What is the area of the shaded figure?

A it is. Completing the triangle having 45'45'90 angles.

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)

Regards,

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

Answer is A

Hope it helps
Cheers
J

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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