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A right pyramid 10 cm high has a square base of which the diagonal is 10 cm. The whole surface area is

A) \( \Large 200 cm^{2} \)
B) \( \Large 205 cm^{2} \)
C) \( \Large 210 cm^{2} \)
D) \( \Large 220 cm^{2} \)

Correct Answer:

A) \( \Large 200 cm^{2} \)

Description for Correct answer:

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

= \( \Large \left(4 \times \frac{75}{2}+50\right) cm^{2} \)

= \( \Large 200\ cm^{2} \)

Part of solved Mensuration questions and answers : >> Aptitude >> Mensuration

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