Question

In the given figure, ABCD is a rectangle with length as 2 and breadth as 1. The area of rectangle BDFE is:

• A. $2$
• B. $\sqrt{2}$
• C. $\sqrt{5}$
• D. $1+\sqrt{2}$

Answer 1

The correct option is A $2$

By Pythagoras' theorem, the length of the diagonal BD is $\sqrt{5}$. Now observe that triangles BCD and BCE are similar.
Thus, $\frac{BE}{BC}=\frac{CD}{BD}=\frac{2}{\sqrt{5}}$, and since BC = 1, if follows that $BE=\frac{2}{\sqrt{5}}$. The area of rectangle BDFE is therefore $\sqrt{5}\times \frac{2}{\sqrt{5}}=2$.