$ABCD$ is a parallelogram. If $\operatorname{ar}(ABC) = 42 \, \text{cm}^2$,then find $\operatorname{ar}(ABCD)$ in $\text{cm}^2$.

The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals $12 \, cm$ and $16 \, cm$ is (in $cm^2$)

Prove that the line segment joining the midpoints of two opposite sides of a parallelogram divides the parallelogram into two parallelograms with equal area.

If the medians of a $\Delta ABC$ intersect at $G$,show that $\operatorname{ar}(AGB) = \operatorname{ar}(AGC) = \operatorname{ar}(BGC) = \frac{1}{3} \operatorname{ar}(ABC)$.

Which of the following figures lie on the same base and between the same parallels? Write the common base and the two parallels for the figure for which the answer is affirmative.

$X$ and $Y$ are points on the side $LN$ of the triangle $LMN$ such that $LX = XY = YN$. Through $X$,a line is drawn parallel to $LM$ to meet $MN$ at $Z$ (See figure). Prove that $\operatorname{ar}(LZY) = \operatorname{ar}(MZYX)$.