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.