Question

ABCD is a trapezium with AB||CD and ANGLE BCD=2 ANGLE DAB. if DC=a and BC=b, the length of AB in terms of a and b is____
A) 2a+2b
B) 2a+b
C) a+2b
D) a+b

Answer 1

Given:

ABCD is a trapezium

AB // CD

∠BCD = 2∠DAB

DC = a

BC = b

To find:

The length of AB in terms of a and b

Construction:

Take a point E on AB and join C and E such that CE // DA as shown in the attached figure

Solution:

Let's assume, ∠DAB = y

∴ ∠BCD = 2∠DAB = 2y° .... (i)

We are given,

AB // CD

⇒ AE // CD

⇒ AECD is a parallelogram .... [from construction we have CE//DA]

∴ CD = AE = a ..... (ii) ..... [Opposite facing sides of a parallelogram are equal in length]

We know that → the sum of the angles of two adjacent sides of a trapezium is equal to 180°

∴ ∠B + ∠BCD = 180°

substituting the value of ∠BCD from (i)

⇒ ∠B = (180 - 2y)° ..... (iii)

Also,

∵ CE // DA

⇒ ∠CEB = ∠DAB ..... [alternate angles]

⇒ ∠CEB = ∠DAB = y° .... (iv)

In Δ CBE, using the angle sum property of a triangle, we get

∠ECB + ∠CEB + ∠B = 180°

substituting from (iii) & (iv) we get

⇒ ∠ECB + y° + (180 - 2y)° = 180°

⇒ ∠ECB + y° + 180° - 2y° = 180°

⇒ ∠ECB + y° - 2y° = 0

⇒ ∠ECB - y° = 0

⇒  ∠ECB =  y° ..... (v)

From (iv) & (v), we get

∠CEB = ∠ECB =  y°

∴ BC = BE ...... [sides opposite to equal angles are also equal to each other]

⇒ BC = BE = b ..... (vi) ..... [∵ BE = b (given)]

Now, from the figure, we get

AB = AE + BE

substituting the value of AE & BE from (ii) & (vi), we get

⇒ AB = a + b ← option (D)

Thus, $\boxed{\bold{The\:length\:of\:AB\:in\:terms\:of\:a\:and\:b\:is \:\underline{a+b}}}.$

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