ABCD is a trapezium. Sides AB and CD are parallel to each other. AB = 6 cm, CD = 18 cm, BC = 8 cm and AD = 12 cm. A line parallel to AB divides the trapezium in two parts of equal perimeter. This line cuts BC at E and AD at F. If BE/EC = AF/FD then what is the value of 2BE/EC?
Show answer & explanation
Correct answer: C
Explanation
The correct answer is Option 3 i.e. 8.
Given,
BE/EC = AF/FD
So,
(8 - EC)/EC = (12 - FD)FD (Given, BC = 8 and AD = 12)
So,
EC/FD = 8/12 = 2/3
Let's say, EC = 2k and FD = 3k.
So, AF = (12 - 3k) and BE = (8 - 2k)
According to the question :
Perimeter of ABEF = Perimeter of FECD = (6 + 8 + 18 + 12)/2 = 22 cm
So,
FE + 3k + 2k + 18 = 22
⇒ FE + 12 − 3k + 8 − 2k + 6 = 22
⇒ FE = (5k − 4)
Again,
FE + CD + FD + EC = 22.
⇒ 3k + 2k + 18 + 5k − 4 = 22
⇒ 10k = 8.
⇒ k = 8/10 = 4/5
So,
2BE/EC = 2(8 - 2k)/2k = (8 - 2k)/k = (8 - 8/5)/4/5 = 32/5 × 5/4 = 8