Subject: General Aptitude · Chapter: Volume and Surface Area · Exam: · Marks: · Difficulty:
A pyramid has an equilateral triangle as its base of which each side is 1m. Its slant edge is 3 m. The whole surface area of the pyramid is equal to
(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
Answer (as printed): C
Explanation
Area of base $=\left(\frac{\sqrt{3}}{4} \times 1^{2}\right) \mathrm{m}^{2}=\frac{\sqrt{3}}{4} \mathrm{m}^{2}$. Clearly, the pyramid has 3 triangular faces each with sides 3 m, 3 m and 1 m. So, area of each lateral face $=\sqrt{\frac{7}{2} \times\left(\frac{7}{2}-3\right)\left(\frac{7}{2}-3\right)\left(\frac{7}{2}-1\right)} \mathrm{m}^{2}=\sqrt{\frac{7}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{5}{2}} \mathrm{m}^{2}=\frac{\sqrt{35}}{4} \mathrm{m}^{2}$. ∴ Whole surface area of the pyramid $=\left(\frac{\sqrt{3}}{4}+3 \times \frac{\sqrt{35}}{4}\right) \mathrm{m}^{2}=\frac{\sqrt{3}+3 \sqrt{35}}{4} \mathrm{m}^{2}$.