#20 - 2005 PSAT Form W

Test Preparation SAT Preparation
1

<p>I'm going over problems I missed from the PSAT to prepare for the January SAT. </p>

<p>I missed #20:</p>

<p>The figure above shows a circle with radius r and center P and an arc of length 6. The two radii shown are extended 3 units outside the circle. There is an arc of length x, which is part of a larger circle (not shown) also centered at P. What must x equal?</p>

<p>A) 9
B) 12
C) r + 3
D) pi ((r+3)/3)
E) (6r + 18) / r</p>

<p>The answer is E. I found a way to solve it, but I think the process is too long. Does anyone know a faster method to solve this problem?</p>

<p>I forgot how to calculate arc length, so I looked it up and found that </p>

<p>arc length/circumference = degree measure of arc/360</p>

<p>With this I found the degree measure of arc for the circle.</p>

<p>6/(2 * pi * r) = degrees / 360
degrees = 1080 / pi *r</p>

<p>Now I use degrees to solve for the arc length with radius (r+3):</p>

<p>x/(2 * pi * (r +3)) = (1080/(pi *r)) / 360
x = (6r + 18) / r</p>

2

<p>o_o; I have no clue, Michael. ><; Sowwie! [pokes] xD Guess who is this huh! I bet you don't know who it is. :P!</p>

3

<p>Hmm.. I remember doing this problem.</p>

<p>I believe I set up a proportion like this:</p>

<p>small circle's radius is to small circle's arc length as big circle's radius is to big circle's arc length</p>

<p>So: </p>

<p>r / 6 = (r+3) / x</p>

<p>cross multiply:</p>

<p>rx = 6r + 18</p>

<p>solve for x:</p>

<p>(6r + 18) / r = x</p>

4

<p>lucky you guys. my stupid first period teacher won't give me my scores. the whole freakin school has theirs, but not me.</p>