mahendru1992
If AD = 10\(\sqrt{3}\) and ADC is a right angle, what is the area of triangle ABC?
(1) AC = 20
(2) \(angle BAD=30^{\circ}\)
This was asked in the Kaplan free test, but I think that the answer is wrong.
Am I wrong (most probably I am, lol). If I am, what's wrong with what i did?
Dear
mahendru1992,
I'm happy to respond.
See this blog for some information about what you can and can't assume.
https://magoosh.com/gmat/2012/gmat-trick ... -possible/That article is about assumptions in the PS questions. In the GMAT DS, you can make absolutely no assumptions from the diagram.
One very easy assumption blatantly encouraged by the diagram is that triangle is symmetrical. This erroneous assumption was your starting point. ABC is drawn to appear isosceles, but there is absolutely nothing in the prompt information that guarantees that it is isosceles --- for all we know, triangle ABD and triangle ADC are completely different, and have nothing in common except for a right angle. It is very tempting to assume symmetry, but that is absolutely something you are allowed to assume, even on the GMAT PS. On the GMAT DS, you can assume nothing.
Does all this make sense?
Mike