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What is the area of the shaded figure? [#permalink]
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24 Aug 2009, 09:28
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What is the area of the shaded figure?
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Last edited by Bunuel on 14 Nov 2013, 12:51, edited 1 time in total.
Added the OA.



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Re: What is the area of the shaded figure? [#permalink]
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24 Aug 2009, 10:44
May be most likely this problem is solved through finding squares of two right triangles.
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Re: What is the area of the shaded figure? [#permalink]
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24 Aug 2009, 12:36
My answer is A.
I split the graph in two right triangles. One triangle with angles 454590, another with angles 603090.
In triangle, 454590, the length of the sides are in ratio 1:1:2^(1/2) In triangle, 306090, the length of the sides are in ratio 1:3^(1/2):2. Using this I figured out area of each triangle and then the total.



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Re: What is the area of the shaded figure? [#permalink]
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24 Aug 2009, 17:03
A
lets split the area into 2 triangles
area of larger tr =
as its a isosceles tr , we can find the side of the tr 2 x^2 = hypt^2
we get side as 3/2
area of smaller tr : one angle is 60 , other angle is 90 and last one is 30 . so we know the base and height .
so we cn find the area of smaller tr .



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Re: What is the area of the shaded figure? [#permalink]
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24 Aug 2009, 22:03
LenaA wrote: My answer is A.
I split the graph in two right triangles. One triangle with angles 454590, another with angles 603090.
In triangle, 454590, the length of the sides are in ratio 1:1:2^(1/2) In triangle, 306090, the length of the sides are in ratio 1:3^(1/2):2. Using this I figured out area of each triangle and then the total. Quite Right LenaA! A it is.
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Re: What is the area of the shaded figure? [#permalink]
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02 May 2011, 00:15
A it is. Completing the triangle having 45'45'90 angles.
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Re: What is the area of the shaded figure? [#permalink]
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04 Jun 2012, 05:43
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Hi, The figure can be split into two triangles: tri(ABE) & tri(DCE) Assuming \(CE=DE=x\)(since angle C = 45) \(CD = \sqrt{2}x = \sqrt{2} + \sqrt{2}/2\) or \(x =3/2\) \(area(DCE) = (1/2)*x^2=1/2*3/2*3/2=9/8\) Now, in tri(ABE) \(AE = ADx=\sqrt{3}/2\) ratio of sides opposite to angles \(30:60:90=1:\sqrt{3}:2\) Thus, BE=1/2 \(area(ABE)=1/2*AE*BE=1/2*\sqrt{3}/2*1/2=\sqrt{3}/8\) area of the figure = area(ABE)+area(DCE)=\((9+\sqrt{3})/8\) (A) Regards,
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Re: What is the area of the shaded figure? [#permalink]
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14 Nov 2013, 12:34
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Re: What is the area of the shaded figure? [#permalink]
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31 Jan 2014, 09:20
mendelay wrote: What is the area of the shaded figure? Nice question we need to know both triangles thr 454590 and 306090, these are a MUST Now the first triangle is the 454590 so the hipothenuse is sqrt (2) + sqrt (2)/2 Therefore the side is 3/2 and area is 9/8 Now for the second triangle we need to subtract 3/2 from (3+sqrt (3)/2) and we get that the side opposite to 60 degrees is sqrt 3/2). Therefore side is 1/2 and area is sqrt (3) /8 So total area is sum of both or 9+ sqrt (3) / 8 Answer is A Hope it helps Cheers J



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Re: What is the area of the shaded figure? [#permalink]
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Re: What is the area of the shaded figure? [#permalink]
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07 Nov 2015, 12:34
woah! a tricky and difficult one.
draw a line to make a 454590 triangle and a 306090 triangle. hypothenuse of the 454590 triangle is sqrt 2 + [(sqrt2)/2)] knowing the properties of a 454590 triangle, the sides must have the ratio of xxx(sqrt 2) or x(sqrt2) = sqrt 2 + [(sqrt2)/2)]. we can conclude that the legs are 3/2. Area of the triangle is thus 9/8 now we know that the exterior angle is 120, thus, interior is a 60 angle. after drawing the above mentioned line, we get the upper triangle 306090, and the proportion of sides x x sqrt 3  2x since we know the length of the longer line (3+ sqrt3)/2, we can find the leg of the 306090 triangle, which is sqrt3/2. thus we can find the leg opposite the 30 angle, which is 1/2 knowing the legs, we can find the area: sqrt 3/2 * 1/2 * 1/2 = sqrt 3 / 8
the area of the figure is the area of the 454590 triangle and 306090 triangle and is equal to 9/8 + sqrt 3/8 = (9+ sqrt 3)/8 A.




Re: What is the area of the shaded figure?
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