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805+ (Hard)|   Geometry|      
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KarishmaB
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If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) 11 Mar 2013, 01:35
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wiseman55555
I thought of the problem in a more simple way, but I'm not sure if its the right thought process...

Basically when I assessed statement 2 I took given that the hypotenuse of the triangle was 5. This means that regardless of what the other 2 sides are, both must be less than 5. So even if the sides were let's say 4.9 * 4.9 ...the multiplication of these two values divided by 2 is always going to give a value less than 15 (especially since 5*5= 25/2 is less than 15).

Bad thought process?
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Certainly not! The thought process is perfectly fine. In fact, it's quite good.
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If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) 25 Aug 2014, 10:38
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gmat620
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
(2) The triangle is right

Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not.
Please help me i have my test 4 days away. Thanks.
Show more

Using the area formula derived from matrices,we find the area to be 5A/2.
From (1)if A<3,then Area<15/2 which is <15.Sufficient
From (2) using right angle in a circle we can get the answer.Sufficient.
So,answer should be D.
Guys,pls let me know if I'm going wrong somewhere. :?:
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If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) 25 Aug 2014, 21:20
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DebWenger
gmat620
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
(2) The triangle is right

Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not.
Please help me i have my test 4 days away. Thanks.

Using the area formula derived from matrices,we find the area to be 5A/2.
From (1)if A<3,then Area<15/2 which is <15.Sufficient
From (2) using right angle in a circle we can get the answer.Sufficient.
So,answer should be D.
Guys,pls let me know if I'm going wrong somewhere. :?:
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Statement 1 tells you that A is less than 3. Since A is just the y coordinate of the 3rd vertex, it can be negative, right? Say, A = -100

Area = |5A/2| = 250 (Mind you, area is the absolute value of 5A/2)
In this case, area is more than 15

If A = 1, Area = 2.5
In this case, area is less than 15

Statement 1 is not sufficient alone.
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If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) 05 Nov 2014, 03:21
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gmat620
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

2. The triangle is right

Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not.
Please help me i have my test 4 days away. Thanks.

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 \(5\) would make the right triangle with diameter. Not necessarily sides to be \(3\) and \(4\). For example we can have isosceles right triangle, which would be 45-45-90: and the sides would be \(\frac{5}{\sqrt{2}}\). OR if we have 30-60-90 triangle and hypotenuse is \(5\), sides would be \(2.5\) and \(2.5*\sqrt{3}\). 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 \((x - a)^2 + (y-b)^2 = r^2\), where \((a,b)\) is the center and \(r\) is the radius.

We know:
\(r=2.5\), as the hypotenuse is 5.
\(a=1.5\) and \(b=0\), as the center is on the X-axis, at the point \((1.5, 0)\), half the way between the (-1, 0) and (4, 0).
We need to determine intersection of the circle with Y-axis, or the point \((0, y)\) for the circle.

So we'll have \((0-1.5)^2 + (y-0)^2 =2.5^2\)

\(y^2=4\) --> \(y=2\) and \(y=-2\). The third vertex is either at the point \((0, 2)\) OR \((0,-2)\). In any case \(Area=2*\frac{5}{2}=5\).
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Bunuel,

Lets say base is - (-1,0), (4,0)
and
(0,A): Third vertices

Base =5, Height =A

1/2 X Base X height = 1/2 X 5 X A

IF A <3 Then Area, A < 7.5 This answers the questions. Hence A is the solution.
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If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) 05 Nov 2014, 05:16
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gmat620
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

2. The triangle is right

Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not.
Please help me i have my test 4 days away. Thanks.

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 \(5\) would make the right triangle with diameter. Not necessarily sides to be \(3\) and \(4\). For example we can have isosceles right triangle, which would be 45-45-90: and the sides would be \(\frac{5}{\sqrt{2}}\). OR if we have 30-60-90 triangle and hypotenuse is \(5\), sides would be \(2.5\) and \(2.5*\sqrt{3}\). 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 \((x - a)^2 + (y-b)^2 = r^2\), where \((a,b)\) is the center and \(r\) is the radius.

We know:
\(r=2.5\), as the hypotenuse is 5.
\(a=1.5\) and \(b=0\), as the center is on the X-axis, at the point \((1.5, 0)\), half the way between the (-1, 0) and (4, 0).
We need to determine intersection of the circle with Y-axis, or the point \((0, y)\) for the circle.

So we'll have \((0-1.5)^2 + (y-0)^2 =2.5^2\)

\(y^2=4\) --> \(y=2\) and \(y=-2\). The third vertex is either at the point \((0, 2)\) OR \((0,-2)\). In any case \(Area=2*\frac{5}{2}=5\).

Bunuel,

Lets say base is - (-1,0), (4,0)
and
(0,A): Third vertices

Base =5, Height =A

1/2 X Base X height = 1/2 X 5 X A

IF A <3 Then Area, A < 7.5 This answers the questions. Hence A is the solution.
Show more

Statement (1) by itself is insufficient. If A is close to 0, the area of the triangle is small, but if A is a large negative (for example, A = -1000), the area of the triangle is large.

Please re-read/study solutions provided on previous pages.
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If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) 09 Feb 2015, 01:31
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VeritasPrepKarishma
gmat620
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

2. The triangle is right

Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not.
Please help me i have my test 4 days away. Thanks.

First, your question - No, you cannot assume that the sides are 3 and 4 if you have 5 as hypotenuse. lagomez is right. What if it is 45-45-90 triangle? The sides will not be 3 and 4.
But in this question, you have something more. You know the line on which your third vertex of the triangle will fall.
Attachment:
The attachment Ques2.jpg is no longer available

There will be only 1 such right triangle so you will be able to say whether the area is greater than 15 or not. You don't need to find the triangle but you know this statement alone is sufficient.
Show more

Quote:

Dear Karishma, from the fogure above the triangle ABC is right triangle lets say that A=(-1,0) B=(4,0) C= (0,A) my

question is why we say that AB is the basic and CD is the hight (middle point of AB) while it shuld be in right triangle

that CA is the basic and CB is the hight I really confused here why the basic and the hight are different. For example if

the triangle ABC is 90-60-30 we will say that the basic is AC and the hight is CB.I mean why we try to find the CD

instead of CB Also if I have right triangle and I have both CD and CB what will be the hight in this case.

Dear Karishma, I apologize for the large number of questions but I really like your explanation.
Show more

Any given triangle has 3 sides. Any one of these sides can be the base. The corresponding altitude will be the height. In a right triangle, the altitude is one of the sides, that's all.
Attachment:
Ques3.jpg
Ques3.jpg [ 11.92 KiB | Viewed 4001 times ]
Look at a triangle with corresponding heights of the same color. Each side has its own altitude. The area of the triangle can be found by Area = (1/2)*base*height. You can take any base-height pair. The area will be the same.
Similarly, look at the right triangle. It also has 3 pairs of base-height. Just that the height in 2 cases corresponds to the sides of the triangle. Again, you can find the area using any of the 3 pairs.
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If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) 23 Mar 2017, 05:45
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Bunuel Thank you for the fantastic explanation. But I have a small doubt. Like you mentioned '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', so the 3rd vertex should have been at 2.5, but when we calculated it came out to be 2.Can you please clear on this[/quote]

How did you come up with the highlighted part? That's not true. The radius is 2.5. The center is at (1.5, 0). The distance from the center to all three vertices is 2.5.[/quote]

Sorry but then should not y coordinate be 2.5+1.5 i.e. +4 or -4

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If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) 23 Mar 2017, 06:01
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KARISHMA315

Sorry but then should not y coordinate be 2.5+1.5 i.e. +4 or -4
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Have you tried to draw it? Hope that the below image will help:
Attachment:
Untitled.png
Untitled.png [ 11.25 KiB | Viewed 3343 times ]
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If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) 23 Mar 2017, 08:06
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KARISHMA315

Sorry but then should not y coordinate be 2.5+1.5 i.e. +4 or -4

Have you tried to draw it? Hope that the below image will help:
Attachment:
Untitled.png
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Thanks a lot Bunuel, got it now. Excellent explanation :)
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If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) 28 Aug 2017, 13:19
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VeritasPrepKarishma
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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

2. The triangle is right

Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not.
Please help me i have my test 4 days away. Thanks.

First, your question - No, you cannot assume that the sides are 3 and 4 if you have 5 as hypotenuse. lagomez is right. What if it is 45-45-90 triangle? The sides will not be 3 and 4.
But in this question, you have something more. You know the line on which your third vertex of the triangle will fall.
Attachment:
Ques2.jpg

There will be only 1 such right triangle so you will be able to say whether the area is greater than 15 or not. You don't need to find the triangle but you know this statement alone is sufficient.
Show more

Karishma - could you please explain why there is only 1 such right angle. ..in a semicircle, with the hypotenuse as the diameter of the circle, there are many 90 degree angle ...

Also even if it the vertex was on the y axis + plus the circle with the diameter as 5, why does this mean, there is only 1 triangle with a 90 degree angle
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If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) 22 Feb 2021, 03:09
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VeritasKarishma
gmat620
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

2. The triangle is right

Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not.
Please help me i have my test 4 days away. Thanks.

First, your question - No, you cannot assume that the sides are 3 and 4 if you have 5 as hypotenuse. lagomez is right. What if it is 45-45-90 triangle? The sides will not be 3 and 4.
But in this question, you have something more. You know the line on which your third vertex of the triangle will fall.
Attachment:
Ques2.jpg

There will be only 1 such right triangle so you will be able to say whether the area is greater than 15 or not. You don't need to find the triangle but you know this statement alone is sufficient.
Show more
VeritasKarishma, why is the hypotenuse 5? I can't get this part
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If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) 23 Feb 2021, 01:21
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VeritasKarishma
gmat620
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

2. The triangle is right

Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not.
Please help me i have my test 4 days away. Thanks.

First, your question - No, you cannot assume that the sides are 3 and 4 if you have 5 as hypotenuse. lagomez is right. What if it is 45-45-90 triangle? The sides will not be 3 and 4.
But in this question, you have something more. You know the line on which your third vertex of the triangle will fall.
Attachment:
Ques2.jpg

There will be only 1 such right triangle so you will be able to say whether the area is greater than 15 or not. You don't need to find the triangle but you know this statement alone is sufficient.
VeritasKarishma, why is the hypotenuse 5? I can't get this part
Show more

Look at the diagram here: https://gmatclub.com/forum/if-vertices- ... ml#p825857

Note that the third vertex is on the Y axis. Then, where will you have the right angle? It cannot be at (-1, 0) or (4, 0) because they will form an incline with any point on the Y axis. The right angle must be at the third vertex only then. Then the side opposite to it must be the hypotenuse. So hypotenuse is 5.
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If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) 19 Jun 2024, 00:49
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