Step-by-step explanation:
ABCD is a rectangle and BD is diagonal such that:
[tex]\angle DBC = 30\degree
(given) \\\\
\angle BCD = 90\degree
(\angle\:of\:a\:rectangle) \\\\
\therefore \angle CDB= 60\degree
(remaining\:\angle\:of \:\triangle)
\\\\
\therefore \triangle BCD\:is \:a\:30\degree , \: 60\degree \:\&\:90\degree \:\triangle. \\\\
\therefore CD = \frac{1}{2} \times BD (side\:oppossite \:to \:30\degree)\\\\
\therefore CD = \frac{1}{2} \times 4\\\\
\therefore CD =2\: units\\\\
\&\\\\
BC = \frac{\sqrt 3}{2} \times BD(side\:oppossite \:to \:60\degree)\\\\
\therefore BC = \frac{\sqrt 3}{2} \times 4\\\\
\therefore BC =2{\sqrt 3}\: units\\\\
Now\\\\
A(\triangle BCD) = \frac{1}{2} \times BC\times CD\\\\
\therefore A(\triangle BCD) = \frac {1}{2} \times {2\sqrt 3} \times 2\\\\
\red{\boxed {\therefore A(\triangle BCD) ={2\sqrt 3}\: units^2}} [/tex]
