How to do this math problem [PIC included]

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<p>Couldn't figure this one out...answer is D.</p>

<p>I remember doing this problem. There are two ways to do it- you can look and decipher it from analyzing the picture, it is TO SCALE. Also, you can use right triangles.</p>

<p>Yea, I guessed that it was D. But how would you go about using right triangles? I attempted that method while doing the problem but couldn’t get the answer.</p>

<p>The answer is actually D.</p>

<p>The key is to form a square by using the 4 centres of the large circles with sides 2.</p>

<p>The next step is to extrapolate a diagonal from the bottom leftmost centre to the topmost right centre.</p>

<p>Label the radius of the small circle x.</p>

<p>Therefore length of diagonal = 1+x+x+1= 2x+2</p>

<p>using d squared/ 2 = area Since We know that the area of the overall square is 4r= 4 * 1= 4</p>

<p>then d^2/2= 4</p>

<p>d^2/2 = (2x+2)^2/ 2= 4</p>

<p>(2x+2)^2= 8
(2x+2) = root 8
2x = (root 8) -2</p>

<p>x= (root 8) -2/ 2= 0.414</p>

<p>D) correct answer</p>

<p>Where did you get this question? I have never seen one like this before. It had me for a while</p>

<p>Another way would be to take a portion of the problem, specifically 1/4 of it such that you have one of the large circles inscribed within a square and 1/4 of the minicircle. This gives the square a side length 2. The diagonal across this square is 2rad2. Take half of that, which is rad2. This distance, rad2, covers the radius of the large circle and the radius of the small circle. Therefore, radius of the small circle = rad2 - 1.</p>

<p>Here’s another way:</p>

<p>Draw a line from the center of the small circle to the centers of two adjacent larger circles. Now draw a line connecting the centers of those two large circles. You know have a 45-45-90 triangle. Let r be the radius of the small circle. Since the hypotenuse (1+1=2) of the triangle is sqrt (2) times the other side (1+r)…</p>

<p>2 = sqrt(2) * (1 + r)
2/sqrt(2) = 1 + r
r = 2/sqrt(2) - 1
r = sqrt(2) - 1</p>