Question

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

Answer 1

Whole surface area of the pyramid 

= Lateral surface area + Area of the square base + (Side)² 

= 4 × Area of triangular faces

Each side of the square (a) = \(\frac{diagonal}{\sqrt{2}}\) = \(\frac{10}{\sqrt{2}}\) cm, Height (h) = 10 cm.

EG = height of a triangular face 

AB = base of the triangular face

∴  Height of the triangle = \(\sqrt{(10)^2 +\big(\frac{1}{2}\times \frac{10}{\sqrt{2}}\big)^2}\) 

= \(\sqrt{100+\frac{25}{2}}\)  = \(\sqrt{112.5}\) = 10.6 cm (approx)

∴  Whole surface area = \(4\times \frac{1}{2}\times AB \times EG +(AB)^2\) 

= \(4\times \frac{1}{2}\times \frac{10}{\sqrt{2}} \times 10.6 +\big(\frac{10}{\sqrt{2}}\big)^2\)  

= 149.91 cm² + 50 cm² = 200 cm² (approx).