ABCD is a parallelogram and EFCD …
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Yogita Ingle 5 years ago
Given that:- ABCD is a parallelogram and EFCD is a rectangle.
To prove:- ar(ABCD)=DC×AL
Proof:-
∴AB∥CD.....(1)[∵Opposite sides of parallelogram are equal]
As given that EFCD is a rectangle and we know that rectangle is also a parallelogram.
∴EF∥CD.....(2)
From eqn(1)&(2), we hahve
AB∥EF
Now, ABCD and EFCD are two parallelograms with same base CD and between tha same parallels EB and CD.
As we know that parallelogram with same base and in between same parallels are equal in area.
∴ar(ABCD)=ar(EFCD)
As from the fig.
ar(EFCD)=DC×FC
⇒ar(ABCD)=DC×FC.....(3)
AL⊥CD[Given]
∴AFCL is a also a rectangle.
∴AL=FC.....(4)
From eqn(3)&(4), we get
ar(ABCD)=DC×AL
Hence proved that ar(ABCD)=DC×AL.
3Thank You