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Senior Manager
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What is the area of the shaded figure? [#permalink]
 24 Aug 2009, 09:28
Question Stats:
70% (01:27) correct
30% (00:00) wrong based on 1 sessions
I gather many of you have seen this in the McH book, but it has not been discussed on the forum yet.
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on681353-02p01v01.jpg [ 10.11 KiB | Viewed 2192 times ]
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Re: What is the area of the shaded figure? [#permalink]
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]
24 Aug 2009, 12:36
My answer is A.
I split the graph in two right triangles. One triangle with angles 45-45-90, another with angles 60-30-90.
In triangle, 45-45-90, the length of the sides are in ratio 1:1:2^(1/2)
In triangle, 30-60-90, 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]
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]
24 Aug 2009, 22:03
LenaA wrote: My answer is A.
I split the graph in two right triangles. One triangle with angles 45-45-90, another with angles 60-30-90.
In triangle, 45-45-90, the length of the sides are in ratio 1:1:2^(1/2)
In triangle, 30-60-90, 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]
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]
04 Jun 2012, 05:43
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}/2or x =3/2area(DCE) = (1/2)*x^2=1/2*3/2*3/2=9/8Now, in tri(ABE) AE = AD-x=\sqrt{3}/2ratio of sides opposite to angles 30:60:90=1:\sqrt{3}:2Thus, BE=1/2 area(ABE)=1/2*AE*BE=1/2*\sqrt{3}/2*1/2=\sqrt{3}/8area 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]
04 Jun 2012, 05:43
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