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Vertices of a triangle have coordinates (-1,0), (4,0) and (0,A). Is the area of the triangle bigger than 15 ? (1) A<3 (2) The triangle is right Source: GMAT Club Tests - hardest GMAT questions I cldnt understand the explanation given for second statement. Can anyone explain?
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What explanation for S2 is trying to say is that if we know that the triangle is right then we can find the coordinates of the point (0,A). Thus, we'll have enough info to answer the question. There are only two possible triangles with S2. See the image below. Do you think we need to change the wording of the OE for S2? If yes, please tell us what exactly was confusing. Hope this helps.
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m16-15.png [ 6.6 KiB | Viewed 4688 times ]
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I hate to be too demanding, but what are the steps to find find the coordinates of (0, A)?
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bmillan01 wrote: I hate to be too demanding, but what are the steps to find find the coordinates of (0, A)? Knowing that the triangle is right means that the right angle must be at the point D(0,A), see the picture attached by dzyubam. Next, we know that line segment AD is perpendicular to the line segment BD, hence the slopes of these line segments must be negative reciprocals of each other. Slope AD=\frac{A-0}{0-(-1)}=A; Slope BD=\frac{A-0}{0-4}=\frac{A}{-4}. As they must be negative reciprocals, then: A=-\frac{-4}{A} --> A^2=4 --> A=2 or -2. Hope it's clear.
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Vertices of a triangle have coordinates , and . Is the area of the triangle bigger than 15?
1. A<3
2. The triangle is right
stmt 1: we know that 1 side length will be 5 and other side will be less than 3 so the third side can be as big as 7.999 and as small as 2.1 so it is not sufficient
stmt 2: if the given triangle is right angle then one side which forms the hypotenuse is 5 so the other two sides will be less than 4. so definitely area of triangle has to be less than 15
so ans is B
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ritula wrote: Vertices of a triangle have coordinates (-1,0), (4,0) and (0,A). Is the area of the triangle bigger than 15 ? (1) A<3 (2) The triangle is right Source: GMAT Club Tests - hardest GMAT questions I cldnt understand the explanation given for second statement. Can anyone explain? Someone pl explain why A cannot be the answer? If we map the triangle on a coordinate plane we have the base on X-axis and apex on Y-axis. Going by formula: Area = 1/2 (base x height) we have the base as 5 units and in order to have the area bigger than 15 we need the height to be >6 units. S1 says A<3, so height of triangle is <3 too hence area is less than 15. S1 SUFFICIENTFrom S2 we know the triangle is right hence if the side on X-axes is 5 units the other 2 sides will be 3 & 4 units. Again, area <15. S2 SUFFICIENTIf above are correct then answer is D. Comments welcome.
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ExecMBA2010 wrote: Someone pl explain why A cannot be the answer? If we map the triangle on a coordinate plane we have the base on X-axis and apex on Y-axis. Going by formula: Area = 1/2 (base x height) we have the base as 5 units and in order to have the area bigger than 15 we need the height to be >6 units. S1 says A<3, so height of triangle is <3 too hence area is less than 15. S1 SUFFICIENT - consider A=-10, what is the area then?
From S2 we know the triangle is right hence if the side on X-axes is 5 units the other 2 sides will be 3 & 4 units. Again, area <15. S2 SUFFICIENT
If above are correct then answer is D. Comments welcome. Hope that helps! 
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To get the value of A apply twice Pythagorean Theorem:
1^2+A^2+4^2+A^2=5^2, A=\pm2, area=5<15. Certainly, B.
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The easiest approach to solve this problem is to realise that there is only one single point in each positive and negative y-axis. or |A|, either A or -A. If there is one measure (I mean |A|) we can definitively say YES or NO, so 2) is suffiicient.
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ExecMBA2010 wrote: ritula wrote: Vertices of a triangle have coordinates (-1,0), (4,0) and (0,A). Is the area of the triangle bigger than 15 ? (1) A<3 (2) The triangle is right Source: GMAT Club Tests - hardest GMAT questions I cldnt understand the explanation given for second statement. Can anyone explain? Someone pl explain why A cannot be the answer? If we map the triangle on a coordinate plane we have the base on X-axis and apex on Y-axis. Going by formula: Area = 1/2 (base x height) we have the base as 5 units and in order to have the area bigger than 15 we need the height to be >6 units. S1 says A<3, so height of triangle is <3 too hence area is less than 15. S1 SUFFICIENTFrom S2 we know the triangle is right hence if the side on X-axes is 5 units the other 2 sides will be 3 & 4 units. Again, area <15. S2 SUFFICIENTIf above are correct then answer is D. Comments welcome. A) can't be the answer because when it is given <3. For positive value statement is true, but we can go in negative direction till infinite and while calculating area of the triangle we will take the magnitude of the height of the triangle. You are just thinking about positive value of the cordinate, just draw your triangle taking -20 or -30 which is also less than 3. Is first statement is true???
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Bunuel wrote: bmillan01 wrote: I hate to be too demanding, but what are the steps to find find the coordinates of (0, A)? Knowing that the triangle is right means that the right angle must be at the point D(0,A), see the picture attached by dzyubam. Next, we know that line segment AD is perpendicular to the line segment BD, hence the slopes of these line segments must be negative reciprocals of each other. Slope AD=\frac{A-0}{0-(-1)}=A; Slope BD=\frac{A-0}{0-4}=\frac{A}{-4}. As they must be negative reciprocals, then: A=-\frac{-4}{A} --> A^2=4 --> A=2 or -2. Hope it's clear. Bunnel as usual you are great!!! +1.
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The way I approached this was rather basic, but it got me the right answer.
Prompt 1 allowed for a range of answers, and these could either be less than 15 or much more (say A = -15). So there is no way to difrectly answer the question.
Prompt 2 only allows 2 possible solutions that will have same area. Since we could potentially calculate the area, we can answer the question.
Therefore B alone is sufficient.
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I think the trick in this question is the framing of the question.
when we substitute a in the area equation we MUST use |a| and not a, because area is absolute. So, the question is really asking is |a| >6?
1) INSUFFICIENT because a<3 could mean when a = -7 then |a| = 7 > 6.
2) If its right angled triangle then we know 5 has to be hypotenuse. So, the base and height lengths will be 3 and 4 respectively. 1/2*3*4 = 6 = 1/2*5*|a|; which means |a| = 12/5 = 2.4. SUFFICIENT.
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nice tricky one, we need to know the numerical value of A(whether greater than 6 or less than -6) 1 if A is less than 3, then it could be 2 or -7, hence insufficient 2 if right angled triangle then sum of square of two sides = 5^2 so each side = 5 sqrt 2 ( either positive or negative, which doesn't matter) 1/2 * base * height = 25....Sufficient IMO B is the correct answer
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i broke this down to: A=bh/2 or A=5/2H or is 15 > 5/2H = H > |6|? A) insuff. B) if it's a right triangle, we have two possibilities which are reflections of each other. suff.
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Bunuel wrote: bmillan01 wrote: I hate to be too demanding, but what are the steps to find find the coordinates of (0, A)? Knowing that the triangle is right means that the right angle must be at the point D(0,A), see the picture attached by dzyubam. Next, we know that line segment AD is perpendicular to the line segment BD, hence the slopes of these line segments must be negative reciprocals of each other. Slope AD=\frac{A-0}{0-(-1)}=A; Slope BD=\frac{A-0}{0-4}=\frac{A}{-4}. As they must be negative reciprocals, then: A=-\frac{-4}{A} --> A^2=4 --> A=2 or -2. Hope it's clear. The solution is as all your solutions bulletproof. Although, this may be the most thorough solution to answer (B) - realizing that...... (A) The third vertice has to be on the Y-axis (which eliminates all right triangles that would be possible if the given side can be seen as one leg of the triangle (5-12-13)) and also ensures that the 90-degree angle is at that vertice. (B) The triangle has to be a right triangle with the base 5 (which is 3-4-5) - which has an area of 6 (That's most likely the one calculated via the slopes calculation) ......makes this a good question to skip any calculations and safe some time. On a sidenote, I believe there is a typo during the calculation of slope BD. Shouldn't this be -A/4? (m= y2-y1/x2-x1) However, the typo does not challenge the validity of the solution. Although finding a typo on a solution happens quite often throughout GMATland (ALL GMAT Prep Companies), on your solutions it does not. So, finding one on a post from 2009 is like finding a needle in a haystack. Thanks for all the great work throughout this site!!!!!!!
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myfish wrote: Bunuel wrote: bmillan01 wrote: I hate to be too demanding, but what are the steps to find find the coordinates of (0, A)? Knowing that the triangle is right means that the right angle must be at the point D(0,A), see the picture attached by dzyubam. Next, we know that line segment AD is perpendicular to the line segment BD, hence the slopes of these line segments must be negative reciprocals of each other. Slope AD=\frac{A-0}{0-(-1)}=A; Slope BD=\frac{A-0}{0-4}=\frac{A}{-4}. As they must be negative reciprocals, then: A=-\frac{-4}{A} --> A^2=4 --> A=2 or -2. Hope it's clear. The solution is as all your solutions bulletproof. Although, this may be the most thorough solution to answer (B) - realizing that...... (A) The third vertice has to be on the Y-axis (which eliminates all right triangles that would be possible if the given side can be seen as one leg of the triangle (5-12-13)) and also ensures that the 90-degree angle is at that vertice. (B) The triangle has to be a right triangle with the base 5 (which is 3-4-5) - which has an area of 6 (That's most likely the one calculated via the slopes calculation) ......makes this a good question to skip any calculations and safe some time. On a sidenote, I believe there is a typo during the calculation of slope BD. Shouldn't this be -A/4? (m= y2-y1/x2-x1) However, the typo does not challenge the validity of the solution. Although finding a typo on a solution happens quite often throughout GMATland (ALL GMAT Prep Companies), on your solutions it does not. So, finding one on a post from 2009 is like finding a needle in a haystack. Thanks for all the great work throughout this site!!!!!!! There is no typo: \frac{-A}{4}=\frac{A}{-4}=-\frac{A}{4}. Hope it's clear.
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as always Bunuel's explanation is always wounderful.... i completely understand now why the answer is B and not D
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I still cannot get the point of this question , and still cannot understand why the right angle must be at point (0,A) specifically ? Can anybody please provide a detailed explanation similar to those of OG ? Thanks in advance
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