Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 09 Sep 2010
Posts: 12

Vertices of a traingle are at (2,1) , (3,1) & (x,y) on a [#permalink]
Show Tags
18 Feb 2011, 10:13
1
This post received KUDOS
10
This post was BOOKMARKED
Question Stats:
60% (01:12) correct 40% (01:35) wrong based on 357 sessions
HideShow timer Statistics
Vertices of a traingle are at (2,1) , (3,1) & (x,y) on a coordinate plane. What is the area of the triangle ? (1) y1 =1 (2) Angle at Vertex (x,y) = 90 degrees Source : MGMAT Test Solution given : (A) The explanation given in the solution is that : From Statement (1) either y=0 or y=2, either way base is 5 and height =1. Hence area can be computed Statement (2) is not enough I did not understand this explanation. Please help !! Thanks in advance.
Official Answer and Stats are available only to registered users. Register/ Login.



Math Forum Moderator
Joined: 20 Dec 2010
Posts: 1914

Re: Coordinate Geometry Area of a Triangle [#permalink]
Show Tags
18 Feb 2011, 10:37
2
This post received KUDOS
Area of triangle= 1/2*base*height Consider line segment joining (2,1) and (3,1) as Base. Base=5. It is a line parallel to xaxis that intercepts at y=1. Now; the ycoordinate of (x,y) can't be 1. 1) y1=1 y1=1 y=2 or y1=1 y=0 We know the ycoordinate of the third vertex. It's either 1 or 0. Say y=1; Irrespective of what x is; the distance between base and the vertex will be 1. Distance is always the perpendicular distance. from the base i.e. make a line of infinite length from vertex(x,y) parallel to the base. The perpendicular line from base to this line will be the distance. So; height=1 Area = 1/2*base*height = 1/2*5*1=5/2 Likewise for y=0; base=5; height=1 Sufficient. 2) Knowing that it is a right triangle won't give us the area. We don't know the length of the sides. Neither do we know anything about the other two angles. Not sufficient. Ans: "A"
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Math Expert
Joined: 02 Sep 2009
Posts: 44598

Re: Coordinate Geometry Area of a Triangle [#permalink]
Show Tags
18 Feb 2011, 10:55
sandhyash wrote: Q  Vertices of a traingle are at (2,1) , (3,1) & (x,y) on a coordinate plane. What is the area of the triangle ? 1) y1 =1 2) Angle at Vertex (x,y) = 90 degrees Source : MGMAT Test Solution given : (A) The explanation given in the solution is that : From Statement (1) either y=0 or y=2, either way base is 5 and height =1. Hence area can be computed Statement (2) is not enough I did not understand this explanation. Please help !! Thanks in advance.Below is almost identical question: Vertices of a triangle have coordinates (2, 2), (3, 2), (x, y). What is the area of the triangle?(1) y  2 = 1 (2) angle at the vertex \((x, y)\) equals 90 degrees Given two points A(2,2) and B(3,2). Question: Area ABC=?, where C(x,y). Attachment:
render.php (1) (1).png [ 17.42 KiB  Viewed 8836 times ]
(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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 09 Sep 2010
Posts: 12

Re: Coordinate Geometry Area of a Triangle [#permalink]
Show Tags
18 Feb 2011, 19:51
I was stuck at understanding Statement (1) for a long time. That diagram really clears my doubt. Thanks guys



Manager
Joined: 16 May 2011
Posts: 198
Concentration: Finance, Real Estate
GMAT Date: 12272011
WE: Law (Law)

Re: Coordinate Geometry Area of a Triangle [#permalink]
Show Tags
10 Jun 2011, 03:13
another simple and straightforward from Bunuel. must admit, i enjoy reading your posts.



Manager
Status: And the Prep starts again...
Joined: 03 Aug 2010
Posts: 129

Re: Vertices of a traingle are at (2,1) , (3,1) & (x,y) on a [#permalink]
Show Tags
08 Mar 2012, 08:25
Bunuel you genius... Best explanation as always.
_________________
My First Blog on my GMAT Journey
Arise, Awake and Stop not till the goal is reached



Manager
Joined: 12 Feb 2012
Posts: 125

If vertices of a triangle have coordinates [#permalink]
Show Tags
17 Sep 2012, 15:05
swatirpr wrote: If vertices of a triangle have coordinates \((2, 2), (3, 2), (x, y)\) , what is the area of the triangle?
1. \(y  2 = 1\) 2. angle at the vertex \((x, y)\) equals 90 degrees Sorry to revive this old post. So 1. \(y  2 = 1\) ===>\(y2=1\)or\(y2=1\) ===> two solutions \(y=1\) or\(y=3\). Notice that that regardless of what value of x, the base will always be the distance between \((2, 2) and (3, 2)\) and the height will always be 1. Therefore the area will always be the same. 2) We know the base and that the third vertex will form a right angle. Well if that is the case, the only right angles that can formed for a given line is one in which it is the diameter of a circle. The third point on this line will always form a right angled triangle. All right angled triangles have their hypotenuse as the diameter of a circle and the third vertex a point on the circle. As you can see, while we know that this third, unknown, vertex will be on this circle, we dont't know where. Therefore the right triangle will have different areas as the third point glides across the circle. The area will reach a maximum when the point forms a isoclese right triangle.
Attachments
gmat_delete2.PNG [ 10.16 KiB  Viewed 6134 times ]
gmat_delete.PNG [ 8.9 KiB  Viewed 6132 times ]



Intern
Joined: 06 Apr 2011
Posts: 13

Re: Vertices of a traingle are at (2,1) , (3,1) & (x,y) on a [#permalink]
Show Tags
18 Mar 2013, 14:04
Bunuel that was a simply marvelous No need for any calculations to solve the question



Intern
Joined: 15 Jul 2012
Posts: 35

Vertices of a traingle are at (2,1) , (3,1) & (x,y) on a [#permalink]
Show Tags
03 Aug 2014, 06:34
Bunuel wrote: sandhyash wrote: Q  Vertices of a traingle are at (2,1) , (3,1) & (x,y) on a coordinate plane. What is the area of the triangle ? 1) y1 =1 2) Angle at Vertex (x,y) = 90 degrees Source : MGMAT Test Solution given : (A) The explanation given in the solution is that : From Statement (1) either y=0 or y=2, either way base is 5 and height =1. Hence area can be computed Statement (2) is not enough I did not understand this explanation. Please help !! Thanks in advance.Below is almost identical question: Vertices of a triangle have coordinates (2, 2), (3, 2), (x, y). What is the area of the triangle?(1) y  2 = 1 (2) angle at the vertex \((x, y)\) equals 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 heps. Hey bunuel, brilliant explanation. But, i have a silly doubt in this problem. arent we assuming that the base is fixed at (2, 2), (3, 2). why cant the base be on the unknown point. from 1st statement we get the value of y as 3 or 1. why cant the base be on (x,1) and (3,2) or (x,3) and (3,2)?



Math Expert
Joined: 02 Sep 2009
Posts: 44598

Vertices of a traingle are at (2,1) , (3,1) & (x,y) on a [#permalink]
Show Tags
12 Aug 2014, 02:44
saggii27 wrote: Bunuel wrote: sandhyash wrote: Q  Vertices of a traingle are at (2,1) , (3,1) & (x,y) on a coordinate plane. What is the area of the triangle ? 1) y1 =1 2) Angle at Vertex (x,y) = 90 degrees Source : MGMAT Test Solution given : (A) The explanation given in the solution is that : From Statement (1) either y=0 or y=2, either way base is 5 and height =1. Hence area can be computed Statement (2) is not enough I did not understand this explanation. Please help !! Thanks in advance.Below is almost identical question: Vertices of a triangle have coordinates (2, 2), (3, 2), (x, y). What is the area of the triangle?(1) y  2 = 1 (2) angle at the vertex \((x, y)\) equals 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 heps. Hey bunuel, brilliant explanation. But, i have a silly doubt in this problem. arent we assuming that the base is fixed at (2, 2), (3, 2). why cant the base be on the unknown point. from 1st statement we get the value of y as 3 or 1. why cant the base be on (x,1) and (3,2) or (x,3) and (3,2)? We can consider any side of a triangle to be the base.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 5255
GPA: 3.82

Re: Vertices of a traingle are at (2,1) , (3,1) & (x,y) on a [#permalink]
Show Tags
10 Nov 2015, 11:41
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Vertices of a traingle are at (2,1) , (3,1) & (x,y) on a coordinate plane. What is the area of the triangle ? (1) y1 =1 (2) Angle at Vertex (x,y) = 90 degrees If we modify the question, we only need to know the value of y, as the height is only affected by the value of y. From condition 1, y1=1, y1=1,1, or y=0,2 Both makes the height 1, so this is sufficient and the answer is (A). Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $79 for 3 month Online Course" "Free Resources30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons  try it yourself"



NonHuman User
Joined: 09 Sep 2013
Posts: 6638

Re: Vertices of a traingle are at (2,1) , (3,1) & (x,y) on a [#permalink]
Show Tags
24 Dec 2017, 20:12
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: Vertices of a traingle are at (2,1) , (3,1) & (x,y) on a
[#permalink]
24 Dec 2017, 20:12






