The side of a regular triangle is 12cm. At a distance of 1 cm from the plane of the triangle

The side of a regular triangle is 12cm. At a distance of 1 cm from the plane of the triangle, a point is taken that is equally distant from all its sides. At what distance from the vertices of the triangle is this point?

Since point D is equidistant from the vertices of the triangle ABC, then connecting point D with the vertices of the triangle, we get a regular triangular pyramid. Point D is projected onto the plane of the triangle to point O, which is the point of intersection of the medians and heights.

Let us determine the length of the median AH.

AH = BC * √3 / 2 = 12 * √3 / 2 = 6 * √3 cm.

The medians of a regular triangle, at the point of their intersection, are divided by the ratio of 2/1.

Then, ОА = 2 * АН / 3 = 2 * 6 * √3 / 3 = 4 * √3 cm.

Triangle OAD is rectangular, then AD ^ 2 = OA ^ 2 + OD ^ 2 = 48 + 1 = 49.

AD = BD = CD = 7 cm.

Answer: Point D is 7 cm from the vertices of the triangle.