In the figure above, square AFGE is inside square ABCD such

Question

In the figure below, square AFGE is inside square ABCD such that G is on arc BD, which is centered at C. If DC=8, what is the area of shaded region DEG

A: \(64\sqrt{2} - 32 - 16\pi\)

B: \(8[4\sqrt{2} - (\pi + 2)]\)

C: \(64 - 32\pi\)

D: \(64 - 16\pi\)

E: \(64\sqrt{2} - 8\pi\)

eshan429 · Answer

Area of the square ABCD = 64 Area of sector CBD = \frac{\pi * 8^2}{4} = 16\pi 1.jpg In the diagram above, GC is the radius of arc BD. So, GC = 8 For square GHCI, GC is the diagonal. Therefore, GH = GI = 4\sqrt{2} which makes FG = GE = 8 - 4\sqrt{2} = 4(2 - \sqrt{2}) Now, as the diagonal AC divides both the square ABCD and the sector CBD into equal halves, Area of region DAG = Area of region BAG Also, the same diagonal divides the small square AEGF into two congruent triangles. So, Area of region DEG = Area of region BFG Now, Area of region DAG + Area of region BAG = Area of the square ABCD - Area of sector CBD - Area of square AEGF Area of region DAG = \frac{64 - 16\pi - ^2}{2} = 32 - 8\pi - \frac{16(4 + 2 - 4\sqrt{2})}{2} = 8(4 - \pi - 6 + 4\sqrt{2}) = 8 Hence the answer is B. Hope it's clear.

mikemcgarry · Answer

Dear PareshGmat, That's a great question. I think it is very much like something the GMAT could ask on one of the harder questions. Here's a solution. If we take the big square ABCD (Area = 64) and subtract both the quarter circle and the little square AFGE, then divide the remaining area in half, we will get the shaded area. Very elegant. Of course, big square ABCD = 64. Quarter circle = 16\pi What about the little square? square, circular arc, smaller square.JPG Draw diagonal AC, which has a length of AC = 8sqrt(2) . Of course, GC is just the radius of the circle: GC = 8 Well, AG = AC - GC = 8sqrt(2) - 8 = 8(sqrt(2) - 1) That's the diagonal of the square AFGE. As a shortcut, we can find the area of any square by squaring the diagonal, and then dividing by two. Area of AFGE = \frac{(AG)^2}{2} Area of AFGE = \frac{(8(sqrt(2) - 1))^2}{2} Area of AFGE = 32(sqrt(2) - 1)^2 Area of AFGE = 32(2 - 2sqrt(2) + 1) Area of AFGE = 96 - 64sqrt(2) Twice shaded region = (Square ABCD) - (quarter circle) - (Square AFGE) Twice shaded region = 64 - 16\pi - (96 - 64sqrt(2)) Twice shaded region = 64sqrt(2) - 32 - 16\pi Divide by 2 Shaded region = 32sqrt(2) - 16 - 8\pi Factor out an 8 Shaded region = 8(4sqrt(2) - 2 - \pi) Shaded region = 8(4sqrt(2) - (2 + \pi)) Answer = (B) Great problem! Mike :-)

Bunuel · Answer

Similar question to practice: in-the-figure-above-square-afge-is-inside-square-abcd-such-127345.html

s1d1 · Answer

mikemcgarry

i have a question. did you assume that Diagonal AC passes through G or is it implied (sorry if you find this question silly)

mikemcgarry · Answer

Dear s1d1 , I'm happy to respond. :-)

That's not an assumption: that's a matter of geometric deduction. Think about it this way. Suppose you start at point A and make a 45 degree angle that bisects the right angle DAB. This 45 degree line would have to pass through G and it would have to pass through C. Both of those opposite points have to be on the same 45 degree line. Does all this make sense? Mike :-)

s1d1 · Answer

mikemcgarry

thank you so much for the explanation!

law258 · Answer

B is definitely the correct answer.

(1)   Quarter Circle Area  
1/4(pi)(8^2) = (64pi)/4 = 16pi

(2)   Area Square ABCD  
s^2=8^2=64

(3)   Area Triangle AFGE   (found by creating a triangle within quarter circle, using that it is 45-45-90 & that the hypotenuse will be 8 to find the remaining side values. They will be 4sqrt(2), so value of side of square AFGE is 8-4sqrt(2)).
(8-4sqrt(2))^2 --> simplify however you'd like

Area will be  /2

Fdambro294 · Answer

Answered the same way, thanks Mike!

because the Square is a SUB-Type of Rhombus, we can use the Same way to find the AREA of a Rhombus = (Diagonal 1) * ( Diagonal 2) / 2

but in a Square, the Diagonals are EQUAL, so just Square the Diagonal and divide by 2 to find the Area of that Square.

In the figure above, square AFGE is inside square ABCD such

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In the figure above, square AFGE is inside square ABCD such

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