Solved Problem 06 | Cube

Pythagorean Theorem for triangle BEC
$x^2 = a^2 + (\frac{1}{2}a)^2$

$x^2 = \frac{5}{4}a^2$

$x = \frac{1}{2}\sqrt{5} \, a$
 

Area of section ABCD
$A_{ABCD} = ax$

$A_{ABCD} = a(\frac{1}{2}\sqrt{5} \, a)$

$A_{ABCD} = \frac{1}{2}\sqrt{5} \, a^2$           answer