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ABCD is a rectangle inscribed in a circle. If the length of AB is thr

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annmary
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ABCD is a rectangle inscribed in a circle. If the length of AB is thr [#permalink] New post   17 Apr 2011, 00:04
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ABCD is a rectangle inscribed in a circle.
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D

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144144
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Re: ABCD is a rectangle inscribed in a circle. If the length of AB is thr [#permalink] New post   19 Apr 2011, 05:47
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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 inscribed in a circle. If the length of AB is thr [#permalink] New post   19 Jul 2015, 00:34
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annmary wrote:
ABCD is a rectangle inscribed in a circle.


Rectangle \(ABCD\) can be broken down into 2 congruent triangles, \(\triangle\)\(ABC\) & \(\triangle\)\(CDA\).

area of rectangle \(ABCD\) = 2 * area of \(\triangle\)\(ABC\)

Let's draw a perpendicular from B to AC and denote the length by h.

Also, \(AC = 2r = 2\)

Now, area of \(\triangle\)\(ABC\) \(= (1/2)*(2)*(h) = h\)

area of rectangle \(ABCD = 2h\)

Now maximum value of h can be r and minimum can be tending towards 0.

So, area of rectangle will vary between \(0\) and \(2\).

\(Answer D\)
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Re: ABCD is a rectangle inscribed in a circle. If the length of AB is thr [#permalink] New post   17 Apr 2011, 00:36
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annmary wrote:
hello
i cant solve this geometry question :shock: , plz help!
tanx :oops:


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 inscribed in a circle. If the length of AB is thr [#permalink] New post   17 Apr 2011, 14:20
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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 inscribed in a circle. If the length of AB is thr [#permalink] New post   01 May 2011, 20:10
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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 inscribed in a circle. If the length of AB is thr [#permalink] New post   01 May 2011, 22:29
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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 inscribed in a circle. If the length of AB is thr [#permalink] New post   17 Apr 2011, 06:12
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thank you fluke
your solution is very well.
+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 inscribed in a circle. If the length of AB is thr [#permalink] New post   19 Apr 2011, 06:07
thanks old friend.
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Re: ABCD is a rectangle inscribed in a circle. If the length of AB is thr [#permalink] New post   17 Oct 2015, 00:27
15 seconds solution
the max area of a rectangle could be the area of squire that can be inscribed in circle with vertices on edges of circle , hence max area could be 2*2 / 2 as 2 is the max length of dignol of squire, hence D
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Re: ABCD is a rectangle inscribed in a circle. If the length of AB is thr [#permalink] New post   05 May 2017, 07:26
Trigonometric Approach

Since the radius is equal to 1, the diagonal of rectangle is equal to 2r = 2

Consider angle CAD = x

Hence CD= 2 sin x and AD= 2 cos x

Area of the rectangle ABCD= AD.CD = (2sinx)(2cosx)= 2(2sinx.cosx) = 2 sin 2x

Now sin2x has maximum value 1 & minimum value 0

Hence 0 < area (ABCD) < 2

Therefore OPTION D
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Re: ABCD is a rectangle inscribed in a circle. If the length of AB is thr [#permalink] New post   26 Jul 2021, 08:33
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