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In the rectangular coordinate system above, if OP < PQ, is

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Re: In the rectangular coordinate system above, if OP < PQ, is  [#permalink]

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New post 07 Aug 2018, 06:45
Bunuel wrote:
Image
In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?

(1) The coordinates of point P are (6,8). Now, if OP were equal to PQ then the triangle OPQ would be isosceles and OS would be equal to SQ and the area would be: 1/2*base*height=1/2(OS+SQ)*PS=1/2*(6+6)*8=48. Since OP<PQ then OS<SQ and the base OQ is more than 12, which makes the area more than 48. Sufficient.

(2) The coordinates of point Q are (13,0) --> we know the length of the base (OQ=13) but know nothing about the height (PS), which may be 1 or 100, so the area may or may not be more than 48. Not sufficient.

Answer: A.

Hope it's clear.

Attachment:
Triangle.png


Hi Bunuel,

how do we know that OS<SQ? I mean how do we deduce that from OP<OQ
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Re: In the rectangular coordinate system above, if OP < PQ, is  [#permalink]

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New post 30 Aug 2018, 12:41
eybrj2 wrote:
Image
In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?

(1) The coordinates of point P are (6,8)

(2) The coordinates of point Q are (13,0)

Attachment:
The attachment Triangle.png is no longer available


\(?\,\,\,:\,\,\,{S_\Delta }\,\,\mathop > \limits^? \,\,48\)

\(\left( 1 \right)\,\,{S_\Delta }\,\,\mathop > \limits^{{\text{see}}\,\,{\text{our}}\,\,{\text{figure !}}} \,\,{S_{\Delta OPQ\,dash}} = \frac{{\left( {6 + 6} \right) \cdot 8}}{2} = 48\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)

\(\left( 2 \right)\,\,{\text{Geometric}}\,\,{\text{Bifurcation}}\)

Please note that:

> The first figure (on the right) is "the same" as the one we used in our first statement (see our figure), in which we answered <YES>
> The second figure (on the right-below) shows that the area-focus may be as small as we which (but still positive, of course) , hence <NO>

This solution follows the notations and rationale taught in the GMATH method.
Attachments

30Ago18_7q.gif
30Ago18_7q.gif [ 18.03 KiB | Viewed 279 times ]


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Re: In the rectangular coordinate system above, if OP < PQ, is  [#permalink]

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New post 09 Dec 2018, 04:44
Its such a good trap question !!!

How one can be aware of such trap questions ?
Is practice and seeing the pattern is the solution ?
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Re: In the rectangular coordinate system above, if OP < PQ, is  [#permalink]

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New post 09 Dec 2018, 04:55
ENEM wrote:
Bunuel wrote:
Image
In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?

(1) The coordinates of point P are (6,8). Now, if OP were equal to PQ then the triangle OPQ would be isosceles and OS would be equal to SQ and the area would be: 1/2*base*height=1/2(OS+SQ)*PS=1/2*(6+6)*8=48. Since OP<PQ then OS<SQ and the base OQ is more than 12, which makes the area more than 48. Sufficient.

(2) The coordinates of point Q are (13,0) --> we know the length of the base (OQ=13) but know nothing about the height (PS), which may be 1 or 100, so the area may or may not be more than 48. Not sufficient.

Answer: A.

Hope it's clear.

Attachment:
Triangle.png


Hi Bunuel,

how do we know that OS<SQ? I mean how do we deduce that from OP<OQ



From the question we have OP < OQ .....(a)
also from Pythagoras theorem we have \(OP^2\) = \(PS^2\) +\(OS^2\) and \(OQ^2\) = \(PS^2\) +\(SQ^2\)
Putting these value in equation a we have
\(PS^2\) +\(OS^2\) < \(PS^2\) +\(SQ^2\)
=> OS<SQ

Hope its clear now
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Re: In the rectangular coordinate system above, if OP < PQ, is   [#permalink] 09 Dec 2018, 04:55

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