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Areas of Parallelogram and Triangles : Exercise 9.3 (Mathematics NCERT Class 9th)


Q.1     In figure , E is any point on median AD of a \Delta ABC. Show that ar (ABE) = ar (ACE).

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Sol.

Given : AD is a medium of \Delta ABC and E is any point on AD.
To prove : ar (ABE) = ar (ACE)
Proof :   Since AD is the median of \Delta ABC
Therefore ar (ABD) = ar (ACD) ... (1)
Also , ED is the median of \Delta EBC
Therefore ar (BED) = ar (CED) ... (2)
Subtracting (2) from (1) , we get
ar (ABD) – ar (BED) = ar (ACD) – ar (CED)
 \Rightarrow ar (ABE) = ar (ACE).


Q.2      In a triangle ABC, E is the mid- point of median AD. Show that ar (BED) = {1 \over 4} ar (ABC).
Sol.

Given : A \Delta ABC, E is the mid- point of the median AD.
To prove : ar (BED) = {1 \over 4} ar (ABC) 

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Proof :  Since AD is a median of \Delta ABC and median divides a triangle into two triangles of equal area.
Therefore ar (ABD) = ar (ADC)
 \Rightarrow ar (ABD) = {1 \over 2} ar (ABC) ... (1)
In \Delta ABD, BE is the median
Therefore ar(BED) = ar (BAE) ... (2)
 \Rightarrow ar(BED) = {1 \over 2} ar (ABD)     [ ar(BAE) = {1 \over 2} ar (ABD) ]
 \Rightarrow ar (BED) = {1 \over 2} \times {1 \over 2} ar (ABC)                 [Using (1)]
 \Rightarrow ar (BED) = {1 \over 4} ar (ABC) .


Q.3      Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Sol.

Given : A parallelogram ABCD.
To prove : The diagonals AC and BD divide the || gm ABCD into four triangles of equal area.
Construction : Draw BL  \bot AC.
Proof : Since ABCD is a || gm and so its diagonals AC and BD bisect each other at O.

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Therefore AO = OC and BO = OD
Now, ar (AOB) = {1 \over 2} × AO × BL
ar (OBC) = {1 \over 2} × OC × BL
But AO = OC
Therefore ar (AOB) = ar (OBC)
Similarly, we can show that
ar (OBC) = ar (OCD) ; ar (OCD) = ar(ODA) ;
ar (ODA) = ar (OAB) ; ar (OAB) = ar (OBC)
ar (OCD) = ar (ODA)
Thus , ar (OAB) = ar (OBC) = ar(OCD) = ar (OAD)


Q.4      In figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).
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Sol.

Given : ABC and ABD are two triangles on the same base AB. A line segment CD is bisected by AB at O i.e., OC = OD.
To prove : ar (ABC) = ar (ABD).

Proof : In \Delta ACD, we have
OC = OD [Given]
Therefore AO is the median.
Therefore ar (AOC) = ar (AOD)         [Since median divides a \Delta in two \Delta s of equal area]
Similarly, in \Delta BCD, BO is the median
Therefore ar (BOC) = ar (BOD)
Adding (1) and (2), we get
ar (AOC) + ar (BOC) = ar (AOD) + ar (BOD)
 \Rightarrow ar (ABC) = ar (ABD)


Q.5     D, E and F are respectively the mid- points of the side BC, CA and AB of a \Delta ABC. Show that
                 (i) BDEF is a parallelogram
                 (ii) ar (DEF) = {1 \over 4} ar (ABC)
                 (iii) ar (BDEF) = {1 \over 2} ar (ABC).
Sol.

remain
Given :
D, E and F are the mid- points of the sides BC, CA and AB respectively of \Delta ABC.

To prove : (i) BDEF is a || gm
(ii) ar (DEF) = {1 \over 4} ar (ABC)
(iii) ar (BDEF) = {1 \over 2} ar (ABC)
(i) We have in \Delta ABC,
EF || BC [By mid- point theorem, since E and F are the mid- points AC and AB respectively]
Therefore EF || BD... (1)
Also ED || AB [By mid- point theorem, since E and D are the mid- points AC and BC respectively]
Therefore ED || BF... (2)
From (1) and (2), BDEF is a || gm.
(ii) Similarly, FDCE and AFDE are || gms.

Therefore ar (FBD) = ar (DEF)   [Since FD is a diagonal of || gm BDEF]... (3)
ar (DEC) = ar (DEF)          [Since ED is a diagonal of || gm FDCE]... (4)
and , ar (AFE) = ar (DEF)       [Since FE is a diagonal of || gm AFDE... (5)]
From (3), (4) and (5)

ar (FBD) = ar (DEC) = ar (AFE) = ar (DEF) .... (6)
ar (ABC) = ar (AFE) + ar (FBD) + ar (DEC) + ar (DEF)
from equation (6) -
ar(ABC) = 4ar(DEF)

 \Rightarrow ar (DEF) = {1 \over 4} ar (ABC).
(iii) Also, ar (BDEF) = 2 ar (DEF)
 = 2 \times {1 \over 4}ar\left( {ABC} \right) = {1 \over 2}ar\left( {ABC} \right)


Q.6     In figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that :
           (i) ar (DOC) = ar (AOB)
           (ii) ar (DCB) = ar (ACB)
           (iii) DA || CB or ABCD is a parallelogram.
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Sol.

(i) Draw DN  \bot AC and BM  \bot AC.
In \Delta s DON and BOM
\angle DNO = \angle BMO [Each = 90º]
\angle DON = \angle BOM [Vert. opp \angle s]
OD = OB [Given]

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By AAS criterion of congruent.
\Delta DON\, \cong \,\Delta BOM ... (1)
In \Delta s DCN and BAM
\angle DNC = \angle BMA [Each = 90º]
DC = AB [Given ]
DN = BM [Since \Delta \,DON \cong \Delta BOM\,\, \Rightarrow \,DN = BM]
Therefore by RHS criterion of congruence,
\Delta \,DCN \cong \Delta BAM ... (2)
From (1) and (2), we get
ar (DON) + ar (DCN) = ar (BOM) + ar (BAM)
\, \Rightarrow ar (DOC) = ar (AOB)

(ii) Since ar (DOC) = ar (AOB)
Therefore ar (DOC) + ar (BOC) = ar (AOB) + ar (BOC)
\, \Rightarrow ar (DCB) = ar (ACB)

(iii) \Delta s DCB and ACB have equal areas and have the same base. So, these \Delta s lie between the same parallels.
\, \Rightarrow DA || CB i.e., ABCD is a|| gm


Q.7    D and E are points on sides AB and AC respectively of \Delta ABC such that ar(DBC) = ar(EBC). Prove that DE|| BC.
Sol.      Since \Delta s DBC and EBC are equal in area and have a same base BC. 

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Therefore altitude from D of \Delta s DBC = Altitude from E of \Delta s EBC.
 \Rightarrow \Delta s DBC and EBC are between the same parallels.
 \Rightarrow DE || BC


Q.8     XY is a line parallel to side BC of a triangle ABC. If BE|| AC and CF || AB meet XY at E and F respectively, show that ar (ABE) = ar (ACF).
Sol.

Since XY|| BC and BE || CY
Therefore BCYE is a || gm.
Since \Delta ABE and ||gm BCYE are on the same base BE and between the same parallel lines BE and AC.

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Therefore ar(ABE)  = {1 \over 2} ar(BCYE) ... (1)
Now, CF || AB and XY || BC
 \Rightarrow CF || AB and XF || BC
 \Rightarrow BCFX is a || gm
Since \Delta ACF and || gm BCFX are on the same base CF and between the same parallel AB and FC .
Therefore ar (ACF)  = {1 \over 2} ar (BCFX) ... (2)
But ||gm BCFX and || gm BCYE are on the same base BC and between the same parallels BC and EF.
Therefore ar (BCFX) = ar(BCYE) ... (3)
From (1) , (2) and (3) , we get
ar (\Delta ABE) = ar(\Delta ACF)


Q.9     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 figure). Show that ar(ABCD) = ar (PBQR).

22Sol.

Join AC and PQ. Since AC and PQ are diagonals of || gm ABCD and ||gm BPQR respectively.
Therefore ar(ABC) = {1 \over 2} ar (ABCD) ... (1)
and ar (PBQ) = {1 \over 2} ar (BPRQ) ... (2) 

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Now, \Delta s ACQ and AQP are on the same base AQ and between the same parallels AQ and CP.
Therefore ar(ACQ) = ar(AQP)
 \Rightarrow ar(ACQ) – ar(ABQ) = ar(AQP) – ar(ABQ)             [Subtracting ar(ABQ) from both sides]
 \Rightarrow ar(ABC) = ar(BPQ)
 \Rightarrow {1 \over 2} ar(ABCD) = {1 \over 2} ar(BPRQ)             [Using (1) and (2)]
 \Rightarrow ar(ABCD) = ar (BPRQ).


Q.10    Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar(AOD) = ar (BOC).
Sol.

Diagonals AC and BD of a trapezium ABCD with AB|| DC intersect each other at O.
Therefore \Delta s ABC and ABD are on the same base AB and between the same parallels AB and DC.

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Therefore ar(ABD) = ar(ABC)
 \Rightarrow ar(ABD) – ar(AOB) = ar(ABC) – ar (AOB)  [Subtracting ar (AOB) from both sides]
 \Rightarrow ar (AOD) = ar(BOC)


Q.11    In figure , ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
                 (i) ar (ACB) = ar (ACF)
                 (ii) ar (AEDF) = ar (ABCDE)
25Sol.

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

(ii) Adding ar(ACDE) on both sides we get
ar(ACF) + ar(ACDE) = ar(ACB) + ar(ACDE)
 \Rightarrow ar(AEDF) = ar(ABCDE)


Q.12    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 above proposal with the condition that he should be given 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.
Sol.       

Let ABCD be the quadrilateral plot. Produce BA to meet CD drawn parallel to CA at E. Join EC. 

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Then , \Delta s EAC and ADC lie on the same parallels DE and CA
Therefore ar(EAC) = ar (ADC)
Now ar(ABCD) = ar(ABC) + ar(ACD)
= ar(ABC) + ar(ADC)
= ar (ABC) + ar(EAC) = ar(EBC)
i.e., quad. ABCD = \Delta EBC
Which is the required explain to the suggested proposal.


Q.13   ABCD is a trapezium with AB|| DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
Sol.       ABCD is a trapezium with AB || DC and XY|| AC, is drawn. Join XC.

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ar(ACX) = ar(ACY) ... (1) [Since \Delta s ACX and ACY have same base AC and are between same parallels AC and XY]
But ar (ACX) = ar(ADX) [Since \Delta s ACX and ADX have same base AX and are between same parallels AB and DC]
From (1) and (2), we have
ar(ADX) = ar (ACY).


Q.14    In figure , AP || BQ|| CR . Prove that ar (AQC) = ar(PBR)

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Sol. 

From the figure, we have
ar (AQC) = ar (AQB) + ar (BQC) ... (1)

and ar (PBR) = ar (PBQ) + ar (QBR) ... (2)
But ar (AQB) = ar(PBQ)  [Since these \Delta s are on the same base BQ and between same parallel line AP and BQ]
Also, ar (BQC) = ar (QBR) [Since these \Delta s are on the same base BQ and between same parallel lines BQ and CR]
Using (3) and (4) in (1) and (2) , we get :
ar (AQC) = ar (PBR).


Q.15    Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.
Sol.        Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that

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ar (AOD) = ar (BOC). ... (1)   [Given]
Adding ar (ODC) on both sides , we get
ar(AOD) + ar (ODC) = ar (BOC) + ar (ODC)
 \Rightarrow ar(ADC) = ar (BDC)
 \Rightarrow {1 \over 2} \times DC \times AL = {1 \over 2} \times DC \times BM
 \Rightarrow AL = BM
 \Rightarrow AB || DC
Hence ABCD is a trapezium.


Q.16    In figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
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Sol.

From the figure,
ar (BDP) = ar (ARC)             [Given]
and ar (DPC) = ar (DRC)       [Given]
On subtracting, we get
ar (BDP) – ar (DPC) = ar (ARC) – ar (DRC)
 \Rightarrow ar (BDC) = ar (ADC)
 \Rightarrow DC || AB    [ar(BDC) and ar(ADC) have equal areas and have the same base. So, these \Delta s lie between the same parallels.]
Hence , ABCD is a trapezium.
ar (DRC) = ar (DPC) [Given]
On subtracting ar (DLC) from both sides, we get
ar (DRC) – ar (DLC) = ar (DPC) – ar (DLC)
 \Rightarrow ar (DLR) = ar (CLP)
ON adding ar (RLP) to both sides, we get
ar (DLR) + ar(RLP) = ar(CLP) + ar(RLP)
 \Rightarrow ar (DRP) = ar (CRP)
 \Rightarrow RP || DC
Hence, DCPR is a trapezium.



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