Through the middle E of the hypotenuse AB of the right-angled triangle ABC, a perpendicular EM

Through the middle E of the hypotenuse AB of the right-angled triangle ABC, a perpendicular EM equal to 4√5 is drawn to its plane. AC = BC = 16cm, angle C = 90 degrees. Calculate: a) The distance from point M to the line AC b) the area of the AFM triangle and its projection onto the plane of this triangle. c) Distance between straight lines EM and BC.

So, let’s designate for H – the middle of AC, then EH is the middle line of the triangle ABC, EH is parallel to CB, therefore EH is perpendicular to AC.

ЕН – is the projection of the inclined МН onto the plane ABC, which means that МH is perpendicular to the АС according to the theorem of three perpendiculars.

We conclude that MH is the required distance from point M to line AC.

Find the length of the segment EH

EH = BC / 2 = 16/2 = 8 cm.

Next, we will consider a triangle MEH, in which the angle ∠ MEH = 90 °.

Let’s apply the Pythagorean theorem:

MH = √ (ME ^ 2 + EH ^ 2) = √ (80 + 64) = √144 = 12 cm.