NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles (Ex 9.3) Exercise 9.3

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NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles (Ex 9.3) Exercise 9.3

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Access NCERT solutions for Class 9 Maths Chapter 9 – Areas of Parallelograms and Triangles

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NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3

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Access Other Exercise Solutions of Class 9 Maths Chapter 9 – Area of Parallelograms and Triangles

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CBSE Study Materials for Class 9

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Q.1 In the given figure, E is any point on median AD of a ∆ABC. Show that ar(ABE) = ar(ACE).


Ans

Given : In ΔABC, AD is median and E is any point on AD.To prove : ar(ABE)=ar(ACE)Proof: In ΔABC, AD is median. So, ar(ΔABD)=ar(ΔACD)     ...(i)[Median divides a triangle into two triangles of equal areas.]In ΔEBC, ED is median. So, ar(ΔEBD)=ar(ΔECD)     ...(ii)[Median divides a triangle into two triangles of equal areas.]Subtracting equation(ii) from equation(i), we getar(ΔABD)−ar(ΔEBD)=ar(ΔACD)−ar(ΔECD)⇒             Area (ΔABE)= Area (ΔACE)

Q.2

In a triangle ABC, E is the mid–point of median AD. Showthat arBED=14arABC.

Ans

Given:In​ ΔABC, AD is the median and E is the mid-point of AD.To prove : arBED=14arABCProof : In ΔABC, AD is median. So, ar(ΔABD)=12ar(ΔABC)      ..(i)[Median divides a triangle into two triangles of equal areas.]In ΔBDA, BE is median. So, ar(ΔBED)=12ar(ΔABD)      ...(ii)[Median divides a triangle into two triangles of equal areas.]From equation(i) and equation(ii), we have      ar(ΔBED)=12{12ar(ΔABC)}      ar(ΔBED)=14ar(ΔABC)                                    Hence proved.

Q.3 Show that the diagonals of a parallelogram divide it into four triangles of equal area.

Ans
Given: ABCD is a parallelogram; in which diagonals AC and BD bisect each other at O. To prove: area(DAOB)= area(DBOC)= area(DCOD)= area(DDOA)

Q.4 In the figure shown below, 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).


Ans

Given:ΔABC and ΔADB have same base AB and OC=OD.To prove: ar(ΔABC)=ar(ΔABD)Proof:In ΔADC, AO is median, so        area(ΔAOC)=area(ΔAOD)     ...(i)[Median divides a triangle into two triangles of equal areas.]In ΔBDC, BO is median, so       area(ΔBOC)=area(ΔBOD)       ...(ii)[Median divides a triangle into two triangles of equal areas.]Adding equation(i) and equation(ii), we getarea(ΔAOC)+area(ΔBOC)=area(ΔAOD)+area(ΔBOD)area(ΔABC)=area(ΔDAB)​       Hence proved.

Q.5

D, E and F are respectively the mid–points of the sidesBC, CA and AB of a ΔABC.Show that(i)BDEF is a parallelogram.(ii) arDEF=14arABC(iii) arBDEF=12arABC

Ans

Given:In ΔABC, D, E and F are the mid-points of BC, CA and AB respectively.To prove: (i)BDEF is a parallelogram.             (ii) ar(DEF)=14ar(ABC)            (iii)ar(BDEF)=12ar(ABC)Proof:(i) In ΔABC, F and E are the mid-points of AB and AC              respectively, so by mid-point theorem, we have                 FE=12BC and FE∥BC     ⇒         FE=BD and FE∥BD     [∵D is the mid-point of BC.]BDEF is a parallelogram because one pair of opposite sides is parallel and equal.(ii) Since,     FE=12BC and FE∥BC      So,                   FE=DC and FE∥DCThus, EFDC is a parallelogram. [Since, one pair of opposite sides is equal and parallel.]Similarly, DEAF is a parallelogram.Since, diagonal of a parallelogram divides it into two congruenttriangles. So,ar(ΔAFE)=ar(ΔDFE), ar(ΔBFD)=ar(ΔDFE) and ar(ΔDFE)=ar(ΔDCE)⇒ ar(ΔAFE)=ar(ΔDFE)=ar(ΔBFD)=ar(ΔDFE)⇒ar(ΔDFE)=14ar(ΔABC)[∵ar(ΔABC)=ar(ΔAFE)+ar(ΔDFE)+ar(ΔBFD)+ar(ΔDFE)](iii) ar(BDEF)=ar(ΔBFD)+ar(ΔDFE)                    =14ar(ΔABC)+14ar(ΔABC)                    =12ar(ΔABC)

Q.6 In figure shown below, 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.


Ans

Given:In quadrilateral ABCD, AC and BD are diagonals which             intersect at O, OB = OD and AB = CD.To prove:  (i) ar (DOC)= ar (AOB)               (ii) ar (DCB)= ar (ACB)             (iii) DA||CB or ABCD is a parallelogram.Construction: Draw BM⊥AC and DN⊥AC.Proof: (i)In ΔBOM and ΔDON           ∠BOM=∠DON          [Vertical opposite angles]           ∠BMO=∠DNO          [each 90°]              OB=OD               [Given]       ∴ΔBOM≅ΔDON        [By​ A.A.S]⇒            BM=DN              [By C.P.C.T.]and   ar(ΔBOM)=ar(ΔDON)   ...(i)In ΔBMA and ΔDNC                 BM=DN           [Proved above]          ∠BMA=∠DNC      [each 90°]              AB=CD          [Given]         ΔBMA≅ΔDNC      [By R.H.S.]⇒   ar(ΔBMA)=ar(ΔDNC)   ...(ii)Adding equation(i) and equation(ii), we getar(ΔBOM)+ar(ΔBMA)=ar(ΔDON)+ar(ΔDNC)       ar(ΔAOB)=ar(ΔDOC)⇒     ar(ΔDOC)=ar(ΔAOB)(ii) Since, ar(ΔDOC)=ar(ΔAOB)Adding ar(ΔBOC) both sides, we getar(ΔDOC)+ar(ΔBOC)=ar(ΔAOB)+ar(ΔBOC)⇒                  ar(ΔDCB)=ar(ΔABC)(iii) ar(ΔDCB)=ar(ΔABC) If two triangles having same base and equal area, thenthey lie between same parallels.So, BC∥ADor DA∥BCAs  ΔAOB≅ΔCOD∴  ∠OAB=∠OCD⇒∠CAB=∠ACD⇒AB∥CD          [Alternate angles are equal.]So, ABCD is a parallelogram because opposite sideare parallel.

Q.7 D and E are points on sides AB and AC respectively of ΔABC such that ar(DBC)=ar(EBC). Prove that DE || BC.

Ans

Given : In  ΔABC, D and E are the points on AB and AC            respectively and ar(DBC)=ar(EBC).To prove : DE∥BCProof : Since, ar(DBC)=ar(EBC)     and both triangles have same base BC.So, DE∥BC   [∵If two triangles have equal area and same base, then they lie between same parallels.]

Q.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).

Ans

Given:In ΔABC, XY∥BC, BE∥AC and CF∥AB.To prove: ar(ABE)= ar(ACF)Proof:Since, parallelograms BCYE and BCFX have same base BC and lie between same parallels.So, ar(BCYE)=ar(BCFX)     ...(i)ΔAEB and parallelogram BCYE have same base and lie between same parallels. So,         ar(ΔABE)=12ar(BCYE)    ..(ii)Similarly,         ar(ΔACF)=12ar(BCFX)    ...(ii)From equation(i), we have 12ar(BCYE)=12ar(BCFX)From equation(ii) and equation(iii), we have         ar(ΔABE)=ar(ΔACF)                                     Hence Proved.

Q.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 the figure below). Show that ar (ABCD) = ar (PBQR).


Ans

Given:ABCD and PBQR are parallelogram. AQ∥CP.To prove: ar(ABCD)=ar(PBQR)Construction: Join AC and PQ.Proof: Since, AQ∥CP ΔACP and ΔQPC have same base CP and both triangles are between same parallels. So,        ar(ΔACP)=ar(ΔQPC)Subtracting ar(ΔBCP) from both sides, we getar(ΔACP)−ar(ΔBCP)=ar(ΔQPC)−ar(ΔBCP)        ar(ΔABC)=ar(ΔQBP)⇒      2×ar(ΔABC)=2×ar(ΔQBP)⇒        ar(ABCD)=ar(PBQR)[∵Diagonal divides a parallelogram in two triangles of equal area.]

Q.10 Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar(AOD) = ar(BOC).

Ans

Given:In trapezium ABCD, AB∥CD.To prove: ar (AOD) = ar (BOC)Proof: Area of two triangles is equal if they have same base and lie between same parallels.Here, AB∥CD, ΔADC and ΔBDC have same base.∴    ar(ΔADC)=ar(ΔBDC)⇒     ar(ΔAOD)+ar(ΔDOC)=ar(ΔBDC)+ar(ΔDOC)⇒     ar(ΔAOD)=ar(ΔBDC)

Q.11 In figure shown below, 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)


Ans

Given: ABCDE is a pentagon and BF∥AC.To prove : (i) ar(ACB)=ar(ACF)             (ii) ar(AEDF)=ar(ABCDE)Proof: (i) Since, AC∥BF, ΔACB and ΔACF have same base AC.          So,       ar(ACB)=ar(ACF)[Areas of two triangles are equal if they have same baseand lie between same parallels.](ii) We proved above:                      ar(ACB)=ar(ACF)Adding ar(AEDC) both sides, we getar(ACB)+ar(AEDC)=ar(ACF)+ar(AEDC)⇒           ar(ABCDE)=ar(AEDF)or            ar(AEDF)=ar(ABCDE)                            Hence proved.

Q.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.

Ans

Given:Let ABCD be the shape of the plot of land of Itwaari and AED is his adjacent plot. ED be the side of his adjoining land.        To prove: ar(ΔAOB)=ar(ΔEOD)Construction: The proposal may be implemented as follows. Join diagonal BD and draw a line parallel to BD through point A. Let it meet the extended side CD of ABCD at point E. Join BE and AD. Let them intersect each other at O. Then,portion ΔAOB can be cut from the original field so that the new shape of the field will be ΔBCE.

Proof: Since, ΔEDB and ΔADB have same base and lie between same parallels. So,       ar(ΔEDB)=ar(ΔADB)Subtracting ar(ΔBOD) from both sides, we getar(ΔEDB)−ar(ΔBOD)=ar(ΔADB)−ar(ΔBOD)                ar(ΔEOD)=ar(ΔBOA)Adding ar(BCDO) both sides, we getar(ΔEOD)+ar(BCDO)=ar(ΔBOA)+ar(BCDO)                  ar(ΔBCE)=ar(ABCD)Thus, we can replace land in the shape of ΔBOA by the land in the shape of ΔEOD to implement the proposal of Itwaari.

Q.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).

Ans

Given: ABCD is a trapezium in which AB∥CD and XY∥AC.To prove: ar(ADX)=ar(ACY)Construction:Join CX.Proof:Since, AB∥CD, then ar(ΔADX)=ar(ΔACX)    ... (i)     [Both Δs have same base andlie between same parallels.]and AC∥XY, then ar(ΔACX)=ar(ΔACY)    ... (ii)From equation(i) and equqation(ii), we have ar(ΔADX)=ar(ΔACY)                                 Hence proved.

Q.14 In Fig.9.28, AP||BQ||CR. Prove that ar(AQC) = ar(PBR).


Ans

Given:In given figure, AP||BQ||CR

To prove:ar(AQC)=ar(PBR)Proof:Since,  AP||BQ|| So, ar(ΔBPQ)=ar(ΔABQ)         ...(i)[Both triangles have same base and lie between same parallels.]Since,  BQ∥CRSo,  ar(ΔBQR)=ar(ΔBQC)               ...(ii)[Both triangles have same base and lie between same parallels.]Adding equation(i) and equation (ii), we getar(ΔBPQ)+ar(ΔBQR)=ar(ΔABQ)+ar(ΔBQC)⇒ar(ΔBPR)=ar(ΔAQC)⇒ ar(ΔAQC)=ar(ΔBPR)              Hence proved.

Q.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.

Ans

Given:ABCD is a quadrilateral and its diagonals AC and BD intersect at O and ar(AOD)=ar(BOC).To prove: ABCD is a trapezium.Proof:It​ is given that           ar(AOD)=ar(BOC)Adding ar(AOB) both sides, we getar(AOD)+ar(AOB)=ar(BOC)+ar(AOB)               ar(ABD)=ar(ABC)Since, both triagles have same base AB and having equal area.So they will lie between same parallels.Thus, AB∥CD⇒  ABCD is a trapezium.

Q.16 In the figure shown below, ar (DRC) = ar (DPC) and ar(BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.


Ans

Given:In given figure, ar(DRC)=ar(DPC) and ar(BDP)=ar(ARC).To prove: The quadrilaterals ABCD and DCPR are trapeziums.Proof: It is given that ar(DRC)=ar(DPC)or            ar(DPC)=ar(DRC)     …(i) Since, both triangles have same base and equal area. Then, DC∥RP⇒DCPR is a trapezium.Given that ar(BDP)=ar(ARC)     …(ii)Subtracting equation(i) from equation(ii), we getar(BDP)−ar(DPC)=ar(ARC)−ar(DRC)⇒ ar(BDC)=ar(ACD) Since, both triangles have same base and equal area. Then, AB∥DC⇒ABCD is a trapezium.In this way, we proved that DCPR and ABCD are trapeziums.

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