Triangles on Same Base and Between Same Parallels


 
 
Concept Explanation
 

Triangles on Same Base and Between Same Parallels

Theorem:   Triangles on the same base and between the same parallels are equal in area.

Given    Two triangles ABC and PBC on the same base BC and between the same parallel line BC and AP.

To prove   ar (Delta ABC)=ar(Delta PBC)

Construction Through B,  draw BD parallel CA  intersecting PA  produced in D and through C, draw CQparallel BP, intersecting line AP;in;Q.

Proof    we have,          BDparallel CA             [By construction]

and,         BCparallel DA                     [Given AD is the extension of line AP]

therefore             BCDA is a parallelogram.

Similarly, BCQP is a parallelogram.

Now, paralllograms BCQP and  BCAD  are on the same base BC, and between the same parallels.

therefore            ar(parallel ^{gm}BCQP)=ar(parallel ^{gm}BCAD)                              ...(i)

We know that the diagonals of a parallelogram divides it into two triangles of equal area.

therefore            ar(Delta PBC) =frac{1}{2}ar(parallel ^{gm}BCQP)                                    ...(ii)

and,       ar(Delta ABC) =frac{1}{2}ar(parallel ^{gm}BCAD)                                    ...(iii)

Now,      ar(parallel ^{gm}BCQP)=ar(parallel ^{gm}BCAD)                         [From (i)]

Rightarrow          frac{1}{2}ar(parallel ^{gm}BCAD)=frac{1}{2}ar(parallel ^{gm}BCQP)

Rightarrow          ar(Delta ABC)=ar(Delta PBC)                                          [Using (ii)and(iii)]

Hence,   ar(Delta ABC)=ar(Delta PBC)

 

ILLUSTRATION: The side AB of a parallelogram ABCD is produced to any point P. A line through A parallel to CP meets CB produced at Q then parallelogram PBQR is constructed. Show that ar( ABCD)= ar (PBQR)

Solution:  large DeltaACQ and large DeltaAPQ stand on the same base  AQ and they are between same parallel AQ || CP.

  ar (large Delta ACQ)= ar (large DeltaAPQ)       [Triangles on the same base and beteween same parallel have equal area]

Subtracting common ar (ABQ) from both sides we get

ar (large Delta ACQ)  - ar (ABQ) = ar (large DeltaAPQ) - ar (ABQ)

ar (large Delta ABC)= ar (large DeltaABP)                                  ................(1)

 We know that the diagonal divides a parallelogram in two congruent triangles

 large ar ( Delta ABC)= frac{1}{2}ar (ABCD)      ..................(2)

 large ar ( Delta PBQ)= frac{1}{2}ar (PBQR)      ..................(3)

From Eq (1), (2) and (3) we get

large frac{1}{2}ar (ABCD)= frac{1}{2}ar (PBQR)

ar( ABCD)= ar (PBQR)

Hence Proved.

Sample Questions
(More Questions for each concept available in Login)
Question : 1
If ar(ABC) = ar(ABD), then choose the correct relationship for p and q lines.

 

Right Option : B
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Explanation
Question : 2
In the figure given below, AB parallel CD. Which of the following statements is true ?

 

Right Option : D
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Explanation
Question : 3
If ABCD is quadrilateral and ar (AOD) = ar (BOC), then choose the correct option.

 

Right Option : C
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Explanation
 
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