In the figure,the area of parallelogram $ABCD$ is:

  • A
    $DC \times DL$
  • B
    $AB \times BM$
  • C
    $AD \times DL$
  • D
    $BC \times BN$

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$(1)$ The part of the plane enclosed by a simple closed figure is called a $\ldots \ldots \ldots$
$(2)$ $\ldots \ldots \ldots$ of the planar region corresponding to a closed figure is called its area.

The mid-points of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to:

In a parallelogram $ABCD$,the base $AB$ and the corresponding altitude $DM$ are $16 \, cm$ and $12 \, cm$,then the area of $ABCD =$ ................ $cm^{2}$.

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

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In the given figure,$ABCD$ is a trapezium with $AB \parallel DC$. $E$ is a point on extended $BC$. Prove that $ar(BDE) = ar(ACED)$.

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