In triangle ABC (angle ABC is 90), AC = 41, BC = 40. A perpendicular is drawn through the vertex A to the plane

In triangle ABC (angle ABC is 90), AC = 41, BC = 40. A perpendicular is drawn through the vertex A to the plane of the triangle, equal to 12. Calculate the distance from the ends of this perpendicular to the leg BC.

In the triangle ABC (angle B = 90 °), we calculate the length of the second leg AB according to the Pythagorean theorem:

AB = √ (AC² – BC²) = √ (41² – 40²) = √ (1681 – 1600) = √81 = 9.

Let EA be the perpendicular to the ABC plane.

EA is perpendicular to AB (since EA is perpendicular to the entire plane ABC and to all lines passing through point A).

AB is perpendicular to BC (these are the legs of a right-angled triangle).

According to the three perpendicular theorem, the segment EB will be perpendicular to BC. Hence, EB is the required distance from the end of the perpendicular to the leg BC.

In the triangle EAB (angle A = 90 °, since EA is perpendicular to AB) EA = 12, AB = 9, we calculate the length EB by the Pythagorean theorem:

EB = √ (EA² + AB²) = √ (12² + 9²) = √ (144 + 81) = √225 = 15.

Answer: the distance from the end of the perpendicular to the BC leg is 15.