Question

ABCD is a parallelogram. W, X, Y and Z are points on sides AB, BC, CD, and DA respectively, such that $AW=DY$. If ar$\left(ABCD\right)=400c{m}^2$, what is the area of quadrilateral WXYZ?

• A. $250c{m}^2$
• B. $200c{m}^2$
• C. $50c{m}^2$
• D. $100c{m}^2$

Answer 1

The correct option is B $200c{m}^2$

Construction: Join WY
Given AW = DY
So, AWYD and WBCY are paralleograms
(opposite sides are equal are parallel)
We know that, if a triangle and a paralleogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of parallelogram.

$\therefore$ $ar(\Delta YZW)=\frac{1}{2}ar\left(AWYD\right)$ ... (1)
and $ar(\Delta YXW)=\frac{1}{2}ar\left(WBCY\right)$ ... (2)

On adding (1) and (2) we get,

$ar\left(YZW\right)+ar(\Delta YZW)=\frac{1}{2}\left(ar\right(AWYD)+ar(AWYD\left)\right)$
$\Rightarrow ar\left(WXYZ\right)=\frac{1}{2}\times ar\left(ABCD\right)$$\Rightarrow ar\left(WXYZ\right)=\frac{1}{2}\times 400c{m}^2$
$\Rightarrow ar\left(WXYZ\right)=200c{m}^2$