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M16, #15 [#permalink] New post 30 Mar 2009, 03:20
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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

[Reveal] Spoiler: OA
B

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I cldnt understand the explanation given for second statement. Can anyone explain?
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Re: M16, #15 [#permalink] New post 30 Mar 2009, 04:19
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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Re: M16, #15 [#permalink] New post 18 Jan 2010, 11:36
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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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Re: M16, #15 [#permalink] New post 18 Jan 2010, 12:02
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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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Re: M16, #15 [#permalink] New post 03 Mar 2010, 05:33
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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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Re: M16, #15 [#permalink] New post 03 Mar 2010, 08:37
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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

[Reveal] Spoiler: OA
B

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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 SUFFICIENT

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.
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Re: M16, #15 [#permalink] New post 03 Mar 2010, 09:14
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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! :-D
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Re: M16, #15 [#permalink] New post 03 Mar 2010, 11:22
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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Re: M16, #15 [#permalink] New post 12 Nov 2010, 13:04
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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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Re: M16, #15 [#permalink] New post 08 Mar 2011, 19:55
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

[Reveal] Spoiler: OA
B

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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 SUFFICIENT

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.


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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Re: M16, #15 [#permalink] New post 08 Mar 2011, 19:56
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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Re: M16, #15 [#permalink] New post 09 Mar 2012, 10:28
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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Re: M16, #15 [#permalink] New post 09 Mar 2012, 21:11
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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Re: M16, #15 [#permalink] New post 10 Mar 2012, 04:55
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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Re: M16, #15 [#permalink] New post 10 Mar 2012, 07:42
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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Re: M16, #15 [#permalink] New post 17 Apr 2012, 17:18
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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Re: M16, #15 [#permalink] New post 17 Apr 2012, 20:35
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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!!!!!!!


There is no typo: \frac{-A}{4}=\frac{A}{-4}=-\frac{A}{4}.

Hope it's clear.
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Re: M16, #15 [#permalink] New post 25 Apr 2012, 22:09
as always Bunuel's explanation is always wounderful.... i completely understand now why the answer is B and not D
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Re: M16, #15 [#permalink] New post 12 Mar 2013, 19:29
First of all right triangle with hypotenuse 5, doesn't mean that we have (3, 4, 5) right triangle. If we are told that values of all sides are integers, then yes: the only integer solution for right triangle with hypotenuse 5 would be (3, 4, 5).

To check this: consider the right triangle with hypotenuse 5 inscribed in circle. We know that a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.

So ANY point on circumference of a circle with diameter would make the right triangle with diameter. Not necessarily sides to be and . For example we can have isosceles right triangle, which would be 45-45-90: and the sides would be . OR if we have 30-60-90 triangle and hypotenuse is , sides would be and . Of course there could be many other combinations.

Back to the original question:
If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) , is the area of the triangle greater than 15?
(1) A < 3 --> two vertices are on the X-axis and the third vertex is on the Y-axis, below the point (0,3). The third vertex could be at (0,1) and the area would be less than 15 OR the third vertex could be at (0,-100) and the area would be more than 15. So not sufficient.

(2) The triangle is right. --> Obviously as the third vertex is on the Y-axis, the right angle must be at the third vertex. Which means the hypotenuse is on X-axis and equals to 5. Again if we consider the circle, the radius mus be 2.5 (half of the hypotenuse/diameter) and the third vertex must be one of two intersections of the circle with Y-axis. We'll get the two specific symmetric points for the third vertex, hence the area would be fixed and defined. Which means that it's possible to answer the question whether the area is more than 15, even not calculating actual value. Sufficient.

Answer: B.

If we want to know how the area could be calculated with the help of statement 2, here you go:

One of the approaches:

The equation of a circle is , where is the center and is the radius.

We know:
, as the hypotenuse is 5.
and , as the center is on the X-axis, at the point , half the way between the (-1, 0) and (4, 0).
We need to determine intersection of the circle with Y-axis, or the point for the circle.

So we'll have

--> and . The third vertex is either at the point OR . In any case .



got this explanation from 700%2B GMAT Data Sufficiency Questions With Explanations
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Re: M16, #15 [#permalink] New post 13 Mar 2013, 00:45
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rakeshy10 wrote:
First of all right triangle with hypotenuse 5, doesn't mean that we have (3, 4, 5) right triangle. If we are told that values of all sides are integers, then yes: the only integer solution for right triangle with hypotenuse 5 would be (3, 4, 5).

To check this: consider the right triangle with hypotenuse 5 inscribed in circle. We know that a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.

So ANY point on circumference of a circle with diameter would make the right triangle with diameter. Not necessarily sides to be and . For example we can have isosceles right triangle, which would be 45-45-90: and the sides would be . OR if we have 30-60-90 triangle and hypotenuse is , sides would be and . Of course there could be many other combinations.

Back to the original question:
If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) , is the area of the triangle greater than 15?
(1) A < 3 --> two vertices are on the X-axis and the third vertex is on the Y-axis, below the point (0,3). The third vertex could be at (0,1) and the area would be less than 15 OR the third vertex could be at (0,-100) and the area would be more than 15. So not sufficient.

(2) The triangle is right. --> Obviously as the third vertex is on the Y-axis, the right angle must be at the third vertex. Which means the hypotenuse is on X-axis and equals to 5. Again if we consider the circle, the radius mus be 2.5 (half of the hypotenuse/diameter) and the third vertex must be one of two intersections of the circle with Y-axis. We'll get the two specific symmetric points for the third vertex, hence the area would be fixed and defined. Which means that it's possible to answer the question whether the area is more than 15, even not calculating actual value. Sufficient.

Answer: B.

If we want to know how the area could be calculated with the help of statement 2, here you go:

One of the approaches:

The equation of a circle is , where is the center and is the radius.

We know:
, as the hypotenuse is 5.
and , as the center is on the X-axis, at the point , half the way between the (-1, 0) and (4, 0).
We need to determine intersection of the circle with Y-axis, or the point for the circle.

So we'll have

--> and . The third vertex is either at the point OR . In any case .



got this explanation from 700%2B GMAT Data Sufficiency Questions With Explanations


This is a solution from my post here: if-vertices-of-a-triangle-have-coordinates-87344.html#p656628

Document you are referring is here: 700-gmat-data-sufficiency-questions-with-explanations-100617.html
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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M16, #15   [#permalink] 13 Mar 2013, 00:45
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