Quadrilateral ABCD is a trapezium, where AD is parallel to BC, BC = 5 and AD = 4
Drop a perpendicular from D to line segment BC (Let us name this point as E)
Now, in triangle DEC;
Angle DAC = 90
Angle EDC = 30 .......... (120 - 90)
Angle ECD = 60
Also, triangle DEC is a 30-60-90 triangle. Ratio of sides in 30-60-90 triangle is \(x : \sqrt{3}x : 2x\)
i.e.
opposite to 30 = \(x\)
opposite to 60 = \(\sqrt{3}x\)
opposite to 90 = \(2x\)
Opposite to 90 is side CD = 3
So, \(2x = 3\)
\(x = \frac{3}{2}\)
Therefore, CE = \(\frac{3}{2}\) and DE = \(\frac{3\sqrt{3}}{2}\)
Area of Trapezium = \(\frac{1}{2}\) x (sum of the parallel sides) x Height
= \(\frac{1}{2}\) x (5 + 4) x \(\frac{3\sqrt{3}}{2}\)
= \(\frac{27\sqrt{3}}{4}\)
Hence, option C
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