Question
Download Solution PDFABCD 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 BE/EC?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
As shown in the above figure ABCD is the given trapezium with AB, CD, BC, AD as 6, 18, 8, and 12 cm respectively. A line is also shown in the above figure which divides the trapezium into two equal perimeters. Also BE/EC = AF/FD
Calculation:
Let us assume the value of AF be x cm, FD be (12 – x) cm, also BE be y cm and EC be (8 – y) cm.
Now as given BE/EC = AF/FD Apply the above assumed values into this
⇒ y/(8 – y) = x/(12 – x)
⇒ y(12 – x) = x(8 – y)
⇒ 12y – xy = 8x – xy, 12y = 8x
⇒ x:y = 3:2 ----(1)
Now as given the parametres of parts ABEF and FECD are equal
⇒ AB + BE + EF + AF = FE + EC + CD + FD
⇒ 6 + y + FE + x = FE + 8 – y + 18 + 12 – x
⇒ 2x + 2y = 32 or x + y = 16 ----(2)
⇒ 3/2y + y = 16 {from(1)}
⇒ 5y = 32, y = 32/5
Now put the value of y in (2)
⇒ x = 16 – 32/5 = 48/5
Now the ratio of BE/EC = y/(8 – y)
⇒ (32/5)/(8 – 32/5) = (32/5)/ (8/5) = 4/1
∴ The value of the ratio of BE/EC = 4:1
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