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If vertices of a triangle have coordinates (-1,0), (4,0),  [#permalink]

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### HideShow timer Statistics 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.

Originally posted by gmat620 on 27 Nov 2009, 08:46.
Last edited by gmat620 on 27 Nov 2009, 10:41, edited 1 time in total.
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gmat620 wrote:
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.

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.

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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I appreciate the below solution but to solve in this lengthy method will need GMAC to provide 150 mins for Q instead of 75 mins

The concept to note here is product of slopes of 2 perpendicular lines = -1

Since the triangle can only be right angle at 0,A , then - A/4*A/1 = -1

Solve for A and then find out the max possible area of the triangle

If you like my method , kindly provide comment
##### General Discussion
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gmat620 wrote:
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.

what if you have a 45-45-90 triangle and the hypotenuse if 5

the other sides would be 5 sqrt2/2
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gmat620 wrote:
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.

You are right.
Since the point (0,a) must be in 0y-axis, the hypotenus must be 5, hence the other two sides must be 3 and 4.
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Manager  Joined: 24 Sep 2009
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lagomez wrote:
what if you have a 45-45-90 triangle and the hypotenuse if 5

the other sides would be 5 sqrt2/2

this is true, but it's not the case here because the last vertex must be on y axis, so this is not a isosceles right triangle.
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Great explanation by Bunuel !!
I wish Gmat wouldn't throw such questions on me. Intern  Joined: 21 Aug 2009
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Hi Bunuel,

Please tell me where I am going wrong.
Calculated using the matrix formula to solve the area of the triangle.
$$1/2 [-1 (1-A) + 4 (A-1) +0(1-1)] >15$$
$$A>7$$

Option 1 says A< 3
Hence Statement A is sufficient.

Am I missing something here? Math Expert V
Joined: 02 Sep 2009
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ramgmat wrote:
Hi Bunuel,

Please tell me where I am going wrong.
Calculated using the matrix formula to solve the area of the triangle.
$$1/2 [-1 (1-A) + 4 (A-1) +0(1-1)] >15$$
$$A>7$$

Option 1 says A< 3
Hence Statement A is sufficient.

Am I missing something here? This is a valid approach if you are familiar with the formula which gives the area based on the coordinates of the three vertices of a triangle.

If the vetices of a triangle are: $$A(a_x, a_y)$$, $$B(b_x, b_y)$$ and $$C(c_x,c_y)$$ then the area of ABC is:

$$area=|\frac{a_x(b_y-c_y)+b_x(c_y-a_y)+c_x(a_y-b_y)}{2}|$$.

So if we consider: $$A(-1,0)$$, $$B(4,0)$$, and $$C(0,A)$$ then the area would be: $$area=|\frac{-1(0-A)+4(A-0)+0(0-0)}{2}|$$ --> $$area=|\frac{5A}{2}|$$.

Question: is $$area=|\frac{5A}{2}|>15$$ --> is $$|A|>6$$.

Statement (1) says A>3, which is not sufficient to say whether $$|A|>6$$.

P.S. You made some errors in calculation and also didn't put the area formula in ||.

Hope it helps.
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Thanks a lot Bunuel! I really appreciate this. I need to be careful of both my calculation as well as my silly mistakes
Intern  Joined: 25 Mar 2009
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Bunuel wrote:
ramgmat wrote:
Hi Bunuel,

Please tell me where I am going wrong.
Calculated using the matrix formula to solve the area of the triangle.
$$1/2 [-1 (1-A) + 4 (A-1) +0(1-1)] >15$$
$$A>7$$

Option 1 says A< 3
Hence Statement A is sufficient.

Am I missing something here? This is a valid approach if you are familiar with the formula which gives the area based on the coordinates of the three vertices of a triangle.

If the vetices of a triangle are: $$A(a_x, a_y)$$, $$B(b_x, b_y)$$ and $$C(c_x,c_y)$$ then the area of ABC is:

$$area=|\frac{a_x(b_y-c_y)+b_x(c_y-a_y)+c_x(a_y-b_y)}{2}|$$.

So if we consider: $$A(-1,0)$$, $$B(4,0)$$, and $$C(0,A)$$ then the area would be: $$area=|\frac{-1(0-A)+4(A-0)+0(0-0)}{2}|$$ --> $$area=|\frac{5A}{2}|$$.

Question: is $$area=|\frac{5A}{2}|>15$$ --> is $$|A|>6$$.

Statement (1) says A>3, which is not sufficient to say whether $$|A|>6$$.

P.S. You made some errors in calculation and also didn't put the area formula in ||.

Hope it helps.

$$area=|\frac{a_x(b_y-c_y)+b_x(c_y-a_y)+c_x(a_y-b_y)}{2}|$$. like this formulation, tks
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As in (1) , A<3 , which means A = -3 , -100 , -200 anything --> Not suffcient

From (2) We conclude that the triangle is rt. angled at (0,A)

We can set up the equation as (A-0)/(0-(-1)) * (A-0)/(0-4) = -1 as product of slopes of two perpendicular lines of a rt angled triangle is -1

A= +/- 2 which gives area of the rt angled traingle as 5 sq units < 15 sq units ---> sufficient

Bunuel wrote:
gmat620 wrote:
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.

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.

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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Mikko wrote:
pratikdas007 wrote:

As in (1) , A<3 , which means A = -3 , -100 , -200 anything --> Not suffcient

From (2) We conclude that the triangle is rt. angled at (0,A)

We can set up the equation as (A-0)/(0-(-1)) * (A-0)/(0-4) = -1 as product of slopes of two perpendicular lines of a rt angled triangle is -1

A= +/- 2 which gives area of the rt angled traingle as 5 sq units < 15 sq units ---> sufficient

the bottom line seems to be from slope we can find coordinates of O,A hence find area and compare
The value of A can be +ve or -ve ....doesnt matter as it is length
and because of co-ordinates of 2 given points which are in x axis we can say that right angle is not between them
so seems like this method works...
thanks
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Great explanation Bunuel...
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the biggest take away...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.

Dont calculate....its a yes or no question. Fixed point means we can calculate area some how.
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from the given info. the base of the triangle is 5 i.e the disantce between
-1 and 4.

for the area to be greater than 15 , 0.5*5*hiegt > 15 ==> hiegt > 6

note that the third vertex(0,A) decides the hieght of the triangle.

Stmnt1. A < 3 ==> A could be 2 (hieght = 2 and answer to the question is NO)
or -10 (hieght = 10 and the anser to the qtn is YES)...hence NOT suff.

stmnt2: the triangle is right

VERY IMPORTANT, USEFUL AND GMAT'S FAVORITE property: the angle on the semi circle is 90, means, if two vertices of
the triangle are on the extreme sides of the diameter and the third is on
the semi circle, then at the third vertex the angle is 90.

Using the above property and the stmnt 2 , we can say that the third verthex
(0,A) is on a semi circle having the diameter connected between (-1,0) and
(4,0) ==> radius of the circle is 5 (distance b/w -1 and 4) / 2 = 2.5

hence the trianlge (right angle) in this semicircle would have the maximum
area = 0.5 * 5 * 2.5 (height=radius) = 6.25 that is < 15 ==> the max area is
< 15 ==> any other possible right angle triangle areas would be < 15 ==>
answer to the question is "NO".

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ok, this is simple. the area of our triangle is (!A!*5)/2 where 5 is a hypotenuse of the triangle (!-1-4! = 5)
first option is not sufficient, cuz !A! (absolute value of A) can be any number. 1, 10, 100000000 etc
second option is sufficient. in any right angle triangle, the square of the altitude to the hypotenuse is equal to the product of two sectors it creates on hypotenuse. in our case one sector is 1 and the other is 4. so A*A=1*4, thus A=2. so the area is (2*5)/2. sorry my math english is not that good. tried my best to explain. Ans: B
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gmat620 wrote:
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.

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 [ 9.29 KiB | Viewed 16380 times ]

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.
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VeritasPrepKarishma wrote:
gmat620 wrote:
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.

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.

Little correction here: actually there will be 2 such right triangles, the second one will be the mirror image of the first (urgent-help-required-87344.html#p656628).
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Yes, that's true. But since we are only concerned with area, we can pretty much ignore the mirror images for this statement.
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Veritas Prep GMAT Instructor Re: Urgent help required   [#permalink] 01 Dec 2010, 15:02

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