Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 08 Jul 2009
Posts: 6

In the rectangular coordinate system above, the area of [#permalink]
Show Tags
08 Jul 2009, 17:11
7
This post received KUDOS
36
This post was BOOKMARKED
Question Stats:
72% (02:15) correct
28% (01:12) wrong based on 880 sessions
HideShow timer Statistics
Attachment:
IMAGE PT1.jpg [ 5.67 KiB  Viewed 66772 times ]
In the rectangular coordinate system above, the area of triangular region PQR is (A) 12.5 (B) 14 (C) 10√2 (D) 16 (E) 25
Official Answer and Stats are available only to registered users. Register/ Login.



Current Student
Joined: 03 Aug 2006
Posts: 115
Location: Next to Google
Schools: Haas School of Business

Re: Area of triangular region PQR [#permalink]
Show Tags
08 Jul 2009, 19:13
56
This post received KUDOS
30
This post was BOOKMARKED
The answer is 12.5 If you look carefully, the triangle is enclosed within a rectangle with dimensions 7 x 4. See attached figure. Area of the \(\triangle PQR\) = Area of rectangle  Area of yellow triangle  Area of blue triangle  Area of red triangle \(= (7 \times 4)  (\frac{1}{2} \times 3 \times 4)  (\frac{1}{2} \times 4 \times 3)  (\frac{1}{2} \times 1 \times 7)\) \(= 28  6  6  3.5\) \(= 12.5\)
Attachments
IMAGE PT1.jpg [ 6.58 KiB  Viewed 66456 times ]



GMAT Forum Moderator
Joined: 01 Nov 2007
Posts: 356
Schools: Wharton Class of 2011

Re: Area of triangular region PQR [#permalink]
Show Tags
09 Jul 2009, 13:52
8
This post received KUDOS
3
This post was BOOKMARKED
I came to the same result but my way of solving was a little bit different. Using Pythagorean theorem we can easily find that PQ and PR are 5 each. Now we have to determine is the angle PQR right angle. Yes it is. How we know that. Using Pythagorean theorem again we can determine that QR is\(5\sqrt{2}\) simply by solving \(\sqrt{1^2 + 7^2}\) = \(\sqrt{50}\) = \(5\sqrt{2}\). We know that PQ and QR are both 5 and their base is \(5\sqrt{2}\) and that diagonale of the square is \(a\sqrt{2}\). So, triangle PQR must be half of the square with the base of 5 or 12,5.
_________________
My GMAT experience My retake experience  GMAT dropped from 720 to 690  advice for retakers My TOEFL experience  108
GMAT Club Premium Membership  big benefits and savings



Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2010

Re: Area of triangular region PQR [#permalink]
Show Tags
21 Mar 2011, 11:15
7
This post received KUDOS
1
This post was BOOKMARKED
gmatdone wrote: trangpham wrote: This answer is (A) 12.5. We can easily prove PQR is a right triangle with the length of PQ and PR are 5, and QR is 5\sqrt{2}. And the area of triangle PQR = square of PQ*square of PR/2=12.5.
And Nookway gave a correct result in very good ways. how can we prove that PQR is a rt triangle. how is PQ =PR=5 If the product of slopes of two lines is 1, those two lines are perpendicular. Here; PR is perpendicular to PQ. Point \(P=(x,y)=(4,0)\) Point \(Q=(x,y)=(0,3)\) Point \(R=(x,y)=(7,4)\) Slope of PR\(=\frac{y_2y_1}{x_2x_1}=\frac{40}{74}=\frac{4}{3}\) Slope of PQ\(=\frac{y_2y_1}{x_2x_1}=\frac{30}{04}=\frac{3}{4}==\frac{3}{4}\) Product of slopes \(= \frac{4}{3}*\frac{3}{4}=1\) Hence, \(PQ \perp PR\) and PQR is a right angled triangle. Formula of distance between two points Distance between two points \((x_1,y_1) & (x_2,y_2) = \sqrt{(x_2x_1)^2+(y_2y_1)^2}\) Distance between point P and point Q \(=\sqrt{(30)^2+(04)^2} = \sqrt{9+16} = \sqrt{25} = 5\) Likewise, Distance between point P and point R \(=\sqrt{(40)^2+(74)^2} = \sqrt{16+9} = \sqrt{25} = 5\) Area of a triangle = 1/2*(PQ)*(PR) = 1/2*5*5=12.5 Ans: "A" Please visit the following link for more on coordinate geometry: http://gmatclub.com/forum/mathcoordinategeometry87652.html
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Math Expert
Joined: 02 Sep 2009
Posts: 39673

Re: Area of triangular region PQR [#permalink]
Show Tags
03 Oct 2010, 02:46
4
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
gottabwise wrote: Pathfinder wrote: I came to the same result but my way of solving was a little bit different.
Using Pythagorean theorem we can easily find that PQ and PR are 5 each. Now we have to determine is the angle PQR right angle.
Yes it is. How we know that. Using Pythagorean theorem again we can determine that QR is\(5\sqrt{2}\) simply by solving \(\sqrt{1^2 + 7^2}\) = \(\sqrt{50}\) = \(5\sqrt{2}\).
We know that PQ and QR are both 5 and their base is \(5\sqrt{2}\) and that diagonale of the square is \(a\sqrt{2}\). So, triangle PQR must be half of the square with the base of 5 or 12,5. Nookway...thanks for the colored graph. It's a great visual and helpful reminder of how to be smarter/more efficient. Pathfinder...is it necessary to determine whether PQR is a right triangle? I'm unsure about the relevance. Method  distance formula for PQ & PR (because they're perpendicular), area formula (1/2B*H) = 1/2(PQ)(PR)=1/2(5)(5) PQR just happened to be right triangle, so if we noticed this fact we could use properties of a right triangle to solve the problem (for example the way Pathfinder did). On the other hand solution provided by nookway works no matter whether PQR is right or not, also it requires much less calculations.
_________________
New to the Math Forum? Please read this: 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
Status: Don't worry about finding inspiration. It eventually come >>>>>>>
Joined: 31 May 2011
Posts: 23
Location: Î Ñ D Ï Â

Re: Area of triangular region PQR [#permalink]
Show Tags
06 Jun 2011, 04:59
4
This post received KUDOS
by calculating area of different different region is a good approach but not effective when you just have 2 sec. left. I have generated one effective way to solve these type of problem. here we know that Q has coordinates (0,3) so measure it roughly through your pencil or pen, now we have one task left that is to measure height of a triangle. So from the measured distance we can predict the height of a triangle which comes to be 3.60 approx. in this question thus Area will be 12.6(approx.) hence option A is 99.99% correct. Hopefully you will enjoy to use this approach!



Intern
Joined: 14 Mar 2015
Posts: 35

Re: In the rectangular coordinate system above, the area of [#permalink]
Show Tags
27 Sep 2015, 06:29
4
This post received KUDOS
akriti1 wrote: I have applied the formulas, but still not clear on this problem. please help what formula did you apply. If you calculate  RP = \(\sqrt{(74)^2 + (40) ^2}\) = \(\sqrt{(3)^2 + (4) ^2}\) = 5 similarly, PQ = \(\sqrt{(40)^2 + (03) ^2}\) = \(\sqrt{(4)^2 + (4) ^2}\) = 5 and, QR = \(\sqrt{(07)^2 + (34) ^2}\) = \(\sqrt{(7)^2 + (1) ^2}\) = \(\sqrt{50}\) = 5\(\sqrt{2}\) So, you can see from the length of the sides, that the triangle is right angled at P. So, the area will be  \(\frac{1}{2} (base * height)\) = \(\frac{1}{2}(5 * 5)\) = 12.5
_________________
It ain’t about how hard you hit. It’s about how hard you can get hit and keep moving forward; how much you can take and keep moving forward. That’s how winning is done!



Intern
Joined: 01 Jul 2009
Posts: 3

Re: Area of triangular region PQR [#permalink]
Show Tags
08 Jul 2009, 19:25
1
This post received KUDOS
1
This post was BOOKMARKED
This answer is (A) 12.5. We can easily prove PQR is a right triangle with the length of PQ and PR are 5, and QR is 5\sqrt{2}. And the area of triangle PQR = square of PQ*square of PR/2=12.5.
And Nookway gave a correct result in very good ways.



Manager
Joined: 24 Jul 2009
Posts: 192
Location: Anchorage, AK
Schools: Mellon, USC, MIT, UCLA, NSCU

Re: Area of triangular region PQR [#permalink]
Show Tags
02 Oct 2010, 21:34
1
This post received KUDOS
1
This post was BOOKMARKED
Pathfinder wrote: I came to the same result but my way of solving was a little bit different.
Using Pythagorean theorem we can easily find that PQ and PR are 5 each. Now we have to determine is the angle PQR right angle.
Yes it is. How we know that. Using Pythagorean theorem again we can determine that QR is\(5\sqrt{2}\) simply by solving \(\sqrt{1^2 + 7^2}\) = \(\sqrt{50}\) = \(5\sqrt{2}\).
We know that PQ and QR are both 5 and their base is \(5\sqrt{2}\) and that diagonale of the square is \(a\sqrt{2}\). So, triangle PQR must be half of the square with the base of 5 or 12,5. Nookway...thanks for the colored graph. It's a great visual and helpful reminder of how to be smarter/more efficient. Pathfinder...is it necessary to determine whether PQR is a right triangle? I'm unsure about the relevance. Method  distance formula for PQ & PR (because they're perpendicular), area formula (1/2B*H) = 1/2(PQ)(PR)=1/2(5)(5)
_________________
Reward wisdom with kudos.



Manager
Status: Last few days....Have pressed the throttle
Joined: 20 Jun 2010
Posts: 68
WE 1: 6 years  Consulting

Re: Area of triangular region PQR [#permalink]
Show Tags
23 Nov 2010, 19:38
1
This post received KUDOS
3
This post was BOOKMARKED
Direct formula for finding area of a triangle in coordinate system 1/2 { (x1x2).(y2y3)  (y1y2).(x2x3) } When we have all the coordinate points of vertices, we can directly substitute and get the area => substituting above, we get: 1/2 {(3)(1)  (4).(7)} = 12.5  (A)
_________________
Consider giving Kudos if my post helps in some way



Current Student
Joined: 26 May 2005
Posts: 563

Re: Area of triangular region PQR [#permalink]
Show Tags
11 Jun 2011, 11:14
1
This post received KUDOS
nss123 wrote: See attached diagram.
In the rectangular coordinate system above, the area of triangular region PQR is:
 12.5  14  10[square_root]2  16  25
This question is from the GMAT Prep Test #1 Question bank. I cannot figure out how to approach the problem and what to look at on the coordinate system.
Thanks for your help. i would use the formula; 1/2 * mod [ x1(y2y3)+ x2(y3y1)+x3(y1y2)] 1/2mod(25) 12.5



Manager
Joined: 13 Aug 2012
Posts: 114

Re: In the rectangular coordinate system above, the area of [#permalink]
Show Tags
25 Jun 2013, 22:17
1
This post received KUDOS
I could do the sum easily by using the following formula 1/2[x1(y2y3)+x2(y3y1)+x3(y1y2)]. This is the easiest and the fastest method acc to me



Retired Moderator
Joined: 02 Sep 2010
Posts: 803
Location: London

Re: Area of triangular region PQR [#permalink]
Show Tags
03 Oct 2010, 02:39
gottabwise wrote: Nookway...thanks for the colored graph. It's a great visual and helpful reminder of how to be smarter/more efficient.
Pathfinder...is it necessary to determine whether PQR is a right triangle? I'm unsure about the relevance.
Method  distance formula for PQ & PR (because they're perpendicular), area formula (1/2B*H) = 1/2(PQ)(PR)=1/2(5)(5)
That is relevant so that you can conclude that PQ can act as a height to the base PR in the area formula, i.e, is it the perpendicular from the opposite vertex on to this side.
_________________
Math writeups 1) Algebra101 2) Sequences 3) Set combinatorics 4) 3D geometry
My GMAT story
GMAT Club Premium Membership  big benefits and savings



Senior Manager
Joined: 07 Sep 2010
Posts: 327

Re: In the rectangular coordinate system above, the area of [#permalink]
Show Tags
17 Jun 2013, 05:53
Hi Bunuel, I'm able to answer the question. However, I'm trying to reinforce some other concept on this question and would like to seek your inputs on the same. What I'm trying to use is to find the coordinate of Point P by using Slope of line QR and finding the midpoint of QR. Since this is issoceles right angle triangle, so PQ and PR can be assumed as two sides of the triangle and QR to be the diagonal of the Square. So, this is what I have found. \(PQ = \sqrt{50}\) \(PR = \sqrt{50}\) \(QR = 5\sqrt{2}\) \(Mid Point of QR = \frac{7}{2}, \frac{7}{2}\) Slope of QR =\(\frac{1}{7}\) If we draw perpendicular from Point P on Side QR, then it will be produced at midpoint of QR. Hence, Using Slope of this perpendicular (7) and using the midpoint of QR, I'm trying to verify the coordinates of Point P. But, here I got stuck, I'm not getting the coordiantes as (4,0).. Infact, I'm stuck how to take care of Negative Sign of the slope(7) Please help me how to proceed further. Your help will be appreciated. Regards, imhimanshu
_________________
+1 Kudos me, Help me unlocking GMAT Club Tests



Intern
Joined: 22 Jul 2013
Posts: 17
Concentration: International Business, Marketing
GPA: 3.89

Re: Area of triangular region PQR [#permalink]
Show Tags
10 May 2014, 02:06
MrMicrostrip wrote: by calculating area of different different region is a good approach but not effective when you just have 2 sec. left. I have generated one effective way to solve these type of problem. here we know that Q has coordinates (0,3) so measure it roughly through your pencil or pen, now we have one task left that is to measure height of a triangle. So from the measured distance we can predict the height of a triangle which comes to be 3.60 approx. in this question thus Area will be 12.6(approx.) hence option A is 99.99% correct. Hopefully you will enjoy to use this approach! This is a great way to think about this. Kudos to you thanks so much! I hit this problem within the last minute of my practice exam and had to guess. Will think about this if something like it pops up again!



Intern
Joined: 28 May 2014
Posts: 3

Re: In the rectangular coordinate system above, the area of [#permalink]
Show Tags
28 May 2014, 08:56
I found that the most simple approach to this question to save time is to visualize the shape in your head, alter the triangle position to be in a horizontal way and then estimate the changes that will occur to its dimensions and do the math. i thought that if i did so i will have an approximate result and then i can chose from answer choices in order for the segment QR to be horizontal we have to shift down R by 1/2 and shift up Q by 1/2. now we have an altitude of 3.5 and a base approximately 7, it should be now less than 7 "after the changes" 1/2 Base*altitude > 1/2*7*3.5= 12.25 I don't know however if this approach works all time because shapes are not always drawn on scale. good luck guys!



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15976

Re: In the rectangular coordinate system above, the area of [#permalink]
Show Tags
03 Jun 2015, 15:00
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



Intern
Joined: 08 Sep 2015
Posts: 3

Re: In the rectangular coordinate system above, the area of [#permalink]
Show Tags
25 Sep 2015, 00:55
I have applied the formulas, but still not clear on this problem. please help



Math Expert
Joined: 02 Sep 2009
Posts: 39673

Re: In the rectangular coordinate system above, the area of [#permalink]
Show Tags
25 Sep 2015, 01:08



Manager
Status: tough ... ? Naaahhh !!!!
Joined: 08 Sep 2015
Posts: 67
Location: India
Concentration: Marketing, Strategy
WE: Marketing (Computer Hardware)

GMATPrep PS [#permalink]
Show Tags
21 Dec 2015, 10:03
Not getting the listed answer. Dont know where i am wrong. Pls help !!!
Attachments
Untitled.png [ 17.76 KiB  Viewed 28085 times ]







Go to page
1 2
Next
[ 26 posts ]




