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Question-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.

Solution:
Given: Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas.                                                                                                                                                                                  
To Prove: The perimeter of the parallelogram ABCD is greater than that of rectangle ABEF.

Proof: Let ABCD be the parallelogram and ABEF be the rectangle on the same base AB and between the same parallels AB and FC. Then, perimeter of the parallelogram ABCD = 2(AB + AD) and, perimeter of the rectangle ABEF = 2(AB + AF).
In Δ ADF,
  AFD = 90°
  ADF is an acute angle. (< 90°)               ( Angle sum property of a triangle)

AFD > ADF
 AD > AF                                 (∵ Side opposite to greater angle of a triangle is longer)

 AB + AD > AB + AF
⇒  2(AB + AD) > 2(AB + AF)
Perimeter of the parallelogram ABCD > Perimeter of the rectangle ABEF.

 

Question-2

 In the figure, D and E are two points on BC such that BD = DE = EC. Show that
ar(ABD) = ar(ADE) = ar(AEC). 

Can you now 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?
 

 


Solution:
Given: In figure, D and E are two points on BC such that BD = DE = EC.

To Prove: ar(Δ ABD) = ar(Δ ADE) = ar(Δ AEC).

Proof:  Δ ABD and Δ ADE are on equal bases (BD = DE) and have the same vertex A.

Their altitudes are also the same.

ar(Δ ABD) = ar(Δ ADE)                        …..(1)
Similarly, we can prove that
ar(Δ ADE) = ar(Δ AEC)                            …..(2)
From (1) and (2), we get
ar(Δ ABD) = ar(Δ ADE) = ar(Δ AEC).
Thus the field of Budhia has been actually divides into three parts of equal area.

Question-3

 In the figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ADE) = ar(BCF).
 

 


Solution:

Given: ABCD, DCFE and ABFE are parallelograms.

To Prove: ar(Δ ADE) = ar(Δ BCF).

Proof: ABCD is a parallelogram

AB || DC                           …..(1)            | Opposite sides of a parallelogram are parallel
DCFE is a parallelogram
DC || EF                          .....(2)             | Opposite sides of a parallelogram are parallel

From (1) and (2),

AB || EF                               ….(3)
ABCD is a parallelogram
AD = BC                           ...(4)             | Opposite sides of a parallelogram are equal
Δ ADE and Δ BCF are on equal bases (AD = BC) and between the same parallels AB and EF.
ar(Δ ADE) = ar(Δ BCF).       (∵ Two triangles on the same base(or equal bases) and between the same parallels are equal in areas)         

Question-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).   [Hint: Join AC]
                                                 


 

Solution:
Given: ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. AQ intersects DC at P.

To Prove: ar(Δ BPC) = ar(Δ DPQ).
Construction: Join AC.                                                                                                             
  
Proof: Δ QAC and Δ QDC are on the same base QC and between the same parallels AD and QC.
ar( Δ QAC) = ar( Δ QDC)       ...(1)  (∵ Two triangles on the same base(or equal bases) and between the
                                                                           same parallels are equal in areas)     
ar(Δ QAC) - ar(Δ QPC) = ar(Δ QDC) - ar(Δ QPC)   ( Subtracting the same areas from both sides )                                               

ar(Δ PAC) = ar(Δ QDP)          ...(2)
Δ PAC and Δ PBC are on the same base PC and between the same parallels AB and DC.
ar(Δ PAC) = ar(Δ PBC)           ...(3)   (∵ Two triangles on the same base(or equal bases) and between the same
                                                                                parallels are equal in areas)     

From (2) and (3),
ar(Δ PBC) = ar(Δ QDP)
ar(Δ BPC) = ar(Δ DPQ). 

Question-5

In the 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) = ar(ABC)
(ii) ar(BDE) = ar(BAE)
(iii) ar(ABC) = 2ar(BEC)
(iv) ar(BFE) = ar(AFD)
(v) ar(BFE) = 2ar(FED)
(vi) ar(FED) = ar(AFC)
[Hint: Join EC and AD. Show that BE½½ AC and DE ½½ AB, etc.]


Solution:
                                                                                                                            

Given: ABC and BDE are two equilateral triangles such that D is the mid-point of BC. AE intersects BC at F.


To Prove:

(i) ar(Δ BDE) = ar(Δ ABC)

(ii) ar(Δ BDE) = ar(Δ BAE)

(iii)  ar(D ABC) = 2ar(D BEC)

(iv) ar(D BFE) = ar(D AFD)

(v) ar(D BFE) = 2ar(D FED)

(vi) ar(D FED) =   ar(D AFC)


Construction: Join EC and AD.


Proof: 
(i) ar(Δ BDE) =ar(Δ ABC)

It is given that triangles ABC and BED are equilateral triangles.

Let AB = BC = CA = x.

Then, BD =   

BD = DE = BE = 

We have,

ar (Δ ABC) =  

ar(Δ BDE) =   

ar(Δ BDE) =  . x2


∴ ar(
Δ BDE) =  ar (Δ ABC) 


(ii)  
ar(
Δ BDE)=  ar(Δ BAE) 

It is given that triangles ABC and BED are equilateral triangles.

РACB = DBE = 60° 

 BE || AC 

ar(Δ BAE) =  ar(Δ BCE)  ( Triangles on the same base BC and between the parallel lines BC and AC.) 
ar(Δ BCE) = 2 ar(Δ BDE)  ( D is the median of Δ EBC) 
ar(Δ BDE) =  ar(Δ BCE) 

ar(Δ BDE) =  ar(Δ BAE)                  


(iii)   
ar(Δ ABC) = 2 ar(Δ BEC)

      Since ED is the median of Δ BEC

ar(Δ BEC) = 2 ar(Δ BDE)

              = 2. ar (Δ ABC) (From (i))

    ar(Δ BEC) =   ar (Δ ABC)

 ar(Δ ABC) = 2 ar(Δ BEC)             


 (iv) ar(Δ BFE) = ar(Δ AFD)

 Δ ABC and  Δ BDE are equilateral triangles.

 ∴ ∠ABC = BDE = 60° 

     AB || DE

 ar(Δ BED) = ar(Δ AED)  ( Triangles in the same base ED and between the same parallel lines AB and ED.)

ar(Δ BED) – ar(Δ EFD) = ar(Δ AED)  ar(Δ EFD)   (  Subtract both the sides by equal area.)

 ar(Δ BEF) =   ar(Δ AFD)   


(v)   ar(Δ BFE) = 2 ar(Δ FED)

ar(Δ BFE)  =  .  . FB

              =    .    . 2FD

              = 2×ar(Δ FED)

ar(Δ BFE) = 2 ar(Δ FED)   


(vi) 
ar
(Δ FED)=   ar(Δ AFC).      

 Let the altitude of Δ ABE be h.

 Then, altitude of Δ BED =       ( ar(Δ BDE) = ar(Δ ABE) )

Now, ar(Δ FED) =. FD. =          

          ar(Δ AFC) =. FC . h

                      = (FD + DC) h = (FD + BD) h

                      = (FD + BF + FD) h = (2FD + BF) h
                      = (2FD + 2FD) h             ( ar (Δ BEF) = ar(Δ AFD) and Altitude of Δ BEF = Altitude of Δ AFD)              

        ar(Δ AFC) = 2.FD.h     

  ar(Δ AFC) =  

                    =  ar(Δ FED)                            

      ar(Δ FED) = ar(Δ AFC)

Question-6

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar(APB) ar(CPD) = ar(APD) ar(BPC). 
[Hint: From A and C, draw perpendicular to BD.]

Solution:
 
Given: Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.

To Prove: ar(
Δ APB) × ar(Δ CPD) = ar(Δ APD) × ar(v BPC).

Construction: From A and C, draw perpendiculars AE and CF respectively to BD.

Proof: ar(Δ APB) × ar(Δ CPD) =
                                       =                            …..(1)
ar(Δ APD) × ar(Δ BPC) =
From (1) and (2),
ar(Δ APB) × ar(Δ CPD) = ar(Δ APD) × ar(Δ BPC).

Question-7

P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that
(i) ar(PRQ) = ar(ARC) 
(ii) ar(RQC) = ar(ABC)

(iii) ar(PBQ) = ar(ARC)

Solution:
Given: P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP.

To Prove: (i) ar(Δ PRQ) = ar(Δ ARC)
(ii) ar(
Δ RQC) = ar(Δ ABC)
(iii) ar(Δ PBQ) = ar(Δ ARC).

Construction: Join AQ and CP.                        

Proof: (i) ar(PRQ) = ar(ARC)
ar(PRQ) = ar(APQ)             ( P is a mid point of the side AB)
           = ar(BPQ)             ( ar(APQ) = ar(BPQ) )
           = ar(CPQ)            (  Q is the mid point of the side BC) 
                                        ( ar(CPQ) = ar(BPQ) )
           = .ar(BPC)         (  Q is the midpoint of BC.)
           = ..ar(ABC)     (  ar(BPC) = ar(ABC) )

ar(PRQ) = ar(ABC)                              ….(1)
ar(ARC) =  ar(APC)  ( R is the midpoint of AP.)
              = ar(APC)
              = .ar(ABC)  (  P is the midpoint of AB.)
ar(ARC) = ar(ABC)                           ……(2)

From (1) and (2), we have

ar(PRQ) = ar(ARC)

(ii) ar(RQC) = ar(ABC)
ar(RQC) = ar(RBQ)                (  Q is the midpoint of BC.)
           = ar(PRQ) + ar(BPQ)
           = ar(ABC) + .ar(PBC)
           = ar(ABC) + .ar(ABC)
           = ar(ABC) + ar(ABC)
           = ar(ABC)

(iii) ar(PBQ) = ar(ARC)
     ar(PBQ) = ar(ABC)                  ….(3)          ( From (ii) )
     ar(ARC) = ar(ABC)                      ….(4)     ( From (i))
From (3) and (4),
ar(PBQ) = ar(ARC)

Question-8

In the 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) MBC ABD (ii) ar(BYXD) = 2ar(MBC)
(iii) ar(BYXD) = ar(ABMN) (iv) FCB ACE
(v) ar(CYXE) = 2ar(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.

Solution:
Given: In the figure, ABC is a right angle 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.

To Prove: (i) Δ MBC Δ ABD
(ii) ar(BYXD) = 2 ar(MBC)
(iii) ar(BYXD) = ar(ABMN)
(iv) Δ FCB Δ ACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar(CYXE) = ar(ACFG)
(vii) ar(BCED) = ar(ABMN) + ar(ACFG)

Proof :

(i) In Δ MBC and D ABD,
MB = AB                        ...(1)              (∵ Sides of a square)
BC = BD                        ...(2)                ( Sides of a square)
MBA= CBD                                       (∵ Each = 90°)
MBA + ABC = CBD + ABC
MBC = ABD ...(3)
In view of (1), (2) and (3),                                (∵  By SAS Rule )
Δ MBC Δ ABD.

(ii) ar(BYXD) = 2 ar(Δ ABD) (Parallelogram BYXD and Triangle ABD is on the same base and between the same parallel lines.)
ar(BYXD) = 2ar(Δ MBC)                             (∵ From(i))

(iii) ar(BYXD) = 2ar(Δ ABD)
ar(ABMN) = 2ar(Δ MBC) = 2ar(ΔABD)           (∵ From(i) )
ar(BYXD) = ar(ABMN)
(iv) In Δ FCB and Δ ACE,
FC = AC  |Sides of a square
CB = CE  |Sides of a square
FCA = BCE
FCA + ACB = BCE + ACB
FCB= ACE                                       
∴Δ FCB ACE                                        (∵  SAS Rule)

(v) ar(CYXE) = 2ar(Δ ACE) = 2ar(Δ FCB)        (∵  From (iv) )

(vi) ar(CYXE) = 2ar(Δ ACE) = 2ar(Δ FCB)
      ar(ACFG) = 2ar(ΔFCB) = 2ar(ΔFCB)
     \ ar(CYXE) = ar(ACFG).

(vii) ar(BCED) = ar(CYXE) + ar(BYXD)
                   = ar(ACFG) + ar(ABMN)
                   = ar(ABMN) + ar(ACFG).




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