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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # In the rectangular coordinate system above, if OP < PQ, is

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

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Bunuel wrote: 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.

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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eybrj2 wrote: 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

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

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

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

> 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 [ 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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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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ENEM wrote:
Bunuel wrote: 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.

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

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