# In a regular triangular prism, through the center line of the base at an angle of 60 to the plane of the base

**In a regular triangular prism, through the center line of the base at an angle of 60 to the plane of the base, a plane is drawn that intersects the lateral edge. Find the cross-sectional area if the side of the base is 4 cm.**

Let us denote the prism given by the condition ABCA1B1C1, the prism is regular, at the bases – equilateral triangles.

MK – the middle line, parallel to the BC.

MK = 1/2 * BC = 2 (cm).

P is the point of intersection of the MPK plane and the AA1 lateral edge.

In the triangle ABC we find the height AH (according to the Pythagorean theorem):

AH = √ (AC² – CH²) = √ (16 – 4) = √12 = 2√3 (cm).

RO is the height of the section and the hypotenuse of the RAO triangle.

cos 60 ° = AO / PO → PO = AO / cos 60 ° = √3 / 1/2 = 2√3 (cm).

We find the area of the triangle MRK – the cross-sectional area.

S MPK = 1/2 * MK * PO = 1/2 * 2 * 2√3 = 2√3 (cm²).

Answer: a cross-sectional area of 2√3 cm².