<p>PG 76, #16</p>
<p>Find the circumference of the small circle and add it to half the circumference of the larger circle. so… half the circumference of the shaded circle is 3pi, b/c circumference = r2pi, and the radius is 3. so the other little circle has a perimiter w/ the same value. so together it = 6pi. and then, you find half the circumference of the larger circle, which has a radius of 6 (radius of 3 and 3 added together). so that value is 2(6)pi = 12 pi. add that to the 6 pi. and the answer is 18 pi…but that answer doesn’t show up. so i’m sorry. i’m right.</p>
<p>^thats 14, i need 16</p>
<p>The triangle CPB’s hypotenuse is 3. The leg CP is smaller than the hypotenuse (the hypotenuse is the longest side of a triangle), so it can be 2, but not 4, 5, 7, 8.
The answer is (A) 2</p>
<p>I would answer A. Here is how I did it.</p>
<p>-You know that CP will form a 90 degree angle with side AB (it says that it it perpendicular).</p>
<p>-Now, we have a right angle triangle! BC is equal to 3. Angle CPB is 90 degrees, since CP is perpendicular to AB. BC=3 and is the hypotenuse of the triangle. </p>
<p>-Therefore, CP (a leg of the right angle triangle) must be less than 3. The only option is A (2).</p>
<p>***Could you please give me links to practice manuals/tests like this? I didn’t know collegeboard has these. Do you need to buy something to receive these booklets?</p>
<p>Stopchair, you forgot to divide the second perimeter you found by two because you only want half of the circle. So 14 comes to 12 pi, or D</p>
<p>I agree with crazybandit</p>
<p>Hey Shaqtus where did you get that booklet? Are there anymore of those online from different years?</p>
<p>SamWhich, they’re the same exams as the official online practice tests.</p>
<p>Thanks guys</p>
<p>This question was from the 07-08 Offical SAT prac. test</p>
<p>theres a few more at this link: <a href=“http://talk.collegeconfidential.com/sat-preparation/757034-college-board-sat-test-links.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/757034-college-board-sat-test-links.html</a></p>
<p>Can someone tell me how to do number 17 in section 2. Here’s the link- <a href=“https://satonlinecourse.collegeboard.com/SR/digital_assets/assessment/pdf/F4D31AB0-66B4-CE32-00F7-F5405701F413-F.pdf[/url]”>https://satonlinecourse.collegeboard.com/SR/digital_assets/assessment/pdf/F4D31AB0-66B4-CE32-00F7-F5405701F413-F.pdf</a></p>
<p>^ answer is b</p>
<p>140x = 180x-360</p>
<p>x = 9</p>
<p>The shape that you see has 4 sides. Any polygon with 4 sides has angles whose values add up to 360. The equation for this sum is 180(n-2), where n is the number of sides. You can deduce this by looking at a triangle (3 sides, sum of angles = 180 degrees) and a square (4 sides, sum of angles = 360 degrees). So since x+y=80, the other 2 sides’ values must add up to 280. Since all angles are equal (given), the value of each side of the whole polygon is 140 degrees.</p>
<p>Notice that since the sum = 180(n-2), the measure of EACH angle is 180(n-2) / n</p>
<p>You solve for the equation 180(n-2)/n = 140:</p>
<p>180(n-2)/n=140
(180n-360)/n=140
180n-360=140n
40n=360
n=9</p>
<p>Answer: <a href=“B”>b</a> Nine**</p>
<p>Working backwards, we can see that the sum of the angles is 180(9-2)=180*7=1260. since there are 9 sides, each angle is 1260/9 = 140 degrees. Since each side is 140 degrees, 2 sides is 280. 280+x+y=280+80=360=360</p>
<p>Thanks, I didn’t even think to use the polygon on the top.</p>
<p>How do you do the one with the triangles where you have DFH and you need to find the total area?</p>