Aug 25 09:00 AM PDT  12:00 PM PDT Join a FREE 1day verbal workshop and learn how to ace the Verbal section with the best tips and strategies. Limited for the first 99 registrants. Register today! Aug 25 08:00 PM PDT  11:00 PM PDT Exclusive offer! Get 400+ Practice Questions, 25 Video lessons and 6+ Webinars for FREE. Aug 28 08:00 AM PDT  09:00 AM PDT Join a FREE live webinar with examPAL and Admissionado and learn how to master GMAT Critical Reasoning questions and the 6pointed star of MBA application essay glory. Save your spot today! Aug 30 08:00 PM PDT  11:00 PM PDT We'll be posting questions in DS/PS/SC/CR in competition mode. Detailed and quickest solution will get kudos. Will be collecting new links to all questions in this topic. Here you can also check links to fresh questions posted. Aug 31 07:00 AM PDT  09:00 AM PDT Want to score 90 percentile or higher on GMAT CR? Attend this free webinar to learn how to prethink assumptions and solve the most challenging questions in less than 2 minutes. Sep 01 07:00 AM PDT  09:00 AM PDT Want to solve 700+ level Algebra questions within 2 minutes? Attend this free webinar to learn how to master the most challenging Inequalities and Absolute Values questions in GMAT Sep 02 08:00 PM PDT  11:00 PM PDT Sign Up, Get $49 Exam Pack 2 FREE. Train to be ready for Round 1 Deadlines with EMPOWERgmat's Score Booster Code: EP22019 Ends: September 2nd
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 57255

Re: If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
Show Tags
18 Jun 2013, 10:34
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) y2=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 y2y1/ x2x1 => u get equation as x2  x5= 0 so again NS please correct me if im wrong Yes, you are wrong. Please check here: ifverticesofatrianglehavecoordinates2232and82159.html#p773371Hope it helps.
_________________



Intern
Joined: 26 Jul 2010
Posts: 1

Re: vertices of a triangle
[#permalink]
Show Tags
12 Jul 2013, 07:30
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) y2=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) y2=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 ?



Math Expert
Joined: 02 Sep 2009
Posts: 57255

Re: vertices of a triangle
[#permalink]
Show Tags
12 Jul 2013, 09:01
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) y2=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) y2=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.
_________________



Senior Manager
Joined: 13 May 2013
Posts: 414

Re: If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
Show Tags
20 Jul 2013, 11:39
If vertices of a triangle have coordinates (2,2), (3,2) and (x,y), what is the area of the triangle?
(1) y2=1 Checkpoint=2
y>2 y2=1 y2=1 y=3 Valid
y<2 y2=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)



Manager
Status: Kitchener
Joined: 03 Oct 2013
Posts: 88
Location: Canada
Concentration: Finance, Finance
GPA: 2.9
WE: Education (Education)

Re: If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
Show Tags
30 Jan 2015, 16:34
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) y2=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) y2=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
_________________
Click +1 Kudos if my post helped



Math Expert
Joined: 02 Sep 2009
Posts: 57255

Re: If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
Show Tags
31 Jan 2015, 05:55
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) y2=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) y2=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 ycoordinate 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?
_________________



Manager
Status: Kitchener
Joined: 03 Oct 2013
Posts: 88
Location: Canada
Concentration: Finance, Finance
GPA: 2.9
WE: Education (Education)

Re: If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
Show Tags
31 Jan 2015, 07:38
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) y2=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). (1) y2=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 Rightangled 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).
_________________
Click +1 Kudos if my post helped



Math Expert
Joined: 02 Sep 2009
Posts: 57255

Re: If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
Show Tags
31 Jan 2015, 08:07
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) y2=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). (1) y2=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 Rightangled 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 [ 12 KiB  Viewed 15410 times ]
_________________



Manager
Status: Kitchener
Joined: 03 Oct 2013
Posts: 88
Location: Canada
Concentration: Finance, Finance
GPA: 2.9
WE: Education (Education)

If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
Show Tags
31 Jan 2015, 08:43
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)(Rightangled triangle ABC2 ), AB is still the basic where it should be the height. I mean that in the Rightangled triangle ABC2 the angle B = 90 , AC2 is the hypotenuse, AB is the height and BC is the basic where is the wrong here?
_________________
Click +1 Kudos if my post helped



Math Expert
Joined: 02 Sep 2009
Posts: 57255

Re: If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
Show Tags
02 Feb 2015, 02:02
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)(Rightangled triangle ABC2 ), AB is still the basic where it should be the height.
I mean that in the Rightangled 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?
_________________



Manager
Status: Kitchener
Joined: 03 Oct 2013
Posts: 88
Location: Canada
Concentration: Finance, Finance
GPA: 2.9
WE: Education (Education)

If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
Show Tags
02 Feb 2015, 07:05
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)(Rightangled triangle ABC2 ), AB is still the basic where it should be the height.
I mean that in the Rightangled 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?
_________________
Click +1 Kudos if my post helped



Math Expert
Joined: 02 Sep 2009
Posts: 57255

Re: If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
Show Tags
02 Feb 2015, 07:13
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)(Rightangled triangle ABC2 ), AB is still the basic where it should be the height.
I mean that in the Rightangled 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:
Untitled.png [ 9.23 KiB  Viewed 15380 times ]
Sorry, I cannot explain any better...
_________________



Intern
Joined: 21 Apr 2015
Posts: 39

Re: If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
Show Tags
08 Jul 2015, 16:45
Bunuel! is something else! amazing solution. Is there any other way to do it? By doing some math... maybe..if yes please explain...



CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2967
Location: India
GMAT: INSIGHT
WE: Education (Education)

Re: If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
Show Tags
09 Jul 2015, 00:37
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) y2=1 B) angle at the vertex (x,y) equals to 90 degrees Point to Observe: (2,2) and (3,2) have equal ycoordinate 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*HeightStatement 1: y2=1i.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, SUFFICIENTStatement 2: angle at the vertex (x,y) equals to 90 degreesi.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 SUFFICIENTAnswer: Option A
_________________
Prosper!!!GMATinsightBhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhihttp://www.GMATinsight.com/testimonials.htmlACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2967
Location: India
GMAT: INSIGHT
WE: Education (Education)

Re: If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
Show Tags
18 Oct 2016, 20:08
noboru wrote: If vertices of a triangle have coordinates (2,2), (3,2) and (x,y), what is the area of the triangle?
(1) y2=1 (2) Angle at the vertex (x,y) equals to 90 degrees Please find answer as attached Answer: option A
Attachments
File comment: www.GMATinsight.com
Answer1.jpg [ 135.82 KiB  Viewed 14438 times ]
_________________
Prosper!!!GMATinsightBhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhihttp://www.GMATinsight.com/testimonials.htmlACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



Manager
Joined: 13 Apr 2017
Posts: 53

Re: If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
Show Tags
20 Apr 2017, 10:04
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) y2=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



Intern
Joined: 30 Aug 2016
Posts: 8

Re: If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
Show Tags
07 Aug 2017, 08:47
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) y2=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). (1) y2=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



Manager
Joined: 27 Jan 2016
Posts: 125

Re: If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
Show Tags
11 Aug 2017, 00:03
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.



Manager
Joined: 27 Jan 2016
Posts: 125

Re: If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
Show Tags
11 Aug 2017, 00:08
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) y2=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). (1) y2=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.



NonHuman User
Joined: 09 Sep 2013
Posts: 12092

Re: If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
Show Tags
19 Aug 2018, 10:04
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: If vertices of a triangle have coordinates (2,2), (3,2) and
[#permalink]
19 Aug 2018, 10:04



Go to page
Previous
1 2
[ 40 posts ]



