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If the area of the polygon ABCDE is 90 square inches, then

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kevincan
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If the area of the polygon ABCDE is 90 square inches, then [#permalink] New post   02 Aug 2006, 12:07
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If the area of the polygon ABCDE is 90 square inches, then the length of line segment AB in inches is closest to

(A) 10 (B) 11 (C) 12 (D) 13 (E) 14



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Re: If the area of the polygon ABCDE is 90 square inches, then [#permalink] New post   02 Aug 2006, 15:05
IMO E.
Let |AB| = 2x and |AD|=y
y^2 = (x)^2+(x)^2

90 = 2xy-(1/2)*y^2
90 = 2sqrt(2)x^2-x^2
90 = 2.8x^2-x^2
90 = 1.8x^2
50 = x^2
x = sqrt(50) = 7.xxxx = 7
thus 2x = 14
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Re: If the area of the polygon ABCDE is 90 square inches, then [#permalink] New post   02 Aug 2006, 16:01
I think its E.

Let AB = y
AD = x
we have AD = DE = EC = CB = x and AB = CD = y

Area of polygon ABCED = xy-x^2/2 = 90............eq1
From triangle DEC we have y = x * SQRT(2)
i.e x = y/SQRT(2).......eq2

substitute eq2 in eq1

y^2/SQRT(2) - y^2/4 = 90

y^2 = [90 * 4 * SQRT(2)] / [4-SQRT(2)]
y^2 = 90 * 4 * SQRT(2) * [4-SQRT(2)] / 14
y^2 = 90 * 4 * [4*SQRT(2) + 2] / 14
y^2 = 90 * 2 * [4*SQRT(2) + 2] / 7
y^2 = 13 * 2 * [4*SQRT(2) + 2]
y^2 = 13 * 4 * [2*SQRT(2) + 1]
y^2 = 52 * [2*1.4 + 1]
y^2 = 52 * [2.8 + 1]
y^2 = 52 * [2.8 + 1]
y^2 = 52 * 3.8
y^2 = 200 approx....
y = 14 approx.....
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Re: If the area of the polygon ABCDE is 90 square inches, then [#permalink] New post   02 Aug 2006, 20:16
ps_dahiya wrote:
I think its E.

Let AB = y
AD = x
we have AD = DE = EC = CB = x and AB = CD = y

Area of polygon ABCED = xy-x^2/2 = 90............eq1
From triangle DEC we have y = x * SQRT(2)
i.e x = y/SQRT(2).......eq2

substitute eq2 in eq1

y^2/SQRT(2) - y^2/4 = 90

y^2 = [90 * 4 * SQRT(2)] / [4-SQRT(2)]
y^2 = 90 * 4 * SQRT(2) * [4-SQRT(2)] / 14
y^2 = 90 * 4 * [4*SQRT(2) + 2] / 14
y^2 = 90 * 2 * [4*SQRT(2) + 2] / 7
y^2 = 13 * 2 * [4*SQRT(2) + 2]
y^2 = 13 * 4 * [2*SQRT(2) + 1]
y^2 = 52 * [2*1.4 + 1]
y^2 = 52 * [2.8 + 1]
y^2 = 52 * [2.8 + 1]
y^2 = 52 * 3.8
y^2 = 200 approx....
y = 14 approx.....


A simpler simplification!

AB = y; AD = x. Per the equations above, y = x√2

and √2(x^^2) - (1/2)(x^^2) = 90;
This simplifies to x^^2 = 90/(√2 - 1/2)
or x^^2 = 180/(2√2-1). Now we know √2 = 1.414
So the expressions reduces to x^^2 = 180/1.828 ~ 100
So x ~ 10 and y ~ 10√2 = 10*1.414 = 14.14
So E.
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Re: If the area of the polygon ABCDE is 90 square inches, then [#permalink] New post   02 Aug 2006, 22:37
E

Let Side AB = DC = x
and side AD = DE = y

Area of rectangle - area of right traingle = area of polygon

xy - 1/2 *y^2 = 90

Also x^2 = 2y^2
Hence x= ysqr(2)

sqr(2)x^2/2 - x^2/4 = 90
Sqr(2) x^2 = 360
x = Sqr(180sqr(2))

x ~ 10

y ~14
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Re: If the area of the polygon ABCDE is 90 square inches, then [#permalink] New post   03 Aug 2006, 01:12
Good work everybody!

Also, we can see that A=((2sqrt(2)-1)/2)s^2, where s=length of BC, CD, DE...

So A is approximately 0.9s^2. For A to be 90, s should be approximately 10 and AB approximately 14



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Re: If the area of the polygon ABCDE is 90 square inches, then [#permalink]
03 Aug 2006, 01:12
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