Question : The total surface area of a regular triangular pyramid with each edge of length 1 cm is:
Option 1: $4 \sqrt{3}$ cm2
Option 2: $\frac{4}{3} \sqrt{3}$ cm2
Option 3: $\sqrt{3}$ cm2
Option 4: $4$ cm2
Correct Answer: $\sqrt{3}$ cm 2
Solution :
A regular triangular pyramid, also known as a tetrahedron, has four equilateral triangles as its faces.
The surface area of an equilateral triangle $ = \frac{\sqrt{3}}{4} a^2$
Since a regular tetrahedron has four faces.
The total surface area is four times the surface area of one face $=4 \times \frac{\sqrt{3}}{4} a^2 = \sqrt{3} a^2$
Putting \(a = 1 \, \text{cm}\),
The total surface area of a regular triangular pyramid $= \sqrt{3}$ cm
2
Hence, the correct answer is $ \sqrt{3}$ cm
2
.
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