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  • ABCD  is a parallelogram in which BC is produced to E such that CE = BC (Fig.). AE intersects CD at F. If ar (DFB) = 3 cm^2, find the area of the parallelogram ABCD.

ABCD  is a parallelogram in which BC is produced to E such that CE = BC (Fig.). AE intersects CD at F.
If ar (DFB) = 3cm^{2}, find the area of the parallelogram ABCD.

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Answer: \left [ ar(ABCD)=12cm^{2} \right ]

Solution.

Given: ABCD is a parallelogram.
CE = BC
The lines BC and AF when produced, meet at E.
ar\triangle DFB=3cm^{2}
Now,
BC = EC & AB \parallel FC   (given)
\therefore FC=\frac{1}{2}AB               (By mid-point theorem)
Also, AB = DC (opposite sides of a parallelogram are equal)
\therefore FC=\frac{1}{2}DC
\Rightarrow FC=DF=\frac{1}{2}DC
Now, let perpendicular distance between parallel lines AB and CD be h
ar(\triangle DFB)=\frac{1}{2}(Base)(Height)
                        =\frac{1}{2}(DF)(h)
                       =\frac{1}{2}\left ( \frac{1}{2}CD \right )(h)
                       =\frac{1}{4}(CD)(h)             …(i)
Area of \parallel ABCD = (Base) (corresponding altitude)
= (CD) (h) …(ii)
From (i) and (ii)
Area of \parallel ABCD = 4 (ar (\triangle DFB))
Area of \parallel ABCD = 4 (3 cm^{2}) = 12 cm^{2}


        

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