Question

The diagonals AC and BD of a parallelogram ABCD intersect each other at O. PQ is a line through O which meets BC at P and AD at Q. If ar(quad. ABPQ)$=$k ar(Parallelogram ABCD), then k$=$.

• A. $\frac{1}{2}$
• B. $4$
• C. $3$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

$\frac{1}{2}$

In a parallelogram, Opposite sides are congruent $(AB=DC)$ and Opposite angels are also congruent

$\angle A=\angle C$ and $\angle B=\angle D$

And $PQ$ is common side for both quadrilateral $ABPQ$ and $CDPQ$

$\therefore ar(quad.ABPQ)=ar(quad.CDPQ)\dots eqn\left(1\right)$

and from the figure,

$ar\left(parallelogramABCD\right)=ar(quad.ABPQ)+ar(quad.CDPQ)$

Therefore from the equation (1),

$ar(quad.ABCD)=2\times ar(quad.ABPQ)$

$ar(quad.ABPQ)=\frac{1}{2}ar(quad.ABCD)$

$\therefore k=\frac{1}{2}$