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# ABCD is a rectangle ...

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Manager
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ABCD is a rectangle ... [#permalink]  16 Apr 2011, 23:04
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Difficulty:

(N/A)

Question Stats:

40% (03:27) correct 60% (01:00) wrong based on 5 sessions
hello
i cant solve this geometry question , plz help!
tanx
[Reveal] Spoiler: OA

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Re: ABCD is a rectangle ... [#permalink]  16 Apr 2011, 23:36
3
KUDOS
annmary wrote:
hello
i cant solve this geometry question , plz help!
tanx

Sol:
Let l=length
w=wdith
d=diagonal=2
Area=l*w=l*\sqrt{d^2-l^2}

l*\sqrt{2^2-l^2}

l*\sqrt{4-l^2}=Area

Squaring both sides;
l^2*(4-l^2)=(Area)^2
4l^2-l^4=(Area)^2
-l^4+4l^2-(Area)^2=0

Let \hspace{2} l^2=x

-x^2+4x-(Area)^2=0

D=\sqrt{b^2-4*a*c}

D=\sqrt{4^2-4*(-1)*-(Area)^2}

D=\sqrt{16-4*(Area)^2}

In inside of the root must be greater or equal to 0 to provide valid roots of the equation.
16-4*(Area)^2 \ge 0

4*(Area)^2 \le 16

(Area)^2 \le 4

(Area) \le 2

Only I and II are less than 2.

Ans: "D"
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Re: ABCD is a rectangle ... [#permalink]  17 Apr 2011, 05:12
thank you fluke
+5 kudos , but i just can give you +1 kudos
other 4 kudos :+1,+1,+1,+1
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Re: ABCD is a rectangle ... [#permalink]  17 Apr 2011, 13:20
1
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Area cannot be 3.2 because area of circle is pi r^2 i.e. 3.14, so rule out III
Even if the length is very close to the diameter i.e. 1 approx, width has to be 1/100 to make the area .01 - possible
similarly 1.9 is also possible.
=>D
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Re: ABCD is a rectangle ... [#permalink]  19 Apr 2011, 04:47
3
KUDOS
i did it a bit different.

Max size for a rectangle is a square.
and minimum is when the width or length = to 0 (or almost 0)

so lets check the max and than we know that everything below the max is a possibility.

so if we know the radii is 1 - we know that if it was a square its diagonals will be 2 each.
so the area of the rectangle will be < 2*2/2 = 2

so - I, II is below 2, above 0 - than - they are ok.
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Re: ABCD is a rectangle ... [#permalink]  19 Apr 2011, 05:07
thanks old friend.
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Re: ABCD is a rectangle ... [#permalink]  01 May 2011, 19:10
1
KUDOS
area of rectangle = l*b

diagonal is 2
also d (diagonal) = sqrt (l^2 + b^2)
l*b = l * sqrt (2 -l^2)
area is positive
therefore (2-l^2) >0
i.e sqrt2>l
similarly sqrt2>b
therefore area = l.b <sqrt 2.sqrt2 = 1.4 *1.4 = 1.96
therefore D is correct.
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Re: ABCD is a rectangle ... [#permalink]  01 May 2011, 21:29
1
KUDOS
Max area of the rectangle possible when diagonal = diameter = 2
Hence, l^2 + b^2 = 4. Implies l and b < 1.5 each.
Also, area of circle = 3.14 * 1^2 = 3.14. hence options C and E POE.
Now,
0<l <1.5 and 0<b<1.5. Thus area can be 0.01 and 1.9 both.
Hence D.
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Re: ABCD is a rectangle ...   [#permalink] 01 May 2011, 21:29
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# ABCD is a rectangle ...

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