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In PQS above, if PQ =3 and PS = 4, then PR = (A) 9/4 (B)

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Director
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In PQS above, if PQ =3 and PS = 4, then PR = (A) 9/4 (B) [#permalink]

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New post 02 Aug 2008, 18:57
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

20. In ΔPQS above, if PQ =3 and PS = 4, then PR =
(A) 9/4
(B) 12/5
(C) 16/5
(D) 15/4
(E) 20/3
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Last edited by rao on 02 Aug 2008, 22:43, edited 1 time in total.

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Re: geometery PS [#permalink]

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New post 02 Aug 2008, 19:22
looks like you missed the last part of the question..

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Re: Tricky geometery [#permalink]

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New post 03 Aug 2008, 04:58
B.

Let angle Q = x. Then angle S=90-x
Also QS=5

Now triangle QRP and QPS are similar triangles because Angle RQP=PQS=x and angle RPQ=PQS=90-x and QRP=QPS=90

so you have PS/QS = PM/PQ
4/5=PM/3 => PM =12/5

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Re: Tricky geometery [#permalink]

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New post 03 Aug 2008, 06:43
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THE ANSWER IS B.

THERE IS A FORMULA FOR THIS.

PQ*PS=PR*QS (ALTITUDE*HYPOTENUSE)

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Director
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Re: Tricky geometery [#permalink]

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New post 03 Aug 2008, 07:28
bhushangiri wrote:
B.

Let angle Q = x. Then angle S=90-x
Also QS=5

Now triangle QRP and QPS are similar triangles because Angle RQP=PQS=x and angle RPQ=PQS=90-x and QRP=QPS=90

so you have PS/QS = PM/PQ
4/5=PM/3 => PM =12/5



For some reason OA is E. But I think its wrong. I compared the similar triangles PQS and PRS and got the same result (B).

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Re: Tricky geometery [#permalink]

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New post 03 Aug 2008, 07:30
prateek11587 wrote:
THE ANSWER IS B.

THERE IS A FORMULA FOR THIS.

PQ*PS=PR*QS (ALTITUDE*HYPOTENUSE)


This is giving 15/4 i.e D

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Re: Tricky geometery [#permalink]

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New post 03 Aug 2008, 07:46
method 1 : similar traingle approach : already explained above

method 2 : area of triangle = 1/2 * base * height
1/2 * 3 * 4 = 1/2 * QS * PR
12 = 5 * PR
PR = 12/5

method 3 : equation of line PR = x/3 + y/4 - 1 = 0
4x+3y-12=0
PR = length of perpendiculare from P (0,0) = |0*4 + 0*3 -12 | / sqrt(3^2 + 4^2)
= 12/5

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Re: Tricky geometery [#permalink]

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New post 03 Aug 2008, 07:54
If the OA is E, then there is something wrong wtih the OA or with the question.

If you think of the hypotnuse QS as the base, then QS * PR * 0.5 = area of the triangle.

Also QP * PS * 0.5 = Area of the triangle.

So

QP * PS * 0.5 = QS * PR * 0.5

3 * 4 * 0.5 = 5 * PR * 0.5

6 = 5/2 * PR

6 * 2/5 = PR = 12/5
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Re: Tricky geometery [#permalink]

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New post 03 Aug 2008, 07:57
PQ = 3
PS = 4
QS = 5

3 * 4 = PR * 5

12 = PR * 5
12/5 = PR

I think you must have gotten mixed up with which numbers go where.

rao_1857 wrote:
prateek11587 wrote:
THE ANSWER IS B.

THERE IS A FORMULA FOR THIS.

PQ*PS=PR*QS (ALTITUDE*HYPOTENUSE)


This is giving 15/4 i.e D

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Re: Tricky geometery [#permalink]

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New post 06 Aug 2008, 11:48
How come PQS and PRS similar triangle??

Can any one explain.
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Re: Tricky geometery   [#permalink] 06 Aug 2008, 11:48
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In PQS above, if PQ =3 and PS = 4, then PR = (A) 9/4 (B)

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