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I'm trying to solve the following problem :

$ABCD$ is a square of side $4$ units. Find the area of the shaded region as shown in the figure.

Figure

The area of square is obviously $16$ , but what after that?

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Do you know a formula for the area of a parallelogram? If so, you've got two of them, whose intersection is a little square, so then you just have to find a side of the square. Alternately, there are 4 identical triangles that are not shaded ---- if you can figure out the height of a triangle, you can get the number you want. –  Gerry Myerson Sep 9 '13 at 10:09
    
@GerryMyerson Is the diagonal of the little square 1 ? ( I don't know how , I'm just guessing ) If so then height of each triangle is 1.5 units , is that right? I don't know how to find base and height of parallelograms. –  A Googler Sep 9 '13 at 10:14
    
But the base and height of the parallelograms are trivial. The base, for example, can be taken to be 1. As for the little square, I don't know what the diagonal is, but I wouldn't try guessing. –  Gerry Myerson Sep 9 '13 at 10:18
    
@GerryMyerson Oh ! so you're talking about the shaded parallelograms ! Then the base is 1 and height is 4 so the area is 4 , for the other parallelogram area is also 4 but now I have to subtract the area of the little square , how can I do that? –  A Googler Sep 9 '13 at 10:24
    
That's a little harder. I got you started --- you take it from here! –  Gerry Myerson Sep 9 '13 at 10:27
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3 Answers

The area of the outer square is indeed $16$. The area of the two right triangles on either side of a shaded bar are each $(1/2) (3)(4)=6$, so that the area of one shaded bar is $16-12=4$. The area of the shaded region is thus $4+4-$ the area of the square center. The side of that square is the width of a shaded bar, which is the sine of the right triangle $4/5$, so that the shaded area is

$$4+4-\left ( \frac{4}{5}\right)^2 = 8 - \frac{16}{25} = \frac{184}{25}$$

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A, D, and C are corners of a $4\times4$ square, right? So the area of the triangle ADC is 8, not 6, and what does it have to do with the shaded bar? –  Gerry Myerson Sep 9 '13 at 10:21
    
@GerryMyerson: I mislabeled. The triangles I care about are $4 \times 3$. –  Ron Gordon Sep 9 '13 at 10:25
    
How do you know that sine is 4/5? What is the angle? –  A Googler Sep 9 '13 at 10:38
    
@AGoogler: the right triangles I reference are 3-4-5 triangles. –  Ron Gordon Sep 9 '13 at 10:43
    
Got it, Thank you. –  A Googler Sep 9 '13 at 10:51
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White triangles are 3:4:5 triangle scaled down so that the longest side is $3$. Therefore, each one has area $\left(\frac{3}{5}\right)^2 \times \frac 12 \times 3 \times 4 = \frac{54}{25}$. The area of the whole square is $16$. So the area of the shaded region is $\frac{400}{25} - \frac{216}{25} = \frac{184}{25}$.

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Cool ! But can you expand on why you multiplied by (3/5)^2? –  A Googler Sep 9 '13 at 10:35
    
The triangle is scaled by the factor of $\frac 35$, so each side that is used in the computation of area ($3$ and $4$) has to be multiplied by $\frac 35$. –  Tunococ Sep 9 '13 at 23:42
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We can compute the area of the white triangle as follows:

  • There is one black/white boundary from $(0,3)$ to $(4,0)$. Call the equation of this line $f(x)$. We can solve for $$f(x)=3-\frac{3}{4}x.$$

  • There is one black/white boundary from $(1,0)$ to $(4,4)$. Call the equation of this line $g(x)$. We can solve for $$g(x)=\frac{4}{3}x-\frac{4}{3}.$$

  • The $y$-coordinate of the intersection of $f(x)$ and $g(x)$ is the height $h$ of the white triangle. We set $g(x)=f(x)$ and solve for $x$ to obtain the $x$-coordinate of the intersection as $$\frac{52}{25}.$$ We substitute this $x$-coordinate into $f(x)$ to obtain the $y$-coordinate $$h=\frac{36}{25}.$$

The area of the white triangle is $\frac{1}{2} 3h$.

The total area in black is $$4^2-4 \times \frac{3}{2}h=\frac{184}{25}$$ since they have equal area.

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If the height were 2, wouldn't the top of the bottom triangle be at the same height as the bottom of the top triangle? (since the height of the square is 4). Doesn't seem right.... –  Gerry Myerson Sep 9 '13 at 10:23
    
@GerryMyerson Shouldn't the height be the x coordinate of the intersection ? –  A Googler Sep 9 '13 at 10:28
    
I filled in the details; it'd just be some arithmetic problem. This answer matches Ron Gordon's. –  Rebecca Sep 9 '13 at 10:29
    
Googler, I guess Wolfram alpha is broken. Or, you didn't put the two equations in correctly. Hard to tell, from this distance. –  Gerry Myerson Sep 9 '13 at 10:32
    
@GerryMyerson I realized my mistake , you're right. –  A Googler Sep 9 '13 at 10:34
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