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# In the diagram to the right, triangle PQR has a right angle

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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 re-post 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]
Attachments

triangle.jpg [ 6.39 KiB | Viewed 2496 times ]

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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.

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13 Apr 2007, 14:21
I have used the long way to solve this. Somebody please suggest some quick approach.

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13 Apr 2007, 16:40
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

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13 Apr 2007, 20:18

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14 Apr 2007, 10:37
1
KUDOS
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 25-9. so 12*16
QRS base=9 height=12...

now PQS/QRS=16/9

Hope this helps...

jaspetrovic wrote:

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14 Apr 2007, 14:32
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 25-9. so 12*16
QRS base=9 height=12...

now PQS/QRS=16/9

Hope this helps...

jaspetrovic wrote:

Very well explained

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14 Apr 2007, 18:51
Thanks fresinha12! The easiest way so far......

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15 Apr 2007, 11:35
How do we know they are 3-4-5 triangles?

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15 Apr 2007, 23:06
Tuneman wrote:
How do we know they are 3-4-5 triangles?

Assuming it to be 3-4-5 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.

15-20-25 (luckily it was 3-4-5 triangle)
_________________

"Education is what remains when one has forgotten everything he learned in school."

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16 Apr 2007, 14:29
Where did you get XY=12Z ? Is there an theorm or eqn outthere. Please shed some light.

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16 Apr 2007, 14:43
Swagatalakshmi wrote:
Tuneman wrote:
How do we know they are 3-4-5 triangles?

Assuming it to be 3-4-5 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.

15-20-25 (luckily it was 3-4-5 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 4-5 minute problem

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17 Apr 2007, 01:49
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?...

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17 Apr 2007, 12:34
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

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17 Apr 2007, 14:18
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

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29 Apr 2007, 01:47
[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 25-9. 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) ?

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29 Apr 2007, 09:09
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

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29 Apr 2007, 10:12
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...

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29 Apr 2007, 10:12
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# In the diagram to the right, triangle PQR has a right angle

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