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 :

Answer: Option C

Area of base :
$$\begin{array}{rl}& =\left(\frac{\sqrt{3}}{4}\times 1^2\right)m^2\\ & =\frac{\sqrt{3}}{4}{m}^2\end{array}$$
Clearly, the pyramid has 3 triangular faces each with sides 3m, 3m and 1 m
So, area of each lateral face :
$$\begin{array}{rl}& =\sqrt{\frac{7}{2}\times \left(\frac{7}{2}-3\right)\left(\frac{7}{2}-3\right)\left(\frac{7}{2}-1\right)}{m}^2\\ & \left[\because s=\frac{3+3+1}{2}=\frac{7}{2}\right]\\ & =\sqrt{\frac{7}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{5}{2}}{m}^2\\ & =\frac{\sqrt{35}}{4}{m}^2\end{array}$$
∴ Whole surface area of the pyramid :
$$\begin{array}{rl}& =\left(\frac{\sqrt{3}}{4}+3\times \frac{\sqrt{35}}{4}\right)m^2\\ & =\frac{\sqrt{3}+3\sqrt{35}}{4}{m}^2\end{array}$$