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If AD is 6 and ADC is a right angle, what is the area of triangular re

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

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New post 16 Oct 2017, 02:46
zanaik89 wrote:
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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.


Hi Bunuel

Cant we assume that that AD bisects BC into two?

Please clear my doubt thanks[/quote]


So u mean that since it isnt given that triangle ABC is an isoceles triangle or equilateral triangle, we can not consider AD as a perpendicular bisector right?
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Re: If AD is 6 and ADC is a right angle, what is the area of triangular re [#permalink]

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New post 16 Oct 2017, 02:53
zanaik89 wrote:

So u mean that since it isnt given that triangle ABC is an isoceles triangle or equilateral triangle, we can not consider AD as a perpendicular bisector right?


Yes, we know that AD is perpendicular to BC but we don't know whether it bisects BC.
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Re: If AD is 6 and ADC is a right angle, what is the area of triangular re   [#permalink] 16 Oct 2017, 02:53

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