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Intern
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In the diagram to the right, triangle PQR has a right angle [#permalink]
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13 Apr 2007, 10:12
This topic is locked. If you want to discuss this question please repost it in the respective forum. In the diagram to the right, 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?
see attached JPEG
3/2
7/4
15/8
16/9
2
[/img]
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triangle.jpg [ 6.39 KiB  Viewed 2250 times ]



Director
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Re: 700 level Geo Question [#permalink]
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13 Apr 2007, 12:53
Witchiegrlie wrote: In the diagram to the right, 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?
see attached JPEG
3/2 7/4 15/8 16/9 2
I donot know the best way;
b = 15
l = 20
h = 25.
solving for the areas of the triangles, the ratio = 16/9.



Director
Joined: 14 Jan 2007
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My answer is 'D'.
I have used the long way to solve this. Somebody please suggest some quick approach.



Director
Joined: 18 Jul 2006
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Got D.
After solving, we get PQ=20, PR=25 and QR=15, PS=16, RS=9
(Area PQS)/(Area PRS) = PS/RS = 16/9



Intern
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Anybody please explain your solution to this problem.



Current Student
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OK..you need to memorize the famous triangles..right angle triangles, usually come in the size 3:4:5, 6:8:10 5:12:13..check it for yourself..
so we know the perimeter: is 60
3X+4x+5x=60; x= 5...
OK..so we know that PQ=20, PR=25 and QR=15.
Ok...lets see the second triangle PQS is again 3:4:5, cause we know that QR=15, the other line is 12, so we know the other line is 9.
alright..now area of triangle = 0.5 base*height.. well get rid of the 0.5 cause they will cancel out in the ratio.
so we have...PQS=base=12 heigth=16; i.e 259. so 12*16
QRS base=9 height=12...
now PQS/QRS=16/9
Hope this helps...
jaspetrovic wrote: Anybody please explain your solution to this problem.



VP
Joined: 03 Apr 2007
Posts: 1342

fresinha12 wrote: OK..you need to memorize the famous triangles..right angle triangles, usually come in the size 3:4:5, 6:8:10 5:12:13..check it for yourself.. so we know the perimeter: is 60 3X+4x+5x=60; x= 5... OK..so we know that PQ=20, PR=25 and QR=15. Ok...lets see the second triangle PQS is again 3:4:5, cause we know that QR=15, the other line is 12, so we know the other line is 9. alright..now area of triangle = 0.5 base*height.. well get rid of the 0.5 cause they will cancel out in the ratio. so we have...PQS=base=12 heigth=16; i.e 259. so 12*16 QRS base=9 height=12... now PQS/QRS=16/9 Hope this helps... jaspetrovic wrote: Anybody please explain your solution to this problem.
Very well explained



Intern
Joined: 10 Feb 2007
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Thanks fresinha12! The easiest way so far......



Manager
Joined: 12 Feb 2007
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How do we know they are 345 triangles?



Director
Joined: 24 Aug 2006
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Location: Dallas, Texas

Tuneman wrote: How do we know they are 345 triangles?
Assuming it to be 345 triangle can be a fatal mistake.
You need to solve three equations:
xy=12z
x^2+y^2=z^2
x+y+z=60
This will give you individual length of the arms.
152025 (luckily it was 345 triangle)
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Manager
Joined: 28 Aug 2006
Posts: 160

Where did you get XY=12Z ? Is there an theorm or eqn outthere. Please shed some light.



Manager
Joined: 12 Feb 2007
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Swagatalakshmi wrote: Tuneman wrote: How do we know they are 345 triangles? Assuming it to be 345 triangle can be a fatal mistake. You need to solve three equations: xy=12z x^2+y^2=z^2 x+y+z=60 This will give you individual length of the arms. 152025 (luckily it was 345 triangle)
yes I was looking for 3 eqns also, but like the above poster mentioned, where did you get that xy=12z?
This seems to be a 45 minute problem



Intern
Joined: 11 Apr 2007
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the area of the triangle is 1/2*x*y
the area of the two small triangles are 1/2*z*12
therefore, 1/2*x*y=1/2*z*12
x*y=12*z
got it?...



Director
Joined: 09 Aug 2006
Posts: 520

chris743 wrote: the area of the triangle is 1/2*x*y the area of the two small triangles are 1/2*z*12
therefore, 1/2*x*y=1/2*z*12
x*y=12*z
got it?...
There is a flaw in u r argument.. The two smaller triangles are not same.. so both of them can't be 1/2*z*12



Manager
Joined: 17 Apr 2007
Posts: 90

Amit05 wrote: chris743 wrote: the area of the triangle is 1/2*x*y the area of the two small triangles are 1/2*z*12
therefore, 1/2*x*y=1/2*z*12
x*y=12*z
got it?... There is a flaw in u r argument.. The two smaller triangles are not same.. so both of them can't be 1/2*z*12
area of triangle PQR can also be found using 1/2*PR*QS (QS is perpendicular to PR) 1/2*z*12



Manager
Joined: 11 Nov 2006
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[quote="fresinha12"]OK..you need to memorize the famous triangles..right angle triangles, usually come in the size 3:4:5, 6:8:10 5:12:13..check it for yourself..
so we know the perimeter: is 60
3X+4x+5x=60; x= 5...
OK..so we know that PQ=20, PR=25 and QR=15.
Ok...lets see the second triangle PQS is again 3:4:5, cause we know that QR=15, the other line is 12, so we know the other line is 9.
alright..now area of triangle = 0.5 base*height.. well get rid of the 0.5 cause they will cancel out in the ratio.
so we have...PQS=base=12 heigth=16; i.e 259. so 12*16
QRS base=9 height=12...
now PQS/QRS=16/9
Hope this helps...
Hi Fresinha, could you please explain what was your criteria to select the right triangle 3:4:5? (..3x+4x+5x=60) ?



Manager
Joined: 30 Mar 2007
Posts: 215

The greatest common divisor of the three numbers 3, 4, and 5 is 1.
Pythagorean triples with this property are called primitive.
From primitive Pythagorean triples, you can get other, imprimitive ones, by multiplying each of a, b, and c by any positive whole number d > 1. This is because
a^2 + b^2 = c^2
if and only if (da)^2 + (db)^2 = (dc)^2.
Thus (a,b,c) is a Pythagorean triple if and only if (da,db,dc) is. For example, (6,8,10) and (9,12,15) are imprimitive Pythagorean triples.
Guess we should know few of the primitive triplets
And then hit and trail
Few of starting the primitive triplets
3 4 5
5 12 13
15 8 17
7 24 25
21 20 29
9 40 41
35 12 37
11 60 61
45 28 53
Please let me know if this makes sense



Manager
Joined: 11 Mar 2007
Posts: 69

I found these thereoms...
 If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other
 In a right triangle, the length of the altitude from the right angle to the hypotenuse is the geometric mean of the lengths of the two segments of the hypotenuse
 In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments. Each leg of the right triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg
 The areas of two similar triangles are proportional to the squares of any two homologous sides.
However, I still can't solve the question. I just run around in circles. Maybe these can help jar something loose in someone else...










