techiesam wrote:

Find the area of trapezium ABCD.

Given angle ADB=45,BCN=30 and side AB=4 and BC=10.

A.5/2(9+5√3)

B.5√3/2(13+5√3)

C. 13(13+2√3)

D.2/5(13+2√3)

E. None of these

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BC = 10

So, we know the hypotenuse of triangle BNC

This is a 30-60-90 degrees triangle

So, L(Side BN) = 1/2 x hypotenuse = 1/2 x 10 = 5 units

L(Side NC) = \sqrt{3}/2 x hypotenuse = \sqrt{3}/2 x 10 = 5\sqrt{3}

Since sides AB and CD are parallel and sides AM and BN are parallel, AB = MN = 4 units

Also, AM = BN = 5

Triangle AMD is a 45-45-90 triangle

which makes AM = MD = 5 units

Area of trapezium ABCD = 1/2 x sum of parallel side AB and CD x height

= 1/2 x (AB + MD + MN + NC) x 5

= 5/2 x (4 + 5 + 4 + 5\sqrt{3} ) = 5/2 x ( 13 + 5\sqrt{3})

E. None of these