar (BPQ)
= 1/8 (ar \Delta ABC) + ar( \Delta BPQ) [from eq (vi) & eq (A) in part (i)]
= 1/8 (ar \Delta ABC) + 1/2 (ar \Delta PBC) [ since PQ is the median of \Delta BPC]
= 1/8 (ar \Delta ABC) + (1/2).(1/2)(ar \Delta ABC) [CP is the medain of \Delta ABC]
= 3/8 (ar \Delta ABC)

Hence proved.

Q7 (iii) P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that

ar (PBQ) = ar (ARC)

Answer:

15958708595231595870857364

QP is the median of \Delta ABQ
Therefore, ar( \Delta PBQ) = 1/2. (ar \Delta ABQ)

= (1/2). (1/2) (ar \Delta ABC) [since AQ is the median of \Delta ABC
= 1/4 (ar \Delta ABC)
= ar ( \Delta ARC) [from eq (A) of part (i)]

Hence proved.

Q8 (i) In Fig. \small 9.34 , ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment \small AX\perp DE meets BC at Y. Show that: \small \Delta MBC\cong \Delta ABD

1640236844952

Answer:

15958709530011595870949664
We have, a \Delta ABC such that BCED, ACFG and ABMN are squares on its side BC, CA and AB respec. Line segment \small AX\perp DE meets BC at Y

(i) \angle CBD = \angle MBA [each 90]
Adding \angle ABC on both sides, we get
\angle ABC + \angle CBD = \angle MBA + \angle ABC
\Rightarrow \angle ABD = \angle MBC
In \Delta ABD and \Delta MBC, we have
AB = MB
BD = BC
\angle ABD = \angle MBC
Therefore, By SAS congruency
\Delta ABD \cong \Delta MBC

Hence proved.

Q8 (ii) In Fig. \small 9.34 , ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment \small AX\perp DE meets BC at Y. Show that: \small ar(BYXD)=2ar(MBC)

Answer:

15958709644551595870961735
SInce ||gm BYXD and \Delta ABD are on the same base BD and between same parallels BD and AX
Therefore, ar( \Delta ABD) = 1/2. ar(||gm BYXD)..........(i)

But, \Delta ABD \cong \Delta MBC (proved in 1st part)
Since congruent triangles have equal areas.
Therefore, By using equation (i) we get
\small ar(BYXD)=2ar(MBC)
Hence proved.

Q8 (iii) In Fig. \small 9.34 , ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment \small AX\perp DE meets BC at Y. Show that: ar(BYXD)=ar(ABMN)

Answer:

15958709721991595870971040
Since, ar (||gm BYXD) = 2 .ar ( \Delta MBC) ..........(i) [already proved in 2nd part]
and, ar (sq. ABMN) = 2. ar ( \Delta MBC)............(ii)
[Since ABMN and AMBC are on the same base MB and between same parallels MB and NC]
From eq(i) and eq (ii), we get

\small ar(BYXD)=ar(ABMN)

Q8 (iv) In Fig. \small 9.34 , ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment \small AX\perp DE meets BC at Y. Show that: \small \Delta FCB\cong \Delta ACE

Answer:

15958709809471595870978105
\angle FCA = \angle BCE [both 90]
By adding \angle ABC on both sides we get
\angle ABC + \angle FCA = \angle ABC + \angle BCE
\angle FCB = \angle ACE

In \Delta FCB and \Delta ACE
FC = AC [sides of square]
BC = AC [sides of square]
\angle FCB = \angle ACE
\Rightarrow
\Delta FCB \cong \Delta ACE

Hence proved.

Q8 (v) In Fig. \small 9.34 , ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment \small AX\perp DE meets BC at Y. Show that: \small ar(CYXE)=2ar(FCB)

Answer:

15958709885241595870987887
Since ||gm CYXE and \Delta ACE lie on the same base CE and between the same parallels CE and AX.
Therefore, 2. ar( \Delta ACE) = ar (||gm CYXE)
B
ut, \Delta FCB \cong \Delta ACE (in iv part)
Since the congruent triangle has equal areas. So,
\small ar(CYXE)=2ar(FCB)
Hence proved.

Q8 (vi) In Fig. \small 9.34 , ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment \small AX\perp DE meets BC at Y. Show that: \small ar(CYXE)=ar(ACFG)

Answer:

15958710006701595870997995
Since ar( ||gm CYXE) = 2. ar( \Delta ACE) {in part (v)}...................(i)
Also, \Delta FCB and quadrilateral ACFG lie on the same base FC and between the same parallels FC and BG.
Therefore, ar (quad. ACFG) = 2.ar( \Delta FCB ) ................(ii)

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

\small ar(CYXE)=ar(ACFG)
Hence proved.

Q8 (vii) In Fig. \small 9.34 , ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment \small AX\perp DE meets BC at Y. Show that: \small ar(BCED)=ar(ABMN)+ar(ACFG)

Answer:

15958710082891595871007517
We have,
ar(quad. BCDE) = ar(quad, CYXE) + ar(quad. BYXD)

= ar(quad, CYXE) + ar (quad. ABMN) [already proved in part (iii)]
Thus, \small ar(BCED)=ar(ABMN)+ar(ACFG)

Hence proved.

Summary Of Class 9 Maths Chapter 9 NCERT Solutions

  • Figures on the Same Base and Between the Same Parallels
  • Parallelograms on the same Base and Between the same Parallels
  • Triangles on the same Base and between the same Parallels

Maths chapter 9 class 9 - Important Points

  • The area of a parallelogram is equal to the product of its base and the corresponding altitude.
  • The altitude of a parallelogram is the perpendicular distance between the two parallel sides.
  • The area of a triangle is half the product of its base and the corresponding altitude.
  • The altitude of a triangle is the perpendicular distance between the base and the opposite vertex.
  • If two parallelograms are on the same base and between the same parallels, then their areas are equal.
  • If a parallelogram and a triangle are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
  • If two triangles have the same base and lie between the same parallels, then their areas are equal.
  • The sum of the areas of two triangles formed by drawing a diagonal of a parallelogram is equal to the area of the parallelogram.
  • The sum of the areas of four triangles formed by drawing the diagonals of a parallelogram is equal to twice the area of the parallelogram.
  • The area of a trapezium is half the product of the sum of its parallel sides and the corresponding altitude.

Interested students can practice class 9 maths ch 9 question answer using the following links

NCERT Solutions For Class 9 Maths - Chapter Wise

Chapter No. Chapter Name
Chapter 1 Number Systems
Chapter 2 Polynomials
Chapter 3 Coordinate Geometry
Chapter 4 Linear Equations In Two Variables
Chapter 5 Introduction to Euclid's Geometry
Chapter 6 Lines And Angles
Chapter 7 Triangles
Chapter 8 Quadrilaterals
Chapter 9 Areas of Parallelograms and Triangles
Chapter 10 Circles
Chapter 11 Constructions
Chapter 12 Heron’s Formula
Chapter 13 Surface Area and Volumes
Chapter 14 Statistics
Chapter 15 Probability

NCERT Solutions For Class 9 - Subject Wise

How To Use NCERT Solutions For Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles

  • Before coming to this, please cover the basic properties related to triangles and parallelograms as well as the previous chapter.
  • Learn some formulas and properties given in the NCERT textbook.
  • Understand the method of solution through some examples.
  • Practice the approach to which solutions are solved in the practice problems given.
  • While doing practice problems, if you get stuck anywhere then assist yourself with NCERT solutions for class 9 maths chapter 9 Areas Of Parallelograms and Triangles.

NCERT Books and NCERT Syllabus

Keep working hard and happy learning!

Frequently Asked Question (FAQs)

1. What are the important topics in chapter ch 9 maths class 9?

Areas of Parallelograms and Triangles, figures with the same base and between parallel lines, are the important topics of this chapter. Students can practice NCERT solutions for class 9 maths to command these concepts which are very important for the exam. 

2. How does the NCERT solutions are helpful ?

NCERT solutions are not only helpful for the students if they stuck while solving NCERT problems but also, these solutions are provided in a very detailed manner which will give them conceptual clarity.

3. Where can I find the complete area of parallelogram and triangle class 9 ?

Here you will get the detailed NCERT solutions for class 9 maths  by clicking on the link. you can practice these ncert solutions for class 9 maths chapter 9 to command the concepts which give the confidence during exams.

4. Do we need to master all the exercises included in the NCERT Solutions for Class 9 Maths Chapter 9?

In order to be able to tackle various types of questions that appear in exams, it is crucial to study and practice all the questions included in the NCERT Solutions for Class 9 Maths Chapter 9. The solutions are presented in a student-friendly language that facilitates easier comprehension of complex problems. These solutions are designed by Careers360 experts who possess extensive knowledge of mathematics.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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A career as a Finance Executive requires one to be responsible for monitoring an organisation's income, investments and expenses to create and evaluate financial reports. His or her role involves performing audits, invoices, and budget preparations. He or she manages accounting activities, bank reconciliations, and payable and receivable accounts.  

3 Jobs Available
Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available
Credit Manager

Credit Management refers to the process of granting credit, setting the terms it’s granted on, recovering the credit when it’s due, and confirming compliance with the organization's credit policy, among other credit-related operations. Individuals who opt for a career as Credit Manager should have hands-on experience with accounting software, a solid understanding of lending procedures, excellent analytical skills with the ability to create and process financial spreadsheets, negotiation skills, and a bachelor’s or master’s degree in a field relevant to finance or accounting. Ultimately, Credit Management job is to help organizations minimize bad debts and increase revenues from the loan.

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
Transportation Planner

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
Construction Manager

Individuals who opt for a career as construction managers have a senior-level management role offered in construction firms. Responsibilities in the construction management career path are assigning tasks to workers, inspecting their work, and coordinating with other professionals including architects, subcontractors, and building services engineers.

2 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
Naval Architect

A Naval Architect is a professional who designs, produces and repairs safe and sea-worthy surfaces or underwater structures. A Naval Architect stays involved in creating and designing ships, ferries, submarines and yachts with implementation of various principles such as gravity, ideal hull form, buoyancy and stability. 

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
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
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
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
Veterinary Doctor

A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

5 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
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
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
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
Healthcare Social Worker

Healthcare social workers help patients to access services and information about health-related issues. He or she assists people with everything from locating medical treatment to assisting with the cost of care to recover from an illness or injury. A career as Healthcare Social Worker requires working with groups of people, individuals, and families in various healthcare settings such as hospitals, mental health clinics, child welfare, schools, human service agencies, nursing homes, private practices, and other healthcare settings.  

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
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
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
Fashion Blogger

Fashion bloggers use multiple social media platforms to recommend or share ideas related to fashion. A fashion blogger is a person who writes about fashion, publishes pictures of outfits, jewellery, accessories. Fashion blogger works as a model, journalist, and a stylist in the fashion industry. In current fashion times, these bloggers have crossed into becoming a star in fashion magazines, commercials, or campaigns. 

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
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
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
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
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
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
Advertising Manager

Advertising managers consult with the financial department to plan a marketing strategy schedule and cost estimates. We often see advertisements that attract us a lot, not every advertisement is just to promote a business but some of them provide a social message as well. There was an advertisement for a washing machine brand that implies a story that even a man can do household activities. And of course, how could we even forget those jingles which we often sing while working?

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
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
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
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
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
Procurement Manager

The procurement Manager is also known as  Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness. 

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
Merchandiser
2 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
QA Manager
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
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
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
ITSM Manager
3 Jobs Available
.NET Developer

.NET Developer Job Description: A .NET Developer is a professional responsible for producing code using .NET languages. He or she is a software developer who uses the .NET technologies platform to create various applications. Dot NET Developer job comes with the responsibility of  creating, designing and developing applications using .NET languages such as VB and C#. 

2 Jobs Available
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