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

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ABCD is a rectangle inscribed in a circle. If the length of AB is thr  [#permalink]

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16 Apr 2011, 23:04
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85% (hard)

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53% (02:15) correct 47% (02:48) wrong based on 151 sessions

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ABCD is a rectangle inscribed in a circle.

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Re: ABCD is a rectangle inscribed in a circle. If the length of AB is thr  [#permalink]

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19 Apr 2011, 04:47
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1
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]

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16 Apr 2011, 23:36
4
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 inscribed in a circle. If the length of AB is thr  [#permalink]

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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 inscribed in a circle. If the length of AB is thr  [#permalink]

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17 Apr 2011, 13:20
1
1
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]

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19 Apr 2011, 05:07
thanks old friend.
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Re: ABCD is a rectangle inscribed in a circle. If the length of AB is thr  [#permalink]

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01 May 2011, 19:10
1
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]

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01 May 2011, 21:29
1
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]

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18 Jul 2015, 23:34
4
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]

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16 Oct 2015, 23:27
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]

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05 May 2017, 06:26
Trigonometric Approach

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

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]

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08 Jun 2018, 03:54
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Re: ABCD is a rectangle inscribed in a circle. If the length of AB is thr &nbs [#permalink] 08 Jun 2018, 03:54
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