# ABCA1B1C1 is a straight triangular prism. the length of each edge of which is 8 cm.

ABCA1B1C1 is a straight triangular prism. the length of each edge of which is 8 cm. Point E, F are the midpoints of segments B1A, BC1, respectively. Calculate the perimeter of the EFC triangle.

The length of the segment EC is half the length of the diagonal CB1.

CB1 ^ 2 = BC ^ 2 + BB1 ^ 2 = 64 + 64.

CB1 = 8 * √2 cm.

Then CE = 8 * √2 / 2 = 4 * √2 cm.

The length of the segment EF is defined as its projection onto the plane of the base, E1F1 = EF = AC / 2 = 8/2 = 4 cm, since E1F1 is the middle line of the triangle ABC.

Quadrangle CEFA is an isosceles trapezoid.

Let’s build the height FH. The length of the segment AH = (AC – EF) / 2 = (8 – 4) / 2 = 2 cm.

Then FH ^ 2 = AF ^ 2 – AH ^ 2 = 32 – 4 = 28.

CH = AC – AH = 8 – 2 = 6 cm.

Then CF ^ 2 = CH ^ 2 + FH ^ 2 = 36 + 28 = 64.

CF = 8 cm.

Then the perimeter of the triangle EFC will be: P = EF + CF + CE = 4 + 8 + 4 * √2 = 12 + 4 * √2 = 4 * (3 + √2) cm.

Answer: The perimeter of the triangle is 4 * (3 + √2) cm. One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.