two math questions from blue book

Test Preparation
1

<h1>20, pg 738</h1>

<h1>16, pg 795</h1>

2

<p>If they're math problems (and not pictures), could you post the problems? I don't own a blue book since I'm already done with my SATs</p>

3

<p>no they are pictures. thanx for the offer though.
Anyone else? plu-essse?</p>

4

<h1>20, 738</h1>

<p>This problem requires you to use the information they give you to figure out what you don't know. It seems confusing, but it's very simple if you reason it properly and don't allow yourself to get confused.</p>

<p>You're given that arc SBT is one quarter of a circle with radius 6, so the length of SBT is 12pi/4 = 3pi. Right away you can eliminate options D and E that don't have 3pi in them. </p>

<p>The perimeter of the entire figure is 12 + 3pi. now you need to find the perimeter of the shaded area, so subtract what isn't shaded:</p>

<p>You are told that the length + width of ABCR = 8. Subtact 8 from 12 + 3pi to get 4 + 3pi</p>

<p>You should notice, visually, that since R is the center of the circle and the distance given from R to the point on the circle S is 6, the length of the hypotenuse of triangle RAC (which is also the diagonal of ABCR) is 6 as well. So add 6 to 4+ 3pi to get 10 + 3pi</p>

<p>The answer is B, 10 + 3pi.</p>

<h1>16, 795</h1>

<p>When you see simple problems with unlabeled triangles like this, create variables for the angles you don't know and use simple algebra and what you know about triangles (exterior angle = sum of remote interior, sum of triangle's angles = 180, etc...): to create a true equation and solve it.
Like this:
z = 65 + a
y = 65 + b
y + z = 130 + a + b
a + b = 180 - 65 = 115</p>

<p>y + z = 130 + 115 = 245</p>

<p>The answer is E, 245. </p>

<p>Hope this helps.</p>

5

<p>Thanx so much Obsessed andre. You are my hero!!!</p>

6

<p>Answers </p>

<hr>

<h1>20, 738</h1>

<p>"
You should notice, visually, that since R is the center of the circle and the distance given from R to the point on the circle S is 6, the length of the hypotenuse of triangle RAC (which is also the diagonal of ABCR) is 6 as well. So add 6 to 4+ 3pi to get 10 + 3pi"</p>

<p>sorry, i don't really catch how you reasoned that the diagonal of ABCR is 6.</p>

7

<p>Because the diagonal of ABCR is a radius of the circle (which we are told is 6)</p>

<p>If you don't see this, draw a straight line from R to B.</p>

8

<p>Sorry but I don't get the explanation for #16 795 could someone rephrase it please?</p>