# Triangle ABC – rectangular, angle C = 90, AC = 8 cm, BC = 6 cm segment CD

**Triangle ABC – rectangular, angle C = 90, AC = 8 cm, BC = 6 cm segment CD perpendicular to the plane ABC find CD if the distance from point D to side AB is 5 cm**

Let’s designate the distance from point D to AB – DK – perpendicular.

CК – projection of DK on ABC, perpendicular to AB (theorem about three perpendiculars).

By the Pythagorean theorem, we find the hypotenuse AB

AB = √ (AC² + BC²) = √100 = 10 (cm).

SK is the height drawn from the top of the right angle.

AK = AC² / AB = 64/10 = 6.4 (cm).

In the triangle ASK, we find SK by the Pythagorean theorem:

CK = √ (AC² – AK²) = √ (64 – 40.96) = √23.04 = 4.8 (cm).

In the triangle DCK, we find DC by the Pythagorean theorem:

DC = √ (DK² – CK²) = √ (25 – 23.04) = √1.96 = 1.4 (cm).

Answer: the distance from point D to side AB is 1.4 cm.