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In the diagram, triangle PQR has a right angle at Q and a [#permalink]
05 Feb 2012, 15:54
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Difficulty:
95% (hard)
Question Stats:
50% (04:53) correct
50% (03:34) wrong based on 354 sessions
Attachment:
Triangle PQR.GIF [ 2.52 KiB | Viewed 35926 times ]
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?
Re: In the diagram, triangle PQR has a right angle at Q and a [#permalink]
05 Feb 2012, 16:43
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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 [ 2.52 KiB | Viewed 35770 times ]
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) \((x+y)^2-2xy=z^2\) --> as from (i) \(x+y=60-z\) and from (ii) \(xy=12z\) then (\(60-z)^2-2*12z=z^2\) --> \(3600-120z+z^2-24z=z^2\) --> \(3600=144z\) --> \(z=25\);
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, \(\frac{x^2}{y^2}=\frac{AREA}{area}\) --> \(\frac{AREA}{area}=\frac{400}{225}=\frac{16}{9}\)
Re: In the diagram, triangle PQR has a right angle at Q and a [#permalink]
01 Mar 2012, 18:18
Hi,
Can u explain in a simpler way, I didn't get it. This is what I tried, since QS is perpendicular to PR, it divides PR in 2 equal parts ie, PS=SR. Also now we are given perimeter is 60 for the given right triangle, hence I tried to use 3-4-5 rule, so now length of the sides are PQ=20, QR=15, PR=25 since its given PQ greater than QR. So now, area of PQS= 1/2 PS * QS =1/2 * 25/2* 12 =75 area of RQS = 1/2*SR*QS = 1/2*25/2*12= 75 I know I'm wrong, could you please correct me.
Re: In the diagram, triangle PQR has a right angle at Q and a [#permalink]
01 Mar 2012, 20:23
Expert's post
priyalr wrote:
Hi,
Can u explain in a simpler way, I didn't get it. This is what I tried, since QS is perpendicular to PR, it divides PR in 2 equal parts ie, PS=SR. Also now we are given perimeter is 60 for the given right triangle, hence I tried to use 3-4-5 rule, so now length of the sides are PQ=20, QR=15, PR=25 since its given PQ greater than QR. So now, area of PQS= 1/2 PS * QS =1/2 * 25/2* 12 =75 area of RQS = 1/2*SR*QS = 1/2*25/2*12= 75 I know I'm wrong, could you please correct me.
Thnx.
Perpendicular to hypotenuse divides it in half if and only the right triangle is isosceles, so when PQ=QR but it's given that PQ>QR, so it's not the case.
Re: In the diagram, triangle PQR has a right angle at Q and a [#permalink]
22 Jul 2012, 20:07
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) \((x+y)^2-2xy=z^2\) --> as from (i) \(x+y=60-z\) and from (ii) \(xy=12z\) then (\(60-z)^2-2*12z=z^2\) --> \(3600-120z+z^2-24z=z^2\) --> \(3600=144z\) --> \(z=25\);
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, PQR 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.
Hey Bunuel,
I am aware that if a right triangle is spliced from the vertex of the 90 degree by a line that is perpendicular to the opposite side, then the two smaller triangles (within the larger triangle) are similar. I have a couple of questions regarding this property:
1) Can we make any inference on how the vertex with 90* is spliced (45*-45* , 60*-30*, 20*-70*)? 2) Does this property hold for other triangles? ( I can think of equaliteral triangles for example, in which I can splice one vertex from the middle at 30-30 and extend the a perpenduular line to the opposite side to form two similar triangles). Can we have a triangle in which we extend a line from that vertex to the opposite perpendicular and not create two smaller triangles? 3) I noticed in your answer that you used this formula:
\(\frac{AREA}{area}=\frac{S^2}{s^2}\)
In the question above the two sides similar, but they also share the same side of 12. How does this change things? the sides of similar triangles are usually multiples of each other (3-4-5 and 6-8-10 are similar because the same similar angles, albeit different size lengths). But in this case the two similar triangles share a common side, they are no longer multiples of one another.
I mean why couldn't we simply pick the side of the two triangles that both triangles each share with another and say \(\frac{AREA}{area}=\frac{12^2}{12^2}=1\)
Re: In the diagram, triangle PQR has a right angle at Q and a [#permalink]
26 Jul 2012, 07:24
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) \((x+y)^2-2xy=z^2\) --> as from (i) \(x+y=60-z\) and from (ii) \(xy=12z\) then (\(60-z)^2-2*12z=z^2\) --> \(3600-120z+z^2-24z=z^2\) --> \(3600=144z\) --> \(z=25\);
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, PQR 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.
Hi Bunnuel
I don't get why you equate both areas with xy=12z With this you are specifying that both triangles have the same area, and the whole purpose of the excercise is to prove this? What am I missing?
Re: In the diagram, triangle PQR has a right angle at Q and a [#permalink]
01 Aug 2012, 17:46
Hey Bunuel,
I am aware that if a right triangle is spliced from the vertex of the 90 degree by a line that is perpendicular to the opposite side, then the two smaller triangles (within the larger triangle) are similar. I have a couple of questions regarding this property:
1) Can we make any inference on how the vertex with 90* is spliced (45*-45* , 60*-30*, 20*-70*)? 2) Does this property hold for other triangles? ( I can think of equaliteral triangles for example, in which I can splice one vertex from the middle at 30-30 and extend the a perpenduular line to the opposite side to form two similar triangles). Can we have a triangle in which we extend a line from that vertex to the opposite perpendicular and not create two smaller triangles? 3) I noticed in your answer that you used this formula:
\(\frac{AREA}{area}=\frac{S^2}{s^2}\)
In the question above the two sides similar, but they also share the same side of 12. How does this change things? the sides of similar triangles are usually multiples of each other (3-4-5 and 6-8-10 are similar because the same similar angles, albeit different size lengths). But in this case the two similar triangles share a common side, they are no longer multiples of one another.
I mean why couldn't we simply pick the side of the two triangles that both triangles each share with another and say \(\frac{AREA}{area}=\frac{12^2}{12^2}=1\)
Re: In the diagram, triangle PQR has a right angle at Q and a [#permalink]
10 Aug 2012, 13:03
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Bunuel,
What about this methodology?
1) Triangle PQR is a right triangle w/ perimeter of 60. We know PQ is the longer side. Using the right triangle ratio 3:4:5, we get QR=15, PQ=20, and PR=25.
2) Triangle QRS has side QS=12 (given) and QR=15 (from 1). Using 3:4:5, SR=9.
3) Triangle PQS has side QS=12 (given) and PQ=20 (from 1). Using 3:4:5, PS=16 (also PR-RS=25-9=16).
Re: In the diagram, triangle PQR has a right angle at Q and a [#permalink]
15 Dec 2012, 00:33
3
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@Marcab, thanks, I didn't see it was that easy and btw I did find a 2minute approach to this problem. Unfortunately I don't have 5 posts yet on this forum to share the link, nevertheless I would say simply Google for "In the diagram, triangle PQR" and click the MGMAT link wrt this problem and you would find what your looking for there. Hth.
P.S.: Please don't forget to click on Kudos if you think my post helped.
Re: In the diagram, triangle PQR has a right angle at Q and a [#permalink]
15 Dec 2012, 01:34
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Thanks Boddhisattva. +1 to you.
Here is the best approach to the problem. Quoting ChristianCRyan from MGMAT:
I personally think that the trick to this particular problem is to switch gears at a certain point -- after you write down the equations describing all the relationships, you might realize (as I myself did on first trying this problem) that it would take a long time to derive the solution using algebra. So then you need to say, let me try some common right triangles. 3-4-5 becomes the first candidate -- especially since the perpendicular is 12, which is divisible by both 3 and 4. So a side of 12 (in the two smaller triangles) is easy to "scale" to -- if it's the shortest side (the "3" side), then the longer sides are 16 and 20; likewise, if it's the middle side, the other two sides are 9 and 15. Magically this fits the perimeter constraint (the big triangle's perimeter=60), and you're done. _________________
Re: In the diagram, triangle PQR has a right angle at Q and a [#permalink]
01 Sep 2013, 14:50
1
This post received KUDOS
Here is another approach I don't think anyone has mention that takes 10's.
We know whatever the answer is going to be it is going to satisfy this equation. A1/A2=S1^2/S2^2=(S1/S2)^2. Now even after ratio S1/S2 is reduced to smallest possible fraction. The answer choices have to satify the condition that the numbers are squares. The only answer choice in this problem where the answer choices are squares is 16/9=4^2/3^2.
Here is my question for Bunuel. How did find the solution for x+y=35 & xy=300. I know you could just substitute and solve. But you get an ugly quadratic. I don't know how to find all the factors of 300 in less than 30 secs. Besides plugging number was there a shortcut you used?
Re: In the diagram, triangle PQR has a right angle at Q and a [#permalink]
10 Sep 2013, 23:00
Ok.. so my idea is not foolproof but it got me the right answer within 15 secs...
By a property (QS)^2 = PS*SR. PS*SR = 12^2 = 144 Also, the traingles are similar so the ratio of the area can be given by ratio of side PS and SR.
In exam I could reverse it to affirm the answer (16+9 = 25 i.e. hyp of pqr is 25. If I take the triplet (3*5,4*5,5*5). I can see perimeter = 60 and so on Basically find a ratio which multiplies to 144
Re: In the diagram, triangle PQR has a right angle at Q and a [#permalink]
07 Oct 2013, 06:40
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I solved it this way:
we are given:
QS = 12 PQ>QR Perimeter = 60
Fist off, we are dealing with a right triangle, it is glaring, and the first thing that pops up in my mind is: Pythagorean Triples. Let's just recall the basic one because any other is just a multiple of the basic one. 3+4+5 = 12 not really close to 60. 15+20+25= 60 there we go!
Now we know that PQ= 20, QR= 15 and PR=25.
Since QS is perpendicular to PR the two smaller triangles are also right triangles. Let's figure out the length of the sides of PSQ. Once again the smaller cathetus (QS) turns out to be our fundamental Pythagorean triplet multiplied by 4. Once realized this we can quickly gauge the length of the remaining cathetus and hypothenuse. 12:16:20
Now we have all the elements that we need to find out our answer.
Area PQS- Area PQS = 54 (=Area QSR) and our ratio is going to be 16/9 _________________
learn the rules of the game, then play better than anyone else.
Re: In the diagram, triangle PQR has a right angle at Q and a [#permalink]
23 Nov 2013, 18:16
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I'm assuming this is WAAAY too intense to actually appear on the Gmat. Look at Bunuel's solution; even if you KNOW that, which there is virtually no way most people would, it would still take about 4 minutes to solve that out.
Re: In the diagram, triangle PQR has a right angle at Q and a [#permalink]
13 Dec 2013, 09:23
Bunuel,
I solved a very similar problem to this yesterday, where we were given x+y = 15 and that x*y =28. I tried to solve the way you solved this problem and got an incorrect answer. How would I know to solve like that for this problem but not for the other one? Would I know because no factors of 28 would add up to 15?
Re: In the diagram, triangle PQR has a right angle at Q and a [#permalink]
03 Apr 2014, 21:10
1
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The question seems to ask for "ratio of the area of triangle PQS to the area of triangle RQS" but Mr. Bunuel, you have provided for the ratio of areas of triangle PQR and SQR, is it a typing error?? Please confirm!!
gmatclubot
Re: In the diagram, triangle PQR has a right angle at Q and a
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03 Apr 2014, 21:10
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