Question - 4 : -
In Fig. 9.24, ABC and ABD are two triangles on the same base AB. Ifline- segment CD is bisected by AB at O, show that: ar(ABC) = ar(ABD).

Answer - 4 : -

In ΔABC, AO is themedian. (CD is bisected by AB at O)

∴ar(AOC) = ar(AOD) —(i)

also,

ΔBCD, BO is themedian. (CD is bisected by AB at O)

∴ar(BOC) = ar(BOD) —(ii)

Adding (i) and (ii),

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 ABof a ΔABC.
Show that
(i) BDEF is a parallelogram.
(ii) ar(DEF) = ¼ ar(ABC)
(iii) ar (BDEF) = ½ ar(ABC)

Answer - 5 : -

(i) In ΔABC,

EF || BC and EF = ½ BC(by mid point theorem)

also,

BD = ½ BC (D is themid point)

So, BD = EF

also,

BF and DE are paralleland equal to each other.

∴, the pair oppositesides are equal in length and parallel to each other.

∴ BDEF is aparallelogram.

(ii) Proceeding fromthe result of (i),

BDEF, DCEF, AFDE are parallelograms.

Diagonal of aparallelogram divides it into two triangles of equal area.

∴ar(ΔBFD) = ar(ΔDEF)(For parallelogram BDEF) — (i)

also,

ar(ΔAFE) = ar(ΔDEF)(For parallelogram DCEF) — (ii)

ar(ΔCDE) = ar(ΔDEF)(For parallelogram AFDE) — (iii)

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

ar(ΔBFD) = ar(ΔAFE) =ar(ΔCDE) = ar(ΔDEF)

⇒ ar(ΔBFD)+ar(ΔAFE) +ar(ΔCDE) +ar(ΔDEF) = ar(ΔABC)

⇒ 4 ar(ΔDEF) =ar(ΔABC)

⇒ ar(DEF) = ¼ar(ABC)

(iii) Area(parallelogram BDEF) = ar(ΔDEF) +ar(ΔBDE)

⇒ ar(parallelogramBDEF) = ar(ΔDEF) +ar(ΔDEF)

⇒ ar(parallelogramBDEF) = 2× ar(ΔDEF)

⇒ ar(parallelogramBDEF) = 2× ¼ ar(ΔABC)

⇒ ar(parallelogramBDEF) = ½ ar(ΔABC)

Question - 6 : -
In Fig. 9.25, diagonals AC and BD of quadrilateral ABCD intersect at Osuch 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.
[Hint : From D and B, draw perpendiculars to AC.]

Answer - 6 : -

Given,

OB = OD and AB = CD

Construction,

DE ⊥ AC and BF ⊥ AC are drawn.

Proof:

(i) In ΔDOE and ΔBOF,

∠DEO = ∠BFO (Perpendiculars)

∠DOE = ∠BOF (Vertically opposite angles)

OD = OB (Given)

∴, ΔDOE ≅ ΔBOF by AAS congruence condition.

∴, DE = BF (By CPCT) —(i)

also, ar(ΔDOE) =ar(ΔBOF) (Congruent triangles) — (ii)

Now,

In ΔDEC and ΔBFA,

∠DEC = ∠BFA (Perpendiculars)

CD = AB (Given)

DE = BF (From i)

∴, ΔDEC ≅ ΔBFA by RHS congruence condition.

∴, ar(ΔDEC) = ar(ΔBFA)(Congruent triangles) — (iii)

Adding (ii) and (iii),

ar(ΔDOE) + ar(ΔDEC) =ar(ΔBOF) + ar(ΔBFA)

⇒ ar (DOC) = ar(AOB)

(ii) ar(ΔDOC) =ar(ΔAOB)

Adding ar(ΔOCB) in LHSand RHS, we get,

⇒ar(ΔDOC) + ar(ΔOCB) =ar(ΔAOB) + ar(ΔOCB)

⇒ ar(ΔDCB) =ar(ΔACB)

(iii) When twotriangles have same base and equal areas, the triangles will be in between thesame parallel lines

ar(ΔDCB) = ar(ΔACB)

DA || BC — (iv)

For quadrilateral ABCD,one pair of opposite sides are equal (AB = CD) and other pair of opposite sidesare parallel.

∴, ABCD isparallelogram.

Question - 7 : -
D and E are points onsides AB and AC respectively of ΔABC such that
ar(DBC) = ar (EBC). Prove that DE || BC.

Answer - 7 : -

Since ΔBCE and ΔBCD arelying on a common base BC and also have equal areas, ΔBCE and ΔBCD will liebetween the same parallel lines.

∴ DE || BC

Question - 8 : -
XY is a line parallel toside BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and Erespectively, show that
ar(ABE) = ar (ACF)

Answer - 8 : -

It is given that

XY || BC ⇒EY || BC

BE || AC ⇒BE || CY

Therefore, EBCY is aparallelogram.

It is given that

XY || BC ⇒XF || BC

FC || AB ⇒FC || XB

Therefore, BCFX is aparallelogram.

Parallelograms EBCY andBCFX are on the same base BC and between the same parallels BC and EF.

∴Area (EBCY) = Area (BCFX) … (1)

Consider parallelogramEBCY and ΔAEB

These lie on the same baseBE and are between the same parallels BE and AC.

∴Area (ΔABE) = Area (EBCY) … (2)

Also, parallelogram BCFX andΔACF are on the same base CF and between the same parallels CF and AB.

∴Area (ΔACF) = Area (BCFX) … (3)

From equations (1), (2),and (3), we obtain

Area(ΔABE) = Area (ΔACF)

Question - 9 : -
The side AB of aparallelogram ABCD is produced to any point P. A line through A and parallel toCP meets CB produced at Q and then parallelogram PBQR is completed (see thefollowing figure). Show that
ar (ABCD) = ar (PBQR).
[Hint:Join AC and PQ. Now compare area (ACQ) and area (APQ)]

Answer - 9 : -

Let us join AC and PQ.

ΔACQ and ΔAQP are on thesame base AQ and between the same parallels AQ and CP.

∴Area (ΔACQ) = Area (ΔAPQ)

⇒Area (ΔACQ) − Area (ΔABQ) = Area (ΔAPQ) − Area (ΔABQ)

⇒Area (ΔABC) = Area (ΔQBP) … (1)

Since AC and PQ arediagonals of parallelograms ABCD and PBQR respectively,

∴Area (ΔABC) = Area (ABCD) … (2)

Area (ΔQBP) = Area (PBQR) … (3)

From equations (1), (2),and (3), we obtain

Area (ABCD) = Area (PBQR)

Area(ABCD) = Area (PBQR)

Question - 10 : -
DiagonalsAC and BD of a trapezium ABCD with AB || DC intersect each other at O. Provethat ar (AOD) = ar (BOC).

Answer - 10 : -

It can be observed thatΔDAC and ΔDBC lie on the same base DC and between the same parallels AB and CD.

∴Area (ΔDAC) = Area (ΔDBC)

⇒Area (ΔDAC) − Area (ΔDOC) = Area (ΔDBC) − Area (ΔDOC)

⇒ Area (ΔAOD) = Area (ΔBOC)

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