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Re: If AD is 6, and ADC is a right angle, what is the area of [#permalink]
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sm0k3rz wrote:

If AD is 6, and ADC is a right angle, what is the area of triangular region ABC?

(1) Angle ABD = 60°

(2) AC = 12

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-ad-is-6-and-adc-is-a-right-angle-what-is-the-area-of-101200.html



If AD is 6 and ADC is a right angle, what is the area of triangular region ABC?

Given: \(AD=6\). Question: \(area_{ABC}=\frac{1}{2}*AD*BC=\frac{1}{2}*6*(BD+DC)=3(BD+DC)=?\)

(1) Angle ABD = 60 --> triangle ABD is 30-60-90 triangle, so the sides are in ratio \(1:\sqrt{3}:2\) --> as AD=6 (larger leg opposite 60 degrees angle) then \(BD=\frac{6}{\sqrt{3}}\) (smallest leg opposite 30 degrees angle). But we still don't know DC. Not sufficient.

(2) AC=12, we can find DC. But we still don't know BD. Not sufficient.

(1)+(2) We know both BD and DC, hence we can find area. Sufficient.

Answer: C.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-ad-is-6-and-adc-is-a-right-angle-what-is-the-area-of-101200.html



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Re: If AD is 6, and ADC is a right angle, what is the area of [#permalink]
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