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.

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NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles
Areas of Parallelograms and Triangles Class 9 Solutions - Important Formulae
Areas of Parallelograms and Triangles Class 9 NCERT Solutions (Intext Questions and Exercise)
Summary Of Class 9 Maths Chapter 9 NCERT Solutions
NCERT Solutions For Class 9 Maths - Chapter Wise
NCERT Solutions For Class 9 - Subject Wise
NCERT Books and NCERT Syllabus

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 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 Questions And Answers PDF Free Download

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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

Answer:

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.
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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:

1595868720095
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:

1595868779714
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.]

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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:

1595868888044
(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.
1595868950842
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. $\small ar(ABE)=ar(ACE)$

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Answer:
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.
1595869100838
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:

1595869168733
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 $\small ar(ABC)=ar(ABD)$ .

1595869232559
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.
1595869300862
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:

1595869319420
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:

1595869344427
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
1595869408243
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:

1595869418016
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:

1595869431927
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)

1595869522642
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.
1595869753492
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:

1595869859506

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:

1595870011551
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 $\small ar(AQC)=ar(PBR)$ .

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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:

1595870246292
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.

1595870351963
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:

1595870426562

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 $\small ar(ABD)=ar(ADE)=ar(AEC)$ .

1640236583768 Answer:

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,
1595870559171
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:

1595870619379
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:

1595870673567
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:

1595870700069
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:

1595870707470
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:

1595870716157
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:

1595870785823
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:

1595870842857
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:

1595870850022
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:

1595870859523

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:

1595870953001
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:

1595870964455
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:

1595870972199
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:

1595870980947
$\angle FCA = \angle BCE$ [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:

1595870988524
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:

1595871000670
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:

1595871008289
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.

Also Read

NCERT Class 9 Maths Notes - Download Chapter Wise PDFs

NCERT Class 9 Maths Chapter 8 Notes Quadrilaterals - Download PDF

NCERT Class 9 Maths Chapter 2 Notes Polynomials - Download PDF

NCERT Class 9 Maths Chapter 1 Notes Number System - Download Free PDF

NCERT Class 9 Maths Chapter 5 Notes Introduction To Euclid’s Geometry - Download PDF

NCERT Class 9 Maths Chapter 6 Notes Lines and Angles - Download PDF

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NCERT Exemplar Class 9 Maths Solutions - Chapter Wise Problems with Solutions

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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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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

Areas of Parallelograms and Triangles Class 9 Solutions - Important Formulae

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

Summary Of Class 9 Maths Chapter 9 NCERT Solutions

NCERT Solutions For Class 9 Maths - Chapter Wise

NCERT Solutions For Class 9 - Subject Wise

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