Yes, we can assume. But we sud understand the reason.
Both the traingles ADB and ADC are cogruent. (RHS property, one angle is right angle and one side of same length. In this case the ommon side AD.). SInce both the traingles are congruent, BD = DC and hence AD is perpendicular bisector.

Answer sud be D.

I get A.

If you know a side and an angle on 30-60-90 right triangle, you can get the other two sides. Statement 1 tells us that this is a 30-60-90. Thus, you can find the other two sides using identities. Dont solve all the way through - just know that you can!

Statement 2 doesnt give us a special right triangle relationship. You cant trust the figure - it may not be drawn to scale.

Can we not assume that AD is a perpendicular bisector of BC?

We ca only assume that provided ABD and ADC are congruent... Unless and untill you can establish that , AD is not the bisector of BC

Statement 1 is insufficient because we can only establish values for AB and BD from this ... For area we need the value of DC and we have no clue how long CD is
Statement 2 is insufficient again because from this we can find the value of CD ..Nothing beyond that ..
Hence answer is C

stmt 1:
abd=60 degree
so by 30-60-90 rule we can find sides as we know ad=6

stmt 2:
ac=12
here we dont know the angle of other sides it can be 45-45-90 or 30-60-90 and so we cant find sides.

so IMO: A

IMO C

Whats the OA?

AD cannot be assumed as a perpendicular bisector as it hasn't been mentioned that the triangle is isosceles or equilateral ( Perpendicular from the vertix to the base bisects the base only in case the triangle is Isosceles or Equilateral)

Statement 1: Not sufficient as with the angle ABD = 60 , we can find the value of AB and BD , which is not sufficient to answer

Statement 2: Not sufficient either, citing the same reason as above

Statement 1 and Statement 2 can together give us the area , Hence C

Won't the answer be D? Each statement by itself is sufficient?

Area of a triangle is 1/2BH.

H is given as 6 (AD). ABD angle is 60 so this is a 30-60-90 triangle. We can find Base by finding BD and then multiplying by 2. Therefore Statement 1 is sufficient.

For statement 2, can't we use Pythagoras Theorem? For a right angle triangle, we are given two sides.
I.E. 12^2=6^2+x^2 and then solve for x where x is DC? Multiply DC by 2 and you have the Base.

I believe the answer should be D. What do you guys think?

I'm in the "I think its (C) camp"

We cannot assume AD to be the bisector. You can easily draw a triangle where AB and BD would be much greater than AC and CD respectively.

Stm1 helps us get BD (using 30-60-90)

Stm2 helps us get CD (using 30-60-90)

Combining 1 and 2 gives us BC, which in turn gives us the area.

ru2008 wrote:Can we not assume that AD is a perpendicular bisector of BC?

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

1.You definitely can not assume that it is bisector . Why would you do so ? There are many traiangles possible with perpendicular and not being bisector .
2 . OA should be A

Explanation :
1.
ABD = 60 ; that makes left hand triangle 3à,60 and 90 traingle.If you know one side (which we know here) , you get the left hand side traingle area. RIght hand triangle is also 30,60 and 90 triangle so you get the area of this triangle too. Hence you can get total area.

2. Statement 2 gives you the right hand triangle sides but you have no way to know left hand side. Because sides of right hand traingle do not give any special relationship either.

GMATMadeEasy wrote:

ru2008 wrote:Can we not assume that AD is a perpendicular bisector of BC?

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

1.You definitely can not assume that it is bisector . Why would you do so ? There are many traiangles possible with perpendicular and not being bisector .
2 . OA should be A

Explanation :
1.
ABD = 60 ; that makes left hand triangle 30, 60 and 90 triangle. If you know one side (which we know here) , you get the left hand side traingle area. RIght hand triangle is also 30, 60 and 90 triangle so you get the area of this triangle too. Hence you can get total area.

But you just did when you make the assumption that the left side triangle is equal to the right side triangle.

There's nothing in the stimulus that says this is an "equilateral triangle", so there is no evidence that AD bisects BC. So I would think the answer should be C as well, unless this is a test of some obscure triangle knowledge, in which if left side triangle is 30/60/90, then the right must also be the same, and that this triangle must be an equilateral triangle.

Btw, if you punched in the number, you can see that this is definitely not an equilateral triangle.

30/60/90 = x/x√3/2x

So if x√3 = 6, then 2x must be 4√3.

so AB is not the same as AC, which is 12.

Also, if your reasoning is that A is sufficient because left side triangle is the same as right side triangle, then you must assume all measure is the same on both sides. Therefore if AC = 12, and AD is 6, we can easily find DC, and since both triangles are the same, the same assumption of (1), then we could do the same for ABD.

So I think the answer would either be D or C depending on your reasoning, but it shouldn't be A or B by itself.

Agreed , I made an error. It should be C .


I ,by mistake,assumed one of the angles as 30 in right side triangle .. sorry.. I didnt mean equilateral triangle at all.

Don't be sorry man, it's a discussion! We're all here to learn and to improve.

Any way to find out the actual answer to the question??

magic of Google: https://www.manhattangmat.com/forums/post3967.html

Lol, Harish Dorai drew the triangle I was going to draw and post here.