NCERT Solutions for Exercise 9.3 Class 9 Maths Chapter 9 - Areas Of Parallelograms And Triangles

NCERT Solutions for Exercise 9.3 Class 9 Maths Chapter 9 - Areas Of Parallelograms And Triangles

The NCERT Solutions for Class 9 Maths exercise 9.3 deal with triangles rather than parallelograms. In Class 9 Maths chapter 9 exercise 9.3 we deal with triangles that have a common base and are in the same parallels.

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The important concepts related to NCERT solutions for Class 9 Maths chapter 9 exercise 9.3 in order to solve this exercise are:

  • Triangles that have the same base or the equal base and are between the same parallel lines have equivalent areas. This is on the grounds that we realize that the space of the triangle is half of the product of base and tallness since the height (distance between the parallels) remains the same for the two triangles and they have the same bases thus their areas are equal.

The figure shown below is a set of triangles with the same base and between the same parallel

The triangles ABC and BCD have the same area.

Along with NCERT book exercise 9.3 Class 9 Maths, the following exercises are also present.

  • Areas Of Parallelograms And Triangles Exercise 9.1

  • Areas Of Parallelograms And Triangles Exercise 9.2

  • Areas Of Parallelograms And Triangles Exercise 9.4

Areas Of Parallelograms And Triangles Class 9 Chapter 9 Exercise: 9.3

Q1 In Fig. , E is any point on median AD of a . Show that.

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

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

On subtracting eq(ii) from eq(i), we get
ar( ABD) - ar( BED) =

Hence proved.

Q2 In a triangle ABC, E is the mid-point of median AD. Show that .

Answer:

We have a triangle ABC and AD is a median. Join B and E.

Since the median divides the triangle into two triangles of equal area.
ar( ABD) = ar ( ACD) = 1/2 ar( ABC)..............(i)
Now, in triangle ABD,
BE is the median [since E is the midpoint of AD]
ar ( BED) = 1/2 ar( ABD)........(ii)

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

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

Hence proved.

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

Answer:


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

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

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

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

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

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

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


Answer:

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

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

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

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

Hence proved.

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

Answer:

We have a triangle ABC such that D, E and F are the midpoints of the sides BC, CA and AB respectively.

Now, in 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.
EF || BC or EF || BD

also, EF = 1/2 (BC)
[ 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 . Show that

Answer:


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
Ar (BDEF) = Ar (DCEF)
Ar( BDF) = Ar ( DEF) .............(i)

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

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

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

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

Thus, Ar ( ABC) = Ar( BDF) + Ar(DCE) + Ar( AEF) + Ar ( DEF)
Ar ( ABC) = 4 . Ar( DEF)

Hence proved.

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


Answer:


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

So, ar(||gm BDEF) = ar( BDF) + ar ( DEF)
= 2 . ar( DEF) [from equation (i)]

Hence proved.

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

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

Answer:
We have ABCD is quadrilateral whose diagonals AC and BD intersect at O. And OB = OD, AB = CD
Draw DE AC and FB AC

In DEO and BFO
DOE = BOF [vertically opposite angle]
OED = BFO [each ]
OB = OD [given]

Therefore, by AAS congruency
DEO BFO
DE = FB [by CPCT]

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

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

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

Hence proved.

Q6 (ii) In Fig. , diagonals AC and BD of quadrilateral ABCD intersect at O such that . If , then show that:

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


Answer:


We already proved that,

Now, add on both sides we get


Hence proved.

Q6 (iii) In Fig. , diagonals AC and BD of quadrilateral ABCD intersect at O such that . If , then show that:

or ABCD is a parallelogram.

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


Answer:


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

Q7 D and E are points on sides AB and AC respectively of such that . Prove that .

Answer:

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


Since DBC and EBC are on the same base BC and having the same area.
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 and meet XY at E and F respectively, show that

Answer:

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

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

ar( ABE) = ar( 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. ). Show that . [ Hint: Join AC and PQ. Now compare and .]

Answer:

Join the AC and PQ.

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

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

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

Now, subtracting ABQ from both sides we get,

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

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

Hence proved.

Q10 Diagonals AC and BD of a trapezium ABCD with intersect each other at O. Prove that .

Answer:

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

Now, subtracting AOB from both sides we get

ar ( AOD) = ar ( BOC)

Hence proved.

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

(i)
(ii)

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

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

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

ar( ACB) + ar(AEDC) = ar ( ACF) + ar(AEDC)

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:


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

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

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

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

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

Q13 ABCD is a trapezium with . A line parallel to AC intersects AB at X and BC at Y. Prove that . [ Hint: Join CX.]

Answer:


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

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

ar( ADX) = ar ( ACY)

Hence proved.

Q14 In Fig. , . Prove that .


Answer:

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

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

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

ar ( AQC) = ar ( PBR)

Hence proved.

Q15 Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that . Prove that ABCD is a trapezium.

Answer:


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

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

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

Hence ABCD is a trapezium.

Q16 In Fig. , and . Show that both the quadrilaterals ABCD and DCPR are trapeziums.


Answer:

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

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

Hence quadrilateral DCPR is a trapezium

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

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

Hence ABCD is a trapezium.

More About NCERT Solutions for Class 9 Maths Exercise 9.3

From NCERT solutions for Class 9 Maths exercise 9.3, we get to know another concept (theorem) which deals with the converse of the above concept. We get to know that if two triangles have the same base (or if they have an equal base) and if they have equal areas then they lie on the same parallel line. Thus, a quick thing to remember from NCERT solutions for Class 9 Maths exercises 9.3 is that if the triangle has a median (a line that divides the base into two parts), then the median divides the triangle into two parts both on each side.

Also Read| Areas Of Parallelograms And Triangles Class 9 Notes

Benefits of NCERT Solutions for Class 9 Maths Exercise 9.3

  • Exercise 9.3 Class 9 Maths, is based on the AREAS OF PARALLELOGRAMS AND TRIANGLES by providing us with a general (much simpler) approach to calculate the area of triangles and parallelograms and concepts of figures in the same base and same parallel lines.

  • NCERT syllabus Class 9 Maths chapter 9 exercise 9.3 introduces us to a new concept of triangles on the same base and between the same parallels and its properties, like calculating the area and height.

  • Understanding the concepts from Class 9 Maths chapter 9 exercise 9.3 will help us to compare the areas in multiple-choice questions easily and thus save time for a particular question.

Also, See

  • NCERT Solutions for Class 9 Maths Chapter 9

  • NCERT Exemplar Solutions Class 9 Maths Chapter 9

NCERT Solutions of Class 10 Subject Wise

  • NCERT Solutions for Class 9 Maths

  • NCERT Solutions for Class 9 Science

Subject Wise NCERT Exemplar Solutions

  • NCERT Exemplar Class 9 Maths
  • NCERT Exemplar Class 9 Science