Answer 1

It is given that ABCD is a square.

∴ AB = BC = CD = DA = x (say)

According to Pythagoras theorem, in ∆ABD

\[{\text{AB}}^2 + {\text{AD}}^2 = {\text{BD}}^2 \]
\[ \Rightarrow x^2 + x^2 = \left( 10\sqrt{2} \right)^2 \]
\[ \Rightarrow 2 x^2 = 200\]
\[ \Rightarrow x^2 = 100\]
\[ \Rightarrow x = \sqrt{100}\]
\[ \Rightarrow x = 10 \text{cm}\]

Hence, the side of the square is 10 cm.

Now,
Perimeter of a square = \[4 \times \left( side \right)\]

=\[4 \times x\]

=\[4 \times 10\]

=\[40\]

Hence, the correct option is 40 cm.