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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of
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04 Apr 2014, 01:32



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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of
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23 Jun 2014, 01:14
Bunuel wrote: enigma123 wrote: In the diagram, triangle PQR has a right angle at Q and a perimeter of 60. Line segment QS is perpendicular to PR and has a length of 12. PQ > QR. What is the ratio of the area of triangle PQS to the area of triangle RQS?
A)3/2 B)7/4 C)15/8 D) 16/9 E)2
Guys any idea how to solve this. I neither have an OA nor an explanation to this. Attachment: Triangle PQR.GIF Let \(PQ=x\), \(QR=y\) and \(PR=z\). Given: \(x+y+z=60\) (i); Equate the areas: \(\frac{1}{2}*xy=\frac{1}{2}*QS*z\) (area of PQR can be calculated by 1/2*leg*leg and 1/2* perpendicular to hypotenuse*hypotenuse) > \(xy=12z\) (ii); Aslo \(x^2+y^2=z^2\) (iii); So, we have: (i) \(x+y+z=60\); (ii) \(xy=12z\); (iii) \(x^2+y^2=z^2\). From (iii) \((x+y)^22xy=z^2\) > as from (i) \(x+y=60z\) and from (ii) \(xy=12z\) then (\(60z)^22*12z=z^2\) > \(3600120z+z^224z=z^2\) > \(3600=144z\) > \(z=25\); From (i) \(x+y=35\) and from (ii) \(xy=300\) > solving for \(x\) and \(y\) > \(x=20\) and \(y=15\) (as given that \(x>y\)). Next, perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. So, PQS and SQR are similar. In two similar triangles, the ratio of their areas is the square of the ratio of their sides: \(\frac{AREA}{area}=\frac{S^2}{s^2}\).So, \(\frac{x^2}{y^2}=\frac{AREA}{area}\) > \(\frac{AREA}{area}=\frac{400}{225}=\frac{16}{9}\) Answer: D. Given the perimeter is a smallish number, wouldn't plugging in numbers be a smarter approach on a question like this? Guessing numbers here to fit the conditions is rather straight forward in this case There could be multiple sets of possible lengths, but then every length should result in the same ratio for areas of the similar triangles Bunnel?



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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of
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23 Jun 2014, 02:03
Kconfused wrote: Bunuel wrote: enigma123 wrote: In the diagram, triangle PQR has a right angle at Q and a perimeter of 60. Line segment QS is perpendicular to PR and has a length of 12. PQ > QR. What is the ratio of the area of triangle PQS to the area of triangle RQS?
A)3/2 B)7/4 C)15/8 D) 16/9 E)2
Guys any idea how to solve this. I neither have an OA nor an explanation to this. Attachment: Triangle PQR.GIF Let \(PQ=x\), \(QR=y\) and \(PR=z\). Given: \(x+y+z=60\) (i); Equate the areas: \(\frac{1}{2}*xy=\frac{1}{2}*QS*z\) (area of PQR can be calculated by 1/2*leg*leg and 1/2* perpendicular to hypotenuse*hypotenuse) > \(xy=12z\) (ii); Aslo \(x^2+y^2=z^2\) (iii); So, we have: (i) \(x+y+z=60\); (ii) \(xy=12z\); (iii) \(x^2+y^2=z^2\). From (iii) \((x+y)^22xy=z^2\) > as from (i) \(x+y=60z\) and from (ii) \(xy=12z\) then (\(60z)^22*12z=z^2\) > \(3600120z+z^224z=z^2\) > \(3600=144z\) > \(z=25\); From (i) \(x+y=35\) and from (ii) \(xy=300\) > solving for \(x\) and \(y\) > \(x=20\) and \(y=15\) (as given that \(x>y\)). Next, perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. So, PQS and SQR are similar. In two similar triangles, the ratio of their areas is the square of the ratio of their sides: \(\frac{AREA}{area}=\frac{S^2}{s^2}\).So, \(\frac{x^2}{y^2}=\frac{AREA}{area}\) > \(\frac{AREA}{area}=\frac{400}{225}=\frac{16}{9}\) Answer: D. Given the perimeter is a smallish number, wouldn't plugging in numbers be a smarter approach on a question like this? Guessing numbers here to fit the conditions is rather straight forward in this case There could be multiple sets of possible lengths, but then every length should result in the same ratio for areas of the similar triangles Bunnel? Whatever suits you best.
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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of
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05 Aug 2014, 04:50
I think one approach not covered so far is the approach required when ideas are not striking you fast enough and you are kind of stuck staring at the screen. And time is running out!
In such a situation, eliminating/ solvingthroughoptions approach comes handy.
Since the two triangles have a common side, the ratio of the two areas will be the ratio of PS and SR. Here, PS and SR add up to form the side PR. And we have perimeter given as 60. So going through options, by trialanderror, we need to find which value of PR give us Pythagorean Triplets.
1) 3/2 > 3+2=5, too small to be hypotenuse since the remaining two sides will have to add up to 55... so not possible 2)7/4 > 7+4=11... too small to be hypotenuse 3)15/8 > 15+8=23 > doesn't form triplets (can apply trial and error to test just in case you are not sure) 4)16/9 > 16+9=25 > common triplet 5)2 >2+1=3.. too small



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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of
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06 Aug 2014, 01:04
enigma123 wrote: Attachment: Triangle PQR.GIF In the diagram, triangle PQR has a right angle at Q and a perimeter of 60. Line segment QS is perpendicular to PR and has a length of 12. PQ > QR. What is the ratio of the area of triangle PQS to the area of triangle RQS? A. 3/2 B. 7/4 C. 15/8 D. 16/9 E. 2 If a,b,c are not integers, the solution will take a lot of time. That is not kind of GMAT question, if it takes like forever to solve the problem For a right triangle, the first thing happened in my mind was a:b:c = 3:4:5. Maybe a = 3k, b = 4k, c = 5k so I have the equation: 3k + 4k + 5k = 60 > 12k = 60 > k = 5 > a = 15, b = 20, c = 25 (b>a) Try whether a*b/2 = QS*25/2 > 15*20/2 = 12*25/2 > correct Area of PQS/ Area of RQS = b^2/a^2 = 16/9 Answer is D
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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of
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26 Jul 2016, 20:25
There is a easier way for this but it involves some guess work(Smart work) QS^2=PS*SR =>12^2=PS*SR=>144=PS*SR Now try to break PS and SR such that the multiplication will yield you 144. The immediate answer that follows is 16*9 Now, PR=PS+SR=16+9=25 Go to the Pythagoras triplets. One possible value is 15, 20, 25. Bingo !!! Perfect match Now, PQ > QR (Given) There fore PQ=20, QR=15  1 Now to get the ratio of the area of triangle PQS to the area of triangle RQS we just need to take the ratio of the square of the sides as they will be Similar. (HOPE YOU KNOW WHY THEY WILL BE SIMILAR) Area(PQS)/Area RQS=PQ^2/QR^2=15^2/20^2 [From 1]=9/16



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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of
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26 Jul 2016, 20:31
Marcab wrote: boddhisattva wrote: Can anyone please explain how do we arrive at the values for x and y using xy = 300 and x + y = 35.
I tried using (x+y)^2 = x^2 + y^2 + 2xy formulae as below:
35^2 = x^2 + y^2 + 2*300 , how to proceed from here to get values of x and y?
Thank you. \(x=300/y\) Put this in equation: \(x+y=35\), you will get the equation \(y^235y+300=0\) You may write it down as : \(y^20y15y300=0\) OR \(y(y20)15(y20)=0> (y20)(y15)\) Hence y=20 or 15. Thereby x=15 or 20 But since its given that PQ>QR, therefore x=20 and y=15. Now my question is: Is there a 2 minute approach to this question. This will take less than a min. All Mental Game There is a easier way for this but it involves some guess work(Smart work) QS^2=PS*SR =>12^2=PS*SR=>144=PS*SR Now try to break PS and SR such that the multiplication will yield you 144. The immediate answer that follows is 16*9 Now, PR=PS+SR=16+9=25 Go to the Pythagoras triplets. One possible value is 15, 20, 25. Bingo !!! Perfect match Now, PQ > QR (Given) There fore PQ=20, QR=15  1 Now to get the ratio of the area of triangle PQS to the area of triangle RQS we just need to take the ratio of the square of the sides as they will be Similar. (HOPE YOU KNOW WHY THEY WILL BE SIMILAR) Area(PQS)/Area RQS=PQ^2/QR^2=15^2/20^2 [From 1]=9/16



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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of
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02 Apr 2017, 07:22
Well, a traditional mathematical solution will consume a lot of time. So here are two ways, which i used to solve this question 1. ar (PQS)/ar(RQS)= 0.5 * QS * PS/ 0.5 * QS * SR = PS/SR  So we need the values of PS and SR. Consider this as Statement 1Now we know, QS ^ 2 = PS * SR ==> 12^2 = PS * SR ==> 144 = PS * SR Look at the options now. Only D can give me 144 (16*9). Hence PS = 16 and SR = 9 Therefore, from Statement 1 ratio of the area has to be 16/9 Second Method: When we talk about a Pythagoras triplet, the first that comes to our mind is 3,4,5 If this is the measurement of a triangle then the perimeter will be 3+4+5= 12 In the question the given parameter (60) is 5 times of 12 Hence the side should also be 5 times  3*5, 4*5, 5*5 = 15,20,25 Adding them will give me 15+20+25 = 60 ( which is the given parameter in the question) Now hypotenuse is 25. ar (PQS)/ar(RQS)= 0.5 * QS * PS/ 0.5 * QS * SR = PS/SR  So we need the values of PS and SR. Consider this as Statement 1PS + SR = 25 (hypotenuse ) Look at the options. None of the options will add up to 25 but option D will add up to 16+9 = 25. Hence PS = 16 and SR = 9 Therefore, from Statement 1 ratio of the area has to be 16/9 I feel both of them can be mentally solved in less than a min



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In the diagram, triangle PQR has a right angle at Q and a perimeter of
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07 Mar 2018, 08:49
Bunuel wrote: enigma123 wrote: In the diagram, triangle PQR has a right angle at Q and a perimeter of 60. Line segment QS is perpendicular to PR and has a length of 12. PQ > QR. What is the ratio of the area of triangle PQS to the area of triangle RQS?
A)3/2 B)7/4 C)15/8 D) 16/9 E)2
Guys any idea how to solve this. I neither have an OA nor an explanation to this. Let \(PQ=x\), \(QR=y\) and \(PR=z\). Given: \(x+y+z=60\) (i); Equate the areas: \(\frac{1}{2}*xy=\frac{1}{2}*QS*z\) (area of PQR can be calculated by 1/2*leg*leg and 1/2* perpendicular to hypotenuse*hypotenuse) > \(xy=12z\) (ii); Aslo \(x^2+y^2=z^2\) (iii); So, we have: (i) \(x+y+z=60\); (ii) \(xy=12z\); (iii) \(x^2+y^2=z^2\). From (iii) \((x+y)^22xy=z^2\) > as from (i) \(x+y=60z\) and from (ii) \(xy=12z\) then (\(60z)^22*12z=z^2\) > \(3600120z+z^224z=z^2\) > \(3600=144z\) > \(z=25\); From (i) \(x+y=35\) and from (ii) \(xy=300\) > solving for \(x\) and \(y\) > \(x=20\) and \(y=15\) (as given that \(x>y\)). Next, perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. So, PQS and SQR are similar. In two similar triangles, the ratio of their areas is the square of the ratio of their sides: \(\frac{AREA}{area}=\frac{S^2}{s^2}\).So, \(\frac{x^2}{y^2}=\frac{AREA}{area}\) > \(\frac{AREA}{area}=\frac{400}{225}=\frac{16}{9}\) Answer: D. can u explain how you deduced eqn iii; bcoz if pr is 12, how come it is z...plz explain . Thanks



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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of
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07 Mar 2018, 08:57
shringi87 wrote: Bunuel wrote: enigma123 wrote: In the diagram, triangle PQR has a right angle at Q and a perimeter of 60. Line segment QS is perpendicular to PR and has a length of 12. PQ > QR. What is the ratio of the area of triangle PQS to the area of triangle RQS?
A)3/2 B)7/4 C)15/8 D) 16/9 E)2
Guys any idea how to solve this. I neither have an OA nor an explanation to this. Let \(PQ=x\), \(QR=y\) and \(PR=z\). Given: \(x+y+z=60\) (i); Equate the areas: \(\frac{1}{2}*xy=\frac{1}{2}*QS*z\) (area of PQR can be calculated by 1/2*leg*leg and 1/2* perpendicular to hypotenuse*hypotenuse) > \(xy=12z\) (ii); Aslo \(x^2+y^2=z^2\) (iii); So, we have: (i) \(x+y+z=60\); (ii) \(xy=12z\); (iii) \(x^2+y^2=z^2\).From (iii) \((x+y)^22xy=z^2\) > as from (i) \(x+y=60z\) and from (ii) \(xy=12z\) then (\(60z)^22*12z=z^2\) > \(3600120z+z^224z=z^2\) > \(3600=144z\) > \(z=25\); From (i) \(x+y=35\) and from (ii) \(xy=300\) > solving for \(x\) and \(y\) > \(x=20\) and \(y=15\) (as given that \(x>y\)). Next, perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. So, PQS and SQR are similar. In two similar triangles, the ratio of their areas is the square of the ratio of their sides: \(\frac{AREA}{area}=\frac{S^2}{s^2}\).So, \(\frac{x^2}{y^2}=\frac{AREA}{area}\) > \(\frac{AREA}{area}=\frac{400}{225}=\frac{16}{9}\) Answer: D. can u explain how you deduced eqn iii; bcoz if pr is 12, how come it is z...plz explain . Thanks Not clear what you mean. Equation (iii) is Pythagoras's theorem: x and y are lengths of the legs and z is the length of the hypotenuse.
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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.
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