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
If a right triangle has
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- fiza gupta
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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).
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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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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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.
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.