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

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

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.

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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:

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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:

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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:

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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:

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(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.
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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.
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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)

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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:

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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:

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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 Questions (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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