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If vertices of a triangle have coordinates (-2,2), (3,2) and

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Re: If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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New post 18 Jun 2013, 09:34
krrish wrote:
noboru wrote:
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

A) |y-2|=1
B) angle at the vertex (x,y) equals to 90 degrees



dude here's my take and this contradicts ur QA :(

s1: this gives two values for Y i.e. 3 and -1 so its insufficient as we dont know any info other than this we need to know either value or x or the nature of triange.
s2: now given angle is 90, here if the angle is 90 then we get a equation from slope, two variables and one equation so again NS

now consider s1 and s2, we have y two values and each of which will yield 2 different values for x : for example

use slope form y2-y1/ x2-x1 => u get equation as x2 - x-5= 0 :(
so again NS :(
please correct me if im wrong


Yes, you are wrong. Please check here: if-vertices-of-a-triangle-have-coordinates-2-2-3-2-and-82159.html#p773371

Hope it helps.
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Re: vertices of a triangle  [#permalink]

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New post 12 Jul 2013, 06:30
Bunuel wrote:
noboru wrote:
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

(1) |y-2|=1
(2) angle at the vertex (x,y) equals to 90 degrees


Given two points A(-2,2) and B(3,2). Question: Area ABC=?, where C(x,y).
Attachment:
render.php (1).png


(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.

(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.

Answer: A.

Hope it heps.


Hi,

From the 2nd statemen, the hypotenuse is 5 units in length since the triangle is a right angle triangle that is 90 degree at point C.. then that basically means that the other 2 sides(base and height of the triangle) will be be either 3 and 4 since 3,4 and 5 forms a Pythagorean triple.. that means area which is half * base * height will always be (3*4)/2 .. That would mean that 2nd statement is also sufficient.. I do get your point about drawing a circle but how would you justify the 3 lengths according to Pythagoras theorem ?
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Re: vertices of a triangle  [#permalink]

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New post 12 Jul 2013, 08:01
debshali wrote:
Bunuel wrote:
noboru wrote:
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

(1) |y-2|=1
(2) angle at the vertex (x,y) equals to 90 degrees


Given two points A(-2,2) and B(3,2). Question: Area ABC=?, where C(x,y).
Attachment:
render.php (1).png


(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.

(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.

Answer: A.

Hope it heps.


Hi,

From the 2nd statemen, the hypotenuse is 5 units in length since the triangle is a right angle triangle that is 90 degree at point C.. then that basically means that the other 2 sides(base and height of the triangle) will be be either 3 and 4 since 3,4 and 5 forms a Pythagorean triple.. that means area which is half * base * height will always be (3*4)/2 .. That would mean that 2nd statement is also sufficient.. I do get your point about drawing a circle but how would you justify the 3 lengths according to Pythagoras theorem ?


You assume with no ground for it that the lengths of the sides are integers. Knowing that hypotenuse equals to 5 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 3:4:5. Or in other words: if \(a^2+b^2=5^2\) DOES NOT mean that \(a=3\) and \(b=4\), certainly this is one of the possibilities but definitely not the only one. In fact \(a^2+b^2=5^2\) has infinitely many solutions for \(a\) and \(b\) and only one of them is \(a=3\) and \(b=4\).

For example: \(a=1\) and \(b=\sqrt{24}\) or \(a=2\) and \(b=\sqrt{21}\) ...

Hope it's clear.
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Re: If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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New post 20 Jul 2013, 10:39
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

(1) |y-2|=1
Checkpoint=2

y>2
|y-2|=1
y-2=1
y=3 Valid

y<2
|y-2|=1
-y+2=1
-y=-1
y=1 Valid

With this triangle, the base will always be five units long and the height (whether it is above or below the base) will always be one. Therefore, regardless of where the x coordinate is on the height the area will always be the same.
SUFFICIENT

(2) Angle at the vertex (x,y) equals to 90 degrees
This might be difficult to solve with the tools allowed on the GMAT, but I say this is sufficient simply because the 90 degree vertex can only form at a specific height with a fixed base. If the base were to change length then the height of the 90 degree vertex would scale accordingly. We could solve by drawing out and getting a rough height estimate. I believe, and correct me if I am wrong, but given graph paper with square sectors, the vertices will only form a right angle when both legs perfectly intersect the squares on the graph paper (i.e. cut them perfectly and diagonally down the middle)
SUFFICIENT

I never thought to look at AB as the diameter of a circle. As stated by Bunuel, (x,y) can be any point on the edge of that circle and it will form a right angle with AB. Therefore, the height of (x,y) could change depending on where it is located on the circle and therefore, the solution for formula a=1/2 (b*h) could also change.
INSUFFICIENT

(A)
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Re: If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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New post 30 Jan 2015, 15:34
Bunuel wrote:
noboru wrote:
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

(1) |y-2|=1
(2) angle at the vertex (x,y) equals to 90 degrees


Given two points A(-2,2) and B(3,2). Question: Area ABC=?, where C(x,y).
Attachment:
render.php (1).png


(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.

(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.

Answer: A.

Hope it heps.


Dear Bunuel, based on the your graphic above, you said that C could be anywhere on the blue line y=3 or anywhere

on the red line y=1
but if C2 move to the right side (on red line) and become equal to (5,2) in this case the height

will be equal to 2 not 1
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Re: If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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New post 31 Jan 2015, 04:55
23a2012 wrote:
Bunuel wrote:
noboru wrote:
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

(1) |y-2|=1
(2) angle at the vertex (x,y) equals to 90 degrees


Given two points A(-2,2) and B(3,2). Question: Area ABC=?, where C(x,y).
Attachment:
render.php (1).png


(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.

(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.

Answer: A.

Hope it heps.


Dear Bunuel, based on the your graphic above, you said that C could be anywhere on the blue line y=3 or anywhere

on the red line y=1
but if C2 move to the right side (on red line) and become equal to (5,2) in this case the height

will be equal to 2 not 1


That does not make sense. First of all how can C2 be (5,2) if it's on line y=1? The y-coordinate of C must be 1 (or 3 if C is on y=3). Next, even if the coordinates of C were (5,2), then how is the height 2? In this case would not all three point be on a line (y=2), and we wouldn't have a triangle at all?
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Re: If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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New post 31 Jan 2015, 06:38
Bunuel wrote:
noboru wrote:
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

(1) |y-2|=1
(2) angle at the vertex (x,y) equals to 90 degrees


Given two points A(-2,2) and B(3,2). Question: Area ABC=?, where C(x,y).
Image

(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.

(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.

Answer: A.

Hope it helps.

Attachment:
render.php (1).png



Sorry Bunuel, I was mean point (5,1). Where I see here in the figure above that when point C2 become equal to (3,1)

we will have Right-angled triangle ABC2 where AB is the height, BC2 is the basic and AC2 is the hypotenuse. Now

when point C2 moves to right side to become (5,1) the basic of the triangle is (BC2) will become longer than when C2

was equal to (3,1). :? :?
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Re: If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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New post 31 Jan 2015, 07:07
23a2012 wrote:
Bunuel wrote:
noboru wrote:
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

(1) |y-2|=1
(2) angle at the vertex (x,y) equals to 90 degrees


Given two points A(-2,2) and B(3,2). Question: Area ABC=?, where C(x,y).
Image

(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.

(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.

Answer: A.

Hope it helps.

Attachment:
The attachment render.php (1).png is no longer available



Sorry Bunuel, I was mean point (5,1). Where I see here in the figure above that when point C2 become equal to (3,1)

we will have Right-angled triangle ABC2 where AB is the height, BC2 is the basic and AC2 is the hypotenuse. Now

when point C2 moves to right side to become (5,1) the basic of the triangle is (BC2) will become longer than when C2

was equal to (3,1). :? :?


It seems that you don't understand the solution...

The base of the green triangle below is 5 and the height is 1 --> area = 1/2*5*1.
Attachments

Untitled.png
Untitled.png [ 12 KiB | Viewed 14223 times ]


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If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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New post 31 Jan 2015, 07:43
Dear Bunuel, I really appreciate your patience. However, could you explain more how we can say that when the point

C2 become equal to (3,1)(Right-angled triangle ABC2 ), AB is still the basic where it should be the height.

I mean that in the Right-angled triangle ABC2 the angle B = 90 , AC2 is the hypotenuse, AB is the height and BC is the

basic where is the wrong here?
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Re: If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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New post 02 Feb 2015, 01:02
23a2012 wrote:
Dear Bunuel, I really appreciate your patience. However, could you explain more how we can say that when the point

C2 become equal to (3,1)(Right-angled triangle ABC2 ), AB is still the basic where it should be the height.

I mean that in the Right-angled triangle ABC2 the angle B = 90 , AC2 is the hypotenuse, AB is the height and BC is the

basic where is the wrong here?


Any side can be consider as the base of a triangle... Sorry, but I don't understand what you mean... Are you saying that the area for (2) can be something other that 5/2?
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If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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New post 02 Feb 2015, 06:05
Bunuel wrote:
23a2012 wrote:
Dear Bunuel, I really appreciate your patience. However, could you explain more how we can say that when the point

C2 become equal to (3,1)(Right-angled triangle ABC2 ), AB is still the basic where it should be the height.

I mean that in the Right-angled triangle ABC2 the angle B = 90 , AC2 is the hypotenuse, AB is the height and BC is the

basic where is the wrong here?


Any side can be consider as the base of a triangle... Sorry, but I don't understand what you mean... Are you saying that the area for (2) can be something other that 5/2?


Yes, where the question did not mention that the AB is the basic and as you said that any side can

be consider as the base of a triangle
so we will have different area. For example if point C2 become equal to

(9,1) in this case we will have AB=5, BC2=sqrt 37, and AC2=sqrt 58 so if we say that the height is =1 what will be the

basic ? can I say the basic is BC2 or AC2 because the question did not mention that the AB is the basic and any side

can be consider as the base of a triangleand. And if the basic is still the AB can you tell me why we can not consider

the other sides to be the basic?
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Re: If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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New post 02 Feb 2015, 06:13
23a2012 wrote:
Bunuel wrote:
23a2012 wrote:
Dear Bunuel, I really appreciate your patience. However, could you explain more how we can say that when the point

C2 become equal to (3,1)(Right-angled triangle ABC2 ), AB is still the basic where it should be the height.

I mean that in the Right-angled triangle ABC2 the angle B = 90 , AC2 is the hypotenuse, AB is the height and BC is the

basic where is the wrong here?


Any side can be consider as the base of a triangle... Sorry, but I don't understand what you mean... Are you saying that the area for (2) can be something other that 5/2?


Yes, where the question did not mention that the AB is the basic and as you said that any side can

be consider as the base of a triangle
so we will have different area. For example if point C2 become equal to

(9,1) in this case we will have AB=5, BC2=sqrt 37, and AC2=sqrt 58 so if we say that the height is =1 what will be the

basic ? can I say the basic is BC2 or AC2 because the question did not mention that the AB is the basic and any side

can be consider as the base of a triangleand. And if the basic is still the AB can you tell me why?


The area still would be 1/2*5*1 = 5/2.
Attachment:
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Untitled.png [ 9.23 KiB | Viewed 14196 times ]

Sorry, I cannot explain any better...
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Re: If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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New post 08 Jul 2015, 15:45
Bunuel! is something else! amazing solution.
Is there any other way to do it? By doing some math... maybe..if yes please explain...
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Re: If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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New post 08 Jul 2015, 23:37
Shreks1190 wrote:
Bunuel! is something else! amazing solution.
Is there any other way to do it? By doing some math... maybe..if yes please explain...


Here it is...

Quote:
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

A) |y-2|=1
B) angle at the vertex (x,y) equals to 90 degrees


Point to Observe: (-2,2) and (3,2) have equal y-co-ordinate i.e. they are on the same horizontal line which can be considered as base of triangle [Base=3-(-2)=5 units] for calculation of area using Area=(1/2)Base*Height

Statement 1: |y-2|=1

i.e. y = 1 or 3
i.e. This point will be 1 unit above both other points or 1 unit below both other points
In each case the Height will be 1 unit and hence will result in equal area. Hence,
SUFFICIENT

Statement 2: angle at the vertex (x,y) equals to 90 degrees

i.e. Three points are making a right angle triangle with hypotenuse = 5 which has various possibilities of lengths of other two sides therefore, Various various values of Area of Triangle. Hence,
NOT SUFFICIENT

Answer: Option A
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Re: If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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New post 18 Oct 2016, 19:08
noboru wrote:
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

(1) |y-2|=1
(2) Angle at the vertex (x,y) equals to 90 degrees



Please find answer as attached

Answer: option A
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Re: If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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New post 20 Apr 2017, 09:04
noboru wrote:
yezz wrote:
noboru wrote:
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

A) |y-2|=1
B) angle at the vertex (x,y) equals to 90 degrees

PS: 1 day to go!!!


first: good luck and wish you a 750+ :)

from 1

suff either ways if we took the distance bet (-2,2),(3,2) as the base the height is either ways 1 unit

from 2


though i am not quite sure , id say insuff

A


hi, thanks!

OK, lets do this one:

I agree: from 1 is suff since you know the height.

But, from 2: I think that there´s only one triangle you can draw that, with 2 points fixed (-2,2 and 3,2), the other vertex has 90 degrees...

you are right, with 2 fixed points and a right triangle => 3 points lie on a circle in which, 2 of 3 creates diameter
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Re: If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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New post 07 Aug 2017, 07:47
Dear Bunuel,

I want to thank you. Your explanations are so flawless and simple.


Bunuel wrote:
noboru wrote:
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

(1) |y-2|=1
(2) angle at the vertex (x,y) equals to 90 degrees


Given two points A(-2,2) and B(3,2). Question: Area ABC=?, where C(x,y).
Image

(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.

(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.

Answer: A.

Hope it helps.

Attachment:
render.php (1).png
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Re: If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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New post 10 Aug 2017, 23:03
Guys,

I've seen many of you questioning, why is B insuff as we can deduce the side lengths as 3,4,5, since we know that the triangle is right angled and the hypotenuse is 5.

But notice that 3,4,5 is not the only possible case. Any of these sets - 1,sqrt(24),5 ; 4,sqrt(21),5 etc..., satisfy the given two conditions, and in each case the area of the triangle is different.

Hence option B is insuff.
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Re: If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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New post 10 Aug 2017, 23:08
anooshehka wrote:
Dear Bunuel,

I want to thank you. Your explanations are so flawless and simple.


Bunuel wrote:
noboru wrote:
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

(1) |y-2|=1
(2) angle at the vertex (x,y) equals to 90 degrees


Given two points A(-2,2) and B(3,2). Question: Area ABC=?, where C(x,y).
Image

(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.

(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.

Answer: A.

Hope it helps.

Attachment:
render.php (1).png


Alternately, We can eliminate option B based on pythagoreous theorem as well. You can Refer to my explanation above.
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