If AD is 6root 3, and ADC is a right angle, what is the area

[ms100]

If AD is 6root 3, and ADC is a right angle, what is the area of triangular region ABC?

(1) Angle ABD = 60°

(2) AC = 12

I don't get why the answer is not d since we know that in 1) abdis 60degrees then th angle DAC is also 30 because the angle ADC is 90 degrees. I've already figured out the alue for DC which is 6 by the information given in statement 1.

Statement 2 says AC is 12 and from the question stem we know AD is 6 root 3 then the degrees correspond to a 30-60-90 and DC is also 6.

Thanks,
Mariela

[StaceyKoprince]

(For our students: if you've done this one before, it is titled "Triangle ABC" so you can look it up. There's a diagram that goes with it - can't post it here.)

We're asked for the area of the larger triangle. Formula for area of triangle = 1/2 (bh). We're given the height (6SQRT3) but not the base, BC.

Don't forget that D requires each statement to work BY ITSELF.

Statement 1 allows us to calculate BD, which is part of BC but not the whole thing. We cannot calculate DC, so we can't find the entire length of BC. Eliminate answer choices A and D (because statement 1 does not work BY ITSELF, and this is required for A or D to be correct).

Statement 2 allows us to calculate that DC = 6, but we cannot figure out the value of BD, which means we can't find the value of BC. Eliminate answer choice B (because statement 2 does not work BY ITSELF, and this is required for B to be correct).

Now, when we combine the two statements, we can figure out the length of BC. Answer is C.

[Guest]

I think the answer should be D bcos of the principle "Angles opposite to equal sides are equal" and vice versa.

1 . Since I know angle ABD it should be equal to angle ACD, hence one should be enough to answer the question, that is the area

2. Since we know length of AC, one can calculate angle ACD which is sinx = AD/AC = sin 60 = angle ABD, so 2 again we can calculate the area

I would like to know where I am wrong

[StaceyKoprince]

1) The problem stem + statement 1 does not give us enough information to say that angle ABD is equal to angle ACD. You cannot assume this just because the angles look similar in the picture - they could be very close to the same angle but still different.

2) Once again, the problem stem + statement 2 does not give us enough information to say that the two angles are equal. You are assuming either based on the picture or based on the fact that, once you've done the work, you see that they are, in fact, equal. But if you don't have the two pieces of info together, you don't actually know that they are equal.

[Guest]

As I stated in my response I am not assuming that the angles are equal based on the diagram but on the principle which I remember from my school says

"Angles opposite to equal sides are equal" and sides oppsite to equal angles are equal"

[tmmyc]

"Angles opposite to equal sides are equal"
This is true.

"Sides opposite to equal angles are equal"
This is false.

Take a simple example like any two isosceles right triangle.

The angles are always 45-45-90, but the sides could be 1, 1, (1*sqrt(2)) or 100, 100, (100*sqrt(2)).

[Guest]

If Angles opposite to equal sides are equal then its corollary Sides opposite to equal angles should equal

lets say ABC is a right isosceles triangle so say angle BAC = angle ACB = 45 and angle ABC=90, then the ratio of sides AB:BC:AC is x:x:x sqrt(2)

which means AB and BC = X, which mean equal sides

All I am saying is side opposite to angle BAC is BC and angle opposite to angle ACB is AB and they will be equal

[tmmyc]

I think the answer should be D bcos of the principle "Angles opposite to equal sides are equal" and vice versa.

1 . Since I know angle ABD it should be equal to angle ACD, hence one should be enough to answer the question, that is the area

2. Since we know length of AC, one can calculate angle ACD which is sinx = AD/AC = sin 60 = angle ABD, so 2 again we can calculate the area

I would like to know where I am wrong

Here is the original thread to which I believe Stacey was referring: http://www.manhattangmat.com/forums/post3967.html

If you go to fourth post, you can see a more exaggerated sample triangle that shows how ABD and ACD are not necessarily equal.


Ignore my previous post because I misunderstood what you were saying originally.

[Guest]

Thanks for the link tmmyc!

So the rule Angles opposite to equal sides are equal and its corollary Sides opposite to equal angles are equal apply only to a single given triangle.

[RonPurewal]

So the rule Angles opposite to equal sides are equal and its corollary Sides opposite to equal angles are equal apply only to a single given triangle.

correct.

if you have a diagram in which two or more triangles are stuck together (via a common side, or vertical angles, or whatever else), you can only apply this rule to one triangle at a time. if you need to build inequalities that compare sides from different triangles, then you should construct those inequalities by using the common side/angle.

[beakdas]

What i fail to understand here is that in a triangle when the perpendicular is dropped on the base it divides the base into equal halves.

Statement gives the value of DC=6.We are told that altitude is 6 root 3. Why can't we conclude that BC is also 6.

Similarly for statement 1?

[cbjohn1]

The text format of this one really threw me for a loop. Using Chrome on Windows 8, the question looks like it says 6^sqrt(3), not 6*sqrt(3). Could someone look into that?

[RonPurewal]

You will never see an irrational number as an exponent on the gmat exam. So, that one is settled.

Remember"”The GMAT will never use a math concept that's not defined by the time you finish first-year high-school algebra and geometry.
So, you might see integer exponents, and you might see fractional exponents (which are the same as roots), but definitely not √3 as an exponent. (It's impossible even to define what that means without calculus.)

In any case, I'll pass this on to the tech team.