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# What is the area of the shaded figure?

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24 Aug 2009, 09:28
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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, 12:51, edited 1 time in total.

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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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24 Aug 2009, 12:36

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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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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24 Aug 2009, 22:03
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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02 May 2011, 00:15
A it is. Completing the triangle having 45'45'90 angles.
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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)
$$AE = AD-x=\sqrt{3}/2$$
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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14 Nov 2013, 12:34
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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 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
Cheers
J

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07 Feb 2015, 06:45
Hello from the GMAT Club BumpBot!

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07 Nov 2015, 12:34
woah! a tricky and difficult one.

draw a line to make a 45-45-90 triangle and a 30-60-90 triangle.
hypothenuse of the 45-45-90 triangle is sqrt 2 + [(sqrt2)/2)]
knowing the properties of a 45-45-90 triangle, the sides must have the ratio of x-x-x(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 30-60-90, 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 30-60-90 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 45-45-90 triangle and 30-60-90 triangle and is equal to
9/8 + sqrt 3/8 = (9+ sqrt 3)/8
A.

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Re: What is the area of the shaded figure?   [#permalink] 07 Nov 2015, 12:34
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