
1.Number System
14
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2.Polynomials
10
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3.Coordinate Geometry
8
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4.Linear Equations
8
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5.Euclid's Geometry
7
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6.Lines and Angles
10
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7.Triangles
11
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8.Quadrilaterals
13
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9.Area of Parallelogram
11
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10.Constructions
7
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11.Circles
11
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12.Heron's Formula
8
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13.Surface Area and Volume
16
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14.Statistics
15
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15.Probability
8
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Lecture15.8

NCERT Solutions – Area of Parallelogram Exercise 9.1 – 9.4
Exercise 9.1
Q.1 Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.
Sol.
The figures mentioned above lie on the same base and between the same parallels as indicated against them :
(i) Base DC, parallels DC and AB.
(ii) Base QR, parallels QR and PS.
(iii) Base AD, parallels AD and BQ.
Exercise 9.2
Q.1 In figure ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
Sol.
We have,
Area of a  gm = Base × Height
Therefore, Area of  gm ABCD = AB × AE
=(16×8)cm2=128cm2 …(1)
Also area of  gm ABCD = AD × CF
=(AD×10)cm2 … (2)
From (1) and (2) we get
128 = AD × 10
⇒ AD=12810cm=12.8cm
Q.2 If E, F, G and H are respectively the mid points of the sides of a parallelogram ABCD, show that ar(EFGH) =12ar(ABCD)
Sol. Δ HGF and  gm HDCF stand on the same base HF and lie between the same parallels HF and DC.
Therefore, ar(ΔHGF)=12ar(HDCF) … (1)
Similarly, Δ HEF and gm ABFH stand on the same base HF and lie between the same parallels
HF and AB.
Therefore ar(ΔHEF)=12ar(ABFH) … (2)
Therefore Adding (1) and (2), we get
ar(ΔHGF)+ar(ΔHEF)=12ar(HDCF)+ar(ABFH)
⇒ ar(EFGH)=12ar(ABCD)
Q.3 P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar(APB) = ar(BQC).
Sol. Δ APB and  gm ABCD stand on the same base AB and lie between the same parallels AB and DC.
Therefore ar(ΔAPB)=12ar(ABCD)…. (1)
Similarly , Δ BQC and  gm ABCD stand on the same base BC and lie between the same parallels BC and AD.
Therefore ar(ΔBQC)=12ar(ABCD) …. (2)
From (1) and (2) , we have
ar(ΔAPB) = ar (ΔBQC)
Q.4 In figure, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar (APB) + ar (PCD) = 12ar(ABCD)
(ii) ar (APD) + ar(PBC) = ar (APB) + ar (PCD).
Sol. Draw EPF parallel to AB or DC and GPH parallel to AD or BC.
Now AGHD is a gm
[Since GH  DA and AG DH]
Similarly, HCBG, EFCD and ABFE are parallelograms.
(i) Δ APB and  gm ABFE stand on the same base AB and lie between the same parallels AB and EF.
Therefore ar(APB)=12ar(ABFE) … (1)
Similarly, ar(PCD)=12ar(EFCD) … (2)
Adding (1) and (2) , we get
ar(APB)+ar(PCD)=12[ar(ABFE)+ar(EFCD)]
ar(APB)+ar(PCD)=12ar(ABCD)....(3)
(ii) Δ APD and  gm AGHD are on the same base AD and lie between the same parallels AD and HG.
Therefore ar(APD)=12ar(AGHD)....(4)
Similarly, ar(PCB)=12ar(GBCH)....(5)
Adding (4) and (5), we get
ar(APD)+ar(PCB)=12[ar(AGHD)+ar(GBCH)]
ar(APD)+ar(PCB)=12ar(ABCD)...(6)
From (3) and (6) we get
ar (APD) + ar (PBC) = ar (APB) + ar(PCD).
Q.5 In figure PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) =12 ar (PQRS).
(i)  gm PQRS and  gm ABRS stand on the same base RS and lie between the same parallels SR and PAQB.
Therefore ar(PQRS) = ar (ABRS) … (1)
(ii) Δ AXS and  gm ABRS stand on the same base AS and lie between the same parallels AS and RB.
Therefore ar (AXS) =12 ar (ABRS)
⇒ ar (AXS) =12 ar (PQRS) [Using (1)]
Q.6 A farmer was having a field in the form of a parallelogram PQRS. He took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should he do it?
Sol. Clearly, the field i.e.,  gm PQRS is divided into 3 parts. Each part is of the shape of triangle.
Since Δ APQ and  gm PQRS stand on the same base PQ and lie between the same parallels PQ and SR.
Therefore ar (APQ) =12 ar (PQRS) … (1)
Clearly, ar (APS) + ar (AQR) = ar (PQRS) – ar (APQ)
= ar (PQRS) −12ar(PQRS) [Using (1)]
=12ar(PQRS)....(2)
From (1) and (2), we get
ar (APS) + ar (AQR) = ar (APQ)
Thus , the farmer should sow wheat and pulses either as [(Δs APS and AQR) or Δ APQ] or as
[Δ APQ or (Δs APS and AQR )]
Exercise 9.3
Q.1 In figure , E is any point on median AD of a Δ ABC. Show that ar (ABE) = ar (ACE).
Given : AD is a medium of Δ ABC and E is any point on AD.
To prove : ar (ABE) = ar (ACE)
Proof : Since AD is the median of Δ ABC
Therefore ar (ABD) = ar (ACD) … (1)
Also , ED is the median of Δ EBC
Therefore ar (BED) = ar (CED) … (2)
Subtracting (2) from (1) , we get
ar (ABD) – ar (BED) = ar (ACD) – ar (CED)
⇒ ar (ABE) = ar (ACE).
Q.2 In a triangle ABC, E is the mid point of median AD. Show that ar (BED) = 14 ar (ABC).
Sol.
Given : A Δ ABC, E is the mid point of the median AD.
To prove : ar (BED) = 14 ar (ABC)
Proof : Since AD is a median of Δ ABC and median divides a triangle into two triangles of equal area.
Therefore ar (ABD) = ar (ADC)
⇒ ar (ABD) = 12 ar (ABC) … (1)
In Δ ABD, BE is the median
Therefore ar(BED) = ar (BAE) … (2)
⇒ ar(BED)=12ar(ABD) [ar(BAE)=12ar(ABD) ]
⇒ ar (BED) = 12×12 ar (ABC) [Using (1)]
⇒ ar (BED) = 14 ar (ABC) .
Q.3 Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Sol.
Given : A parallelogram ABCD.
To prove : The diagonals AC and BD divide the  gm ABCD into four triangles of equal area.
Construction : Draw BL ⊥ AC.
Proof : Since ABCD is a  gm and so its diagonals AC and BD bisect each other at O.
Therefore AO = OC and BO = OD
Now, ar (AOB) = 12 × AO × BL
ar (OBC) = 12 × OC × BL
But AO = OC
Therefore ar (AOB) = ar (OBC)
Similarly, we can show that
ar (OBC) = ar (OCD) ; ar (OCD) = ar(ODA) ;
ar (ODA) = ar (OAB) ; ar (OAB) = ar (OBC)
ar (OCD) = ar (ODA)
Thus , ar (OAB) = ar (OBC) = ar(OCD) = ar (OAD)
Q.4 In 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 (ABC) = ar (ABD).
Sol.
Given : ABC and ABD are two triangles on the same base AB. A line segment CD is bisected by AB at O i.e., OC = OD.
To prove : ar (ABC) = ar (ABD).
Proof : In Δ ACD, we have
OC = OD [Given]
Therefore AO is the median.
Therefore ar (AOC) = ar (AOD) [Since median divides a Δ in two Δs of equal area]
Similarly, in Δ BCD, BO is the median
Therefore ar (BOC) = ar (BOD)
Adding (1) and (2), we get
ar (AOC) + ar (BOC) = ar (AOD) + ar (BOD)
⇒ ar (ABC) = ar (ABD)
Q.5 D, E and F are respectively the mid points of the side BC, CA and AB of a Δ ABC. Show that
(i) BDEF is a parallelogram
(ii) ar (DEF) = 14 ar (ABC)
(iii) ar (BDEF) = 12 ar (ABC).
Sol.
Given : D, E and F are the mid points of the sides BC, CA and AB respectively of Δ ABC.
To prove : (i) BDEF is a  gm
(ii) ar (DEF) = 14 ar (ABC)
(iii) ar (BDEF) = 12 ar (ABC)
(i) We have in Δ ABC,
EF  BC [By mid point theorem, since E and F are the mid points AC and AB respectively]
Therefore EF  BD… (1)
Also ED  AB [By mid point theorem, since E and D are the mid points AC and BC respectively]
Therefore ED  BF… (2)
From (1) and (2), BDEF is a  gm.
(ii) Similarly, FDCE and AFDE are  gms.
Therefore ar (FBD) = ar (DEF) [Since FD is a diagonal of  gm BDEF]… (3)
ar (DEC) = ar (DEF) [Since ED is a diagonal of  gm FDCE]… (4)
and , ar (AFE) = ar (DEF) [Since FE is a diagonal of  gm AFDE… (5)]
From (3), (4) and (5)
ar (FBD) = ar (DEC) = ar (AFE) = ar (DEF) …. (6)
ar (ABC) = ar (AFE) + ar (FBD) + ar (DEC) + ar (DEF)
from equation (6) –
ar(ABC) = 4ar(DEF)
⇒ ar (DEF) = 14 ar (ABC).
(iii) Also, ar (BDEF) = 2 ar (DEF)
=2×14ar(ABC)=12ar(ABC)
Q.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 (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA  CB or ABCD is a parallelogram.
Sol.
(i) Draw DN ⊥ AC and BM ⊥ AC.
In Δs DON and BOM
∠DNO=∠BMO [Each = 90º]
∠DON=∠BOM [Vert. opp ∠s]
OD = OB [Given]
By AAS criterion of congruent.
ΔDON≅ΔBOM … (1)
In Δs DCN and BAM
∠DNC=∠BMA [Each = 90º]
DC = AB [Given ]
DN = BM [Since ΔDON≅ΔBOM⇒DN=BM]
Therefore by RHS criterion of congruence,
ΔDCN≅ΔBAM … (2)
From (1) and (2), we get
ar (DON) + ar (DCN) = ar (BOM) + ar (BAM)
⇒ ar (DOC) = ar (AOB)
(ii) Since ar (DOC) = ar (AOB)
Therefore ar (DOC) + ar (BOC) = ar (AOB) + ar (BOC)
⇒ ar (DCB) = ar (ACB)
(iii) Δs DCB and ACB have equal areas and have the same base. So, these Δs lie between the same parallels.
⇒ DA  CB i.e., ABCD is a gm
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Q.7 D and E are points on sides AB and AC respectively of Δ ABC such that ar(DBC) = ar(EBC). Prove that DE BC.
Sol. Since Δs DBC and EBC are equal in area and have a same base BC.
Therefore altitude from D of Δs DBC = Altitude from E of Δs EBC.
⇒ Δs DBC and EBC are between the same parallels.
⇒ DE  BC
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).
Sol.
Since XY BC and BE  CY
Therefore BCYE is a  gm.
Since Δ ABE and gm BCYE are on the same base BE and between the same parallel lines BE and AC.
Therefore ar(ABE) =12 ar(BCYE) … (1)
Now, CF  AB and XY  BC
⇒ CF  AB and XF  BC
⇒ BCFX is a  gm
Since Δ ACF and  gm BCFX are on the same base CF and between the same parallel AB and FC .
Therefore ar (ACF) =12 ar (BCFX) … (2)
But gm BCFX and  gm BCYE are on the same base BC and between the same parallels BC and EF.
Therefore ar (BCFX) = ar(BCYE) … (3)
From (1) , (2) and (3) , we get
ar (Δ ABE) = ar(Δ ACF)
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 figure). Show that ar(ABCD) = ar (PBQR).
Join AC and PQ. Since AC and PQ are diagonals of  gm ABCD and gm BPQR respectively.
Therefore ar(ABC) = 12 ar (ABCD) … (1)
and ar (PBQ) = 12 ar (BPRQ) … (2)
Now, Δs ACQ and AQP are on the same base AQ and between the same parallels AQ and CP.
Therefore ar(ACQ) = ar(AQP)
⇒ ar(ACQ) – ar(ABQ) = ar(AQP) – ar(ABQ) [Subtracting ar(ABQ) from both sides]
⇒ ar(ABC) = ar(BPQ)
⇒ 12 ar(ABCD) = 12 ar(BPRQ) [Using (1) and (2)]
⇒ ar(ABCD) = ar (BPRQ).
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Q.10 Diagonals AC and BD of a trapezium ABCD with AB  DC intersect each other at O. Prove that ar(AOD) = ar (BOC).
Sol.
Diagonals AC and BD of a trapezium ABCD with AB DC intersect each other at O.
Therefore Δs ABC and ABD are on the same base AB and between the same parallels AB and DC.
Therefore ar(ABD) = ar(ABC)
⇒ ar(ABD) – ar(AOB) = ar(ABC) – ar (AOB) [Subtracting ar (AOB) from both sides]
⇒ ar (AOD) = ar(BOC)
Q.11 In figure , 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)
Sol.
(i) Since Δ s ACB and ACF are on the same base AC and between the same parallels AC and BF.
Therefore ar(ACB) = ar (ACF)
(ii) Adding ar(ACDE) on both sides we get
ar(ACF) + ar(ACDE) = ar(ACB) + ar(ACDE)
⇒ ar(AEDF) = ar(ABCDE)
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 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.
Sol.
Let ABCD be the quadrilateral plot. Produce BA to meet CD drawn parallel to CA at E. Join EC.
Then , Δ s EAC and ADC lie on the same parallels DE and CA
Therefore ar(EAC) = ar (ADC)
Now ar(ABCD) = ar(ABC) + ar(ACD)
= ar(ABC) + ar(ADC)
= ar (ABC) + ar(EAC) = ar(EBC)
i.e., quad. ABCD = Δ EBC
Which is the required explain to the suggested proposal.
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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).
Sol. ABCD is a trapezium with AB  DC and XY AC, is drawn. Join XC.
ar(ACX) = ar(ACY) … (1) [Since Δ s ACX and ACY have same base AC and are between same parallels AC and XY]
But ar (ACX) = ar(ADX) [Since Δs ACX and ADX have same base AX and are between same parallels AB and DC]
From (1) and (2), we have
ar(ADX) = ar (ACY).
Q.14 In figure , AP  BQ CR . Prove that ar (AQC) = ar(PBR)
From the figure, we have
ar (AQC) = ar (AQB) + ar (BQC) … (1)
and ar (PBR) = ar (PBQ) + ar (QBR) … (2)
But ar (AQB) = ar(PBQ) [Since these Δs are on the same base BQ and between same parallel line AP and BQ]
Also, ar (BQC) = ar (QBR) [Since these Δs are on the same base BQ and between same parallel lines BQ and CR]
Using (3) and (4) in (1) and (2) , we get :
ar (AQC) = ar (PBR).
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.
Sol. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that
ar (AOD) = ar (BOC). … (1) [Given]
Adding ar (ODC) on both sides , we get
ar(AOD) + ar (ODC) = ar (BOC) + ar (ODC)
⇒ ar(ADC) = ar (BDC)
⇒ 12×DC×AL=12×DC×BM
⇒ AL = BM
⇒ AB  DC
Hence ABCD is a trapezium.
Q.16 In figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
Sol.
From the figure,
ar (BDP) = ar (ARC) [Given]
and ar (DPC) = ar (DRC) [Given]
On subtracting, we get
ar (BDP) – ar (DPC) = ar (ARC) – ar (DRC)
⇒ ar (BDC) = ar (ADC)
⇒ DC  AB [ar(BDC) and ar(ADC) have equal areas and have the same base. So, these Δs lie between the same parallels.]
Hence , ABCD is a trapezium.
ar (DRC) = ar (DPC) [Given]
On subtracting ar (DLC) from both sides, we get
ar (DRC) – ar (DLC) = ar (DPC) – ar (DLC)
⇒ ar (DLR) = ar (CLP)
ON adding ar (RLP) to both sides, we get
ar (DLR) + ar(RLP) = ar(CLP) + ar(RLP)
⇒ ar (DRP) = ar (CRP)
⇒ RP  DC
Hence, DCPR is a trapezium.
Exercise 9.4
Q.1 Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
Sol.
Given : A  gm ABCD and a rectangle ABEF with same base AB and equal areas.
To prove : Perimeter of  gm ABCD > Perimeter of rectangle ABEF.
Proof : Since opposite sides of a gm and rectangle are equal.
Therefore AB = DC [Since ABCD is a  gm]
and, AB = EF [Since ABEF is a rectangle]
Therefore DC = EF … (1)
⇒ AB + DC = AB + EF (Add AB in both sides) … (2)
Since, of all the segments that can be drawn to a given line from a point not lying on it, the perpendicular segment is the shortest.
Therefore BE < BC and AF < AD
⇒ BC > BE and AD > AF
⇒ BC + AD > BE + AF … (3)
Adding (2) and (3), we get
AB + DC + BC + AD > AB + EF + BE + AF
⇒ AB + BC + CD + DA > AB + BE + EF + FA
⇒ perimeter of  gm ABCD > perimeter of rectangle ABEF.
Hence,the perimeter of the parallelogram is greater than that of the rectangle.
Q.2 In figure , D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).
Can you know answer the question that you have left in the ‘Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?
Let AL be perpendicular to BC. So, AL is the height of Δs ABD, ADE and, AEC.
Therefore ar (ABD) =12 × BD × AL
ar (ADE) =12 ×DE × AL
and, ar (AEC) =12 × EC × AL
Since BD = DE = EC
Therefore ar (ABD) = ar (ADE) = ar(AEC)
Yes , altitudes of all triangles are same. Budhia has use the result of this question in dividing her land in three equal parts.
Q.3 In figure , ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF)
Since opposite sides of a gm are equal
Therefore , AD = DC [Since ABCD is a  gm]
DE = CF [Since DCFE is a  gm]
and AE = BF [Since ABFE is a  gm]
Consider Δs ADE and BCF, in which AE = BF , AD = BC and DE = CF
Therefore by SSS criterion of congruence
ΔADE≅ΔBCF
⇒ ar(ADE) = ar (BCF)
Q.4 In figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ).
Join AC.
Since Δs APC and BPC are on the same base PC and between the same parallels PC and AB.
Therefore , ar(APC) = ar (BPC) … (1)
Since AD = CQ
and , AD  CQ [Given]
Therefore in the quadrilateral ADQC, one pair of opposite sides is equal and parallel.
Therefore ADQC is a parallelogram.
⇒ AP = PQ and CP = DP [Since diagonals of a gm bisect each other]
In Δs APC and DPQ we have
AP = PQ [Proved above]
∠APC=∠DPQ [Vertically opp. ∠s]
and , PC = PD [Proved above]
Therefore by SAS criterion of congruence,
ΔAPC≅ΔDPQ
⇒ ar (APC) = ar (DPQ) … (2)
[Since congruent Δs have equal area]
Therefore ar (BPC) = ar (DPQ) [From (1)]
Hence , ar (BPC) = ar (DPQ)
Q.5 In figure, ABC and BDE are two equilateral triangles such that D is the mid point of BC. If AE intersects BC at F, Show that
(i) ar (BDE) = 14 ar (ABC)
(ii) ar (BDE) = 12 ar (BAE)
(iii) ar (ABC) = 2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED) = 18 ar (AFC).
Sol.
Join EC and AD. Let a be the side of Δ ABC. then,
ar(ABC)=3√4a2=Δ(say)
(i) ar(BDE)=3√4(a2)2
[SinceBD=12BC=a2]
=3√16a2=Δ4
⇒ ar(BDE)=14ar(ABC)
(ii) We have , ar (BDE) =12 ar (BEC) … (1)
[Since DE is a median of Δ BEC and each median divides a triangle
in two other Δs of equal area]
Now, ∠EBC=60o
and ∠BCA=60o
⇒ ∠EBC=∠BCA
But these are alterante angles with respect to the line segments BE and CA and their transversal BC.
Hence BE  AC.
Now, Δs BEC and BAE stand on the same base BE and lie between the same parallels BE and AC.
Therefore ar(BEC) = ar(BAE)
Therefore From (1) ar (Δ BDE) = 12 ar (BAE).
(iii) Since ED is a median of Δ BEC and we know that each median divides a triangle in two other Δs of equal area.
Therefore ar(BDE) = 12 ar (BEC)
From part (i),
ar (BDE) = 14 ar (ABC)
Combining these results , we get
14 ar (ABC) = 12 ar (BEC)
⇒ ar (ABC) = 2 ar (BEC)
(iv) Now,∠ABD=∠BDE=60∘ [Given]
But ∠ABDand∠BDE are alternate angles with respect to the line segment BA and DE and their transversal BD.
Hence BA  ED.
Now, Δs BDE and AED stand on the same base ED and lie between the same parallels BA and DE
Therefore ar (BDE) = ar (AED)
⇒ ar(BDE) – ar(FED) = ar (AED) – ar (FED)
⇒ ar (BEF) = ar (AFD)
(v) In Δ ABC, AD2=AB2−BD2
=(a)2−(a2)2
=a2−a24=3a24
⇒ AD=3√2a
In Δ BED , EL2=DE2−DL2
=(a2)2−(a4)2=a24−a216=3a216 [EL is median ofΔ BED]
⇒ EL=3√a4
Therefore ar(AFD) = 12 × FD × AD
= 12 × FD × 3√2a … (1)
and ar(EFD) = 12 × FD × EL
= 12 × FD = 3√4a … (2)
From (1) and (2), we have ar (AFD) = 2 ar (EFD)
Combining this result with part (iv).
We have ar (BFE) = ar (AFD) = 2ar (EFD)
(vi) From part (i)
ar (BDE) = 14 ar (ABC)
⇒ ar (BEF) + ar (FED) = 14 × 2ar (ADC)
⇒ 2ar (FED) + ar (FED) = 12 (ar(AFC) – ar (AFD) [Using part (v)]
⇒ 3ar (FED) = 12 ar (AFC) – 12 × 2 ar (FED)
⇒ 4ar (FED) = 12 ar (AFC)
⇒ ar (FED) = 18 ar (AFC)
Q.6 Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
Sol. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.
Draw AM ⊥ BD and CN ⊥ BD.
Now, ar (APB) × ar(CPD)
=(12×BP×AM)×(12×DP×CN)
=14 × BP × DP × AM × CN
and , ar (APD) × ar (BPC)
=(12×DP×AM)×(12×BP×CN)
=14 × BP × DP × AM × CN
From (1) and (2), we have
ar (APB) × ar (CPD) = ar (APD) × ar (BPC)
Q.7 P and Q are respectively the mid points of sides AB and BC of of a triangle ABC and R is the mid point of AP, show that
(i) ar (PRQ) = 12 ar (ARC)
(ii) ar (RQC) = 38 ar (ABC)
(iii) ar (PBQ) = ar (ARC)
Sol.
P and Q are respectively the mid points of sides AB and BC of Δ ABC and R is the mid point of AP.
Join AQ and PC.
(i) We have,
ar (PQR) = 12 ar (APQ) [Since QR is a median of Δ APQ and it divides the Δ into two other Δs of equal area]
=12×12 ar (ABQ) [Since QP is a median of Δ ABQ]
=14 ar (ABQ) = 14×12 ar (ABC)
[Since AQ is a median of Δ ABC]
=18 ar (ABC) … (1)
Again , ar(ARC) = 12 ar (APC) [Since CR is a median of Δ APC]
=12×12 ar (ABC) [Since CP is a median of Δ ABC]
=14 ar (ABC) … (2)
From (1) and (2) , we get
ar (PQR) = 18 ar (ABC) = 12×14 ar (ABC)
=12 ar (ARC).
(ii) We have,
ar (RQC) = ar(RQA) + ar (AQC) – ar (ARC) … (3)
Now, ar (Δ RQA) = 12 ar (PQA) [Since RQ is a median of Δ PQA]
=12×12 ar (AQB) [Since PQ is a median of Δ AQB]
=14 ar (AQB)
=14×12 ar(ABC) [Since AQ is a median of Δ ABC]
=18 ar (ABC) … (4)
ar (AQC) = 12 ar (ABC) … (5) [Since AQ is a median of Δ ABC]
ar (Δ ARC) = 12 ar (APC) [Since CR is a median of Δ APC]
=12×12 ar(ABC) [Since CP is a median of Δ ABC]
=14 ar (ABC) … (6)
From (3), (4) , (5) and (6) we have
ar (RQC) = 18 ar (ABC) + 12 ar (ABC) – 14 ar (ABC)
=(18+12−14) ar (ABC)
=38 ar (ABC)
(iii) We have,
ar (PBQ) = 12 ar (ABQ) [Since PQ is a median of Δ ABQ]
=12×12ar(ABC) [Since AQ is a median of Δ ABC]
=14 ar (ABC)
= ar(ARC) [Using (6)]
Q.8 In figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that :
(i) ΔMBC≅ΔABD
(ii) ar (BYXD) = 2 ar (MBC)
(iii) ar (BYXD) = ar (ABMN)
(iv) ΔFCB≅ΔACE
(v) ar (CYXE) = 2 ar (FCB)
(vi) ar (CYXE) = ar (ACFG)
(vii) ar (BCED) = ar (ABMN) + ar(ACFG)
Note : Result (vii) is the famous theorem of pythagoras. You shall learn a simpler proof of this theorem in class X.
Sol.
(i) In Δs MBC and ABD, we have
BC = BD [Sides of the square BCED]
MB = AB [Sides of the square ABMN]
∠MBC=∠ABD [Since Each = 90º + ∠ABC]
Therefore by SAS criterion of congruence, we have
ΔMBC≅ΔABD
(ii) Δ ABD and square BYXD have the same base BD and are between the same parallels BD and AX.
Therefore ar(ABD) =12 ar (BYXD)
But ΔMBC≅ΔABD [Proved in part (i)]
⇒ ar(MBC) = ar (ABD)
Therefore ar(MBC) = ar (ABD) = 12 ar (BYXD)
⇒ ar(BYXD) = 2 ar (MBC).
(iii) Square ABMN and Δ MBC have the same base MB and are between same parallels MB and NAC.
Therefore ar(MBC) = 12 ar (ABMN)
⇒ ar(ABMN) = 2ar (MBC)
= ar (BYXD) [Using part (ii)]
(iv) In Δs ACE and BCF, we have
CE = BC [Sides of the square BCED]
AC = CF [Sides of the square ACFG]
and ∠ACE=∠BCF [Since Each = 90º + ∠BCA]
Therefore by SAS criterion of congruence,
ΔACE≅ΔBCF
(v) Δ ACE and square CYXE have the same base CE and are between same parallels CE and AYX.
Therefore ar(ACE) = 12 ar (CYXE)
⇒ ar(FCB) = 12 ar (CYXE) [Since ΔACE≅ΔBCF, part (iv)]
⇒ ar(CYXE) =2 ar (FCB) .
(vi) Square ACFG and Δ BCF have the same base CF and are between same parallels CF and BAG.
Therefore ar(BCF) = 12 ar(ACFG)
⇒ 12 ar(CYXE) = 12 ar (ACFG) [Using part (v)]
⇒ ar(CYXE) = ar (ACFG)
(vii) From part (iii) and (vi) we have
ar (BYXD) = ar (ABMN)
and ar (CYXE) = ar (ACFG)
On adding we get
ar (BYXD) + ar(CYXE) = ar (ABMN) + ar(ACFG)
ar (BCED) = ar (ABMN) + ar(ACFG)