SAT Math Problem

Test Preparation SAT Preparation
1

<p>This is from the Big Blue book 2e, test 3 section 2 question 17 on page 518.</p>

<p>Semicircular arcs AB, AC, BD, and CD divide the circle above into regions. The points shown along the diameter AD divide it into 6 equal parts. If AD = 6, what is the total area of the shaded regions?</p>

<p>A)4pi
B)5pi
C)6pi
D)12pi
E)24pi</p>

<p>Here is a picture:</p>

<p>

<a href="http://i.imgur.com/duDIL.jpg%5B/img%5D"&gt;http://i.imgur.com/duDIL.jpg

</a></p>

<p>D is the point on the opposite side of A.</p>

<p>Much appreciated !</p>

2

<p>Go to the website for the study guide explanations, I think I did for the same problem last week and it helped.</p>

<p>You have to find the area of the smallest circle (pir^2), Subtract that from the area of the medium white circle to find the amount you need to subtract from the total circle area. Each segment from one point to another is 1 because all together they’re 6. then you should get your answer.</p>

3

<p>Check your diagram.</p>

<p>Is there supposed to be a blue semicircle from a to b? If so, the answer is C, 6pi.</p>

<p>pi(3^2 - 2^2 +1^2).</p>

<p>If your diagram is correct, the answer is 5.5 * pi</p>

<p>pi(3^2 -2^2 + 1^2/2)</p>

4

<p>The diagram is off. There is a semicircle that is shaded for arc AB.</p>

5

<p>Another way to get the answer is to …</p>

<p>take the top half of the circle</p>

<p>reflect it such that arc AB forms a circle, arc AC forms a circle, and arc AD forms a circle.</p>

<p>you now have three circle, each a different size. Now you find the area using A=pi x r^2</p>

<p>because you only want the shaded region, the equation should look like this</p>

<p>A= pi x 3^2 - pi x 2^2 + pi x 1^2
A= 9pi - 4pi + pi
A= 9pi - 3pi
A= 6pi</p>

<p>*the “pi x 3^2” is the area of the largest circle with diameter AD
*the “pi x 2^2” is the area of the not shaded circle with diameter AC
*the “pi x 1^2” is the area of the smallest shaded circle with diameter AB</p>

<p>The answer is C, 6pi</p>

6

<p>Trust me this is the easiest solution for this problem that you may find:
1- Look at the figure and you see that the circle is made of 3 regions.
2- Find the total area: d=6 then r=3 . A= 9 pi .
3- The shaded region to non-shaded region is 2/3 .
4- (2/3)*9 pi= 6 pi
The answere is 6 pi</p>

7

@Qaradaghi‌ I swear ur answer was the only one that made sense to me lol.

8

One should pay closer attention to the answer of Merupan: Here’s an older post with a diagram:

http://talk.collegeconfidential.com/sat-preparation/1290303-cant-understand-this-math-problem.html

9

If the diameter is 6, then that means that each segment on the diameter of the circle is 1 and there’s a radius of 3.
Whole circle = 9 pi

Then you find the area of the non-shaded portions:

Area of the non-shaded semicircle from A to C is 4 pi divided by 2, so = 2 pi.
Area of the other non-shaded semicircle from B to D is also 2 pi, but then you look at the small blue semicircle from C to D which has an area to be 1 pi. Thus, the non-shaded semicircle from B to D is 2 pi - 1 pi = 1 pi.

2 pi + 1pi = 3 pi
Subtract 3 pi from the whole circle 9 pi to get 6 pi, which is letter ©.

10

Trust me this is the easiest solution for this problem that you may find:
1- Look at the figure and you see that the circle is made of 3 regions.
2- Find the total area: d=6 then r=3 . A= 9 pi .

3- The shaded region to non-shaded region is 2/3 . >>>>>>>>>>>>>>>>>> Check that part. How do you determine that and how accurate is this?

4- (2/3)*9 pi= 6 pi
The answere is 6 pi