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Re: What is the area of the shaded figure? [#permalink]

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04 Jun 2012, 05:43

2

This post received 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)=\((9+\sqrt{3})/8\)

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 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.

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 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. _________________

GMAT offended me. Now, its my turn! Will do anything for Kudos! Please feel free to give one.

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 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

Re: What is the area of the shaded figure? [#permalink]

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07 Feb 2015, 06:45

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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 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.

gmatclubot

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

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