From the center O of the correct Δ ABC, the perpendicular OM is drawn to its plane. Find the area Δ ABM

From the center O of the correct Δ ABC, the perpendicular OM is drawn to its plane. Find the area Δ ABM, if AB = 6√3, OM = 4.

Since point O is the center of the regular triangle ABC, it is also the center of the inscribed and circumscribed circle around the triangle ABC.

The radius of the inscribed circle in a regular triangle is: OH = r = AB * √3 / 6 = 6 * √3 * √3 / 6 = 3 cm.

In a right-angled triangle OMH, according to the Pythagorean theorem, MH ^ 2 = OH ^ 2 + OM ^ 2 = 9 + 16 = 25.

MH = 5 cm.

MH is the height of the triangle ABM.

Determine the area of the triangle ABM.

Savm = AB * MH / 2 = 6 * √3 * 5/2 = 15 * √3 cm2.

Answer: The area of the triangle ABM is 15 * √3 cm2.