Question

ABCD is a trapezium in which AB is parallel to CD. The sides AD and BC when extended, intersect at point E. If AB = 2 cm, CD = 1 cm, and perimeter of ABCD is 6 cm, then the perimeter, in cm, of ∆AEB is

• A. 10
• B. 9
• C. 8
• D. 7

Answer 1

Answer: (C)

8

Answer 2

Let AD = x and BC = y.

Given, AB = 2 cm, CD = 1 cm, and perimeter of ABCD = 6 cm. So, 2 + 1 + x + y = 6 ⇒ x + y = 3

Now, let's consider triangles ABE and CDE. These triangles are similar. So, the ratio of their sides is equal.

$\frac{AE}{CE} = \frac{AB}{CD} = \frac{2}{1}$

Let AE = 2k and CE = k.

Now, AD = AE + ED = 2k + k = 3k = x BC = BE + EC = 2k + k = 3k = y

Therefore, x = y = 3k.

Since x + y = 3, we get 6k = 3, or k = 0.5.

So, AE = 2k = 1, BE = 2k + 1 = 2, and CE = k = 0.5.

The perimeter of triangle AEB = AE + BE + AB = 1 + 2 + 2 = 5.

Therefore, the perimeter of triangle AEB is 5 cm.