Given a trapezoid with bases of 5 cm and 14 cm, divide one of the sides into 3 equal
Given a trapezoid with bases of 5 cm and 14 cm, divide one of the sides into 3 equal parts and from the dividing points draw segments parallel to the base to the other side, find the lengths of these segments.
Consider a trapezoid ABCD with bases AD = 14 and BC = 5.
Let points M and N divide AB into 3 equal parts and through them there are straight lines parallel to the base of AD that intersect the lateral side of CD at points K and L. Point M is closer to the vertex B
You need to find the lengths MK and NL.
Let us write the solution in general form. Let AD = a, BC = b, MK = c, NL = d.
Consider 2 trapezoids: NBCL and AMKD. The middle lines of these trapezoids MK and NL are:
c = (b + d) / 2, d = (a + c) / 2,
2 * d = (a + (b + d) / 2) = (2 * a + b) / 2 + d / 2, 3/2 * d = (2 * a + b) / 2,
d = (2 * a + b) / 3.
c = (b + (2 * a + b) / 3) / 2 = (3 * b + 2 * a + b) / 6 = (2 * b + a) / 3.
Substitute a = 14 and u = 5:
d = (2 * 14 + 5) / 3 = 11,
c = (2 * 5 + 14) / 3 = 8.