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
In the given figure \(BD=6\) and \(DC=4\). If angle \(BAD\)
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- Jay@ManhattanReview
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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.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
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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