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Exercise 9.3

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

Sol. Construction: Draw AL ⊥ BC, which meets BC at L.
Proof :  $ar(ABD) = {1 \over 2} \times BD \times AL$                      …..(i)
$ar(ACD) = {1 \over 2} \times DC \times AL$                                   ……(ii)

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

ar(ABD) = ar(ACD)                                                                          ….(iv)

Similarly, in triangle BEC, ED is the median, proceeding as above,
we have

∴           ar(BED) = ar(CED) …                                                                ……(v)
From (iv) and (v), we get

ar(ABD) – ar(BED) =  ar(ACD) – ar(CED)
∴            ar(ABE) = ar(ACE).

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

$ar(BED) = {1 \over 2}ar(ABC)$

Sol. AD is median of triangle ABC.

[Median of a triangle divides the triangle into two parts of equal area.]

$\Rightarrow ar(ABD) = {1 \over 2}ar(ABC)$

Also, BE is median of triangle ABD.

$\Rightarrow ar(BED) = {1 \over 2}ar(ABC)$           [Reason same Es above]

$\Rightarrow ar(BED) = {1 \over 2} \times {1 \over 2}ar(ABC)$

$\Rightarrow ar(BED) = {1 \over 4}ar(ABC)$

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

Sol. Construction: Draw DL perpendicular to AC.
Proof: We know that diagonals of a parallelogram bisect each other.

i.e., OA = OC and OB = OD                                   …(i)

$ar(AOD) = {1 \over 2} \times AO \times DL$                                         ….(ii)

ar(DOC) =  ${1 \over 2}$   × OC × DL
From (i), (ii), (iii), we get
ar(AOD) = ar(DOC)                                                                                             …(iv)
Similarly, by drawing perpendiculars from C and B on BD and AC respectively, we can show that
ar(DOC) = ar(COB)                                                                                               …(v)
and ar(BOC) = ar(AOB)                                                                                       …(vi)
From (iv), (v) and (vi), we get
ar(AOB) = ar(BOC) = ar(COD) = ar(AOD).

4. In figure, ABC and ABD are two triangles on the at 0, show that same base AB. If line-segment CD is bisected by AB at O , show that
ar(ABC) = ar(ABD).

Sol. Construction: Draw perpendiculars CL and DM on AB.
Proof: Consider triangles CLO and DMO.

OC = OD                                                                                  …(i) [Given]

∠CLO = ∠DMO                                                                        …(ii) [90° each]

∠COL = ∠DOM                                                                       …(iii) [Vertically opposite angles]

From (i), (ii) and (iii),

ΔCLO ≅  ΔDMO                                                                       [ASS]

∴       CL = DM                                                                        ….(iv)  [CPCT]

Now, ar(ABC) =  ${1 \over 2}$ × AB × CL                  ……(v)

and    ar (ADB) =${1 \over 2}$ × AB ×DM             …….(vi)

∴    ar(ABC)  = ar(ABD)                                                    [From (iv), (v),(vi)]

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

Sol.
(i) Proof: Since E and F are respectively the mid-points of AC and AB of ΔABC.

∴   $EF\parallel BC$     and EF =${1 \over 2}$BC                                 [Mid-point theorem]
Also,    BD = ${1 \over 2}$ BC                         [ D is mid-point of BC]

∴     $EF\parallel BD$  and EF = BD.

⇒ BDEF is a parallelogram.

[In a quadrilateral, if a pair of opposite sides is equal and parallel, then it is a parallelogram.]
(ii) DF is diagonal.
∴     ar(BFD) = ar(DEF)                                                                 ..(i)
[Diagonal of a parallelogram divides it into two triangles of equal area.]
Similarly, we can show that
ar(DEF) = ar(CDE) = ar(AFE)                                                     ….(ii)
From equations (i) and (ii),
$ar(ABC) = 4ar(DEF) \Rightarrow ar(DEF) = {1 \over 4}ar(ABC)$
(iii)  ar(BDEF) = ar(BDF) + ar(DEF) = 2ar(DEF)       [From (i)]

$= 2 \times {1 \over 4}ar(ABC) = {1 \over 2}ar(ABC)$                  [Using result of part (ii)]

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)  $DA\parallel CB$  or ABCD is a parallelogram.
[Hint: From D and B, draw perpendiculars to AC.]

Sol. (i) Construction: Draw DL and BM perpendiculars to AC.
Consider triangles DOL and BOM,

OB = OD                                                                            [Given)

∠DLO = ∠BMO                                                               [Construction]

∠DOL = ∠BUM                                                               [Vertically opposite angles]

∴   ΔDLO  ≅  ΔBMO                                                        [RAS]

∴   DL =  BM                                                          …(i)[CPCT]

Consider triangles CLD and AMB,
DL = BM                                                                           [Proved above] [From (i)]

DC = AB                                                                            [Given]

∠DLC = ∠BMA                                                                [90° each]

∴                 ΔDCL  ≅   ΔBAM                                                            [RHS]

∴               ∠DCL = ∠BAM                                             …(ii) [CPCT]

Consider triangles COD and AOB,

OD = OB                                                                          [Given]

∠COD = ∠AOB                                                              [Vertically opposite angles]

∠DCO = ∠BAO                                                              [From (ii)]

∴    ΔCOD ≅  ΔAOB                                                      [AAS]

⇒    ar(COD) =  ar(AOB)

(ii) From result of part (i),

ar(AOB) = ar(COD)

⇒        ar(AOB) + ar(BOC) = ar(COD) + ar(BOC)

⇒        ar(ABC) = ar(DCB)
(iii) AB = CD and ∠DCO = ∠BAO                                                          [Given and from result (ii)]

⇒                  AB = CD and $AB\parallel CD$.

Therefore, ABCD is a parallelogram.
[In a quadrilateral, if a pair of opposite sides is equal and parallel, then it is a parallelogram.]
∴   $AD\parallel BC$

7.  D and E are points on sides AB and AC respectively of ΔABC such that ar(DBC) = ar(EBC). Prove that $DE\parallel BC$ .

Sol.
Construction: Draw DL and EM
perpendicular to BC.

Proof : ar(DBC) = ar(EBC)              [Given]

$\Rightarrow {1 \over 2} \times BC \times DL = {1 \over 2} \times BC \times EM$

⇒    DL = EM

As distance between two given lines BC and DE is same at different points.

∴                         $DE\parallel BC$

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

Sol. Triangle ABE and parallelogram BCQE are on the same base BE and between the same parallels BE and AC.

$ar(ABE) = {1 \over 2}ar(BCQE)$                                                        …..(i)

Similarly, we can show that

ar(ACF) = ${1 \over 2}$ar(BCFP)                                                        …..(ii)

and ar(BCQE) = ar(BCFP)                                                                     …..(iii)

From (i), (ii), (iii), we get
ar(ABE) = ar(ACF).

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).

[Hint: Join AC and PQ. Now compare ar(ACQ) and ar(APQ).]

Sol. Construction: Draw diagonals AC and PQ.
Proof: $AQ\parallel CP$

Triangles ACQ and APQ are on the same base AQ and between the same parallels AQ and PC.

∴         ar(ACQ) = ar(APQ)

⇒   ar(ACB) + ar(ABQ) = ar(ABQ) + ar(PBQ)

⇒  ar(ACB) = ar(PBQ)

ar(ACB) =${1 \over 2}$ ar(ABCD)

[Diagonals of a parallelogram divide the parallelogram into two parts of equal area.]

and ar(PBQ) = ${1 \over 2}$ ar(PBQR)                                             …(iii) [Reason same as above]

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

${1 \over 2}$ar(ABCD) =${1 \over 2}$ar(PBQR)

⇒   ar(ABCD) = ar(PBQR).

10. Diagonals AC and BD of a trapezium ABCD with $AB\parallel DC$ intersect each other at 0. Prove that ar(AOD) = ar(BOC).

Sol. Triangles DAB and CAB are on the same base AB and between the same parallels AB and DC.

∴    ar(DAB) =  ar(CAB)
⇒   ar(DOA) + ar(OAB) = ar(OAB) + ar(COB)
⇒   ar(AOD) = ar(BOC).

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. Triangles ABC and AFC are on the same base AC and between the same parallels AC and BF.
(i)  ∴               ar(ACB) = ar(ACF).
(ii) Adding ar(AEDC) to both sides, we get
ar(ACB) + ar(AEDC) = ar(ACF) + ar(AEDC)
⇒         ar(ABCDE) = ar(AFCDE)
or        ar(ABCDE) = ar(AEDF)

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.

Sol. Let the villager’s plot be quadrilateral ABCD. Let the corner ABM be taken over to construct the Health Centre in a way that AM produced meets the side DC produced at E and is such that diagonal AC is parallel to BE as shown in the adjoining figure.

Now, ΔACE and ΔACB are on the same base AC and between the same parallels AC and BE.
Therefore, ar(ACE) = ar(ACB)
⇒  ar(ACM) + ar(CEM) = ar(ACM) + ar(ABM)
ar(CEM) = ar(ABM)                                                                      …….(i)
Thus, the land CEM is provided to the villager according to the condition.
Equation (i) shows that the area of the Health Centre is equal to the area of the land provided to the villager.

13. ABCD is a trapezium with $AB\parallel DC$. A line parallel to AC intersects AB at X and BC at Y Prove that ar(ADX) = ar(ACY).
[Hint: Join CX.]

Sol. Join CX. Triangles ADX and ACX are on the same base AX and between the same parallels AB and CD.

Triangles ACX and ACY are on the same base AC and between the same parallels AC and XY.
∴    ar(ACX) = ar(ACY)
From (i) and (ii), we get

14. In figure, $AP\parallel BQ\parallel CR$. Prove that ar(AQC) = ar(PBR).

Sol. Triangles ABQ and PBQ are in the same base BQ and between the same parallels AP and BQ.
∴         ar(ABQ) = ar(PBQ)

⇒  ar(AXB) + ar(BXQ) = ar(BXQ) + ar(PXQ)
⇒  ar(AXB) = ar(PXQ)                                               …..(i)
Similarly, we can show that
ar(BYC) = ar(QYR)                                                   ……(ii)
Adding (i) and (ii), we get
ar(AXB) + ar(BYC) = ar(PXQ) + ar(QYR)
Adding ar(BXQY) to both sides, we get
ar(AXB) + ar(BYC) + ar(BXQY)
=   ar(PXQ) + ar(QYR) + ar(BXQY)
⇒   ar(AQC) = ar(PBR).
Alternative Method:
ΔBAQ and ΔBPQ are on the same base BQ and between same parallels AP and BQ.

∴   ar(BAQ) = ar(BPQ)                                  …(i)
Also, ABCQ and ABRQ are on the same base BQ and between the same parallels CR and BQ.
∴          ar(BCQ) = ar(BRQ)                            ……(ii)
Adding equations (i) and (ii), we have
ar(BAQ) + ar(BCQ) = ar(BPQ) + ar(BRQ)
⇒               ar(AQC) = ar(PBR).

15. Diagonals AC and BD of a quadrilateral ABCD intersect at 0 in such a way that ar(AOD) = ar(BOC). Prove that ABCD is a trapezium.

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

Adding ar(AOB) to both sides, we get
ar(AOD) + ar(AOB) = ar(AOB) + ar(BOC)
Triangles ADB and ACB are on the same base AB and as their areas are equal, so these must be in the same parallels.
∴     $AB\parallel CD$
Hence, ABCD is a trapezium.

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

Sol. ar(DRC) = ar(DPC)                                  [Given]

As triangles DRC and DPC are on the same base DC and their areas are same, hence these must be in the same parallels.

∴       $DC\parallel RP$.         Hence, DCPR is trapezium.

Now,       ar(BDP) = ar(ARC)                                           [Given]

Also, ar(DPC) = ar(DRC)                                                 [Given]

∴   ar(BDP) – ar(DPC) = ar(ARC) – ar(DRC)

∴   $AB\parallel DC$. Hence, ABCD is a trapezium.