help with math problem

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<p><a href=“http://i35.■■■■■■■.com/2r6jf46.jpg[/url]”>http://i35.■■■■■■■.com/2r6jf46.jpg</a></p>

<p>i just made this on paint. heres the queston</p>

<li>In the figure above, a smaller circle is inscribedin a square, which is nscribed in a larger circle. If a point on the figure is chose at random, what is the probability that the point is in the shaded area?</li>
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<p>A- 1/4pi
B 4-pi / 2pi
C pi-1/2pi
D pi-2/pi
E 2/pi</p>

<p>Thanks a lot.</p>

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<p>^ Answer should be B.</p>

<p>Here is the long way:</p>

<p>Assume bigger circle radius is 8
We know the area is 64pi^2 (bigger circle)</p>

<p>Using the same 8 radius, we know that the hypotenuse of 1 small triangle in the square is 8 and the full hypotenuse of the square is 16. Since the diagonal of the square bisect the angle, we know it is a 45-45-90 triangle and can conclude that the side is x rt(2) = 8 rt2.</p>

<p>Since the circle is inscribed in the square we know half the side of the square is the radius of the circle so the radius MUST be 4 rt 2.</p>

<p>Plug into equation</p>

<p>(area of square) - (area of circle) / area of big circle</p>

<p>128 - 32pi / 64 pi</p>

<p>= 4-pi/2pi</p>

<p>Of course, choosing 8 or whatever number is arbitrary. You could've easily put in 1 and then multiplied the fractions by 8/8 as well.</p>

<p>Faster way: </p>

<p>We know that the denominator must be the area of the circle.
We know that the square's area will not have a pi in it.
We know that the small circle's area will have a pi in it.</p>

<p>We can delete A and E because it does not have pi in its numerator which should be the smaller circle's area.</p>

<p>From here, you go two ways depending on how safe you want to play it:</p>

<p>I would see that pi-1 and pi-2 are unlikely because the square is bigger than the circle so it should be square MINUS circle not the other way around. Of course, there are a few exceptions to this but it is still unlikely to happen. I would pick B immediately without any calculation..</p>

<p>However, if you want to play it safe, following this notion, you should check it which should not take more than 10 seconds!</p>

<p>You see that 2pi is on the bottom and this must be the larger circle's area:
Right away you know that if it is the right answer, the radius should be rt 2.</p>

<p>You know 45-45-90 so the side of the square should be 2.</p>

<p>Square's area 2x2 = 4 -- you should know it's the correct answer by now.
You can go even further by see that the smaller circle's area is the length of the square divided by 2 = 1. </p>

<p>^ Hope these two ways helped. In these type of problems, you can get the answer without even attempting to solve it at all.</p>

<p>Total amount of time with fastest way = app 10 seconds.</p>

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<p>wow. awesome explanation. I was trying the first method but I got lost somewhere. Second one just amazing. I never would have thought about the pi or no pi thing and how fast it could be done. Thanks a lot.</p>

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<p>^ where did you get lost?</p>

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<p>This was in a real timed practice test so time was a problem. So with like less than a min left, I tried to plug in 4 for diameter. So with 4pi as my denom, I tried finding squares area. I tried doing it but what I was about to do was pytgagorean backwards. Thats when time ran out. For this kind of problem and method, is there a quicker way of finding the squares area than pythagorean?</p>

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<p>If 4 is the diameter, the bigger circle's area would be 4pi^2. To find the square's area you could make the diameter a pyth triple as well (not 3-4-5 cause then you'd have 2.5 as your radius). </p>

<p>Also, doing pyth backwards isn't that hard. Since it's a square you already know </p>

<p>x^2 + x^2 = 2x^2 = hypotenuse^2</p>

<p>**Edit: I know a lot of SAT Math books tell you not to look at the answer choices first because it may confuse you or something...I know Princeton advocates this with its Joe Bloggs method etc. etc. This is why I don't like Kaplan/Princeton because they seem to give innocuous advice which in actuality, harms your speed significantly. The rule of thumb for HARD math problems involving geometry is to look at the choices first and eliminate accordingly with logic and common sense.</p>

<p>Remember, if your method is convoluted (Example: You start dividing angles and drawing lines everywhere), it's probably not the method the SAT wants you to use.</p>

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<p>Oh right. But after looking at your easier method, I keep doing it subconsciously for this problem. lol. </p>

<p>I actually look at answer choices a lot even for mediums. I usually plug back in answers or find some way of getting the right one (mostly). But I think I'll try it specifically for geometry now. That and probability always get me annoyed. </p>

<p>And yeah. Ill keep that in mind. Thanks a lot again.</p>

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<p>

Round pi to 3 and evaluate B, C, and D:
B 4-pi / 2pi = 1/6
C pi-1/2pi = 1/3
D pi-2/pi = 1/3</p>

<p>Shaded area is definitely less then 1/3 of the larger circle area.
Thus B.</p>

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<p>gcf- lmao. i did that exact thing too when i couldnt get it since it didnt specify not to scale. I was stuck between A and B and put A cuz I couldnt tell if it was 1/6 or 1/12</p>