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Maths / Areas Of Parallelograms And Triangles / Relation between Area of Parallelogram and Triangle
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Relation between Area of Parallelogram and Triangle

Theorem:     If  a triangle and a parallelogram are on the same base and between the same parallels the area of the triangle is equal to half of the parallelogram.

Given     Delta ABC   and  aparallel ^{gm} BCDE   on the same base BC and between the same parallels BC and AD

To prove      ar(Delta ABC)=frac{1}{2}ar(parallel ^{gm}BCDE)

Construction     Draw AL perp BC;;and;;DM perp BC, meeting  BC in M

Proof   Since A, E and D are collinear and BC || AD

therefore          AL=DM              ...(i)   [becauseDistance between parallel lines is always same]

Now,         ar(Delta ABC)=frac{1}{2}(BCtimes AL)

Rightarrow ar(Delta ABC)=frac{1}{2}(BCtimes DM)                  ............(1)              [because AL=DM (from(i))]

large ar(parallel ^{gm}BCDE)= BC ;times;DM     ...............(2)

From Eq (1) and Eq (2)

Rightarrow ar(Delta ABC)=frac{1}{2}ar(parallel ^{gm}BCDE)

ILLUSTRATION   P and Q are any two points lying on the side DC and AD respectively of a parallelogram ABCD. Show that ar(APB)= ar(BQC)

Solution: Now large parallel ^{gm}ABCD and large DeltaABP lie on the same base AB and between the same parallel AB || CD

Rightarrow ar(Delta ABP)=frac{1}{2}ar(parallel ^{gm}ABCD)          .................. (1)

Similarly large parallel ^{gm}ABCD and large DeltaBCQ  lie on the same base BC and between the same parallel BC || AD

Rightarrow ar(Delta BCQ)=frac{1}{2}ar(parallel ^{gm}ABCD)          .................. (2)

From Eq (1) and Eq (2)

ar(Delta ABP)=ar(Delta BCQ)

Hence Proved

 


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