In an isosceles triangle ABC, the base is BC = 12 cm, and the lateral side is 10 cm. From the vertex A

In an isosceles triangle ABC, the base is BC = 12 cm, and the lateral side is 10 cm. From the vertex A, a segment AM = 15 cm is drawn, perpendicular to the plane of triangle ABC. Find the distance from point M to side BC.

From the vertex A of the isosceles triangle, we draw the height AH, which is also the median of the triangle, then in the right-angled triangle ACH CH = BC / 2 = 12/2 = 6 cm.

By the Pythagorean theorem, AH ^ 2 = AC ^ 2 – CH ^ 2 = 100 – 36 = 64.

AH = 8 cm.

According to the three perpendicular theorem, the ML is perpendicular to the BC and is the desired distance.

From a right-angled triangle MAH MH ^ 2 = AM ^ 2 + AH ^ 2 = 225 + 64 = 289.

MH = 17 cm.

Answer: The distance from point M to side BC is 17 cm.