# 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).

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)

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.

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).

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.

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.
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]

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.

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.

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).

Sol.

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)

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.

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)
Sol.

(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.

Then , $\Delta$ s EAC and ADC lie on the same parallels DE and CA
Now ar(ABCD) = ar(ABC) + ar(ACD)
= 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.

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

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

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

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.

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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