Triangle AB = AC = 20 BC = 24 AM perpendicular to triangle ABC = 12. find the distance from

Triangle AB = AC = 20 BC = 24 AM perpendicular to triangle ABC = 12. find the distance from point M to the side of the BC.

Let’s build the height АН of the isosceles triangle ABC, which is also its median, and then СН = ВС / 2 = 24/2 = 12 cm, and the triangle АСН is rectangular.

Then, by the Pythagorean theorem, AH ^ 2 = AC ^ 2 – CH ^ 2 = 400 = 144 = 256.

AH = 16 cm.

Let’s connect points M and H.

The height АН is the projection of the inclined МН onto the plane of the triangle ABC, then the inclined МН is perpendicular to the side ВС, and therefore there is the required distance.

By the Pythagorean theorem, in a right-angled triangle AMН, MH ^ 2 = AM ^ 2 + AH ^ 2 = 144 + 256 = 400.

MH = 20 cm.

Answer: From point M to straight BC 20 cm.