Correct Option: A

Volume of the tetrahedron =	1	Area of Base × height
	3

∴ Area of the base =	√3	× 12 × 12 = 36√3 sq cm
	4

A regular tetrahedron is made up of 4 equilateral triangles.
One is the base triangles and other are the 3 faces.

In ∆DBC, draw DF ⊥ BC. ∆DBC is are equilateral triangle. DF (perpendicular) [DF 1 AF]
= √DC² - FC² = √12² - 6²
= √18 × 6
= 6√3 = AF [altitude of ∆ABC]
[∆ABC is also an equilateral ∆ with side 12cm].

FE =	1	× 6√3 = 2√3 cm.
	100

[E is the centroid].
∴ AE = √AF² - FE²
= √(6√3)² - (2√3)²
= √108 - 12 = √96 = 4√6 cm

∴ Required volume =	1	× 36√3 = 4√3
	3

= 144√2 cu.cm.