Point K is equidistant from the vertices of a regular triangle with sides 6 cm and 8 cm away from

Point K is equidistant from the vertices of a regular triangle with sides 6 cm and 8 cm away from the plane of the triangle Find the distance from point K to the apex of the triangle.

Let’s connect all the points and get a regular triangular KAВС pyramid. According to the condition of the problem, point K is equidistant from the vertices of a regular triangle, which means that the base of the height will be point O – the center of the circle circumscribed about the triangle.
Consider a right-angled triangle KOA, in it:
KO – leg – height, 8 cm;
OA – leg – the radius of the circumscribed circle. We find by the formula:
ОА = a / √3 = 6 / √3 = 2√3 (cm).
By the Pythagorean theorem, we find the hypotenuse AK – the distance from point K to the apex of the triangle:
AK = √ (KO² + OA²) = √ (64 + 12) = √76 = 8.717797 ≈ 8.7 (cm).
Answer: the distance is 8.7 cm.