There is a trapezoid ABCD whose side AD and BC are parallel as showing in the figure below.
E is a point on the side BC. AB and DE are parallel.
This trapezoid is divided into five triangle P, Q, R, S, and T with AC, DE, and BF.
The area of P is 32 cm2.
The area of Q is larger than the area of S by 8 cm2.
(1) Answer triangles whose area is equal among P, Q, R, S, and T.
(2) When the height of trapezoid ABCD is 16cm and an area is 280cm2, find the length of the EC.
Answer
(1) R, T
(2) 15 cm
Solution
(1) As quadrangle ADEB is parallelogram, P + R = Q. As triangle ACD = ABD, P + T = Q.
Thus R = T.
(2) According to 280 × 2 / 16 = 35, AD + BC = 35 cm.
P = 32 (cm2),
As P +R = Q, 32 + R = Q.
R = Q - 32.
As Q = S + 8 and Q = S - 8, R = S - 24 and T = R = S - 24.
Thus, 32 + S + 8 + S - 24 + S -24 = 280.
4S = 288
S = 288/4 = 72
R = 72 - 24 = 48
BE : CE = 48 : 72 = 2 : 3
AD : BE : EC = 2 : 2 : 3
Thus EC = 35 × 3/ (2+2+3) = 15 cm.
