Breaking A Trapezium Apart

$ABCD$ has 4 sides, it is a quadrilateral. Two of its sides are $AB$ and $CD$ are parallel. Therefore, $ABCD$ is a trapezium/trapezoid.

We drop perpendiculars to $CD$ i.e. $AE \perp BC$ and $BF \perp BC$ . $E$ and $F$ are points on side $CD$ . Then in triangle $AED$ ,

$\cos30^\circ = \dfrac{ED}{AD} = \dfrac{ED}2 \Rightarrow \dfrac{\sqrt3}2 = \dfrac{ED}2 \Rightarrow ED = \sqrt3 \text{ cm} \; .$

By Pythagorean theorem , $AE=1 \text{ cm}$ . Clearly $AE \perp BF$ (Perpendicular distances between parallel lines are equal).

Then $BF=1\text{ cm}$ . Following to this, in triangle $BFC$ ,

$\tan60^\circ = \dfrac{BF}{FC} = \dfrac1{FC} \Rightarrow \sqrt3 = \dfrac1{FC} \Rightarrow FC = \dfrac1{\sqrt3} \text{ cm} \; .$

Now, $CD=ED+FE+FC=\sqrt3 + 4 + \dfrac1{\sqrt3}$ .

Area of a trapezium/trapezoid is

$\dfrac12 \times (\text{ sum of parallel sides }) \times (\text{ height }) = \dfrac12 (AB +CD) \times 1 = \dfrac12 \left (4 + \sqrt3 + 4 + \dfrac1{\sqrt3} \right)\; .$

Solving it and evaluating it correct to 3 decimal places, we get $\boxed{5.155}$ as the answer.