Question

If the length of each edge of a regular tetrahedron is 'a', then its surface area is
(a) $\sqrt{3}{a}^2$ sq. units
(b) $\frac{\sqrt{3}}{2}{a}^2$ sq. units
(c) $\sqrt{\frac{2}{3}}{a}^2$ sq. units
(d) $\sqrt{6}{a}^2$ sq. units

Answer 1

Given: The length of each edge of a regular tetrahedron is 'a' units.
We know, the surface area of tetrahedron = Sum of area's of 4 equilateral triangles.
Thus, Surface Area=$4\times \frac{\sqrt{3}}{4}\times a^2$
$=\sqrt{3}{a}^2$ sq. units
Hence, the correct option is (a).