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# In the diagram, triangle PQR has a right angle at Q and a perimeter of

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In the diagram, triangle PQR has a right angle at Q and a perimeter of  [#permalink]

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Updated on: 20 Feb 2019, 00:43
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75
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95% (hard)

Question Stats:

47% (03:16) correct 53% (03:14) wrong based on 370 sessions

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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

Attachment:

Triangle PQR.GIF [ 2.52 KiB | Viewed 238251 times ]

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Originally posted by enigma123 on 05 Feb 2012, 16:54.
Last edited by Bunuel on 20 Feb 2019, 00:43, edited 1 time in total.
Updated.
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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of  [#permalink]

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05 Feb 2012, 17:43
18
18
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

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, $$\frac{x^2}{y^2}=\frac{AREA}{area}$$ --> $$\frac{AREA}{area}=\frac{400}{225}=\frac{16}{9}$$

Attachment:

Triangle PQR.GIF [ 2.52 KiB | Viewed 205727 times ]

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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of  [#permalink]

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10 Aug 2012, 14:03
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6
Bunuel,

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).

4) Area of PQS is 1/2*12*16=96

5) Area of QRS is 1/2*12*9=54

6) Ratio is 96/54=16/9
##### General Discussion
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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of  [#permalink]

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01 Mar 2012, 19:18
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.

Thnx.
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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of  [#permalink]

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01 Mar 2012, 21:23
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.

Hope it's clear.
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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of  [#permalink]

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14 Dec 2012, 22:17
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.
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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of  [#permalink]

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15 Dec 2012, 00:23
1
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 $$y^2-35y+300=0$$

You may write it down as :
$$y^-20y-15y-300=0$$

OR

$$y(y-20)-15(y-20)=0---------> (y-20)(y-15)$$
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.
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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of  [#permalink]

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15 Dec 2012, 01:33
5
@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.
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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of  [#permalink]

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15 Dec 2012, 02:34
5
2
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.
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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of  [#permalink]

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01 Sep 2013, 15:50
1
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?
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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of  [#permalink]

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11 Sep 2013, 00:00
1
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
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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of  [#permalink]

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07 Oct 2013, 07:40
7
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
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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of  [#permalink]

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06 Aug 2014, 01:04
1
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

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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of  [#permalink]

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22 Oct 2018, 07:50
Given the time constraints, I tried to solve by using the similarity properties of the two triangles formed by splitting the right angle using QS.

To frame the question, ratio of area PQS:RSQ = ratio of PS:SR.

Note that SR:QS = QS:PS; QS = 12.
Therefore, PS:SR = QS^2 = 12^2 = 144 = (3 x 4)^2 = (3^2 x 2^4).
There are many ways to assign the prime values of 144 but only 1 answer fits a suitable ratio i.e. 2^4 : 3^2 = 16:9.
Therefore, answer is D. No other answer allows the allocation of the prime factors available. This is another way to solve it. I was pressed for time and not able to do the detailed method Bunuel did, so I tried my luck and got the answer. Should there be another answer that fits the available ratio, then I would have compared it to the given information in the stem and try to figure out from there.

Hope this works.
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Re: In the diagram, triangle PQR has a right angle at Q and a perimeter of  [#permalink]

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24 Oct 2018, 08:25
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

Attachment:
Triangle PQR.GIF

Ans:

Since its not a equilateral right triangle, the sides must be in ration of 3:4:5. Using this

1) Triangle PQR is a right triangle w/ perimeter of 60, 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).

4) Area of PQS is 1/2*12*16=96

5) Area of QRS is 1/2*12*9=54

6) Ratio is 96/54=16/9
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