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If a right triangle has

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by fiza gupta » Tue Nov 22, 2016 9:46 am
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?

a) 1
b) 2
c) 3
d) 4
e) 5

OA:D
Fiza Gupta
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by Matt@VeritasPrep » Fri Nov 25, 2016 2:54 pm
We've got:

a + b + c = 19

c > 9

a² + b² = c²

and we want ab/2.

Let's start with a little rejiggering:

a² + b² = c²

(a + b)² - 2ab = c²

(a + b + c - c)² - 2ab = c²

(19 - c)² - 2ab = c²

361 - 38c = 2ab

We want ab/2, so we'll divide both sides by 4:

90.25 - 9.5c = ab/2

Since c > 9, we know the left hand side is less than 90.25 - 9.5*9, or less than 4.75.

So any value between 0 and 4.75 should be possible, giving us FOUR integer areas (1, 2, 3, and 4).
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by Matt@VeritasPrep » Fri Nov 25, 2016 2:56 pm
For anyone reading this in search of questions, this problem was a lot of fun, but not at all a question you'd see on the GMAT - it's too hard.
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by Matt@VeritasPrep » Fri Nov 25, 2016 3:01 pm
Another approach just occurred to me too.

Let's say that c = 9. Since c > 9, whatever we get with c = 9 should give us the max.

If we have a + b + c = 19 and a² + b² = c² and c = 9, then we've got

a + b = 10

a² + b² = 81

Then we can say

a² + b² is (a + b)² - 2ab

so

81 = 10² - 2ab

19 = 2ab

4.75 = ab/2

So as before, we can have any area less than 4.75.
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