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# What is the area of the shaded figure?

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What is the area of the shaded figure? [#permalink]  24 Aug 2009, 08:28
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What is the area of the shaded figure?
[Reveal] Spoiler: OA

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Last edited by Bunuel on 14 Nov 2013, 11:51, edited 1 time in total.
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Re: What is the area of the shaded figure? [#permalink]  24 Aug 2009, 09:44
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, 11:36

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, 16:03
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, 21:03
LenaA wrote:

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]  01 May 2011, 23:15
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, 04:43
2
KUDOS
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)
$$AE = AD-x=\sqrt{3}/2$$
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)

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Re: What is the area of the shaded figure? [#permalink]  14 Nov 2013, 11:34
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Re: What is the area of the shaded figure? [#permalink]  31 Jan 2014, 08:20
mendelay wrote:
What is the area of the shaded figure?

Nice question we need to know both triangles thr 45-45-90 and 30-60-90, these are a MUST

Now the first triangle is the 45-45-90 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

Hope it helps
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Re: What is the area of the shaded figure? [#permalink]  07 Feb 2015, 05:45
Hello from the GMAT Club BumpBot!

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Re: What is the area of the shaded figure?   [#permalink] 07 Feb 2015, 05:45
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