## Saturday, August 11, 2007

### SAT challenge

Note: Figure not drawn to scale.

The circle shown above has center O and a radius of length 5. If the area of the shaded region is 20 pi, what is the value of x?

18
36
45
54
72

I like this challenging SAT problem because it demonstrates once again that background knowledge is needed to deploy critical and creative thinking. This should give pause to "critical thinking" advocates who disparage knowledge (known as "content" nowadays).

Here is the reasoning, based on background knowledge, offered by SAT (don't read past this point if you want to solve this problem yourself):

In order to find the value of x, you should first determine the measure of the angle that is located at point O in the right triangle. To determine this angle, you must calculate what fraction of the circle’s area is unshaded. The radius r of the circle is 5 and its area is pi r^2, or 25 pi. The area of the shaded region is 20 pi, so the area of the unshaded region must be 5 pi. Therefore, the fraction of the circle’s area that is unshaded is 5 pi/25 pi, or 1/5. A circle contains a total of 360 degrees of arc, which means that 1/5 of 360 degrees, or 72 degrees, is the measure of the angle at point O in the unshaded region. Since you now know that two of the three angles in the triangle measure 72 degrees and 90 degrees and that the sum of the measures of the three angles is always 180 degrees, the third angle must measure 18 degrees. Therefore, x = 18.

Anonymous said...

http://apps.collegeboard.com/qotd/question.do?src=O&questionId=180

Anonymous said...

I like the ratio between the angle(Q) and area(A) of a sector:
Q/360 = A/(pi*r^2)
With a little thought and a nice diagram, its very obvious.

So, since the whole area of the circle is pi*r^2=25*pi, then the area of the non-shaded sector(inside the triangle) is 25*pi-20*pi=5*pi. So, the center angle is Q=360*(5*pi)/(25*pi) = 360*1/5 = 72.

So, x=90-72 = 18.

rightwingprof said...

What button?

Instructivist said...

RWP,

Buttons makes sense in the original context (the SAT site) where you really can click on one of the choices of this MC test question.

Rob Kremer said...

Boy they aren't kidding when they say the diagram is not drawn to scale! I came to the same answer of 18 degrees, and then started questioning it because the scale is so badly off. The non-shaded portion of the circle isn't even close to being the 1/5 of the circle that the math requires.

NYC Math Teacher said...

No kidding, dude! That angle is hardly 73 degrees as drawn. Still, the covered their tuchases with the caveat.

NYC Math Teacher said...

Ummm...that's 72 degrees. Typo there.

Instructivist said...

I think it's not drawn to scale on purpose. Otherwise some test-takers might be tempted to circumvent the problem by simply measuring the angle.

Cal said...

They can't measure it; it's not as if they take the test with a protractor. It is deliberately not drawn to scale because the test has to test certain geometry facts without allowing the student to use spatial references.

As for the ratio, it's a basic fact tested on all standardized math tests.

All standardized general math tests address the following circle facts:

1) Area formula
2) Circumference formula
3) d = 2r
4) 360 degrees
5) the ratio

The SAT also tests the "tangent to a radius is perpendicular", but that one isn't universal.

All test prep books or courses covers this.

The actual ratio as taught:

center angle/360 = arc/circumference = sector/area.

Or, a pizza slice takes out the same percentage of the crust and the pie.

Anonymous said...

Here is my 2-cents on why discovery learning/critical thinking "advocates" should love this problem as well. Sorry it is so long...The key is how did you learn the fundamentals needed to solve this problem. Did you memorize the formulas, or did you use experiential learning to discover them. If you discovered them you will be better prepared to handle these types of problems.

Take two classes of high school students. Class #1 is told that Area of a circle= pi*r^2 by the teacher and they do 50-100 problems to memorize this equation and apply it in various ways . Meanwhile, class #2 is told that they will do some experiments to discover the mathematical relationships between Area and the size of circles. Class #2 starts out by cutting out and weighing the mass of 1cmx1cm, 2cmx2cm and 3cmx3cm squares. Students gaph the mass versus area for the 3-squares and see that mass of paper is directly proportional to area, and they can calculate the mass per cm^2 of paper from the slope. Next the class is told to make 5 different circles with radi from 1-5cm. They cut and weigh the mass of each circle, and using the slope from their previous data they can determine the area for each circle in cm^2. Students plot area versus radius and find that it is non-linear. They can experiment with different equations to get a linear fit...until they find out that a plot of Area versus r^2 is linear, and the slope of this graph is pi! Now class #2 only has enough time to do 10-20 problems because they spent so much time "discovering" what class#1 had to memorize. Hopefully to them the equation now makes sense and is not another formula to memorize and apply in a given situation.

Students from both classes are taking the SAT and encounter your question. Which group will do better? The memorizer-high repetition class#1 or the "discovery" class#2? Guided inquiry research would suggest that the best students in each class will do equally well, but more of the average to below average students from class#2 will be able to answer the question correct!

Barry Garelick said...

Class #2 starts out by cutting out and weighing the mass of 1cmx1cm, 2cmx2cm and 3cmx3cm squares.

Uh, exuse me? Is Class 2 the only alternative to Class 1? And why do you presume that Class 1 does not learn about sector area? In high school, I learned about the relationship between ratio of sector area to area of a circle being proportional to the ratio of the measure of the arc of the sector to 360 degrees (or 2 pi). I didn't learn it by cutting things out; we read about it in a book and it took 5 to 10 minutes. Then we did problems. I remembered it well enough years later that I solved the SAT problem. I don't recall the class being memorizer/high repetition. In fact, we used the SMSG geometry book (this was mid-60's) which was criticized as being too rigorous and mathematical for high school students.

Anonymous said...

My 10 year old - a Singapore math student - was able to solve this without too much difficulty. She easily found the area of the circle, figured that the unshaded portion was 1/5, calculated 1/5 of 360 and then did a little addition to get the final answer. Took her less than 5 minutes.

RSydnor55 said...

I found a nice site that provides great information on the SAT and interesting strategies.

http://www.macunderground.org/satfaq.php

RSydnor55 said...

I found a nice site that provides great information on the SAT and interesting strategies.

http://www.macunderground.org/satfaq.php

Anonymous said...

Ideally a child of about eleven should be able to solve this without writing anything down.

Anonymous said...

this shit is pretty easy man there are some challenging problems on the sats that are way more complicate than that b