Bihar Board Class 9th Maths Solutions Chapter 9 Areas of Parallelograms and Triangles Ex 9.3 Textbook Questions and Answers.

## BSEB Bihar Board Class 9th Maths Solutions Chapter 9 Areas of Parallelograms and Triangles Ex 9.3

Question 1.

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

Solution:

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

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

Proof:

∵ AD is the median of ∆ ABC

∴ ar(ABD) = ar(ACD) … (1)

Also, ED is the median of ∆ EBC

∴ ar(BED) = ar(CED) … (2)

Subtracting (2) from (1), we get

ar(ABD) – ar(BED) = ar(ACD) – ar(CED)

⇒ ar(ABE) = ar(ACE).

Question 2.

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

Solution:

Given : ∆ ABC, E is the mid-point of the median AD.

To prove : ar(BED) = \(\frac { 1 }{ 4 }\) ar(ABC).

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

∴ ar(ABD) = ar(ADC)

⇒ ar(ABD) = \(\frac { 1 }{ 2 }\) ar(ABC) … (1)

In ∆ ABD, BE is the median

∴ ar(BED) = \(\frac { 1 }{ 2 }\) ar(BAE)

⇒ ar(BED) = \(\frac { 1 }{ 2 }\) ar(ABD) … (2)

⇒ ar(BED) = \(\frac { 1 }{ 2 }\) x \(\frac { 1 }{ 2 }\)ar(ABC) [Using (1)]

⇒ ar(BED) = \(\frac { 1 }{ 2 }\) ar(ABC).

Question 3.

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

Solution:

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.

∴ AO = OC and BO = OD

Now, ar(AOB) = \(\frac { 1 }{ 2 }\) x AO x BL

ar(OBC) = \(\frac { 1 }{ 2 }\) x OC x BL

But AO = OC

∴ 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).

Question 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).

Solution:

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]

∴ AO is the median.

ar(AOC) = ar(AOD) …(1)

[∵ Median divides a ∆ in two As of equal area]

Similarly, in ∆ BCD, BO is the median

∴ ar(BOC) = ar(BOD) … (2)

Adding (1) and (2), we get

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

⇒ ar(ABC) = ar(ABD).

Question 5.

D, E and F are respectively the mid-points of the sides BC, CA and AB of a ∆ ABC. Show that

(i) BDEF is a parallelogram

(ii) ar(DEF) = \(\frac { 1 }{ 4 }\) ar(ABC)

(iii) ar(BDEF) = \(\frac { 1 }{ 4 }\) ar(ABC).

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) = \(\frac { 1 }{ 4 }\) ar(ABC)

(iii) ar(BDEF) = \(\frac { 1 }{ 2 }\) ar(ABC)

Solution:

(i) We have in ∆ ABC,

EF || BC [By mid-point theorem, since E and F are the mid-points of AC and AB respectively]

∴ EF || BD … (1)

Also, ED || AB [By mid-point theorem, since E and D are the mid-points of AC and BC respectively]

∴ ED || AF

From (1) and (2), BDEF is a ||gm.

(ii) Similarly, FDCE and AFDE are ||gms.

∴ ar(FBD) = ar(DEF) [∵ FD is a diagonal of ||gm BDEF]

ar(DEC) = ar(DEF) [∵ ED is a diagonal of ||gm FDCE]

and, ar(AFE) = ar(DEF) [∵ FE is a diagonal of ||gm AFDE]

∴ ar(FBD) = ar(DEC) = ar(AFE) = ar(DEF) …(3)

⇒ ar(DEF) = \(\frac { 1 }{ 4 }\) ar(ABC).

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

= 2 x \(\frac { 1 }{ 4 }\) ar(ABC) = \(\frac { 1 }{ 2 }\)ar(ABC).

Question 6.

In figure, diagonals AC and BD of quadrilateral ABCD interest at O such that OB = OD. If AB = CD, then show that:

(i) ar(DOC) = ar(AOB)

(ii) ar(DCB) = ar(ACB) C

(iii) DA || CB or ABCD is a parallelogram.

Solution:

(i) Draw DN ⊥ AC and BM ⊥ AC.

In ∆s DON and BOM

∠DNO = ∠BMO [Each = 90°)

∠DON = ∠BOM [Vert. opp. ∠s]

OD = OB [Given]

By AAS criterion of congruent Q

∆ DON ≅ ∆ BOM … (1)

In ∆s DCN and PAM

∠DNC = ∠BMA [Each = 90°]

DC = AB [Given]

DN = BM [∵ A DON ≅ A BOM ⇒ DN = BM]

∴ By RHS criterion of congruence,

∆ DCN ≅ ∆ BAM

From (1) and (2), we get

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

⇒ ar(DOC) = ar(AOB).

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

∴ ar(DOC) + ar{BOC) = ar(AOB) + ar(BOC)

⇒ ar(DCB) = ar(ACB).

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

⇒ DA || CB i.e., ABCD is a ||gm.

Question 7.

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

Solution:

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

∴ Altitude from D of ∆ DBC = Altitude from E of ∆ EBC.

⇒ As DBC and EBC are between the same parallels.

⇒ DE || BC.

Question 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).

Solution:

Since XY || BC and BE || CY.

∴ BCYE is a ||gm.

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

∴ ar(ABE) = \(\frac { 1 }{ 2 }\)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

∴ ar(ACF) = \(\frac { 1 }{ 2 }\)ar(BCFX) …(2)

But ||gm BCFX and ||gm BCYE are on the same base BC and between the same parallels BC and EF.

∴ ar(BCFX) = ar(BCYE) … (3)

From (1), (2) and (3), we got

∴ ar(∆ ABE) = ar(∆ ACF).

Question 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).

Solution:

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

∴ ar(ABC) = \(\frac { 1 }{ 2 }\) ar(ABCD)… (1)

and, ar(PBQ) = \(\frac { 1 }{ 2 }\) ar(BPRQ) … (2)

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

∴ ar(ACQ) = ar(AQP)

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

⇒ ar(ABC) = ar(BPQ)

⇒ \(\frac { 1 }{ 2 }\) ar(ABCD) = \(\frac { 1 }{ 2 }\) ar(BPRQ) [Using (1) and (2)]

⇒ ar(ABCD) = ar(BPRQ).

Question 10.

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

Solution:

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O.

∴ ∆s ABC and ABD are on the same base and between the same parallels. A

∴ ar(ABD) = ar(ABC)

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

⇒ ar(AOD) = ar(BOC).

Question 11.

In figure, ABCDE is A B 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).

Solution:

(i) Since ∆s ACB and ACF are on the same base AC and between the same parallels AC and BF.

∴ 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).

Question 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 the 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.

Solution:

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

Then, ∆s EAC and ADC lie on the same parallels DE and CA

∴ 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.

Question 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).

Solution:

ABCD is a trapezium with AB || DC and XY || AC, is drawn. Join XC.

ar(ACX) = ar(ACY) … (1) [∵ As ACX and ACY have same base AC and are between same parallels AC and XY]

But ar(ACX) = ar(ADX) … (2) [∵ As 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).

Question 14.

In figure, AP || BQ || CR. Prove that

ar(AQC) = ar(PBR).

Solution:

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) …(3)

[∵ These As are on the same base BQ and between same parallel lines AP and BQ]

Also, ar(BQC) = ar(QBR) …(4)

[∵ These As 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).

Question 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.

Solution:

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

ar(AOD) = ar(BOC) … (1)

Adding ar (ODC) on both sides, we get

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

⇒ ar(ADC) = ar(BDC)

⇒ \(\frac { 1 }{ 2 }\) x DC x AL = \(\frac { 1 }{ 2 }\) x DC x BM

⇒ AL = BM

⇒ AB || DC

Hence, ABCD is a trapezium.

Question 16.

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

Solution:

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

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.