# ABCD is a regular tetrahedron, the edge length of which is 4 cm. Through point O – the middle of the edge BC

**ABCD is a regular tetrahedron, the edge length of which is 4 cm. Through point O – the middle of the edge BC – a plane is drawn perpendicular to the straight line CD. Calculate the perimeter of the resulting section.**

1. Building a section.

From point O we lower the perpendicular to the side CD to point H. Then we draw the perpendicular HN from point N, which belongs to side AC, to point H. Connect points O and N. The resulting section is the triangle NOH.

2. By the condition of the problem, the edges are 4. Therefore, DB = DC = BC = 4 cm. The CBD triangle is equilateral, all of its angles are 60 degrees. And since O is the middle of BC, then BO = OC = 2 cm.

3. Consider a right-angled triangle OHC. Angle C = 60 degrees, then angle O = 90 – 60 = 30 degrees. and the HC leg is 2 times smaller than the OC hypotenuse (from the properties of a right-angled triangle). HC = OC / 2 = 1 cm.

By the Pythagorean theorem, we find the side OH = (OC ^ 2 – HC ^ 2) ^ (1/2) = 3 ^ (1/2).

4. Consider a triangle CHN – rectangular. Similar to point 3, angle C = 60 degrees.

tg (C) = NH / CH.

NH = tan (C) * CH = tan (60) * 1 = (3) ^ (1/2) cm.

5. By the Pythagorean theorem, CN = (HN ^ 2 + CH ^ 2) ^ (1/2) = 2 cm. Therefore, the triangle CON is equilateral, since the angle NCO = 60 degrees, and CN = CO = 2 cm. Then NO = 2 cm.

6. Perimeter section = 2 + (3) ^ (1/2) + (3) ^ (1/2) = 2 + 2 * (3) ^ (1/2) cm.

Answer: 2 + 2 * (3) ^ (1/2) cm.