There are surface, namely OAB, OBC, OCD, ODA and ABCD
where \( \Large \triangle OAB \) = \( \Large \triangle OBC \) = \( \Large \triangle OCD \) = \( \Large \triangle ODA \) .
Therefore, Total surface area = \( \Large \triangle \)
\( \Large 4 \times surface\ area\ of\ \triangle OAB + surface\ area\ of\ ABCD \) ...(i)
Now \( \Large BD = 10 cm \)
=> \( \Large x^{2}+x^{2}=10^{2} \)
=> \( \Large x = 5\sqrt{2} \)
Therefore Area of ABCD = \( \Large x^{2} = 50 \) ...(ii)
Again, MP = \( \Large \frac{1}{2}BC=\frac{1}{2}x=\frac{5}{\sqrt{2}} \)
Therefore, \( \Large OM^{2} = MP^{2} + OP^{2} \)
= \( \Large \frac{25}{2}+100 = \frac{225}{2}\)
=> \( \Large OM = \frac{15}{\sqrt{2}} \)
Therefore, Area of \( \Large \triangle OAB=\frac{1}{2} \times AB \times OM \)
= \( \Large \frac{1}{2} \times 5\sqrt{2} \times \frac{15}{\sqrt{2}} cm^{2} \) ...(iii)
= \( \Large \frac{75}{2} cm^{2} \)
From equation (i), (ii) and (iii),
total surface area of the pyramid
= \( \Large \left(4 \times \frac{75}{2}+50\right) cm^{2} \)
= \( \Large 200\ cm^{2} \)