If a right triangle has a perimeter of 19 and a hypotenuse that is gre

Question

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

Bunuel · Accepted Answer

Let the hypotenuse be c and the legs a and b . Then area=\frac{ab}{2} . Given: a+b+c=19 and c>9 . Also as the length of any side of a triangle must be smaller than the sum of the other two sides then a+b>c . So must be true that: 9 9.5^2-c^2<2ab<10^2-c^2 ; As the highest limit of c^2 is 9.5^2 and the lowest is 9^2=81 then: 0<2ab<19 --> divide by 4 --> 0<\frac{ab}{2}<\frac{19}{4} --> 0

IanStewart · Answer

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 /4 < AREA < /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.

prashantbacchewar · Answer

Great solution bunuel +1 to you
+1 to Ian as well

prashantbacchewar · Answer

Bunuel ..Can we expect such problem on GMAT it is to difficult

HH · Answer

Thanks! In my opinion It is a very difficult one, let's see what say the experts

Bunuel · Answer

As Ian Stewart said the question is far beyond the scope of the GMAT and I completely agree.

EBITDA · Answer

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

Could someone please help me with the following question?

Thank you so much.

Kurtosis · Answer

Applying Pythagoras Theorem:

a^2 + b^2 > 9^2
(a + b)^2 - 2ab > 9^2
-2ab > 9^2 - (a + b)^2
2ab < (a + b)^2 - 9^2
ab < ((a + b)^2 - 9^2)/2

a + b + c = 19
c > 9
a + b < 10

Assume that a + b = 10 (approx)

ab < (10^2 - 9^2)/2
ab < 9.5
Area = ab/2 < 4.75

There can be 4 integral values for the area.

Answer: D

Senthil7 · Answer

Hi from where did you get this question?
The principle of Triangle is that sum of any 2 sides must always be greater than the third side and difference of 2 sides must be less than the third side. The values stated in your question does not fall under valid triangular parameters. Wonder how to answer this?

Kurtosis · Answer

Hi,

Its not stated in the question that the sides have integer values.

For example, consider a normal triangle where the sum of two sides is 9.9 and the other side can be 9.1. So a triangle can exist in such a circumstance.

Senthil7 · Answer

Hi Vyshak,

But the question is how many integer values for area is possible so i do not think any decimal or integesr x root values will give (bxh) / 2 an integer value. Can you please clarify with an example that works?

EBITDA · Answer

Hi Vyshak,

How do you go from this part:

"Area = ab/2 < 4.75"

To conclude that:

"Its area can be 4 positive integers."

Please explain the link between both. Can you plug in the examples of the numbers that would fit in?

Going back to the beginning. What is the question asking? I understand that it asks how many different positive integer areas can this triangle take. Am I right?

Thank you so much.

chetan2u · Answer

Hi,

My approach would be..
Perimeter = a+b+h = 19...... 
so a+b = 19-h...... 
square both sides....... . (a+b)^2=(19-h)^2....................a^2+b^2+2ab = 19^2+h^2-2*19*h

Now a^2+b^2 = h^2,... so ..2ab=19^2-38h=19(19-2h).................\frac{1}{2}ab=19\frac{(19-2h)}{4}.....

Since area is an integer, 19\frac{(19-2h)}{4} should be an integer
so the value of h has to be between 9 and 9.5, which can make 19-2h div by 4..

Now h has to have a multiple of 19 in the denominator....
so lets try 38 in denominator as 2h will cancel out 2 in 38 and 19 outside the BRACKET will cancel out 19 in 38..
 9 \frac{3}{38}, 9 \frac{7}{38}, 9 \frac{11}{38} and 9 \frac{15}{38} ...

the second denominator can be 19*4, but there will be 2 still remaining as ONLY one of the two in denominator will get cancelled out by 2 in 2h..

ans 4

Edited the answer

mikemcgarry · Answer

Dear chetan2u , My friend, I respectfully disagree with you. I'm going to rewrite your fraction in non-fraction form 19(19 - 2h) = 4A where h is the hypotenuse and A is the area, which should be an integer. We know that when h = 9, 19(19 - 2h) = 19(19 - 18) = 19 which, of course, is not a multiple of 4. Well, we could increase h by a tiny fraction, such as 3/38. This would be just enough to bring 19(19 - 2h) down to 16, which would lead to an integer area of A = 4. h = 9\frac{3}{38} leads to 19(19 - 2h) = 16, so that A = 4 h = 9\frac{7}{38} leads to 19(19 - 2h) = 12, so that A = 3 h = 9\frac{11}{38} leads to 19(19 - 2h) = 8, so that A = 2 h = 9\frac{15}{38} leads to 19(19 - 2h) = 4, so that A = 1 Four possibilities. OA = (D) Mike :-)

chetan2u · Answer

Thanks Mike,
You disagree and I respectfully agree you are correct here...
Although I was sure that the method has to be on thesE lines and something in me told me to check for values of h between 9&9.5, I think I somehow went against that...

EBITDA · Answer

Dear Mike,

Thank you for your response. Please note that you have made some errors on your calculations.

If:

A=1, h=9 23/38
A=2, h=9 27/38
A=3, h=9 31/38
A=4, h=9 35/38

Cheers.

mikemcgarry · Answer

Dear EBITDA, Thank you. :-)

You are perfectly correct that I made the purely boneheaded mistake of putting 19 as the integer part instead of 9 . That was a silly mistake and I have corrected it my post above. Thank you for pointing this out. I will say, though, I believe the fraction parts are correct. Let's look at your values. If h = 9 \frac{35}{38} , then 19(19 - 2h) = 19* = 19* = 19* = 19*(38/38 - 70/38) = 19*(-32/38) = -32/2 = -16 That would correspond to an area of negative four. I infer that you swapped a sign in your calculations some place, and that the values you found are for areas of -1, -2, -3, and -4. The fraction values I found were for the correct positive values. Because I've already made one mistake, I invite you and anyone else reading this to double-check my calculations with the fraction values given in my post above. Mike :-)

mvictor · Answer

I understand the approach, though...how can a triangle, with area equal to 1 (leg 1 * leg 2 / 2 = 1), have legs the sum of which is greater than 9.
same approach for 2 -> leg 1* leg 2 /2 = 2-> leg1 * leg 2 = 4. 
any ideas?
it is a property that does not need to be ignored - sum of any two sides must be greater than the third one. 
who can come with examples of values for leg 1 and leg 2, so as leg 1* leg 2 = 2, but leg 1 + leg 2 > third side - this case, >9 at least...

mikemcgarry · Answer

Dear mvictor , I'm happy to respond. :-)

Think about it this way. The longer and skinnier a geometry shape, the more perimeter it gets with respect to its area. Suppose we have a right triangle JKL, with right angle at K. Let one leg JK = 500, and let the other leg KL = 1/250 = 0.004. This would be a triangle that has an area of one, because A = \frac{1}{2}bh = \frac{1}{2}*500*\frac{1}{250} = 1 What is the perimeter of this triangle? Well, the hypotenuse given by JL^2 = JK^2 + JL^2 = 500^2 + 0.004^2 We don't need an exact calculation. We can see that JL would be number slightly larger than 500. Therefore, this triangle has a perimeter of over 1000, yet it has an area of 1. Does all this make sense? Mike :-)

If a right triangle has a perimeter of 19 and a hypotenuse that is gre

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