# 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 ABC. Show that ar (ABE) = ar (ACE).**

Given : AD is a medium of ABC and E is any point on AD.

To prove : ar (ABE) = ar (ACE)

Proof : Since AD is the median of ABC

Therefore ar (ABD) = ar (ACD) ... (1)

Also , ED is the median of EBC

Therefore ar (BED) = ar (CED) ... (2)

Subtracting (2) from (1) , we get

ar (ABD) – ar (BED) = ar (ACD) – ar (CED)

ar (ABE) = ar (ACE).

**Q.2 In a triangle ABC, E is the mid- point of median AD. Show that ar (BED) = ar (ABC).
**

**Sol.**Given : A ABC, E is the mid- point of the median AD.

To prove : ar (BED) = ar (ABC)

Proof : Since AD is a median of ABC and median divides a triangle into two triangles of equal area.

Therefore ar (ABD) = ar (ADC)

ar (ABD) = ar (ABC) ... (1)

In ABD, BE is the median

Therefore ar(BED) = ar (BAE) ... (2)

[ ]

ar (BED) = ar (ABC) [Using (1)]

ar (BED) = 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 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) = × AO × BL

ar (OBC) = × 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 ACD, we have

OC = OD [Given]

Therefore AO is the median.

Therefore ar (AOC) = ar (AOD) [Since median divides a in two s of equal area]

Similarly, in BCD, BO is the median

Therefore ar (BOC) = ar (BOD)

Adding (1) and (2), we get

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

ar (ABC) = ar (ABD)

**Q.5 D, E and F are respectively the mid- points of the side BC, CA and AB of a ABC. Show that **

**(i) BDEF is a parallelogram**

**(ii) ar (DEF) = ar (ABC) **

**(iii) ar (BDEF) = ar (ABC). **

**Sol.**

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

To prove : (i) BDEF is a || gm

(ii) ar (DEF) = ar (ABC)

(iii) ar (BDEF) = ar (ABC)

**(i)** We have in 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)

ar (DEF) = ar (ABC).

** (iii)** Also, ar (BDEF) = 2 ar (DEF)

**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 AC and BM AC.

In s DON and BOM

[Each = 90º]

[Vert. opp ]

OD = OB [Given]

By AAS criterion of congruent.

... (1)

In DCN and BAM

[Each = 90º]

DC = AB [Given ]

DN = BM [Since ]

Therefore by RHS criterion of congruence,

... (2)

From (1) and (2), we get

ar (DON) + ar (DCN) = ar (BOM) + ar (BAM)

ar (DOC) = ar (AOB)

** (ii)** Since ar (DOC) = ar (AOB)

Therefore ar (DOC) + ar (BOC) = ar (AOB) + ar (BOC)

ar (DCB) = ar (ACB)

** (iii)** DCB and ACB have equal areas and have the same base. So, these lie between the same parallels.

DA || CB i.e., ABCD is a|| gm

**Q.7 D and E are points on sides AB and AC respectively of ABC such that ar(DBC) = ar(EBC). Prove that DE|| BC.**

**Sol.**** **Since DBC and EBC are equal in area and have a same base BC.

Therefore altitude from D of DBC = Altitude from E of EBC.

DBC and EBC are between the same parallels.

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 ABE and ||gm BCYE are on the same base BE and between the same parallel lines BE and AC.

Therefore ar(ABE) ar(BCYE) ... (1)

Now, CF || AB and XY || BC

CF || AB and XF || BC

BCFX is a || gm

Since ACF and || gm BCFX are on the same base CF and between the same parallel AB and FC .

Therefore ar (ACF) 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 ( ABE) = ar( 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). **

Join AC and PQ. Since AC and PQ are diagonals of || gm ABCD and ||gm BPQR respectively.

Therefore ar(ABC) = ar (ABCD) ... (1)

and ar (PBQ) = ar (BPRQ) ... (2)

Now, s ACQ and AQP are on the same base AQ and between the same parallels AQ and CP.

Therefore ar(ACQ) = ar(AQP)

ar(ACQ) – ar(ABQ) = ar(AQP) – ar(ABQ) [Subtracting ar(ABQ) from both sides]

ar(ABC) = ar(BPQ)

ar(ABCD) = ar(BPRQ) [Using (1) and (2)]

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 s ABC and ABD are on the same base AB and between the same parallels AB and DC.

Therefore ar(ABD) = ar(ABC)

ar(ABD) – ar(AOB) = ar(ABC) – ar (AOB) [Subtracting ar (AOB) from both sides]

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 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)

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 , 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 = 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 s ACX and ACY have same base AC and are between same parallels AC and XY]

But ar (ACX) = ar(ADX) [Since 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)**

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 s are on the same base BQ and between same parallel line AP and BQ]

Also, ar (BQC) = ar (QBR) [Since these 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.
**

*Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that*

**Sol.**

ar (AOD) = ar (BOC). ... (1) [Given]

Adding ar (ODC) on both sides , we get

ar(AOD) + ar (ODC) = ar (BOC) + ar (ODC)

ar(ADC) = ar (BDC)

AL = BM

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)

ar (BDC) = ar (ADC)

DC || AB [ar(BDC) and ar(ADC) have equal areas and have the same base. So, these 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)

ar (DLR) = ar (CLP)

ON adding ar (RLP) to both sides, we get

ar (DLR) + ar(RLP) = ar(CLP) + ar(RLP)

ar (DRP) = ar (CRP)

RP || DC

Hence, DCPR is a trapezium.

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