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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

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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

by AAPL » Thu Jul 01, 2021 10:04 am

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

by gmatbyexample » Thu Jul 01, 2021 4:40 pm
In these kind of problems, I recommend redrawing the triangle after deriving some preliminary information.

Two properties we will use here:

1) Angles on a line are supplementary (their sum is 180)
2) Sum of angles of the triangle is 180

Since Angle ADC is 120 => angle ADB = 60 (sum must be 180). Since Angle BAD is 30, the angle ABD (or angle ABC) = 90.
beat_1.png
Now that we know that angle ABC is 90 - lets redraw the triangle to make it look better.
beat_2.png
To get the area of triangle ADC, we just need to figure out "h" and the area would be 1/2 * h * 4

Finding h is very easy for the 30, 60, 90 , h = 6* sqrt{3}

Area = 1/2 * 4 * 6 * sqrt{3}

Area = 12 * sqrt{3}

(D)
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