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1. A pyramid has an equilateral triangle as its base of which each side is 1 m. Its slant edge is 3 m. The whole surface are of the pyramid is equal to :

A. $\frac{\sqrt{3} + 2 \sqrt{13}}{4} s q . m$

B. $\frac{\sqrt{3} + 3 \sqrt{13}}{4} s q . m$

C. $\frac{\sqrt{3} + 3 \sqrt{35}}{4} s q . m$

D. $\frac{\sqrt{3} + 2 \sqrt{35}}{4} s q . m$

Solution:

Area of base :

\begin{aligned} = (\frac{\sqrt{3}}{4} \times 1^{2}) m^{2} \\ = \frac{\sqrt{3}}{4} m^{2} \end{aligned}

Clearly, the pyramid has 3 triangular faces each with sides 3m, 3m and 1 m
So, area of each lateral face :

\begin{aligned} = \sqrt{\frac{7}{2} \times (\frac{7}{2} - 3) (\frac{7}{2} - 3) (\frac{7}{2} - 1)} m^{2} \\ [∵ s = \frac{3 + 3 + 1}{2} = \frac{7}{2}] \\ = \sqrt{\frac{7}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{5}{2}} m^{2} \\ = \frac{\sqrt{35}}{4} m^{2} \end{aligned}

∴ Whole surface area of the pyramid :

\begin{aligned} = (\frac{\sqrt{3}}{4} + 3 \times \frac{\sqrt{35}}{4}) m^{2} \\ = \frac{\sqrt{3} + 3 \sqrt{35}}{4} m^{2} \end{aligned}

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