Question

EF divides the rectangle ABCD in two equal parts of _____.

Answer 1

$EF$ divides the rectangle ABCD into two trapeziums.

Area of trapezium ABFE
$=\frac{1}{2}\times \left(Sumofparallelsides\right)\times \left(Distancebetweenthem\right)$
$=\frac{1}{2}\times (AE+BF)\times AB$

$=\frac{1}{2}\times (1+2)\times 4$

$=\frac{1}{2}\times 3\times 4=6c{m}^2$

Area of trapezium CDEF $=\frac{1}{2}\times (FC+ED)\times CD$

$=\frac{1}{2}\times (1+2)\times 4$

$=\frac{1}{2}\times 3\times 4=6c{m}^2$

$\text{Area of both the trapeziums are equal.}$

$\therefore$ $EF$ divides the rectangle $ABCD$ in two equal parts and the area of both the parts are $6c{m}^2$.