# A straight line parallel to the bases of trapezoid ABCD intersects its lateral sides AB and CD

**A straight line parallel to the bases of trapezoid ABCD intersects its lateral sides AB and CD at points E and F, respectively. Find the length of the line segment EF if AD = 21 cm, BC = 5 cm, and the area of the trapezoid AEFD refers to the area of the trapezoid EFCB as 45: 7.**

Let the length of the segment EF = X cm.

From points C and F we draw straight lines CK and FP parallel to the side of the AC, as well as straight lines CM and FH perpendicular to EF and AD.

Triangles CKF and FРD are similar in two angles, then: KF / PD = CF / FD. (1)

Rectangular triangles CMF and FНD are also similar in acute angle, then: CM / FH = CF / FD. (2).

In equalities 1 and 2, the right-hand sides are the same, then: KF / PD = CM / FH.

KF = X – BC = (X – 5). RD = AD – AP = AD – X = (21 – X).

Then: (X – 5) / (21 – X) = CM / FH.

The area of the trapezoid EBCF is equal to: S1 = (BC + X) * CM / 2 = (5 + X) * CM / 2.

The area of the trapezoid AEFD is: S2 = (AD + X) * FH / 2 = (21 + X) * FH / 2.

S2 / S1 = 45/7 = ((21 + X) * FH / 2) / ((5 + X) * CM / 2).

45/7 = (21 + X) / (5 + X) * FH / CM = (21 + X) * (21 – X) / (5 + X) * (X – 5).

45/7 = (212 – X ^ 2) / (X ^ 2 – 52).

-7 * X ^ 2 + 3087 = 45 * X ^ 2 – 1125.

52 * X ^ 2 = 4212.

X ^ 2 = 4212/52 = 81.

X = EF = 9 cm.

The length of the segment EF is 9 cm.