a) $$\frac{{\sqrt 3 + 2\sqrt {13} }}{4}sq.m$$
b) $$\frac{{\sqrt 3 + 3\sqrt {13} }}{4}sq.m$$
c) $$\frac{{\sqrt 3 + 3\sqrt {35} }}{4}sq.m$$
d) $$\frac{{\sqrt 3 + 2\sqrt {35} }}{4}sq.m$$

Answer: c
Explanation: Area of base :
$$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} \times {1^2}} \right){m^2} \cr & = \frac{{\sqrt 3 }}{4}{m^2} \cr} $$
Clearly, the pyramid has 3 triangular faces each with sides 3m, 3m and 1 m
So, area of each lateral face :
$$\eqalign{ & = \sqrt {\frac{7}{2} \times \left( {\frac{7}{2} – 3} \right)\left( {\frac{7}{2} – 3} \right)\left( {\frac{7}{2} – 1} \right)} {m^2} \cr & \left[ { s = \frac{{3 + 3 + 1}}{2} = \frac{7}{2}} \right] \cr & = \sqrt {\frac{7}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{5}{2}} {m^2} \cr & = \frac{{\sqrt {35} }}{4}{m^2} \cr} $$
Whole surface area of the pyramid :
$$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} + 3 \times \frac{{\sqrt {35} }}{4}} \right){m^2} \cr & = \frac{{\sqrt 3 + 3\sqrt {35} }}{4}{m^2} \cr} $$