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1. Suff since it's a 5-12-13 triangle 2. Insuff because the height to side 5, which can be calculated to be 12 may not necessarily be the leg.
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Chinese Democracy is misunderstood...at your nearest BestBuy.

Best AWA guide here: http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html

I guess we'll end up with a pretty complex equation - one with \(x^3\) in it. But do we need to solve ? for the purpose of DS all we need to know is that there is a method to solve.

There was a problem in one of the challenges where I knew we could solve and find out an answer using trigonometry, but trigonometry is not required knowledge for GMAT so does that mean that it can't be solved ??? I guess I dont know what is correct here..

I guess we'll end up with a pretty complex equation - one with \(x^3\) in it. But do we need to solve ? for the purpose of DS all we need to know is that there is a method to solve.

There was a problem in one of the challenges where I knew we could solve and find out an answer using trigonometry, but trigonometry is not required knowledge for GMAT so does that mean that it can't be solved ??? I guess I dont know what is correct here..

What is the OA ?

OA should be A.

BC could be 12 or > 12. So only A is suff....
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2) refer to my diagram. Given the area, we can calculate the height h since we know the base (13). Therefore now we have two 90 right triangles. We can find out x using pythagoras ( i am not solving it here since this is a DS problem). We can then find out y by doing 13 - x. Now that we know y and h, we can find out z (or AB), hence this info is sufficient

seems you missed the posts above yours. Read them again as D is incorrect..

buffdaddy wrote:

The answer is infact D, here is how

1) this is sufficient (as agreed by all)

2) refer to my diagram. Given the area, we can calculate the height h since we know the base (13). Therefore now we have two 90 right triangles. We can find out x using pythagoras ( i am not solving it here since this is a DS problem). We can then find out y by doing 13 - x. Now that we know y and h, we can find out z (or AB), hence this info is sufficient

seems you missed the posts above yours. Read them again as D is incorrect..

buffdaddy wrote:

The answer is infact D, here is how

1) this is sufficient (as agreed by all)

2) refer to my diagram. Given the area, we can calculate the height h since we know the base (13). Therefore now we have two 90 right triangles. We can find out x using pythagoras ( i am not solving it here since this is a DS problem). We can then find out y by doing 13 - x. Now that we know y and h, we can find out z (or AB), hence this info is sufficient

I disagree. The answer must be D. S2: Sqrt [S(S-a)(S-b)(S-c)] =30 Where S=(a+b+c)/3. @GMATTIGER, so what you are saying is that you can draw other triangle with two sides 5, and 13, with area = 30. And you are saying you can draw other triangle with the third side greater than 12 but can keep the area and other two sides constant.I think it is not possible. Can you draw a triangle with sides 5, 13, 1million, with area 30? AND at the same time can you draw other triangle with sides 5,13, 20 with area 30? Answer should be D.Please help me if I missed anything.

seems you missed the posts above yours. Read them again as D is incorrect..

buffdaddy wrote:

The answer is infact D, here is how

1) this is sufficient (as agreed by all)

2) refer to my diagram. Given the area, we can calculate the height h since we know the base (13). Therefore now we have two 90 right triangles. We can find out x using pythagoras ( i am not solving it here since this is a DS problem). We can then find out y by doing 13 - x. Now that we know y and h, we can find out z (or AB), hence this info is sufficient

Sorry, answer has to be D.

GMAT TIGER wrote:

also we can draw the trangle with same area, same height, same measures for ab and ac. in this case bc is > 12.

i guess, drawing a perpendicular to a base AB (if that's what you meant) will in fact lead you to the same result while drawing a perpendicular to base AC will tell you nothing.

As to BC=12 or >12 in different instances, again differing results may be due to rounding. Try calculating it in excel. Under buffdaddy's approach you ll get 12.

Ill be happy to be proven wrong.

Cheers

Last edited by NightAlum on 11 May 2008, 22:26, edited 1 time in total.

For everyone doubting the OA, check with Heron's formula. 2*sqrt(61) and 12 will both work for line segment BC. The more interesting question is why ASS is sufficient for triangle congruence. If I had not recognized the triangle, I would've said A is insufficient. Is 90 degrees a special case?

Let imagine AB and AC as sticks bounded in point A by a hinge. So, we can move AB relatively to AC and change angle BAC form 0 to 180. The area of ABC triangle will change form 0 (angle BAC=0) through maximum value 12*5*13=32.5 (angle BAC=90) to 0 (angle BAC=180). If you imagine that, It will be obvious that for 0<area<32.5 we will get two possible value of BC and for area=32.5 we will get only one value of BC. In our case area=30<32.5. Therefore, the second condition is insufficient.
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Thanks all of you guys for all yor explanations. But I still can;t get this straight. It would be helpful if someone could draw any sample triangle that is not 5, 12, 13 and meets condition II. Thanks for your help.

Let imagine AB and AC as sticks bounded in point A by a hinge. So, we can move AB relatively to AC and change angle BAC form 0 to 180. The area of ABC triangle will change form 0 (angle BAC=0) through maximum value 12*5*13=32.5 (angle BAC=90) to 0 (angle BAC=180). If you imagine that, It will be obvious that for 0<area<32.5 we will get two possible value of BC and for area=32.5 we will get only one value of BC. In our case area=30<32.5. Therefore, the second condition is insufficient.

walker, you said "maximum value 12*5*13=32.5" that i did not get. if the sides are 12, 5, and 13, then it is a right angle triangle and should have 30 as area. how did you get 32.5 for a triangle with sides 12, 5, and 13?
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walker, you said "maximum value 12*5*13=32.5" that i did not get. if the sides are 12, 5, and 13, then it is a right angle triangle and should have 30 as area. how did you get 32.5 for a triangle with sides 12, 5, and 13?

Hi, Tiger!

Sorry for typo. It should be 1/2*5*13=32.5 instead of 12*5*13=32.5.

I used formula for area: area=1/2*a*h. if a=5, the maximum height will be h=13
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What a bliss! I was being jerk! For those who did not understand Walker's expln, check out this link. (Interactive triangle) http://www.mathopenref.com/heronsformula.html For there are 2 triangles with the area 30. 1: 5,13,12 (For acute angle) 2: 5,13,15.5... (Not 15.5) ( For obtuse angle) see the attachment. Thank you all

Contradicting statement 2 : We can keep the fact in mind that triangles between two parallel lines have same area.(draw a line ll to AC / BC's value will vary and still satisfy the area being constant at 30 ,as base and height in this case are constant (AC) and distance between ll lines remain the same , which in this case would be the height of the triangle)

Heron's formula makes one jump to a quick conclusion about statement 2 being sufficient but an important point is overlooked by the fact using it to find the value of BC would give some polynomial ( x3 or x4 or whatever ....) and that will lead to multiple values of x (at least 2 in this case ). Hence insufficient.

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