Question

From a square with sides of length $5$ unit, triangular pieces from the four corners are removed to from a regular octagon. Then the area removed is
(correct answer + 1, wrong answer - 0.25)

• A. $25(3+\sqrt{2})\text{sq. units}$
• B. $25(3+2\sqrt{2})\text{sq. units}$
• C. $25(3-2\sqrt{2})\text{sq. units}$
• D. $25(3-\sqrt{2})\text{sq. units}$

Answer 1

The correct option is C $25(3-2\sqrt{2})\text{sq. units}$


From above figure,
$5-2x=\sqrt{2}x$
$\Rightarrow x=\frac{5}{2+\sqrt{2}}=\frac{5}{\sqrt{2}(\sqrt{2}+1)}$
$\therefore \text{Removed area}=4\times \frac{1}{2}{x}^2$
$=2x^2$
$=2\times \frac{25}{2(\sqrt{2}+1{)}^2}=25(3-2\sqrt{2})$


Alternate solution:


The angle subtended at the centre by a side is $\frac{2\pi }{8}=\frac{\pi }{4}$ and $OP=\frac{5}{2}.$
Therefore, side of the regular octagon is
$x=2\cdot \frac{5}{2}\cdot \tan \frac{\pi }{8}$
$\Rightarrow x=5(\sqrt{2}-1)$

$\therefore \text{Removed area}=4\times \frac{1}{2}\times \frac{x}{\sqrt{2}}\times \frac{x}{\sqrt{2}}$
$=x^2$
$=\left[5\right(\sqrt{2}-1){]}^2=25(3-2\sqrt{2})\text{sq. units}$