Triangle ABC is equilateral with a side equal to 6cm. Point M is 4cm away from the vertices

Triangle ABC is equilateral with a side equal to 6cm. Point M is 4cm away from the vertices of the triangle. Find the distance from point M to plane ABC.

Connecting point M with the vertices ABC, we get a regular pyramid.

Since all the side edges of the pyramid are equal, its center is projected to the center of the circle described about the triangle ABC, that is, to the intersection point of the mid-perpendiculars, and since ABC is equilateral, then to the point of intersection of heights and medians, O.

The height of an equilateral triangle is:

AH = BC * √3 / 2 = 6 * √3 / 2 = 3 * √3 cm.

Point O divides the median and height AH in the ratio 2/1, starting from the top A, then AO = 2 * AH / 3 = 2 * √3 cm.

The AOD triangle is rectangular, then OM^2 = AM^2 – AO^2 = 16 – 12 = 4.

OM = 2 cm.

Answer: From point M to plane ABC 2 cm.