AAPL wrote:e-GMAT



In the given figure, \(BD=6\) and \(DC=4\). If angle \(BAD=30^o\) and angle \(ADC=120^o\), find the area of the triangle \(ADC\).

A. \(6\sqrt{3} \text{ cm}^{2}\)

B. \(8\sqrt{3} \text{ cm}^{2}\)

C. \(10\sqrt{3} \text{ cm}^{2}\)

D. \(12\sqrt{3} \text{ cm}^{2}\)

E. \(18\sqrt{3} \text{ cm}^{2}\)

OA D

Given /_ADC = 120, we have /_ADB = 180 - 120 = 60. Again, given /_BAD = 30 and /_ADB = 60, we have /_ABD = 90. Thus, ∆ABD is a 30-60-90 rightangled triangle.

For the 30-60-90 the rightangled triangle ∆ABD, given that BD = 6, we have AB = 6√3

Thus, area of ∆ADC = 1/2 * AB * DC = 1/2 * 6√3 * 4 = 12√3

The correct answer: D

Hope this helps!

-Jay
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