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Question-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.


Solution:
(i) The figures (quadrilaterals APCD and ABCD), and (quadrilaterals PBCD and ABCD), lie on the same base DC and between the same parallels DC and AB.

(iii) The figures (
ΔTRQ and parallelogram SRQP), (quadrilaterals TPQR and parallelogram SRQP), (quadrilateral STQR and SRQP), lie on the same base RQ and between the same parallels RQ and SP.

(v) Quadrilaterals APCD and ABQD lie on the same base AD and between the same parallels AD and BQ.

Question-2

In the figure, ABCD is a parallelogram, AE DC And CF AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Solution:

Area of parallelogram ABCD = CD × AE = 16 × 8 cm2         ( AB = CD, ABCD is a parallelogram)
                                                   = 128 cm2       ….(1)
Also area of parallelogram ABCD= AD × CF = AD × 10 cm2 ….(2)

From (1) and (2), we get
AD × 10 = 128
AD = AD = 12.8 cm

Question-3

 E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD. Show that ar (EFGH) = ar(ABCD). 
  
Solution:

Given: E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD.

To Prove: Area(EFGH) = Area(ABCD)
 
Construction:
Join OF, OG, OH and OE. Also, join diagonals AC and BD to intersect at O.

Proof: In
ΔBCD,
F and G are the mid-points of BC and DC respectively.
FG || BD …..(1)       ( In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side)  
In Δ BAD,

E and H are the mid-points of AB and AD respectively.

EH || BD    ..…..(2)    (∵  In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side) 
From (1) and (2),
EH || BD || FG

Hence EH || FG..........(3)

Similarly, we can prove that
EF || HG ...........(4)
From (3) and (4),
Quadrilateral EFGH is a parallelogram  (  Opposite sides are parallel)

A quadrilateral is a parallelogram if its opposite sides are equal

F is the mid-point of CB and O is the mid-point of CA

FO || BA                 

FO || CG   .............(5)  (  BA || CD, Opposite sides of a parallelogram are parallel)
BA || CG

and FO = BA  CD          ( Opposite sides of a parallelogram are equal)   
           = CG                 ...(6)   ( G is the mid-point of CD )
In view of (5) and (6),

Quadrilateral OFCG is a parallelogram
(  A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length)
OP = PC              | Diagonals of a parallelogram bisect each other

Δ OPF and Δ CPF have equal bases (  OP = PC) and have a common vertex F
Their altitudes are also the same Area(Δ OPF) = Area(Δ CPF)


Similarly, Area(
Δ OQF) = Area(Δ BQF)

Adding, we get


Ar(Δ OPF) + Ar(Δ OQF) = Ar(Δ CPF) + Ar(Δ BQF)
Area of parallelogram OQFP = Area(Δ CPF) + Area(Δ BQF)  ...(7)
Similarly, Area of parallelogram OPGS = Area(Δ GPC) + Area(Δ DSG) ...(8)

Area of parallelogram OSHR = Area(Δ DSH) + Area(Δ HAR)       ...(9)

Area of parallelogram OREQ = Area(Δ ARE) + Area(Δ EQB)     ...(10)

Adding the corresponding sides of (7), (8), (9) and (10), we get

Area(parallelogram EFGH) = {Area(Δ CPF) + Area(Δ GPC)} + {Ar(Δ DSG) + ArΔ DSH)} + {Area(Δ HAR) + Area(Δ ARE)} 

                                                + {Area(Δ BQF) +Area(Δ EQB)}

                     = Area(
Δ FCG) +Area(Δ GDH) + Area(Δ HAE) + Area(Δ EBF)

                     = Area(parallelogram ABCD) - Area(parallelogram EFGH)
 
 2 × Area(parallelogram EFGH) = Area(parallelogram ABCD)

  Area of parallelogram EFGH = × Area of parallelogram ABCD

Question-4

P and Q are any two points lying on the sides DC and AD respectively of a Parallelogram ABCD. Show that
area(APB) = area(BQC).

Solution:
                                                                               
Given: P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.

To Prove: Area(Δ APB) = Area(Δ BQC).

Proof:  Δ APB and parallelogram ABCD are on the same base AB and between the same parallels AB and DC.

Area(Δ APB)=  Area of parallelogram ABCD           ...(1)

Δ BQC and parallelogram ABCD are on the same base BC and between the same parallels BC and AD.

 
Area(Δ BQC) = Area( parallelogram ABCD)          ...(2)
From (1) and (2),
Area(Δ APB) = Area(Δ BQC).

Question-5

 In the figure P is a point in the interior of a parallelogram ABCD. Show that
 

(i) area(APB) + area(PCD) =  area(ABCD)
(ii) area(APD) + area(PBC) = area(APB) + area(PCD).

[Hint: through P, draw a line parallel to AB]

 


Solution:
Given: P is a point in the interior of a parallelogram ABCD.

To Prove: (i) Area(APB) + Area(PCD) = 
Area(ABCD)



(ii) Area(APD) + Area(PBC) = Area(APB) + Area(PCD)

Construction: Through P, draw a line EF parallel to AB.

Proof: (i) EF || AB                            ...(1) ( by construction )

AD || BC          ( Opposite sides of a parallelogram are parallel )
AE || BF                                      ...(2)

In view of (1) and (2),
Quadrilateral ABFE is a parallelogram    ( A quadrilateral is a parallelogram if its opposite sides are parallel )
Similarly, quadrilateral CDEF is a parallelogram

Δ APB and parallelogram ABFE are on the same base AB and between the same parallels AB and EF.

Area(Δ APB)=  Area(parallelogram ABFE)         ...(3)

Δ PCD and parallelogram CDEF are on the same base DC and between the same parallels DC and EF.

Area(Δ PCD) =   Area(parallelogram CDEF)        ...(4)
Adding (3) and (4), we get
Area(Δ APB) + Area(Δ PCD) =   Area(parallelogram ABFE) +Area(parallelogram CDEF)

                                     =   [Area(parallelogram ABFE) +Area(parallelogram CDEF)]
                                     =  Area(parallelogram ABCD)                        
Area(Δ APB) +Area(Δ PCD) =  Area(parallelogram ABCD) ...(5)

(ii) Area(Δ APD) + Area(Δ PBC) = Area(parallelogram ABCD) - Area ΔAPB + Area ΔPCD)]

                                         = 2 [Area(Δ APB) + Area(Δ PCD)] - [AreaΔ APB + AreaΔ PCD]     ( from (5) )

                                         = Area(Δ APB) + Area(Δ PCD) 
 Area(Δ APD) + Area(Δ PBC) = Area(Δ APB) +Area(Δ PCD)

Question-6

 In the figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) area(PQRS) = area(ABRS)

(ii) area(AXS) = 
 area(PQRS)
 
 

 


Solution:
Given: PQRS and ABRS are parallelograms and X is any point on side BR.

To Prove:

(i) area(PQRS) = area(ABRS)
(ii) area(AXS) =  area(PQRS)

Proof: (i) In Δ PSA and Δ QRB,
SPA = RQB             ...(1) | Corresponding angles from PS || QR and transversal PB
PAS = QBR             ...(2) | Corresponding angles from AS || BR and transversal PB
PSA = QRB             ...(3) | Angle sum property of a triangle
Also, PS = QR               ...(4) | Opposite sides of parallelogram PQRS

In view of (1), (3) and (4),

Δ PSA Δ QRB               ...(5) | By ASA Rule
area(Δ PSA) = area(Δ QRB) ...(6) | Congruent figures have equal areas
Now, area(PQRS) = area(Δ PSA) + area(AQRS)

                              = area(Δ QRB) + area(AQRS)         | Using (6)

                        = ar(ABRS)

(ii)
Δ AXS and parallelogram ABRS are on the same base AS and between the same parallels AS and BR.
area(Δ AXS) =  area(parallelogram ABRS)

                    = 
{area(AQRS) + area(Δ QRB)}
                    =  {area(AQRS) + area(Δ PSA)}           | Using (6)
                    =  area(parallelogram PQRS).

Question-7

A farmer has 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 can the fields be 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 ?

Solution:
The field is divided into three triangles. The three triangles are:
(i) Δ APS

(ii)
Δ APQ

(iii)
Δ AQR
If the farmer wants to sow wheat and pulses in equal portions of the field separately, then A must be taken such that
SA = SR. Also, B must be taken such that
PB = . Let us assume that SA = SR and PB = .
Now, we are going to prove area(Δ ASP) = area(parallelogram ARBP) = area(Δ BQR),
so that the farmer can sow wheat and pulses in equal portions of the field seperately.

    
 
   SR || PQ                            ( PQRS is a parallelogram )
AR || PB               …..(1)
Again AR =               (by assumption SA = SR )
PB =                        ( by assumption PB = )
SR = PQ                            (  Opposite sides of a parallelogram are equal )

AR = PB                   …..(2)
In view of (1) and (2),
ARBP is a parallelogram.    ( A quadrilateral is a parallelogram if one of its pair of opposite sides is parallel and of equal length.)

Δ ASP, parallelogram ARBP and Δ BQR are in between the same parallels PQ and SR.
Their altitudes are equal. Let it be x.

Now, Area of Δ ASP = =                (by assumption)

                                          =                                                ….(3)

Area of parallelogram ARBP = (AR)(x) =        ...(4)   (by assumption

Area of Δ BQR = =                        (by assumption

                                             =
                                    =                                                .…(5)
In view of (3), (4) and (5)
area(Δ ASP) = area(parallelogram ARBP) = area(Δ BQR)

Question-8

In the given figure, E is any point on the median AD of a ΔABC. Show that area(ABE) = area(ACE).
 

 


Solution:
Given: E is any point on median AD of a Δ ABC.

To Prove: area(Δ ABC) = area(Δ ACE).

Proof: In Δ ABC,
AD is a median.

area(Δ ABD) = area(Δ ACD) ….(1)
In Δ EBC,
ED is a median.

area(Δ EBD) = area(Δ ECD) …..(2)
Subtracting (2) from (1), we get
area(Δ ABD) – area(Δ EBD) = area(Δ ACD) – area(Δ ECD)
area(Δ ABE) = area(Δ ACE). 

Question-9

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

Solution:

 Given: In a triangle ABC, E is the mid-point of median AD.

To Prove: area(
Δ BED) = area(Δ ABC).

Proof: In Δ ABC,

AD is a median.
area(Δ ABD) = area(Δ ACD) = area(Δ ABC)…..(1)   ( median of a triangle divides it into two triangles of equal areas.)

In Δ ABD,

BE is a median.

area(Δ BED) = area(Δ BEA) =  area(Δ ABD)

area(Δ BED) = area(Δ ABD) =.area(Δ ABC) (From(1) )

                    = area (Δ ABC).

Question-10

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 intersect at O.
Divide it into four trianlges. They are ΔOAB, ΔOBC, ΔOCD and ΔODA.

To Prove: Area(Δ OAB) = Area(Δ OBC) = Area(Δ OCD) = Area(Δ ODA).

Construction: Draw BE AC.

Proof: ABCD is a parallelogram

OA = OC and OB = OD   (  Diagonals of a parallelogram bisect each other)


Now, Area(Δ OAB) = =

and Area(Δ OBC) = =
But OA = OC

Area(Δ OAB) = Ar(Δ OBC)                           .….(1)
Similarly,
Area(Δ OBC) = Ar(Δ OCD)                               …..(2)

and Area(Δ OCD) = Ar(Δ ODA)                        .….(3)
From (1), (2) and (3),
Area(ΔOAB) = Area(ΔOBC) = Area(ΔOCD) = Area(ΔODA).

Question-11

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 Area(ΔABC) = Area(ΔABD). 


Solution:
Given: ABC and DBC are two triangles on the same base AB. Line segment CD is bisected by AB at O.

To Prove: Area(Δ ABC) = Area(Δ ABD).

Proof:  Line segment CD is bisected by AB at O.
 OC = OD
⇒  BO is a median of Δ BCD and AO is a median of Δ ACD.
BO is a median of Δ BCD. 

Area(ΔOBC) = Area(ΔOBD)             …………(1)

AO is a median of Δ ACD

Area(Δ OAC) = Area(Δ OAD)           ………….(2)

Adding (1) and (2), we get
Area(Δ OBC) + Area(Δ OAC) = Area(Δ OBD) + Area(Δ OAD)
Area(Δ ABC) = Area(Δ ABD).

Question-12

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) area(DEF) =  area(ABC)

(iii) area(BDEF) = 
area(ABC)

Solution:


Given: D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC.

To Prove:
(i) BDEF is a parallelogram
(ii) area(
Δ DEF) =area(Δ ABC) 
(iii) area(BDEF) =  area(Δ ABC).

Proof: (i) In Δ ABC,                                                                                
F is the mid-point of side AB and E is the mid-point of side AC.

EF || BC | In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side.
EF || BD                .......... (1)

Similarly, ED || BF       .......... (2)
In view of (1) and (2),
BDEF is a parallelogram.      (A quadrilateral is a parallelogram if opposite sides are parallel)

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

AFDE and FDCE are parallelograms.

FD is a diagonal of the parallelogram BDEF.

area(Δ FBD) = area(Δ DEF)                               ...(3)

Similarly, area(Δ DEF) = area(Δ FAE)                      ...(4)
and, area(Δ DEF) = area(Δ DCE)                            ...(5)
From (3), (4) and (5), we have
area(Δ FBD) = area(Δ DEF) = area(Δ FAE) = area(Δ DCE) ...(6)

Δ ABC is divided into four non-overlapping triangles Δ FBD, Δ DEF, Δ FAE and Δ DCE 

area(Δ ABC) = area(Δ FBD) + area(Δ DEF) + area(Δ FAE) + area(Δ DCE)
                    = 4area(Δ DEF)                          | From (6)
area(Δ DEF) = area(Δ ABC)                               ...(7)

(iii) area(BDEF) = area(Δ FBD) + ar(Δ DEF)            | From (3)
                     = area(Δ DEF) + ar(Δ DEF) 
                     = 2 area(Δ DEF)
                     = 2.area(Δ ABC)                       | From (7)
                     = area(Δ ABC)

Question-13

In the figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i) area(DOC) = area(AOB)
(ii) area(DCB) = area(ACB)
(iii) DA || CB or ABCD is a parallelogram.

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

Solution:
 Given: Diagonals AC and BD of a quadrilateral ABCD intersect at O such that OB = OD.

To Prove: If AB = CD, then

(i) area(Δ DOC) = area(Δ AOB)

(ii) area(
Δ DCB) = area(Δ ACB)

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

Construction: Draw DE
AC and BF AC.



Proof: (iii) In Δ ADB,
AO is a median
area(Δ AOD) = area(Δ AOB)                   ....(1) | A median of a triangle divides it into two triangles of equal areas

In ΔCBD,
CO is a median.
area(Δ COD) = area(Δ COB)                   ....(2) | A median of a triangle divides it into two triangles of equal areas

Adding (1) and (2), we get
area(Δ AOD) + area(Δ COD) = area(Δ AOB) + area(Δ COB)
area(Δ ACD) = area(Δ ACB)

=     ( Area of triangle = Base × Corresponding altitude) 


DE = BF ...(3)
In right as DEC and BFA,
Hyp. DC = Hyp. BA           | given
DE = BF                          | From (3)


Δ DEC Δ BFA              | R. H. S. Rule

DCE = BAF             | CPCT
But these angles form a pair of equal alternate interior angles.
DC || AB                   ...(4)
DC = AB and DC || AB

ABCD is a parallelogram. ( A quadrilateral is a parallelogram if a pair of opposite sides is parallel and equal)
DA || CB                     (Opposite sides of a parallelogram are parallel)
(i) ABCD is a parallelogram
OC = OA        ...(5)      (Diagonals of a parallelogram bisect each other)
area(ΔDOC) =
area(Δ AOB) =
DE = BF          ( From (3) )
and OC = OA       ( From (5) )
area(Δ DOC) = area(Δ AOB).

(ii) From (i), area(
Δ DOC) = area(Δ AOB)
area(Δ DOC) + area(Δ OCB) = area(Δ AOB) + area(Δ OCB)     (Add same area (Δ OCB) on both sides )
area(Δ DCB) = area(Δ ACB)

Question-14

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

Solution:

Given: D and E are points on sides AB and AC respectively of ΔABC such that area(Δ DBC) = area(ΔEBC).

To Prove: DE || BC
Proof: Δ DBC and Δ EBC are on the same base BC and have equal areas.
Their altitudes must be the same.
DE || BC

Question-15

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(DABE) = ar(DACF).

Solution:

Given: XY is a line parallel to side BC of a triangle ABC. BE || AC and CF || AB meet XY at E and F respectively.

To Prove: ar(Δ ABE) = ar(Δ ACF).

Proof: XY || BC (given )
and CF || BX      (  CF || AB (given) )
BCFX is a parallelogram  
BC = XF                             (∵ Opposite sides of a parallelogram are equal )  
BC = XY + YF     …..(1)
Again,
XY || BC                         (   BE || AC )
and BE || CY
BCYE is a parallelogram
BC = YE                      (∵ Opposite sides of a parallelogram are equal )
BC = XY + XE ...(2)
From (1) and (2),
XY + YF = XY + XE

YF = XE
XE = YF

Δ AEX and Δ AYF have equal bases (XE = YF) on the same line EF and have a common vertex A.

Their altitudes are also the same.

ar(Δ AEX) = ar(Δ AFY)     ...(4)

Δ BEX and Δ CFY have equal bases (XE = YF) on the same line EF and are between the same parallels EF and BC (XY || BC).

ar(Δ BEX) = ar(Δ 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(
Δ AEX) + ar(Δ BEX) = ar(Δ AFY) + ar(Δ CFY)
ar(Δ ABE) = ar(Δ ACF).

Question-16

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). 

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

 

 


Solution:
Given: 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.

To Prove: ar(parallelogram ABCD) = ar(parallelogram PBQR).

Construction: Join AC and PQ.
                                                                                                      
Proof: AC is a diagonal of parallelogram ABCD
ar(Δ ABC) = ar(parallelogram ABCD)            …..(1)
PQ is a diagonal of parallelogram BQRP
ar(Δ BPQ) = ar(parallelogram BQRP)          …….(2)
Δ ACQ and Δ APQ are on the same base AQ and between the same parallels AQ and CP.
ar(Δ ACQ) = ar(Δ APQ)         

ar(Δ ACQ) - ar(Δ ABQ) = ar(Δ APQ) - ar(Δ ABQ) (  Subtract the same area ar(Δ ABQ) from both sides )
ar(Δ ABC) = ar(Δ BPQ)
ar(parallelogram ABCD) =ar(parallelogram PBQR)

ar(parallelogram ABCD) = ar(parallelogram PBQR)          | From (1) and (2)

Question-17

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

Solution:
Given: Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O.

To Prove: area(Δ AOD) = area(Δ BOC).

Proof: Δ ABD and Δ ABC are on the same base AB and between the same parallels AB and DC.


ar(Δ ABD) = ar(Δ ABC)   
 

Subtract ar(Δ AOB) both sides,
ar(Δ ABD) - ar(Δ AOB) = ar(Δ ABC) - ar(Δ AOB)     
ar(Δ AOD) = ar(Δ BOC).

Question-18

In the 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)

Solution:
Given: ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F.
To Prove: (i) ar(Δ ACB) = ar(Δ ACF)

               (ii) ar(AEDF) = ar(ABCDE)

                                                                                  
Proof: (i) Δ ACB and ΔACF are on the same base AC and between the same parallels AC and BF.    (  AC || BF (given))

ar(Δ ACB) = ar(Δ ACF) (  Two triangles on the same base (or equal bases) and between the same parallels are equal in area)

∵ ar(Δ ACB) = ar(Δ ACF)
ar(Δ ACB) + ar(AEDC) = ar(Δ ACF) + ar(AEDC)      ( Add ar(AEDC) on both sides )
ar(ABCDE) = ar(AEDF)
ar(AEDF) = ar(ABCDE).

Question-19

A villager Itwaari has a plot of land in 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 center. 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 in the shape of a quadrilateral. Let the portion ADE be taken over by the Gram Panchayat of the village from one corner 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(ΔADE) = ar(ΔPEC).


Proof: ΔDAP and ΔDCP are on the same base DP and between the same parallels DP and AC.
ar(Δ DAP) = ar(Δ DCP) ( Two triangles on the same base (or equal bases) and between the same parallels are equal in area )
ar(Δ DAP) - ar(Δ DEP) = ar(Δ DCP) - ar(Δ DEP) ( Subtract ar(Δ DEP) from both sides )
ar(Δ ADE) = ar(Δ PCE)
ar(Δ ADE ) + ar(ABCE) = ar(Δ PCE) + ar(ABCE) ( Add ar(ABCE) both sides )
ar(ABCD) = ar(Δ ABP)

Question-20

ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X And BC at y. Prove that
area(Δ ADX) = area(Δ 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.

To Prove: ar(
Δ ADX) = ar(Δ ACY)

Construction: Join CX

Proof: Δ ADX and Δ ACX are on the same base AX and between the same parallels AB and DC.


ar(Δ ADX) = ar(Δ ACX)                 ...(1)  
Δ ACX and Δ ACY are on the same base AC and between the same parallels AC and XY.

ar(Δ ACX) = ar(Δ ACY)                ...(2)        

From (1) and (2), we get
ar(Δ ADX) = ar(Δ ACY).

Question-21

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

 

 


Solution:
Given: AP || BQ || CR.
To Prove: ar(Δ AQC) = ar(Δ PBR).
Proof: Δ BAQ and Δ BPQ are on the same base BQ and between the same parallels BQ and AP.
ar(Δ BAQ) = ar(Δ BPQ)          …..(1)      

 Δ BCQ and Δ BQR are on the same base BQ and between the same parallels BQ and CR.
ar(Δ BCQ) = ar(Δ BQR)             .…(2)         

Adding the corresponding sides of (1) and (2), we get
ar(Δ BAQ) + ar(Δ BCQ) = ar(Δ BPQ) + ar(Δ BQR)
ar(Δ AQC) = ar(Δ PBR).

Question-22

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:
                                                                                                             
Given: Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(Δ AOD) = ar(Δ BOC).
To Prove: ABCD is a trapezium.
Proof: ar(Δ AOD) = ar(Δ BOC)
ar(Δ AOD) + ar(Δ AOB) = ar(Δ BOC) + ar(Δ AOB)     ( Adding the same areas on both sides)
ar(Δ ABD) = ar(Δ ABC)
But Δ ABD and Δ ABC are on the same base AB.
Δ ABD and Δ ABC will have equal corresponding altitudes

and Δ ABD and Δ ABC will lie between the same parallels
⇒ AB || DC

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

Question-23

 In figure ar(ΔDRC) = ar(ΔDPC) and ar(ΔBDP) = ar(ΔARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
 

 


Solution:
Given: ar(Δ DRC) = ar(Δ DPC) and ar(Δ BDP) = ar(Δ ARC).

To Prove: Both the quadrilaterals ABCD and DCPR are trapeziums.


Proof: ar(Δ DRC) = ar(Δ DPC)  (given )  ...(1)       
But Δ DRC and Δ DPC are on the same base DC.
Δ DRC and Δ DPC will have equal corresponding altitudes. 
and Δ DRC and Δ DPC will lie between the same parallels.
DC || RP             ( A quadrilateral is a trapezium if exactly one pair of opposite sides is parallel.)  
ž  DCPR is a trapezium.       
Again, ar(Δ BDP) = ar(Δ ARC)

ar(Δ BDC) + ar(Δ DPC) = ar(Δ ADC) + ar(Δ DRC)

ar(Δ BDC) = ar(Δ ADC)       ( Using (1) )
But Δ BDC and Δ ADC are on the same base DC.

Δ BDC and Δ ADC will have equal corresponding altitudes.

and D BDC and Δ ADC will lie between the same parallels.

AB || DC

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




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