Areas of Parallelograms and Triangles (Exercise 9.3)

Areas of Parallelograms and Triangles


Chapter 9


Exercise 9.3


EX 9.3 QUESTION 1.


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

Solution:

Given: In ΔABC, AD is median.

∴ar(ABD) = ar(ACD) — (i)  [ A median of a Δ divides it into two triangles of equal area.]

also, In ΔABC, ED is the median  [Given]

∴ar(EBD) = ar(ECD) — (ii)

Subtracting (ii) from (i),

ar(ABD) – ar(EBD) = ar(ACD) – ar(ECD)

⇒ar(ABE) = ar(ACE)


EX 9.3 QUESTION 2.


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

Solution:

Given: In ΔABC, AD is median.

ar(ABD) = ar(ACD)

ar(ABD) =  ½ ar(ABC)  [ A median of a Δ divides it into two triangles of equal area.]

Salso, In ΔABD, BE is the median.  [E is the midpoint of AD]

ar(BED) =ar(ABE)

ar(BED) = ½ ar(ABD)

ar(BED) = ½[½ ar(ABC)]   [ar(ABD) = ½ ar(ABC) ]

⇒ ar(BED) = ¼ ar(ABC)


EX 9.3 QUESTION 3.


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

Solution:

Diagonals of a parallelogram bisect each other.

∴ AO = OC and DO = OB

O is the midpoint of AC and BD.

In ΔABC, BO is the median.

∴ar(AOB) = ar(BOC) — (i)

Now, In ΔBCD, CO is the median.

∴ar(BOC) = ar(COD) — (ii)

In ΔACD, OD is the median.

∴ar(AOD) = ar(COD) — (iii)

In ΔABD, AO is the median.

∴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, the diagonals of a parallelogram divide it into four triangles of equal area.


EX 9.3 QUESTION 4.


In Fig. 9.24, 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:

In ΔABC, AO is the median.  [CO= OD]

∴ar(AOC) = ar(AOD) — (i)

also,  ΔBCD, BO is the median. [CO= OD]

∴ar(BOC) = ar(BOD) — (ii)

Adding (i) and (ii),

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

⇒ar(ABC) = ar(ABD)


EX 9.3 QUESTION 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) = ¼ ar(ABC)

(iii) ar (BDEF) = ½ ar(ABC)

Solution:

(i) In ΔABC, E and D are the midpoints of BC and CA

FE || BC and EF = ½ BC [midpoint theorem]

⇒ FE || BC  and FE || DC   [D is a midpoint of BC]

⇒ BDEF is a parallelogram.

(ii) BDEF is a parallelogram. [Proved]

BDEF, DCEF, AFDE are parallelograms.

[Diagonal of a parallelogram divides it into two triangles of equal area.]

∴ar(ΔBFD) = ar(ΔDEF) (for BDEF) — (i)

also,ar(ΔAFE) = ar(ΔDEF) (For DCEF) — (ii)

ar(ΔCDE) = ar(ΔDEF) (For  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) = ¼ ar(ABC)

(iii) ar (pBDEF) = ar(ΔDEF) +ar(ΔBDE)

⇒ ar( BDEF) = ar(ΔDEF) +ar(ΔDEF)

⇒ ar(BDEF) = 2× ar(ΔDEF)

⇒ ar(BDEF) = 2× ¼ ar(ΔABC)

⇒ ar(BDEF) = ½ ar(ΔABC)


EX 9.3 QUESTION 6.


 In Fig. 9.25, 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.
[Hint : From D and B, draw perpendiculars to AC.]

Solution:

OB = OD and AB = CD

Construction,

DE ⊥ AC and BF ⊥ AC are drawn.

Proof:

(i) In ΔDOE and ΔBOF,

∠DEO = ∠BFO [Each 90°]

∠DOE = ∠BOF [Vertically opposite angles]

OD = OB [Given]

∴, ΔDOE ≅ ΔBOF  [AAS congruence condition.]

∴, DE = BF [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) on both sides,

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

⇒ ar(ΔDCB) = ar(ΔACB)

(iii) ar(ΔDCB) = ar(ΔACB) are both on the same base CB and having equal areas. They lie between the same parallels CB and DA.

DA || BC — (iv)

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

∴, ABCD is parallelogram.


EX 9.3 QUESTION 7.


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

Solution:

ΔDBC and ΔEBC are on the same base BC and also having equal areas.

∴, DE || BC. [Triangles on the same base and having equal area lie between same parallels.]


EX 9.3 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(ΔABE) = ar(ΔACF)

Solution:

In ∆ABC , XY || BC,

BE || AC and CF || AB.

Since, XY ||BC and BE || CY

∴ BCYE is a parallelogram.
Now, the parallelogram BCYE and ∆ABE are on the same base BE and between the same parallels BE and AC.

∴ ar(∆ABE) =  ½ ar(BCYE)…..(1)  [If a parallelogram and a triangle are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram]

Now, ∆ACF and parallelogram BCFX are on the same base CF and between the same parallels AB and CF.

∴ar(∆ACF) =½ar (BCFX)……….(2)

ar(BCYE) = ar(BCFX)…………(3)      [parallelogram BCFX and parallelogram BCYE are on the same base BC and between the same parallels BC and EF.]

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

ar(∆BE) = ar(∆ACF)


EX 9.3 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 Fig. 9.26). Show that
ar(ABCD) = ar(PBQR).
[Hint : Join AC and PQ. Now compare ar(ACQ) and ar(APQ).]

Solution:

Join AC and PQ.

(△ACQ and △APQ on the same base AQ and between the same parallel lines AQ and CP)

Ar(△ACQ) = ar(△APQ)

Subtracting ar(ABQ) from both the sides

⇒ ar(△ACQ)-ar(△ABQ) = ar(△APQ)-ar(△ABQ)

⇒ ar(△ABC) = ar(△QBP) ⇒ ½ ar(ABCD) = ½ ar(PBQR)

⇒ ar(ABCD) = ar(PBQR)


EX 9.3 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:

∆ABD and ∆ABC are on the same base AB and between the same parallels AB || DC

∴ ar(∆ABD) = ar(∆ABC) [Triangles on the same base and having equal area lie between same parallels.]

Subtracting ar(∆AOB) from both sides, we get

ar(∆ABD) – ar(∆AOB) = ar(∆ABC) – ar(∆AOB)

⇒ ar(∆AOD) = ar(∆BOC)


EX 9.3 QUESTION 11.


In Fig. 9.27, 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)

Solution:

(i)△ACB and △ACF lie on the same base AC and between the same parallels AC || BF.

∴ar(△ACB) = ar(△ ACF) [Triangles on the same base and having equal area lie between same parallels.]

(ii)ar(△ACB) = ar(△ACF)

Adding ar ACDE on the both sides

⇒ ar(△ACB)+ar(△ACDE) = ar(△ACF)+ar(△ACDE)

⇒  ar(ABCDE) = ar(△AEDF)


EX 9.3 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:

The plot in the form of a quadrilateral ABCD.

Let us draw DF || AC and join AF and CF.

Now, ∆DAF and ∆DCF are on the same base DF and between the same parallels AC || DF.

∴ ar(ADAF) = ar(ADCF)

Subtracting ar(∆DEF) from both sides, we get

ar(∆DAF) – ar(∆DEF) = ar(∆DCF) – ar(∆DEF)

⇒ ar(∆ADE) = ar(∆CEF)

The portion of ∆ADE can be taken over by the Gram Panchayat by adding the land (∆CEF) to his (Itwaari) land so as to form a triangular plot,

i.e. ∆ABF.

Let us prove that ar(∆ABF) = ar(quad. ABCD),

ar(ACEF) = ar(AADE) [Proved above]

Adding ar(quad. ABCE) to both sides,

ar(∆CEF) + ar(quad. ABCE) = ar(∆ADE) + ar (quad. ABCE)

⇒ ar(∆ABF) = ar (quad. ABCD)


EX 9.3 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).

[Hint : Join CX.]

Solution:

In trapezium ABCD , AB || DC.

XY || AC meets AB at X and BC at Y. Let us join CX.

∆ADX and ∆ACX are on the same base AX and between the same parallels AX || DC.

∴ ar(∆ADX) = ar(∆ACX) …(1)

∵∆ACX and ∆ACY are on the same base AC and between the same parallels AC || XY.

∴ ar(∆ACX) = ar(∆ACY) …(2)

From (1) and (2),

ar(∆ADX) = ar(∆ACY)


EX 9.3 QUESTION 14.


In Fig.9.28, AP || BQ || CR. Prove that ar(△AQC) = ar(△PBR).

Solution:

AP || BQ || CR

∵ ∆ABQ and ∆PBQ are on the same base BQ and between the same parallels AP and BQ.

∴ ar(∆ABQ) = ar(∆PBQ) …(i)

∵ ∆BCQ and ∆BQR are on the same base BQ and between the same parallels BQ and CR.

∴ ar(∆BCQ) = ar(∆BQR) …(ii)

Adding (i) and (ii),

ar(∆BCQ) + ar(∆ABQ) = ar(∆BQR) + ar(∆PBQ)

⇒ ar(∆AQC) = ar(∆PBR)


EX 9.3 QUESTION 15.


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

Solution:

ar(∆AOD) = ar(∆BOC)   [Given]

Adding ar(∆AOB) to both sides,

ar(∆AOD) + ar(∆AOB) = ar(∆BOC) + ar(∆AOB)

⇒ ar(∆ABD) = ar(∆ABC)

∆ABD and ∆ABC are on the same base AB and ar(∆ABD) = ar(∆ABC).

∴ AB || DC [Triangles on the same base and having equal area lie between same parallels.]

So, ABCD is a trapezium.


EX 9.3 QUESTION 16.


In Fig.9.29, ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Solution:

Given, ar(△DRC) = ar(△DPC) ….(i)

△DRC and△DPC are on the same base DC and ar(△DRC) = ar(△DPC)

DC || RP [Triangles on the same base and having equal area lie between same parallels.]

So, DCPR is a trapezium.

ar(△ARC) = ar(△BDP) ….(ii) [Given]

Subtracting equation (i) from equation (ii)

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

⇒ ar(△ADC) = ar(△BDC)

ar(△BDP) = ar(△ARC)

⇒ ar(△BDP) – ar(△DPC) = ar(△DRC)

⇒  ar(△BDC) = ar(△ADC)

ar(△BDC) = ar(△ADC).

△ADC and△BDC are on the same base DC and ar(△ADC) = ar(△BDC)

∴, AB ∥ CD [Triangles on the same base and having equal area lie between same parallels.]