Triangle ABC is isosceles rectangular with a right angle C and a hypotenuse of 4 cm. The segment CM is perpendicular

Triangle ABC is isosceles rectangular with a right angle C and a hypotenuse of 4 cm. The segment CM is perpendicular to the plane and is equal to 2 cm. Find the distance from point M to line AB.

1. Consider a triangle ABC: angle C = 90 degrees, AB = 4 cm, AC = BC. By the Pythagorean theorem:
AB = √ (AC ^ 2 + BC ^ 2).
Since AC = BC, we denote them as x. Then:
√ (x ^ 2 + x ^ 2) = 4;
√ (2x ^ 2) = 4;
x√2 = 4;
x = 4 / √2 = 4√2 / 2 = 2 / √2.
AC = BC = 2 / √2 cm.
2. Draw from point C the height CH to AB. Since ABC is isosceles, the CH height will also be the median. In a right-angled triangle, the median drawn to the hypotenuse is half the hypotenuse. Then:
CH = AB / 2;
CH = 4/2 = 2 (cm).
3. The distance from point M to line AB will be the perpendicular MH.
Consider the MCH triangle: the MCH angle = 90 degrees (since CM is perpendicular to ABC), CM = 2 cm, CH = 2 cm.Since the MCH angle = 90 degrees, then the MCH triangle is rectangular, then CM and CH are legs, and МН – hypotenuse (since it lies opposite to the angle equal to 90 degrees). By the Pythagorean theorem:
MH = √ (CM ^ 2 + CH ^ 2);
МН = √ (2 ^ 2 + 2 ^ 2) = √ (4 + 4) = √8 = 2√2 (cm).
Answer: MH = 2√2 cm.