It is currently 19 Oct 2017, 17:32

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

Author Message
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1259

Kudos [?]: 335 [0], given: 0

If a right triangle has a perimeter of 19 and a hypotenuse [#permalink]

Show Tags

01 Apr 2007, 10:46
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

Kudos [?]: 335 [0], given: 0

VP
Joined: 06 Feb 2007
Posts: 1020

Kudos [?]: 201 [0], given: 0

Show Tags

01 Apr 2007, 12:24

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

Kudos [?]: 201 [0], given: 0

Director
Joined: 13 Dec 2006
Posts: 506

Kudos [?]: 243 [0], given: 0

Location: Indonesia

Show Tags

01 Apr 2007, 12:38
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

Kudos [?]: 243 [0], given: 0

Director
Joined: 13 Dec 2006
Posts: 506

Kudos [?]: 243 [0], given: 0

Location: Indonesia

Show Tags

01 Apr 2007, 12:41
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

Kudos [?]: 243 [0], given: 0

VP
Joined: 06 Feb 2007
Posts: 1020

Kudos [?]: 201 [0], given: 0

Show Tags

01 Apr 2007, 12:47
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.

Kudos [?]: 201 [0], given: 0

Manager
Joined: 18 Mar 2004
Posts: 50

Kudos [?]: [0], given: 0

Show Tags

01 Apr 2007, 13:41
Kevincan, do you have an answer to this question?

Kudos [?]: [0], given: 0

Director
Joined: 28 Dec 2005
Posts: 917

Kudos [?]: 57 [0], given: 0

Show Tags

01 Apr 2007, 14:24
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.

Kudos [?]: 57 [0], given: 0

GMAT Instructor
Joined: 04 Jul 2006
Posts: 1259

Kudos [?]: 335 [0], given: 0

Re: Area of a Triangle [#permalink]

Show Tags

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

Kudos [?]: 335 [0], given: 0

VP
Joined: 25 Jun 2006
Posts: 1161

Kudos [?]: 190 [0], given: 0

Show Tags

01 Apr 2007, 20:00
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.

Kudos [?]: 190 [0], given: 0

Director
Joined: 13 Dec 2006
Posts: 506

Kudos [?]: 243 [0], given: 0

Location: Indonesia

Show Tags

02 Apr 2007, 01:30
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

Kudos [?]: 243 [0], given: 0

VP
Joined: 25 Jun 2006
Posts: 1161

Kudos [?]: 190 [0], given: 0

Show Tags

02 Apr 2007, 05:38
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.

Kudos [?]: 190 [0], given: 0

Director
Joined: 13 Dec 2006
Posts: 506

Kudos [?]: 243 [0], given: 0

Location: Indonesia

Show Tags

02 Apr 2007, 05:46
Thanks tennisball

regards,

Amardeep

Kudos [?]: 243 [0], given: 0

Senior Manager
Joined: 19 Sep 2004
Posts: 367

Kudos [?]: 6 [0], given: 0

Re: Area of a Triangle [#permalink]

Show Tags

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?

Kudos [?]: 6 [0], given: 0

Senior Manager
Joined: 27 Mar 2007
Posts: 327

Kudos [?]: 41 [0], given: 0

Show Tags

04 Apr 2007, 07:50
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!

Kudos [?]: 41 [0], given: 0

Manager
Joined: 11 Mar 2007
Posts: 82

Kudos [?]: 5 [0], given: 0

Show Tags

04 Apr 2007, 10:51
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 ???

Kudos [?]: 5 [0], given: 0

Manager
Joined: 11 Mar 2007
Posts: 82

Kudos [?]: 5 [0], given: 0

Show Tags

04 Apr 2007, 11:19
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

Kudos [?]: 5 [0], given: 0

Senior Manager
Joined: 04 Mar 2007
Posts: 431

Kudos [?]: 83 [0], given: 0

Show Tags

04 Apr 2007, 14:24
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

D is the correct answer here I think.

Kudos [?]: 83 [0], given: 0

GMAT Instructor
Joined: 04 Jul 2006
Posts: 1259

Kudos [?]: 335 [0], given: 0

Show Tags

04 Apr 2007, 15:54
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!

Kudos [?]: 335 [0], given: 0

Manager
Joined: 28 Dec 2006
Posts: 61

Kudos [?]: [0], given: 0

Show Tags

04 Apr 2007, 16:12

I could not come up with any.

Kudos [?]: [0], given: 0

Manager
Joined: 11 Mar 2007
Posts: 82

Kudos [?]: 5 [0], given: 0

Show Tags

04 Apr 2007, 17:04
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]

Kudos [?]: 5 [0], given: 0

04 Apr 2007, 17:04

Go to page    1   2    Next  [ 29 posts ]

Display posts from previous: Sort by