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Re: What is the area of the shaded figure? [#permalink]
24 Aug 2009, 11:36
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]
24 Aug 2009, 21:03
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]
04 Jun 2012, 04:43
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]
14 Nov 2013, 11:34
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]
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
Re: What is the area of the shaded figure? [#permalink]
07 Feb 2015, 05:45
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).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: What is the area of the shaded figure? [#permalink]
07 Nov 2015, 11:34
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?
[#permalink]
07 Nov 2015, 11:34
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