B encloses a region where the altitude value will give the area range.
B
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Re (2): there are some properties of right triangles which can be useful.
Please, refer to the attached drawing.

All three right triangles (the large one ABC and the two small ones, ABD and ADC) are similar (it can be easily seen that their angles are congruent).

From the similarity of the triangles ABD and CBA we obtain:
\frac{AB}{BC}=\frac{BD}{AB}, from which it follows that AB^2=BD*BC.
Similarly, using the similar triangles ACD and BCA, we get that AC^2=DC*BC.
We can call it the "leg property", as it expresses the leg of the right triangle in terms of its projection to the hypotenuse and the hypotenuse itself.

In our case, using the above property, we can find the two legs of the right triangle being \sqrt{5} = \sqrt{1*5} and 2\sqrt{5}=\sqrt{4*5}.
Then the height of the right triangle is \frac{\sqrt{5}*2\sqrt{5}}{5}=2 <--- leg * leg / hypotenuse.

But there is another useful property of the right triangle, which helps us find directly the height if we know the two projections of the legs to the hypotenuse.
This property was already mentioned by "bellcurve".
Using the similarity of the triangles ABD and CAD, we can write \frac{AD}{DC}=\frac{BD}{AD}, from which AD^2=BD*DC.
So, in our case, the height of the right triangle is \sqrt{1*4}=2.
We can call this the "height property".

Well, we used to call the above properties "The leg Theorem" and "The height Theorem" back then in Romania. Although I have a PhD in Applied Math, I learned the above stuff in junior high. What I have learned during my graduate studies, cannot really help on the GMAT :O)

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For any triangle the max area is can be got by for a equilateral triangle. - This is not a correct statement. You can have a "tiny" equilateral triangle or a "huge" right or any other triangle. The correct statement is "For a given perimeter, the maximum area is that of an equilateral triangle."

The second statement isn't correct either:
For any right angled triangle the max area can be got by forming an isosceles triangle.
I think what you meant is "For a right triangle inscribed in a given circle (which means given constant diameter), the maximum area is that of an isosceles right triangle."

For the given question, in fact we don't have to go the either of the above properties.
Lets' denote by B the point with the coordinates (-1,0) and by C the point with the coordinates (4,0).
Since BC lies on the X-axis and A is on the Y-axis, the area of the triangle ABC is given by Base*Height/2, which in this case is 5*Lenght of OA/2, regardles whether A is above or below the X-axis (A or A_1 in the attached drawing). So, the area of triangle ABC is 5*A/2, where A is the length of OA (consider A>0). Therefore, the question in fact is "Is 5A/2>15?" or"Is A>6?"

(1) Not sufficient, as A_1 can be below -6, in which case A>6, but if A is above 0 and below 3, obviously the area is less than 15.
(2) If the triangle ABC is a right triangle, then necessarily \angle{BAC} is a right angle and BC is the diameter of the circle in which this triangle can be inscribed. Even if the triangle is isosceles, the height AO cannot be greater than the radius of the circle AO_1, which is 5/2. So, A is definitely smaller than 6.
Sufficient.

Answer B.

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Love GMAT Quant questions and running.

ANd for the 2nd one -> The right triangle with the largest area will be an isosceles right triangle (where both the base and height are of equal length).
In our case, not correct. Or rather, not clearly formulated. BC can be only the hypotenuse of the right triangle, as A must be on the Y-axis. So, we cannot have an isosceles right triangle with both legs 5. If A is not on the Y-axis, with no other restrictions, the area can be as large as we wish.
In this case, the right triangle is uniquely determined (except its mirror image), because the diameter is fixed. A must be at the intersection point of the circle with the Y-axis.
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PhD in Applied Mathematics
Love GMAT Quant questions and running.

Another approach to solve this question:
Since the area of the triangle is 1/2( Base * Perpendicular height), therefore the question can be interpreted as IS A>6?
#Base=5,
# questionable area=15
Hence IS \frac{(5*A)}{2}> 15?
or IS A>6?

Statement 1 is clearly insufficient as A can be 1 or -100.

Staement 2 is sufficient because its given that vertex at y-axis forms a right angle. Hence the base is the hypotenuese. Now if we draw a circle in such a way that (0,A) is at the circle, then the radius will be 2.5. Anything beyond 2.5 units from the centre of the circle will not form the right angle. Hence A is always less than 2.5 units from the centre of the circle.
Sufficient.
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Here's a simple way to calculate the area .. in case anybody is interested..

http://www.beatthegmat.com/ds-triangle- ... 89546.html


Press kudos if this helps
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