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In the figure shown, what is the are of the triangular [#permalink]
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29 Oct 2013, 03:14
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In the figure shown, what is the are of the triangular region PRT? (1) The area of rectangular region PQST is 24. (2) The length of line segment RT is 5.
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Re: In the figure shown, what is the are of the triangular [#permalink]
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29 Oct 2013, 04:15
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It statement 1, is it supposed to say "The area of the rectangular region PQST is 24? If so, statement 1 gives you enough information to prove that the area is 12. This is because the area of a quadrilateral is equal to base * height. Because the triangle is inscribed within the rectangle, we know that the base * height of the triangle must also equal 24. But the area of a triangle is equal to 1/2*base*height, so its area must equal 1/2*24 = 12. Try some numbers to prove this to yourself. Say that the base of the rectangle is 24, and its height 1. That makes the base of the triangle 24 and its height 1, and 1/2*24*1 = 12. How about if the base is 6 and the height is 4? 1/2*6*4 = 12. The area of the triangle will always be 12. Statement 2, on the other hand, provides neither the base nor the height, and is this insufficient. This makes the correct answer A. I hope this helps!!!
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Re: In the figure shown, what is the are of the triangular [#permalink]
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29 Oct 2013, 04:21
Thanks a lot for the crystal clear explanation! Never ever question the official answers.
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Re: In the figure shown, what is the are of the triangular [#permalink]
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29 Oct 2013, 17:25
@ bunuel  pls help. Is it always that area of any quadrilateral will be its bxh?



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Re: In the figure shown, what is the are of the triangular [#permalink]
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yuvrajsub wrote: Attachment: image.png In the figure shown, what is the are of the triangular region PRT? (1) The area of rectangular region PQST is 24. (2) The length of line segment RT is 5. [spoiler=]Source: GMATPrep Exam Pack 1 OA: A Theory: The area of any # of trianlge(s) between A.2 fixed parallel lines, and B.With the same base is ALWAYS the same. The first point makes sure that the height of the given triangles is always the same;the second point mandates the base lenght to be the same for all of them. From the given image, try sliding the point R towards Q till the line segment RP coincides with the side QP of the rectangle and thus, RT becomes the diagonal. Now, if we were told to find the area of the newly formed triangle PRT, it would be nothing but \(\frac{1}{2}*24\) Hope this helps.
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Re: In the figure shown, what is the are of the triangular [#permalink]
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Re: In the figure shown, what is the are of the triangular [#permalink]
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24 Nov 2014, 14:32
I'm sorry but I have a bit of a question on this:
(1) The are of PQST is 24 (.5*B*H) so that's the area of the triangular region: sufficient
(2) The length of the line segment RT is 5.
RT is 5 If you drop a perpendicular line down from R to PT that forms a 90 degree triangle. Let's call that point X. Can we then assume a 345 triangle? I got 90 degrees and a 5 right? Then triangle XRT is similar to XRP because they share the same side RX so you can set up proportions to find length PX and XT to get length PT and since you know RX is 4. b*h*.5 is the answer? thus sufficient???



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Re: In the figure shown, what is the are of the triangular [#permalink]
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25 Nov 2014, 04:00
SweetiePi wrote: I'm sorry but I have a bit of a question on this:
(1) The are of PQST is 24 (.5*B*H) so that's the area of the triangular region: sufficient
(2) The length of the line segment RT is 5.
RT is 5 If you drop a perpendicular line down from R to PT that forms a 90 degree triangle. Let's call that point X. Can we then assume a 345 triangle? I got 90 degrees and a 5 right? Then triangle XRT is similar to XRP because they share the same side RX so you can set up proportions to find length PX and XT to get length PT and since you know RX is 4. b*h*.5 is the answer? thus sufficient??? No, the red part is not correct. Hypotenuse of 5 does not necessarily means that the legs must be 3 and 4.
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Re: In the figure shown, what is the are of the triangular [#permalink]
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06 Mar 2015, 00:24
yuvrajsub wrote: Attachment: image.png In the figure shown, what is the are of the triangular region PRT? (1) The area of rectangular region PQST is 24. (2) The length of line segment RT is 5. You may also prove this algebraically. Area of the rectangle is the sum of the all three embedded triangles: (1/2*RQ*ST) + (1/2*RS*ST) + (1/2*PT*ST). But, you will notice that RQ +RS = QS= PT; hence the sum of area of triangle PQR and RST is equal to the area of PRT. Thus, the area of triangle PRT is half of the rectangle.



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Re: In the figure shown, what is the are of the triangular [#permalink]
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yuvrajsub wrote: Attachment: image.png In the figure shown, what is the are of the triangular region PRT? (1) The area of rectangular region PQST is 24. (2) The length of line segment RT is 5. Area of triangle = 1/2bh. Area of rectangle = bh 1) Gives bh and is therefore suff. 2) Cannot get b or h. Insuff. A Kudos if you agree!



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Re: In the figure shown, what is the are of the triangular [#permalink]
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14 May 2017, 15:01
yuvrajsub wrote: Attachment: image.png In the figure shown, what is the are of the triangular region PRT? (1) The area of rectangular region PQST is 24. (2) The length of line segment RT is 5. I picked E because I thought we couldn't assume R was on QS since it is a Data Sufficiency Question and therefore, we don't know the height of the triangle. Just as we couldn't assume PQST was a quadrilateral until statement 1. I'm confused on what we can and cannot assume in these questions. Does anybody have any good advice on this?



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Re: In the figure shown, what is the are of the triangular [#permalink]
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Re: In the figure shown, what is the are of the triangular [#permalink]
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05 Sep 2017, 19:00
yuvrajsub wrote: Attachment: image.png In the figure shown, what is the are of the triangular region PRT? (1) The area of rectangular region PQST is 24. (2) The length of line segment RT is 5. Tricky question ahhh What the trap in this question is getting you to think that you need a particular hypotenuse length in order to calculate the respective areas HOWEVER if you think about the question from a formulaic perspective then you will see that the area of a right triangle 1/2 BH essentially equals the area of the square It doesn't actually matter what the height and length are 6 and 4 or 8 and 3 because they will multiply to 24 which can just be substituted for BH in 1/2 BH And will result in the same answer A




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