This problem tests your ability to work with the two most important circle formulas (both given to you in your SAT test booklet). The circumference of a circle is , and the area of a circle is .  Since you’re given the circumference of the circle as , you can set up an equation to solve for , the radius:

Then you can plug in 6 as the radius in the area formula:

This problem puts a slight twist on the area of a circle formula, . You’re given the area of half a circle as , so to apply the area formula you should double  to account for the missing half of the circle so you can use the classic area formula. Since the area of the entire circle would be  64pi, you can use the area formula to solve for the radius:

So the radius is 8, which means you’re ready to apply the formula for circumference of a circle. But again, you’re only dealing with half the circumference since the problem asks for the arc length of the semicircle, so while the formula for circumference i s, you can cut that in half to find the distance of half the circumference. That means you need to calculate .  With a radius of 8, that means that your answer is .

This problem tests your ability to work with the two most important circle formulas (both given to you in your SAT test booklet). The circumference of a circle is , and the area of a circle is .  Since you’re given the area of the circle as , you can set up an equation to solve for , the radius:

Now you can plug in that radius of 12 into the circumference formula:

Because these triangles are similar, they will have the exact same ratio between their sides, meaning that the sine of Z will be the same as the sine of C. And since sine = opposite/hypotenuse, you can calculate that as , which reduces to .

You probably know SOH-CAH-TOA for sine, cosine, and tangent, which of course is absolutely necessary knowledge for the trigonometry questions on the SAT. The next piece of advanced knowledge about trigonometry that the SAT loves to test is the following set of rules:

You must know these rules to be able to solve advanced SAT trigonometry questions!

Here is another way to think about these rules:

In a right triangle, there are always two smaller angles . 

The sine of one angle = the cosine of the other angle.

The cosine of one angle = the sine of the other angle.

If you understand SOH-CAH-TOA and right triangles, this is logical: From SOH and CAH, you can see that the only difference between the sine and the cosine is that the sine has the Opposite side length in the numerator and the cosine has the Adjacent side length in the numerator. Well, in a right triangle, when you switch from one of the smaller angles to the other one, you are swapping which side is Opposite and which side is Adjacent! (The Hypotenuse always remains the same side.) Based on SOH and CAH, swapping the Opposite and Adjacent sides means the same thing as swapping the sine and cosine values of the angles.

Getting back to this specific question, since ,  and  have to be the two smaller angles of the right triangle. Therefore, based on the rules explained above,  and .

Since  , this means that  also! Thus the correct answer choice is .

This is a classic trick question because the test question writers know that most students will not be expecting the correct answer to be exactly the same as the other value given in the question! So students who don’t know the rules or who aren’t 100% confident in their knowledge will be afraid to choose that answer and will be likely to guess a different answer. This is the challenge that the SAT question writers present to students. You must know these rules very well in order to successfully pass this challenge and choose the correct answer!

First, you have to sort out the corresponding angles in the similar triangles. Clearly since ,  and  are the right angles. Thus we know that within each triangle,  and .  The key extra piece of information in the question about the angles is that . This means that  and . Now we know exactly which angles in the similar triangles are congruent (equal) to each other. 

Now the key step of the question is to use the value  to find the value of . Since we now know that , we know that . Now we can focus only on the triangle with angles , , and .

You probably know SOH-CAH-TOA for sine, cosine, and tangent, which of course is absolutely necessary knowledge for the trigonometry questions on the SAT. The next piece of advanced knowledge about trigonometry that the SAT loves to test is the following set of rules:

You must know these rules to be able to solve advanced SAT trigonometry questions!

Here is another way to think about these rules:

In a right triangle, there are always two smaller angles .

The sine of one angle = the cosine of the other angle.

The cosine of one angle = the sine of the other angle.

If you understand SOH-CAH-TOA and right triangles, this is logical: From SOH and CAH, you can see that the only difference between the sine and the cosine is that the sine has the Opposite side length in the numerator and the cosine has the Adjacent side length in the numerator. Well, in a right triangle, when you switch from one of the smaller angles to the other one, you are swapping which side is Opposite and which side is Adjacent! (The Hypotenuse always remains the same side.) Based on SOH and CAH, swapping the Opposite and Adjacent sides means the same thing as swapping the sine and cosine values of the angles.

Getting back to this specific question, since ,  and  have to be the two smaller angles of the right triangle. Therefore, based on the rules explained above,  and .

Since  , this means that  also! Thus the correct answer choice is 0.8.

This is a classic trick question because the test question writers know that most students will not be expecting the correct answer to be exactly the same as the other value given in the question! So students who don’t know the rules or who aren’t 100% confident in their knowledge will be afraid to choose that answer and will be likely to guess a different answer. This is the challenge that the SAT question writers present to students. You must know these rules very well in order to successfully pass this challenge and choose the correct answer!

First of all, watch out for the trap answer choice “The answer cannot be determined from the information given in the question”!! NEVER GUESS THIS ANSWER CHOICE if you are not completely confident that it MUST be true! Most often the SAT uses this answer choice as a trap for students who don’t know the advanced method that does exist and is necessary to find the correct answer, which can be determined from the information given in the question if you know the correct method. If you don’t understand how to answer the question and you have to guess, guess one of the other three answer choices!

You probably know SOH-CAH-TOA for sine, cosine, and tangent, which of course is absolutely necessary knowledge for the trigonometry questions on the SAT. The next piece of advanced knowledge about trigonometry that the SAT loves to test is the following set of rules:

You must know these rules to be able to solve advanced SAT trigonometry questions!

Here is another way to think about these rules:

In a right triangle, there are always two smaller angles . 

The sine of one angle = the cosine of the other angle.

The cosine of one angle = the sine of the other angle.

If you understand SOH-CAH-TOA and right triangles, this is logical: From SOH and CAH, you can see that the only difference between the sine and the cosine is that the sine has the Opposite side length in the numerator and the cosine has the Adjacent side length in the numerator. Well, in a right triangle, when you switch from one of the smaller angles to the other one, you are swapping which side is Opposite and which side is Adjacent! (The Hypotenuse always remains the same side.) Based on SOH and CAH, swapping the Opposite and Adjacent sides means the same thing as swapping the sine and cosine values of the angles.

Getting back to this specific question,  and  have to be the two smaller angles of the right triangle. Therefore, based on the rules explained above,  and .

Therefore, . Thus the correct answer choice is 0.

You probably know SOH-CAH-TOA for sine, cosine, and tangent, which of course is absolutely necessary knowledge for the trigonometry questions on the SAT. The next piece of advanced knowledge about trigonometry that the SAT loves to test is the following set of rules:

You must know these rules to be able to solve advanced SAT trigonometry questions!

Here is another way to think about these rules:

In a right triangle, there are always two smaller angles . 

The sine of one angle = the cosine of the other angle.

The cosine of one angle = the sine of the other angle.

If you understand SOH-CAH-TOA and right triangles, this is logical: From SOH and CAH, you can see that the only difference between the sine and the cosine is that the sine has the Opposite side length in the numerator and the cosine has the Adjacent side length in the numerator. Well, in a right triangle, when you switch from one of the smaller angles to the other one, you are swapping which side is Opposite and which side is Adjacent! (The Hypotenuse always remains the same side.) Based on SOH and CAH, swapping the Opposite and Adjacent sides means the same thing as swapping the sine and cosine values of the angles.

Getting back to this specific question, we have to think about the rules explained above in another way: Since the rules tell us that , and the question tells us that , we can draw the logical conclusion that . [The question also adds the information that both angles are acute, which means , so we know we are dealing with “normal” angle measures between  and , and not some more complicated angle measures greater than  for example.] Now we can rearrange  in the simpler form .

Now we can simply substitute in the values  and  into the equation . Continuing by adding 25 to both sides, we get . Dividing both sides by 10, we get . Therefore, the correct answer choice is 11.5.

If you mistakenly think  you would get the wrong answer choice 20.5. If you mistakenly think  you would get the wrong answer choice 38.5. This large value 38.5 is included among the answer choices mainly in order to make the more tempting wrong answer choice 20.5 appear to look more “reasonable” by comparison, to trap more students into guessing 20.5 incorrectly.

The SAT likes to pack many different concepts into a single question, to force you to use all of your knowledge of math to answer the question. Here you have to begin with the information that DE is perpendicular to AB, to establish that  is a right angle and so triangle ADE is a right triangle. Further, since DE and BC are parallel, that means  must also be a right angle, and triangle ABC is a right triangle.

Since triangles ADE and ABC also share the same angle A, their third angles  and  must also be equal, because the sum of the three angles of each triangle must add up to . Now that you know all three angles of triangles ADE and ABC are the same, this means they are similar triangles.

The next key piece of information in this question is . Knowing SOH-CAH-TOA, you know that . In these triangles, the adjacent sides to A are AD in the small triangle and AB in the large triangle. The hypotenuses are AE in the small triangle and AC in the large triangle. Therefore you know that .

Now you should recognize 12 and 13 as values in the “Pythagorean triple” of the 5-12-13 right triangle! Recall that these are the lengths of the sides of a right triangle because . However, be careful! The cosine value  only tells you the ratio of the side lengths, not the actual side length measures themselves. They could be 12 and 13, 24 and 26, 120 and 130, 1.2 and 1.3, 0.12 and 0.13, or any other pair of lengths with a ratio of .

The information in the question that  helps us determine the actual side lengths. Since BC must be the shortest side of a 5-12-13 ratio right triangle, you can see that the actual side lengths of triangle ABC must be two times 5, 12, and 13: Thus,  and . Further, the question tells you that . Therefore, subtracting the length of EC from the length of AC, you know that . This means that  and . Since  and , this means that .

Now you finally have all the side lengths of the trapezoid DBCE, and you are ready to calculate its area. You need to know the formula for the area of a trapezoid: . This looks complicated, but the logical way to understand it and remember it is to realize that  is just the average of the two bases of the trapezoid. Always keep in mind that the bases are the two sides that are parallel to each other, regardless of which direction they are oriented in the diagram. Here the two bases are DE and BC. Their lengths are 5 and 10, so their average is 7.5. The height of a trapezoid is the altitude from one base to the other that is perpendicular to both bases. In this case, since  and  are right angles DB is perpendicular to DE and BC, and so DB itself is the height of the trapezoid. Thus the height is 12, and the area of the trapezoid is , so the correct answer choice is 90.

Although the SAT can sometimes present geometry questions like this one without giving you a diagram, often you should draw a diagram yourself to visualize what is going on in the question!

Here the question tells you that LM is perpendicular to MN. Perpendicular lines meet at a right angle, so this is a fancy hidden way to tell you that  is a right angle. Now you know that LMN is a right triangle, so you should draw your own diagram of it:

Next, the question tells you that . Knowing SOH-CAH-TOA, you know that . The opposite side from L is MN, and the adjacent side to L is LM, so now you know that . Technically we do not know the actual side lengths, only the ratio of one side length to the other: They could be 4 and 3, or 8 and 6, or 400 and 300, or 0.8 and 0.6, or any two lengths with a ratio of . But in this particular question that does not matter, since the question is asking you for the value of , which is also a ratio of side lengths just like tan L is. So in this question, you can just keep it simple and assume the simplest case of  and .

You know the 3-4-5 right triangle, so now you can assume .

Again from SOH-CAH-TOA, you know that . Make sure that you switch your perspective now from angle L to angle N! The question asks you for the value of . From the perspective of angle N, the adjacent side is MN and the hypotenuse is LN. So . Therefore the correct answer choice is .