Gujarat Board GSEB Solutions Class 9 Maths Areas of Parallelograms and Triangles Ex 9.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 9 Maths Areas of Parallelograms and Triangles Ex 9.3

Question 1.

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

Solution:

Given: E is any point on median AD of a Î”ABC. In Î”ABC,

âˆ´ AD is a median.

âˆ´ ar(Î”ABD) = ar(Î”ACD) ………(1)

A median of a triangle divides it into two triangles of equal areas.

In Î”EBC,

âˆ´ ED is a median.

âˆ´ ar(Î”EBD) = ar(Î”ECD) ………(2)

A median of a triangle divides it into two triangles of equal areas.

Subtracting (2) from (1), we get

ar(Î”ABD) â‰… ar(Î”EBD)

= ar(Î”ACD) â‰… ar(Î”ECD)

â‡’ ar(Î”ABE) = ar(Î”ACE)

Question 2.

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

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

Solution:

In Î”ABC,

âˆ´ AD is a median.

âˆ´ ar(Î”ABD) = ar(Î”ACD) = \(\frac {1}{4}\) ar(Î”ABC) …….(1)

A median of a triangle divides it into two triangles of equal areas

Similarly, BE is a median in Î”ABD,

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

= \(\frac {1}{2}\) . \(\frac {1}{2}\) ar (Î”ABD) [ From (1)]

\(\frac {1}{4}\) ar (Î”ABD)

Question 3.

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

Solution:

Given: ABCD is a parallelogram whose diagonals AC and BD intersecting at O divide it into four triangles Î”OAB, Î”OBC, Î”OCD and Î”ODA.

To Prove: ar(Î”OAB) = ar(Î”OBC)

= ar(Î”OCD) = ar(Î”ODA)

Construction: Draw BE âŠ¥ AC

Proof: Now,

= \(\frac {(OC) (BE)}{2}\)

But OA = OC

âˆ´ ar(Î”OAB) = ar(Î”OBC) …….(1)

Similarly,

ar(Î”OBC) = ar(Î”OCD) …….(2)

and, ar(Î”OCD) = ar(Î”ODA) …….(3)

From (1), (2) and (3), we get ar(Î”OAB) = ar(Î”OBC)

= ar(Î”OCD) = ar(Î”ODA)

Question 4.

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

Solution:

Line segment CD is bisected by AB at O.

âˆ´ OC = OD

âˆ´ BO is a median of ABCD, and

AO is a median of AACD.

âˆ´ BO is a median of ABCD.

âˆ´ ar(Î”OBC) = ar(Î”OBD) ………..(1)

âˆ´ A median of a triangle divides it into two triangles of equal areas

Similarly, AO is a median of Î”ACD.

âˆ´ ar(Î”OAC) = ar(Î”OAD) ……….(2)

Adding (1) and (2), we get

ar(Î”OBC) + ar(Î”OAC)

= ar(Î”OBD) + ar(Î”OAD)

â‡’ ar(Î”ABC) = ar(Î”ABD)

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(Î”DEF) = \(\frac {1}{4}\) ar(Î”ABC)

(iii) ar(â–¡BDEF) = \(\frac {1}{2}\) ar(Î”ABC)

Solution:

(i) In AABC, F is the mid-point of side AB and E is the mid-point of side AC.

âˆ´ EF || BC

âˆ´ In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side.

â‡’Â EF || BD ………(1)

Similarly, ED || BF ………(2)

In view of (1) and (2),

â–¡BDEF is a parallelogram.

âˆ´ A quadrilateral is a parallelogram if its opposite sides are parallel.

(ii) As in (i), we can prove that

â–¡AFDE and â–¡FDCE are parallelograms,

âˆ´ FD is a diagonal of || gm BDEF.

âˆ´ ar(Î”FBD) = ar(Î”DEF) ……(3)

Similarly, ar(Î”DEF) = ar(Î”FAE) ……(4)

and, ar(Î”DEF) = ar(Î”DCE) ……(5)

From (3), (4) and (5), we have

ar(Î”FBD) = ar(Î”DEF) = ar(Î”FAE) = ar(Î”DCE) ……..(6)

Now, Î”ABC is divided into four non-overlapping triangles Î”FBD, Î”DEF, Î”FAE and Î”DCE.

âˆ´ ar(Î”ABC) = ar(Î”FBD) + ar(Î”DEF) + ar(Î”FAE) + ar(Î”DCE) = 4 ar(Î”DEF) [From (6)]

â‡’ ar(Î”DEF) = \(\frac {1}{4}\) ar(Î”ABC) ……..(7)

(iii) ar(â–¡BDEF)

= ar(Î”FBD) + ar(Î”DEF)

= ar(Î”DEF) + ar(Î”DEF) [From (3)] = 2ar(Î”DEF)

= 2. \(\frac {1}{4}\)ar(Î”ABC) [From (7)]

= \(\frac {1}{2}\) ar(Î”ABC).

Question 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(Î”DOC) = ar(Î”AOB)

(ii) ar(Î”DCB) = ar(Î”ACB)

(iii) DA || CB or ABCD is a parallelogram

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

Solution:

Construction: Draw DE âŠ¥ AC and BF âŠ¥ AC.

(i) In Î”ODE and DOBF,

âˆ OED = âˆ OFB (each 90Â°)

âˆ DOE = âˆ BOF (Vertically opposite angles)

OB = OB (given)

âˆ´ Î”DOE = Î”BOF (By AAS congruency)

â‡’ DE = BF

and OE = OF ……..(1) (CPCT)

Also, ar(Î”DOE) = ar(Î”BOF) ………(2)

Again, in right Î”DEC and Î”BFA,

Hyp. DC = Hyp. BA [Given]

DE = BF (Proved above)

âˆ DEC = âˆ BFA (each 90Â°)

âˆ´ Î”DEC = Î”BFA | RHS rule

âˆ´ CE = AF ………(3)

Adding eqn. (2) and (4),

ar(Î”DOE) + ar(Î”CDE) = ar(Î”BOF) + ar(Î”ABF)

â‡’ ar(Î”DOC) – ar(Î”AOB)

(ii) From (i)

ar(Î”DOC) = ar(Î”AOB)

â‡’ ar(Î”DOC) + ar(Î”OCB) = ar(Î”AOB) + ar(Î”OCB)

Adding equal areas on both sides

â‡’ ar(Î”DCB) = ar(Î”ACB)

(ii) In point (iii), ADCB and Î”ACB having same area lie on the same base AD, therefore, DA || CB

Adding eqn. (1) and (3), we get

OE = CE = OF + AF

â‡’ OC = OA

Also, OB = OD (given)

â‡’ ABCD is a parallelogram

(âˆ´ If diagonals of quadrilateral bisect each other, ten it is a parallelogram)

Question 7.

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

Solution:

Given: D and E are points on side AB and AC respectively of Î”ABC, such that ar(Î”DBC) = ar(Î”EBC)

To prove: DE || BC

Proof: Î”DBC and AEBC are on the same base BC and have equal areas.

âˆ´ Their altitudes must be the same.

âˆ´ DE || BC

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(Î”ABE) = ar(Î”ACF).

Solution:

XF || BC [Given XY || BC]

and CF || BX [âˆ´ CF || AB (Given)]

âˆ´ â–¡BCFX is a ||gm.

A quadrilateral is a parallelogram I if its opposite sides are parallel

âˆ´ BC = XF

Opposite sides of a parallelogram are equal.

â‡’ BC = XY + YF ………(1)

Again,

âˆ´ EY || BC [Given]

and BE || CY [âˆ´ BE || AC (Given)]

âˆ´ Î”BCYE is a parallelogram.

A quadrilateral is a parallelogram I if its opposite sides are parallel.

âˆ´ BC = YE

Opposite sides of a parallelogram are equal.

â‡’ BC = XY + XE ………(2)

From (1) and (2),

XY + YF = XY + XE

â‡’ YF = XE

â‡’ XE = YF ………(3)

âˆ´ Î”AEX and AAYF have equal bases

(âˆ´ XE = YF) on the same line EF and have a common vertex A.

âˆ´ Their altitudes are also the same.

âˆ´ ar(Î”AEX) = ar(Î”AFY) ……..(4)

âˆ´ Î”ABEX and ACFY have equal bases (âˆ´ XE = YF) on the same line EF and are between the same parallels EF and BC (âˆ´ XY || BC).

âˆ´ ar(Î”BEX) = ar(Î”CFY) ………(5)

Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

Adding the corresponding sides of (4) and (5), we get

ar(Î”AEX) + ar(Î”BEX) = ar(Î”AFY) + ar(Î”CFY)

â‡’ ar(Î”ABE) = ar(Î”ACF)

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. Show that ar(|| gm ABCD) = ar(|| gm PBQR)

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

Solution:

To Prove: ar(|| gm ABCD) = ar(|| gm PBQR)

Construction: Join AC and PQ

Proof: AC is a diagonal of || gm ABCD

âˆ´ ar(Î”ABC) = \(\frac {1}{2}\) ar(||gm ABCD) ……..(1)

PCQ is a diagonal of || gm BQRP

ar(Î”BPQ) = \(\frac {1}{2}\) ar(||gm BQRP) ……..(2)

Also, ar(Î”ACQ) = ar(Î”APQ)

Î”ACQ and Î”APQ are on the same base AQ and between the same. parallels (AQ and CP are equal in area

â‡’ ar(Î”ACQ) – ar(Î”ABQ) = ar(Î”APQ) – ar(Î”ABQ)

Subtracting the same areas from both sides

â‡’ ar(Î”ABC) = ar(Î”BPQ)

â‡’ \(\frac {1}{2}\) ar(|| gm ABCD) = \(\frac {1}{2}\) ar(|| gm PBQR) [From (1) and (2)]

â‡’ ar(|| gm ABCD) = ar(|| gm PBQR)

Question 10.

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar(Î”AOD) = ar(Î”BOC)

Solution:

Î”ABD and Î”ABC are on the same base AB and between the same parallels AB and DC.

âˆ´ ar(Î”ABD) = ar(Î”ABC)

Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

ar(Î”ABD) – ar(Î”AOB) = ar(Î”ABC) – ar(Î”AOB)

Subtracting the same areas from both sides.

â‡’ ar(Î”AOD) = ar(Î”BOC)

Question 11.

In figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that:

(i) ar(Î”ACB) = ar(Î”ACF)

(ii) ar(â–¡AEDF) = ar(Î”BCDE)

Solution:

(i) âˆ´ Î”ACB and Î”ACF are on the same base AC and between the same parallels AC and BF.

[âˆ´ AC || BF (Given)]

âˆ´ ar(Î”ACB) = ar(Î”ACF)

Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

(ii) From (i),

ar(Î”ACB) = ar(Î”ACF)

â‡’ ar(Î”ACB) + ar(â–¡AEDC) = ar(Î”ACF) + ar(â–¡AEDC)

Adding the same areas on both sides.

â‡’ ar(ABCDE) = ar(â–¡AEDF)

âˆ´ ar(â–¡AEDF) = ar(ABCDE)

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 comers to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given an 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 the shape of a quadrilateral. Let the portion ADE be taken over by the Gram Panchayat of the village from one comer D to construct a Health Centre. Join AC. Draw a line through D parallel to AC to meet BC produced in P. Then Itwaari must be given the land ECP adjoining his plot so as to form a triangular plot ABP as then

ar(Î”ADE) = ar(Î”PEC)

Proof: âˆ´ Î”DAP and Î”DCP are on the same base DP and between the parallels DP and AC.

âˆ´ ar(Î”DAP) = ar(Î”DCP) ………..(1)

âˆ´ Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

â‡’ ar(Î”DAP) – ar(Î”DEP) = (Î”DCP) – ar(Î”DEP)

Subtracting the same areas from both sides.

â‡’ ar(Î”ADE) = ar(Î”PCE)

â‡’ ar(Î”ADE) + ar(â–¡ABCE) = ar(Î”PCE) + ar(â–¡ABCE)

Adding the same areas to both sides.

â‡’ ar(â–¡ABCD) = ar(Î”ABP)

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(Î”ADX) = ar(Î”ACY).

[Hint: Join CX]

Solution:

Given: ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y.

âˆ´ Î”ADX and Î”ACX are on the same base AX and between the same parallels AB and DC.

âˆ´ ar(Î”ADX) = ar(Î”ACX) …….(1)

Two triangles on the same base (or equal bases) and between the same parallels are equal in area,

âˆ´ 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) …….(2)

Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

From (1) and (2), we get

ar(Î”ADX) = ar(Î”ACY)

Question 14.

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

Solution:

âˆ´ ABAQ and ABPQ are on the same base BQ and between the same parallels BQ and AP.

âˆ´ ar(Î”BAQ) = ar(Î”BPQ) ……..(1)

âˆ´ Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

Similarly, Î”BCQ and Î”BQR are on the same base BQ and between the same parallels BQ and CR.

âˆ´ ar(Î”BCQ) = ar(Î”BQR) ……..(2)

Adding the corresponding sides of (1) and (2), we get

ar(Î”BAQ) + ar(Î”BCQ) = ar(Î”BPQ) + ar(Î”BQR)

âˆ´ ar(Î”AQC) = ar(Î”PBR).

Question 15.

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

Solution:

Given: ar(Î”AOD) = ar(Î”BOC)

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

Adding the same areas on both sides.

â‡’ ar(Î”ABD) = ar(Î”ABC)

But Î”ABD and Î”ABC are on the same base AB.

âˆ´ Î”ABD and Î”ABC will have equal corresponding altitudes.

âˆ´ Î”ABD and Î”ABC will lie between the same parallels.

âˆ´ AB || DC

âˆ´ â–¡ABCD is a trapezium.

A quadrilateral is a trapezium if exactly one pair of opposite sides is parallel.

Question 16.

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

Solution:

Given: ar(Î”DRC) = ar(Î”DPC) ……..(1)

But Î”DRC and Î”DPC are on the same base DC.

Î”DRC and Î”DPC will have equal corresponding altitudes.

âˆ´ Î”DRC and Î”DPC will lie between the same parallels.

âˆ´ DC || RP

âˆ´ â–¡DCPR is a trapezium.

A quadrilateral is a trapezium if exactly one pair of opposite sides are parallel.

Again, ar(Î”BDP) = ar(Î”ARC)

â‡’ ar(Î”BDC) + ar(Î”DPC) = ar(Î”ADC) + ar(Î”DRC)

â‡’ ar(Î”BDC) = ar(Î”ADC) [using (1)]

But Î”BDC and Î”ADC are on the same base DC.

âˆ´ Î”BDC and Î”ADC will have equal corresponding altitudes.

âˆ´ Î”BDC and Î”ADC will lie between the same parallels.

âˆ´ AB || DC

âˆ´ â–¡ABCD is a trapezium

A quadrilateral is a trapezium if exactly one pair of opposite sides is parallel.