Chapter 9 Areas of Parallelograms and Triangles Ex- 9.1 |
Chapter 9 Areas of Parallelograms and Triangles Ex- 9.2 |
Chapter 9 Areas of Parallelograms and Triangles Ex- 9.4 |

In Fig.9.23, E is any point on median AD of a ΔABC. Show that ar (ABE)= ar(ACE).

**Answer
1** :

Given,

AD is median of ΔABC. ∴, it will divide ΔABC into two triangles ofequal area.

∴ar(ABD) = ar(ACD) —(i)

also,

ED is the median ofΔABC.

∴ar(EBD) = ar(ECD) —(ii)

Subtracting (ii) from(i),

ar(ABD) – ar(EBD) =ar(ACD) – ar(ECD)

⇒ar(ABE) = ar(ACE)

In a triangle ABC, E is the mid-point of median AD. Show that ar(BED)= ¼ ar(ABC).

**Answer
2** :

ar(BED) = (1/2)×BD×DE

Since, E is themid-point of AD,

AE = DE

Since, AD is themedian on side BC of triangle ABC,

BD = DC,

DE = (1/2) AD — (i)

BD = (1/2)BC — (ii)

From (i) and (ii), weget,

ar(BED) =(1/2)×(1/2)BC × (1/2)AD

⇒ ar(BED) =(1/2)×(1/2)ar(ABC)

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

**Answer
3** :

O is the mid point ofAC and BD. (diagonals of bisect each other)

In ΔABC, BO is themedian.

∴ar(AOB) = ar(BOC) —(i)

also,

In ΔBCD, CO is themedian.

∴ar(BOC) = ar(COD) —(ii)

In ΔACD, OD is themedian.

∴ar(AOD) = ar(COD) —(iii)

In ΔABD, AO is themedian.

∴ar(AOD) = ar(AOB) —(iv)

From equations (i),(ii), (iii) and (iv), we get,

ar(BOC) = ar(COD) =ar(AOD) = ar(AOB)

Hence, we get, thediagonals of a parallelogram divide it into four triangles of equal area.

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)

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)

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.

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

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)

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).

**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)

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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