All five faces of a regular pyramid with a square base are found to be of the

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All five faces of a regular pyramid with a square base are found to be of the same area. The height of the pyramid is 3 cm. What is the minimum total area (integral) of all its surfaces?

Options:

A . 10

^{2}

sq cm

B . 12

^{2}

sq cm

C . 15

^{2}

sq cm

D . 20

^{2}

sq cm

Answer: Option B
:
B

Altitude of the traingular faces = $\sqrt{\frac{a^{2}}{4} + 9}$
Area of faces =2 x a x $\sqrt{\frac{a^{2}}{4} + 9}$
The total surface area = area of base + lateral surface area
= a $^{2}$ + 2 x a x $\sqrt{\frac{a^{2}}{4} + 9}$
Lets us put a=8 (smallest value whch will yield an integer area, Area = 64 + 2 $\times$ 8 $\times$ 5 =144)

= 12 $^{2}$

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