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# A triangle

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Manager
Joined: 25 Feb 2008
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19 Apr 2008, 01:45
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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A tunnel.doc [30 KiB]

SVP
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19 Apr 2008, 02:04
Capthan wrote:

1. QS = 5, (pitago)
2. triangles QPS and PRS are similar. So PS/QS = PR/QP or 4/5 = PR/3 or PR=12/5
3. B

Anyway, due to my hurry up, correct me if I am wrong!
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Manager
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19 Apr 2008, 05:23
Capthan wrote:

(B) for me.
Can propose another way of solving: area of PQR = 1/2*QP*PS = 1/2*PR*QS
PR*QS = 12, while QS = 5
PR = 12/5
Senior Manager
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02 May 2008, 15:28
Sunchaser20 wrote:
Capthan wrote:

(B) for me.
Can propose another way of solving: area of PQR = 1/2*QP*PS = 1/2*PR*QS
PR*QS = 12, while QS = 5
PR = 12/5

Can someone explain the solution again or may be elaborate on the solutions already given ? I didn't understand any of the above solutions
Manager
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02 May 2008, 16:34
If PQ=3 and PS=4, then QS=5.
The key is to figure out what QR and RS are. QR+RS=QS=5.

QR^2 + PR^2 = 3^2 = 9
RS^2 + PR^2 = 4^2 = 16 ----->sub in RS = 5-QR-------->(5-QR)^2 + PR^2 =16

9 - QR^2 = 16 - (5-QR)^2
9 - QR^2 = 16 - 25 +10QR - QR^2
10QR = 18
QR = 9/5

QR^2 +PR^2 = 9
(9/5)^2 + PR^2 = 9
PR^2 = 144 / 25
PR = 12/5

So the answer is B. Hope this helps. Anyone is welcome to suggest an easier way.
Manager
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04 May 2008, 09:38
Create a mirror image of the triangle along QS and you have a rectangle. Using Pythagorean theorem, QS = 5 (3-4-5). Both line QS and the perpendicular imaginary QS will intersect almost at mid point 2.5

I use elimination process

A. $$\frac{9}{4}$$ = 2.25 (not as close as 2.4)
B. $$\frac{12}{5}$$ = 2.40 (closest to 2.5)
C. $$\frac{16}{5}$$ = 3.20 (too far from 2.5 - eliminate)
D. $$\frac{15}{4}$$ = 3.75 (too far from 2.5 - eliminate)
E. $$\frac{20}{3}$$ = 6.67 (too far from 2.5 - eliminate)

Ans: B
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Jimmy Low, Frankfurt, Germany
Blog: http://mytrainmaster.wordpress.com
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Re: A triangle   [#permalink] 04 May 2008, 09:38
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