A line segment parallel to the second-smallest side of a right triangl

Official Explanation

To understand this question, we will draw a right triangle:



We draw one that is very non-isosceles so that there is no confusion about what the smallest and second-smallest sides are. The hypotenuse, as always is the longest side. The left side can be called the height of the triangle and is the smallest side. The base of the triangle is the second-smallest side. The line that we have to add bisects the smallest side:



The "smaller triangle" that the question refers to must be what's labeled here as ABC; it, too, is a right triangle, since BC is parallel to the base of the larger triangle. We want the area of this smaller triangle. On to the data statements, separately first, as always.

Statement (1) gives the area of the larger triangle. Will that be sufficient to get the area of the smaller triangle? It's clear that the height of the smaller triangle is half the height of the larger triangle. There is a more specific relationship between the two triangles, however, since they have identical angles. They are therefore "similar triangles," meaning they have the same shape. Since the side opposite a given angle scales proportionally with that angle, two similar triangles, with identical angles, will have sides of proportional lengths. The upshot is that, since the height of the smaller triangle is half the height of the larger triangle, the base must be half of the base of the larger triangle. Substituting h --> h/2 and b --> b/2 into the area formula indicates that the area of the smaller triangle is 1/2*h/2*b/2 = 1/4(bh/2) , one quarter of the larger triangle. In our diagram, the smaller triangle doesn't quite look that small--could four of them fit inside the larger triangle? Indeed, if we draw the situation accurately we get:



Four triangles in one! If you aren't certain about this conclusion, you might be able to convince yourself by analysis by cases, in fact. Namely, if you can convince yourself that the case above of 4-in-1 triangle configuration is possible, you can conclude that it must entail bisecting the height and the base, and then you can convince yourself that, working the other way, when the height is bisected by a line parallel to the base, the only possible case is the 4-in-1 case. The area of the smaller triangle is a quarter of the larger area, so it must be 12, and we have answered the question. Sufficient.

A further side note: this is actually connected to a way of thinking about the formula for the area of a triangle. It's no coincidence that the formula for the area of a triangle bh/2 is half the formula of a rectangle, bh, in that any triangle can be doubled and converted into a rectangle. For example:



(It's true also for non-right triangles. You can get out your Tangram set and play around with two congruent triangles.)

Statement (2): Knowing the base is insufficient, because we know nothing about the height of the smaller or bigger triangle. We can imagine two cases, one in which the triangle has a small height and another in which the triangle has a large height. Both cases are allowed by the data and yield different areas, so Statement (2) is insufficient.

The correct answer is (A).