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Re: Area of triangular region PQR [#permalink]
09 Jul 2009, 12:52

5

This post received KUDOS

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. _________________

Re: Area of triangular region PQR [#permalink]
21 Mar 2011, 10:15

4

This post received KUDOS

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

Re: Area of triangular region PQR [#permalink]
03 Oct 2010, 01:46

3

This post received KUDOS

Expert's post

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. _________________

Re: Area of triangular region PQR [#permalink]
08 Jul 2009, 18:25

1

This post received KUDOS

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.

Re: Area of triangular region PQR [#permalink]
02 Oct 2010, 20:34

1

This post received KUDOS

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) _________________

Re: Area of triangular region PQR [#permalink]
06 Jun 2011, 03:59

1

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!

Re: Area of triangular region PQR [#permalink]
11 Jun 2011, 10:14

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(y2-y3)+ x2(y3-y1)+x3(y1-y2)] 1/2mod(-25) 12.5

Re: Area of triangular region PQR [#permalink]
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. _________________

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.

Re: Geormetry problem [#permalink]
11 Dec 2012, 03:11

Expert's post

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. _________________

Re: Geormetry problem [#permalink]
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!

Re: In the rectangular coordinate system above, the area of [#permalink]
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 _________________

Re: Area of triangular region PQR [#permalink]
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!

Re: In the rectangular coordinate system above, the area of [#permalink]
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!

gmatclubot

Re: In the rectangular coordinate system above, the area of
[#permalink]
28 May 2014, 07:56

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