If AD is 6 and ADC is a right angle, what is the area of triangular region ABC?[/b]
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 the area. Sufficient.
Answer: C.
Bunuel - My doubt is if a triangle like this which has a 30-60-90 angles, shouldn't the the triangle ADC should also be 30-60-90 making triangle ABC an equilateral triangle? because an equilateral triangle can be formed by joining two identical triangles, each with 30-60-90 angles.