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In the rectangular coordinate system above, the area of

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In the rectangular coordinate system above, the area of [#permalink] New post 08 Jul 2009, 16:11
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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
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Re: Area of triangular region PQR [#permalink] New post 08 Jul 2009, 18:13
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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

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Re: Area of triangular region PQR [#permalink] New post 08 Jul 2009, 18:25
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.
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Re: Area of triangular region PQR [#permalink] New post 09 Jul 2009, 12:52
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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 is5\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.

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Re: Area of triangular region PQR [#permalink] New post 02 Oct 2010, 20:34
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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 is5\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)

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Re: Area of triangular region PQR [#permalink] New post 03 Oct 2010, 01:39
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.

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Re: Area of triangular region PQR [#permalink] New post 03 Oct 2010, 01:46
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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 is5\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.

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Re: Area of triangular region PQR [#permalink] New post 23 Nov 2010, 18:38
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Direct formula for finding area of a triangle in coordinate system

1/2 { (x1-x2).(y2-y3) - (y1-y2).(x2-x3) }

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)

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Re: Area of triangular region PQR [#permalink] New post 21 Mar 2011, 10:15
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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_2-y_1}{x_2-x_1}=\frac{4-0}{7-4}=\frac{4}{3}
Slope of PQ=\frac{y_2-y_1}{x_2-x_1}=\frac{3-0}{0-4}=\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_2-x_1)^2+(y_2-y_1)^2}

Distance between point P and point Q =\sqrt{(3-0)^2+(0-4)^2} = \sqrt{9+16} = \sqrt{25} = 5
Likewise,
Distance between point P and point R =\sqrt{(4-0)^2+(7-4)^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/math-coordinate-geometry-87652.html

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Re: Area of triangular region PQR [#permalink] New post 06 Jun 2011, 03:59
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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!
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Re: Area of triangular region PQR [#permalink] New post 11 Jun 2011, 10:14
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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(y2-y3)+ x2(y3-y1)+x3(y1-y2)]
1/2mod(-25)
12.5
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Geormetry problem [#permalink] New post 11 Dec 2012, 03:06
Hi everyone!

I took the GMATPrep first CAT yesturday and I couldn't find the answer of the last Quant question. Can someone help me? The image of the problem is enclosed to the post.

Thanks and good luck to you all!

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Re: Geormetry problem [#permalink] New post 11 Dec 2012, 03:11
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AzizaAmri wrote:
Hi everyone!

I took the GMATPrep first CAT yesturday and I couldn't find the answer of the last Quant question. Can someone help me? The image of the problem is enclosed to the post.

Thanks and good luck to you all!


Merging similar topics. Please refer to the solutions above.

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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

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Re: Geormetry problem [#permalink] New post 11 Dec 2012, 03:39
Bunuel wrote:
AzizaAmri wrote:
Hi everyone!

I took the GMATPrep first CAT yesturday and I couldn't find the answer of the last Quant question. Can someone help me? The image of the problem is enclosed to the post.

Thanks and good luck to you all!


Merging similar topics. Please refer to the solutions above.


Thank you very much, I'm not familiar with the forum yet :) and I found my answer!
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Re: In the rectangular coordinate system above, the area of [#permalink] New post 17 Jun 2013, 04:53
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

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Re: In the rectangular coordinate system above, the area of [#permalink] New post 25 Jun 2013, 21:17
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I could do the sum easily by using the following formula 1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]. This is the easiest and the fastest method acc to me
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Re: Area of triangular region PQR [#permalink] New post 10 May 2014, 01:06
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!
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Re: In the rectangular coordinate system above, the area of [#permalink] New post 28 May 2014, 07:56
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!
Re: In the rectangular coordinate system above, the area of   [#permalink] 28 May 2014, 07:56
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