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GMAT Instructor
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If a right triangle has a perimeter of 19 and a hypotenuse [#permalink]
 01 Apr 2007, 10:46
If a right triangle has a perimeter of 19 and a hypotenuse that is greater than 9, its area can be one of how many positive integers?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5
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VP
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I am relatively new to the math forum, so I can be wrong about this one:)
We know that the sum of any two sides on the triangle must be greater than the third side. Therefore, we know that the sum of other two sides of this tringle has to be =10.
I came up with 2 integers that satisfy the condition.
One was with sides 4 & 6: the area of triangle would then be: (4*6)/2=12 and the other with sides 2 & 8: the area is then: (2*8)/2 = 8
Sides 10 & 0 are i think impossible, since the shape would no longer be a trinalge.
Hence, I pick B - 2 integers
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Director
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I would have picked A
lets Hypoteneous be h, perpendicular be p, and base be b.
from pythogorous thereom we have h^2=p^2 + b^2
or h^2 = (p+b)^2 - 2pb
but we know that h+p+b = 19.... given
hence p+b = 19-h
substituting this value we will get h^2 = (19-h)^2 - 2pb
or 2pb = (19-h)^2 - h^2
or pb/2 =( ( 19-h)^2 - h^2)/4..... whereas pb/2= area of triange
Solving this we will have = 19^2 + h^2 -38h-h^2
or (361 - 38h)/4 = area of triangle
so we need h to be more than 9 but such a number that the overall equation can be positive integer.
Also h shoud be lesser than 10, as 38*10 = 380 > 361
if h = 9.5 we have area of triangle = 0
though this doesnt make negative integer, the area of the triangle cannot be 0..... so I am confused!!!!
In test I would have picked A.
regards,
Amardeep
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Director
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Dear Nervousgmat,
Since its a right angle triangel, it should also satisfy the pythogorous equation h^2 = p^2 + b^2
I think we should look for 6*sqrt3 or 7* sqrt 2 as the hypoteneous.
regards,
Amardeep
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VP
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Amardeep Sharma wrote: Dear Nervousgmat,
Since its a right angle triangel, it should also satisfy the pythogorous equation h^2 = p^2 + b^2
I think we should look for 6*sqrt3 or 7* sqrt 2 as the hypoteneous.
regards,
Amardeep
Ooops! Thanks for pointing this out! I solved the problem forgetting the fact that the triangle is right. Now I can't come up with any integers that would satisfy the property of this right traingle.
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Manager
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Kevincan, do you have an answer to this question?
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Director
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Hmm, I would have thought 0, but I dont see that in the answer choice.
19 = a+b+h where h is the hypotynuse.
Property of triangles: h <a> a-b.
If h = 10, and perimeter = 19, then a+b = 9, which makes h > a+b.
Same for all other values of h greater than 9.
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GMAT Instructor
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Re: Area of a Triangle [#permalink]
01 Apr 2007, 15:19
kevincan wrote: If a right triangle has a perimeter of 19 and a hypotenuse that is greater than 9, its area can be one of how many positive integers?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5
If a and b are the legs of this right triangle,
a+b=19-h where h is the hypotenuse
h= sqrt (a^2+b^2)>9
Thus a+b<10>81
We want to know about ab/2
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VP
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I got D.
this is how i get it:
Condition One: a + b + c = 19 and c > 9.
You get a+b < 10
Condition Two:
With a2 + b2 = c2, and c2 > 81.
You get a ^2 + b^2 > 81.
so from Condition One, I get 0 < a^2 + b^2 + 2ab < 100.
From Condition Two, I can get 0 < 2ab < 19.
Then 0 < ab/2 i.e. the area < 19/4 = 4.75.
so ab/2 can be 1, 2, 3, 4. Thus, D.
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Director
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Dear Tennisball,
I couldnt understand the following lines...
so from Condition One, I get 0 < a^2 + b^2 + 2ab < 100.
From Condition Two, I can get 0 < 2ab < 19.
Then 0 < ab/2 i.e. the area < 19/4 = 4.75.
from the equation above we have a^2 + b^2 < 81, suppose its value is just 25, then in that case, 2ab can be greater than 19 but less than 75....
regards,
Amardeep
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VP
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I had a little trouble posting it. I think the forum script interpreted the signs badly.
In my post:
a^2 + b^2 = c^2. and c > 9. so a^2 + b^2 > 81. NOT < 81 as in your post.
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Director
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Thanks tennisball
regards,
Amardeep
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Senior Manager
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Re: Area of a Triangle [#permalink]
02 Apr 2007, 11:44
kevincan wrote: If a right triangle has a perimeter of 19 and a hypotenuse that is greater than 9, its area can be one of how many positive integers?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5
Is Such a Triangle Possible? I don'tthink so because whatever value you pick it should Satisfy a^2 +b^2=c^2
Or am I missing something here?
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Senior Manager
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Since a+b= 9, there are only 4 possibilities to get an area that consists of an integer, i.e. 0.5x5x4 / 0.5x6x3 / 0.5x7x2 and 0.5x8x1, therefore (D) 4. cheers!
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Manager
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Assume a,b,c to be sides of triangle with c being the HYP
thus
a+b+c = 19
a+b = 19 -c
as per rule: sum of the 2 sides of triangle should be greater than or third
so a+b >= 19 - c
now the min value of c = 10
there for
a+b >= 19 - 10
a+b >= 9
(a+b)/2 >= 9/2
area >= 4.5
therefore area = 5
Thus choice E
Thoughts ???
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Manager
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Just a suggestion
whenver some one post's such trivial questions, it would be nice if the person who initiated the post, post the answer too.
This way we discuss how the author derivied a particular answer. Now what's happening is, every body is posting his/her ideas & there is no way to know, what is correct & what's wrong.
Just a thought !!
cheers
ms
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Senior Manager
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msingh wrote: Just a suggestion
whenver some one post's such trivial questions, it would be nice if the person who initiated the post, post the answer too.
This way we discuss how the author derivied a particular answer. Now what's happening is, every body is posting his/her ideas & there is no way to know, what is correct & what's wrong.
Just a thought !!
cheers
ms
Agree with your suggestion.
D is the correct answer here I think.
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GMAT Instructor
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msingh wrote: Just a suggestion
whenver some one post's such trivial questions, it would be nice if the person who initiated the post, post the answer too.
This way we discuss how the author derivied a particular answer. Now what's happening is, every body is posting his/her ideas & there is no way to know, what is correct & what's wrong.
Just a thought !!
cheers
ms
What makes you say that the question is trivial? Tennis-ball's solution is great- I said so myself. You need to judge the solutions posted on their own merits- this is part of the learning process!
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Manager
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OA Please ?
I could not come up with any.
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Manager
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Agreed... i was wrong.. in considering (a+b)/2 as area...
but in the above equation
is (a+b) > 9 ?? i.e sum of the other 2 sides greater than 3rd side ??
is'nt that a rule saying that sum of the 2 side should be greater than or equal to the third side
so solving the above
(a+b) ^2 >= 9
2ab > 81 -(a^2 +b^2)
2ab > 81 - (HYP)^2
ab/2 >= (81 -hyp^2)/4
assume hyp = 10
ab/2 > (81 -100)/4
ab/2 > -19/4
ab/2 > 4.75
Thus area has to 5 ....
& correct choice is E .... is this correct ... can some please tell me what is wrong in the above ??[/quote]
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