HH wrote:
Hi, I do not understand this question? help please
if a right triangle has a perimeter of 19 and a hypotenuse that is greater than 9, its area can be how many positive integers?
1 2 3 4 5
This is a very difficult question - way too hard for the GMAT.
First, say our sides are a, b and c, where c is the hypotenuse. Then, by the triangle inequalities, we know that a + b > c. If c > 9, and if a + b + c = 19, then it must be that a + b > 9.5, and 9 < c < 9.5. We'll need these inequalities later.
Now, we know:
a + b = 19 - c (using the perimeter)
a^2 + b^2 = c^2 (by Pythagoras)
Squaring both sides of the first equation, we have:
a^2 + b^2 + 2ab = 361 - 38c + c^2
Now, substitute c^2 in place of a^2 + b^2 :
c^2 + 2ab = 361 - 38c + c^2
2ab = 361 - 38c
Notice now that ab/2 is the area of our triangle, so dividing by 4 on both sides,
ab/2 = AREA = (361 - 38c)/4
Now, finally using the fact that 9 < c < 9.5 we find that
[361 - 38(9.5)]/4 < AREA < [361 - 38(9)]/4
0 < AREA < 4.75
So the only integer values that are possible for the area are 1, 2, 3 and 4. Now, it's not completely obvious, I don't think, that it's possible to construct triangles satisfying the restrictions in the question with each of these areas. Conceptually, you can imagine first making a right triangle with perimeter 19, and with one side which is very close to 0 in length. That triangle will have an area almost equal to 0. Now, if we gradually increase the length of the shortest side, while decreasing the lengths of the other sides to keep the perimeter constant, the area will gradually increase, up until we get to an isosceles right triangle. If you take this perspective, you can see that it's possible to have each of the four integer areas we found above, though proving this algebraically is actually not at all easy, and this is one of the reasons the question is far beyond the scope of the GMAT.
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