OA Explanation

T.S.A of the Square Pyramid = Area of the Square base + \(4\) x (Area of equilateral faces)

Area of equilateral triangle with each side \(a\) = \(\frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4}6^2 = 9\sqrt{3}\)

Area of Square base with each side \(a\) = \(a^2 = 6^2 = 36\)

Therefore, T.S.A of the Square Pyramid = \(36\) + \(4(9\sqrt{3}\)) = \(36 + 36\sqrt{3}\)

Hence, option B