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If a right triangle has a perimeter of 19 and a hypotenuse

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If a right triangle has a perimeter of 19 and a hypotenuse [#permalink] New post 01 Apr 2007, 09:46
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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 [#permalink] New post 01 Apr 2007, 11:24
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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 [#permalink] New post 01 Apr 2007, 11:38
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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 [#permalink] New post 01 Apr 2007, 11:41
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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 [#permalink] New post 01 Apr 2007, 11:47
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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 [#permalink] New post 01 Apr 2007, 12:41
Kevincan, do you have an answer to this question?
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 [#permalink] New post 01 Apr 2007, 13:24
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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Re: Area of a Triangle [#permalink] New post 01 Apr 2007, 14:19
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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 [#permalink] New post 01 Apr 2007, 19:00
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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 [#permalink] New post 02 Apr 2007, 00:30
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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 [#permalink] New post 02 Apr 2007, 04:38
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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 [#permalink] New post 02 Apr 2007, 04:46
Thanks tennisball

regards,

Amardeep
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Re: Area of a Triangle [#permalink] New post 02 Apr 2007, 10:44
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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 [#permalink] New post 04 Apr 2007, 06:50
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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 [#permalink] New post 04 Apr 2007, 09:51
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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 [#permalink] New post 04 Apr 2007, 10:19
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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 [#permalink] New post 04 Apr 2007, 13:24
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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 [#permalink] New post 04 Apr 2007, 14:54
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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 [#permalink] New post 04 Apr 2007, 15:12
OA Please ?

I could not come up with any.
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 [#permalink] New post 04 Apr 2007, 16:04
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]
  [#permalink] 04 Apr 2007, 16:04

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