Question

Sahana drew a rectangle $ABCD$ of which $AB=4cm$ and $BC=6cm$. I draw a triangle with equal area of that rectangle.

Answer 1

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