Question

Sahana drew a quadrilateral $ABCD$ of which $AB=4cm,BC=5cm,CD=4.8cm,DA=$ $4.2cm$ and diagonal $AC=6cm$. I draw a triangle with equal area of that quadrilateral.

Answer 1

A. Draw the given quadrilateral ABCD.
B. Draw the diagonal $AC$ of quadrilateral $ABCD$.
C. Draw a parallel line through point $D$ to diagonal $AC$ of quadrilateral $ABCD$ which intersects at $F$ produced $BC$.
D. DF $\parallel AC,ABF$ is the required triangle.
Proof:
$△ADC=△ACF$ (on same base $AC$ and between same parallels $AC$ and DF)
$\therefore △ADC=△ACF$
$△ABC+△ADC=△ACF+△ABC$ (adding area of $△ABC$ on both sides)
$\therefore$ quadrilateral $ABCD=△ABF$