Haryana State Board HBSE 9th Class Maths Solutions Chapter 9 Areas of Parallelograms and Triangles Ex 9.3 Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Maths Solutions Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3

Question 1.

In figure 9.23, E is any point on median AD of a Î”ABC. Show that ar (ABE) = ar (ACE).

Solution:

In Î”ABC, AD is a median and median divides a triangle into two triangles of equal areas.

âˆ´ ar (Î”ABD) = ar (Î”ACD) …(i)

Again, in Î”EBC, ED is a median.

âˆ´ ar (Î”EBD) = ar (Î”ECD) …(ii)

Subtracting (ii) from (i), we get

ar (Î”ABD) – ar (Î”EBD) = ar (Î”ACD) – ar (Î”ECD)

â‡’ ar (Î”ABE) = ar (Î”ACE).

Hence proved

Question 2.

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

Solution:

Since AD is a median of AABC and median divides a triangle into two triangles of equal areas.

âˆ´ ar (Î”ABD) = ar (Î”ACD)

â‡’ ar (Î”ABD) = \(\frac{1}{2}\)ar (Î”ABC) …(i)

In Î”ABD, BE is the median.

ar (Î”BED) = ar (Î”BAE)

ar (Î”BED) = \(\frac{1}{2}\)ar (Î”ABD)

ar (Î”BED) = \(\frac{1}{2} \times \frac{1}{2}\)ar (Î”ABC) [using (i)]

ar (Î”BED) = \(\frac{1}{4}\)ar (Î”ABC).

Hence proved

Question 3.

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

Solution:

Given: A parallelogram ABCD in which diagonals AC and BD intersect at O.

Prove : ar (Î”AOB) = ar (Î”BOC) = ar (Î”COD) = ar (Î”DOA).

Proof : Since, diagonals of a parallelogram bisect each other.

OA = OC and OB = OD and a median divides the triangle into two triangles of equal areas.

Now in Î”ABC, BO is the median.

ar (Î”AOB) = ar (Î”BOC) …..(i)

In Î”BCD, CO is the median.

ar (Î”COD) = ar (Î”BOC) ……(ii)

In Î”CDA, DO is the median.

ar (Î”COD) = ar (Î”AOD) …..(iii)

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

we get ar (Î”AOB) = ar (Î”BOC)

= ar (Î”COD) = ar(Î”AOD)

Hence proved

Question 4.

In figure 9.24, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at 0, show that ar (ABC) = ar (ABD).

Solution:

In Î”ACD, we have

OC = OD [âˆµ given line CD is bisected by AB]

âˆ´ AO is the median.

â‡’ ar (AOC) = ar (AOD) …(i)

[median divides a A in two As of equal areas]

Similarly, in Î”BCD, BO is the median.

â‡’ ar(BOC) = ar (BOD) ….(ii)

Adding (i) and (ii), we get

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

â‡’ ar (ABC) = ar (ABD).

Hence proved

Question 5.

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

(i) BDEF is a parallelogram

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

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

Solution:

Given : A triangle ABC in which D, E and Fare the mid points of the sides BC, CA and AB respectively.

To prove : (i) BDEF is a parallelogram.

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

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

Proof : (i) In Î”ABC, F and E are the mid points of the sides AB and AC respectively.

âˆ´ FE || BC and FE = \(\frac{1}{2}\)BC

â‡’ FE || BD and FE = BD [âˆµ D is the mid point of BC, âˆ´ BD = CD]

âˆ´ BDEF is a parallelogram.

(ii) Similarly, F and D are the mid points of the sides AB and BC respectively.

âˆ´ FD || AC and FD = \(\frac{1}{2}\)AC

âˆ´ FD || EC and FD = \(\frac{1}{2}\)AC

â‡’ FD || EC and FD = EC

âˆ´ FDCE is a parallelogram.

Similarly, we can prove that AFDE is a parallelogram.

Since, FD is a diagonal of parallelogram BDEF

âˆ´ ar(FBD) = ar (DEF) …(i)

[âˆµ a diagonal divides a || gm into two Î”s of equal areas]

Similarly, DE is a diagonal of parallelogram FDCE.

âˆ´ ar(ECD) = ar (DEF) …(ii)

and FE is a diagonal of parallelogram AFDE.

âˆ´ ar (AFE) = ar (DEF) …(iii)

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

ar (FBD) = ar (DEF) = ar (ECD)

= ar (AFE) …..(iv)

Now, ar (FBD) + ar (DEF)+ ar (ECD) + ar (AFE) = ar (ABC)

â‡’ ar (DEF) + ar (DEF)+ ar (DEF) + ar (DEF)

= ar (ABC) [using (iv)]

â‡’ 4 ar (DEF) = ar (ABC)

â‡’ ar (DEF) = \(\frac{1}{4}\)ar (ABC)

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

(as proved above)

2 ar (DEF) = \(\frac{2}{4}\)ar (ABC)

â‡’ ar (DEF) + ar (DEF) = \(\frac{1}{2}\)ar (ABC)

â‡’ ar (DEF) + ar (FBD) = \(\frac{1}{2}\)ar (ABC) [using (i)]

â‡’ ar (BDEF) = \(\frac{1}{2}\)ar (ABC).

Hence proved.

Question 6.

In figure 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 Dand B, draw perpendiculars to AC]

Solution:

(i) Draw BM âŠ¥ AC and DN âŠ¥ AC.

In Î”DON and Î”BOM, we have

âˆ DNO = âˆ BMO (Each = 90Â°)

âˆ DON = âˆ BOM

(vertically opposite angles)

and DO = OB (given)

âˆ´ Î”DON â‰… Î”BOM

(By AAS congruecne rule)

â‡’ ar (DON) = ar (BOM) ……(i)

â‡’ DN = BM (CPCT)

In Î”DCN and Î”BAM, we have

âˆ DNC = âˆ BMA (Each = 90Â°)

Hyp. DC = Hyp. AB (given)

and DN = BM (as proved above)

âˆ´ Î”DCN â‰… Î”BAM (By RHS congruence rule)

â‡’ ar (DCN) = ar (BAM) …(ii)

Adding (i) and (ii), we get

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

â‡’ ar (DOC) = ar (AOB).

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

â‡’ ar (DOC) + ar(COB) = ar (AOB) + ar (COB)

(Adding ar (COB) on both sides)

â‡’ ar (DCB) = ar (ACB).

(ii) Since Î”DCB and Î”ACB are on the same base BC and have equal areas.

âˆ´ DA || CB

Now, Î”DON â‰… Î”BOM (as proved above)

âˆ ODN = âˆ OBM(CPCT) … (iii)

Î”DCN â‰… Î”BAM (as proved above)

â‡’ âˆ CDN = âˆ ABM (CPCT) …(iv)

Adding (iii) and (iv), we get

âˆ ODN + âˆ CDN = âˆ OBM + âˆ ABM

â‡’ âˆ ODC = âˆ OBA

But these are alternate interior angles.

âˆ´ CD || AB

âˆ´ ABCD is a parallelogram. Hence proved

Question 7.

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

Solution:

Since Î”DBC and Î”EBC are on the same base BC and have equal areas.

Therefore Î”DBC and Î”EBC lie between the same parallels.

âˆ´ DE || BC. Hence proved

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:

XY || BC and EB || YC [âˆµ BE || AC]

âˆ´ BCYE is a parallelogram.

Again, XY || BC and BX || CF. [âˆµ AB || CF]

BCFX is a parallelogram.

Since Î”ABE and parallelogram BCYE are on the same base EB and between the same parallels EB and AC.

âˆ´ ar (ABE) = \(\frac{1}{2}\)ar (BCYE) …(i)

Similarly, Î”ACF and parallelogram BCFX are on the same base CF and between the same parallels CF and AB.

âˆ´ ar (ACF) = ar (BCFX) ……(ii)

and parallelograms BCYE and BCFX are on the same base BC and between the same parallels BC and EF.

âˆ´ ar(BCYE) = ar (BCFX) …(iii)

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

ar (ABE) = ar (ACF).

Hence proved.

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 9.26). Show that ar (ABCD) = ar (PBQR).

Solution:

Join AC and PQ.

Since Î”AQC and Î”AQP are on the same base AQ and between the same parallels AQ and CP.

ar (Î”AQC) = ar (Î”AQP)

â‡’ ar (Î”AQC) – ar (Î”ABQ) = ar (Î”AQP) – ar (Î”ABQ)

â‡’ ar (Î”ABC) = ar (Î”PBQ)

â‡’ \(\frac{1}{2}\)ar (||gm ABCD) = \(\frac{1}{2}\)ar (||gm PBQR) [âˆµ a diagonal divides the ||gm into two Î”s of equal areas)

â‡’ ar (||gm ABCD) = ar (||gm PBQR).

Hence proved

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:

In a trapezium ABCD,

we have AB || CD

So, Î”ABC and Î”ABD are on the same base AB and between the same parallels AB and CD.

âˆ´ ar (Î”ABC) = ar (Î”BAD) (by theorem 9.2)

â‡’ ar (Î”ABC) – ar (Î”AOB) = ar (Î”BAD) – ar (Î”AOB)

â‡’ ar (Î”BOC) = ar (Î”AOD).

Hence proved.

Question 11.

In figure 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) Since Î”ACB and Î”ACF are on the same base AC and between the same parallels AC and BF.

âˆ´ ar(ACB) = ar (ACF)

(ii) ar (ACB) = ar (ACF) (as proved above)

â‡’ ar (ACB) + ar (ACDE) = ar (ACF) + ar (ACDE)

[Adding ar (ACDE) on both sides]

â‡’ ar (ABCDE) = ar (AEDF)

â‡’ ar (AEDF) = ar (ABCDE). Hence proved

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 plot of land of villager Itwaari and the Gram Panchayat decided to take some portion of land from corner D (say) of the plot ABCD. Since Itwaari wants to have equal amount of land in the lie of land COD we may proceed it as follows:

Join AC and draw a line through D parallel to AC to meet BA produced at E. Join EC intersects AD at O.

Now, Î”ACD and Î”EAC are on the same base AC and between the same parallels AC and DE.

âˆ´ ar (Î”ACD) = ar (Î”EAC)

â‡’ ar (Î”ACD) – ar (Î”AOC) = ar (EAC) – ar (Î”AOC)

â‡’ ar (COD) = ar (Î”AOE) ……(i)

Now, ar (quad. ABCD) = ar (quad. OABC) + ar (Î”COD)

â‡’ ar (quad. ABCD)= ar (quad. OABC) + ar (Î”AOE) [using (i)]

â‡’ ar(quad. ABCD) = ar (ACEB)

So, the Gram Panchayat take area (Î”COD) and provided ar (Î”AOE) to the Itwaari.

Question 13.

ABCD is a trapezium with AB || CD. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).

Solution:

ABCD is a trapezium in which AB || CD and XY || AC. Join XC and XD. Since, Î”ACX and Î”ACY are on the same base AC and between the same parallels AC and XY.

âˆ´ ar (Î”ACX) = ar (Î”ACY) …(i)

Again, Î”ACX and Î”ADX are on the same base AX and between the same parallels AX and CD.

âˆ´ ar (Î”ACX) = ar (Î”ADX) …(i)

From (1) and (2), we get

ar (ADX) = ar (ACY).

Hence proved

Question 14.

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

Solution:

Since Î”ABQ and Î”PQB are on the same base BQ and between the same parallels AP and BQ.

âˆ´ ar (Î”ABQ) = ar (Î”PQB) …(i)

Again Î”BCQ and Î”QRB are on the same base BQ and between the same parallels BQ and CR.

âˆ´ ar (Î”BCQ) = ar (Î”QRB) …(ii)

Adding (i) and (ii), we get

ar (Î”ABQ) + ar (Î”BCQ) = ar (Î”PQB) + ar (Î”QRB)

â‡’ ar (Î”AQC) = ar (Î”PBR).

Hence proved

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:

In a quadrilateral ABCD, diagonals AC and BD intersect at O in such a way that

ar (Î”AOD) = ar (Î”BOC)

â‡’ ar (Î”AOD) + ar (Î”AOB) = ar (Î”BOC) + ar (Î”AOB)

[adding ar (Î”AOB) on both sides]

â‡’ ar (Î”ABD) = ar (Î”ABC)

Thus, Î”ABD and Î”ABC are on the same base AB and having equal area.

âˆ´ AB || CD

Hence ABCD is a trapezium. Proved

Question 16.

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

Solution:

Ð°r (Î”DRC) = ar (Î”DPC) …(1) (given)

So, Î”DRC and Î”DPC are on the same base DC and having equal area.

âˆ´ RP || DC

â‡’ DCPR is a trapezium.

and ar(Î”BDP) = ar (Î”ARC) (given)

ar (Î”ARC) = ar (Î”BDP) …….(2)

Subtracting (1) from (2), we get

ar (Î”ARC) – ar (DRC) = ar (Î”BDP) – ar (Î”DPC)

â‡’ ar (Î”ACD) = ar (Î”BDC)

So, Î”ACD and Î”BDC are on the same base DC and having equal area.

âˆ´ DC || AB

â‡’ ABCD is a trapezium.

Hence, DCPR and ABCD are trapeziums.

Hence proved