Question

Pritam drew a quadrilateral $ABCD$ of which $AB=5cm,BC=6cm,CD=4cm,DA=3cm$ and $\angle ABC={60}^{\circ }$, I draw a triangle with equal area of that quadrilateral.

Answer 1

A. Draw the given quadrilateral ABCD.
B. Draw the diagonal $DB$ of quadrilateral $ABCD$.
C. Draw a parallel line through point $A$ to diagonal $DB$ of quadrilateral $ABCD$ which intersects at $F$ produced $BC$.
D. Join FD, AF $\parallel BD$

$△DFC$ is the required triangle.

Proof:
$△ABD=△BFD$ (on same base $DB$ and between same parallels $DB$ and $AF$ )
$\therefore △ABD=△BFD$
$△DBC+△ABD=△BFD+△DBC$ (adding area of $△DBC$ on both sides)
$\therefore$ quadrilateral $ABCD=△DFC$