﻿ THE INSIDE SCOOP FOR SOLVING PROBLEMS - Important Tactics - SAT SUBJECT TEST MATH LEVEL 1 ﻿

## CHAPTER 1Important Tactics

### THE INSIDE SCOOP FOR SOLVING PROBLEMS

Why do some students do so much better on the Math 1 test than others? Of course, A+ students tend to do better on the test than B+ or A- students. Among students with exactly the same grades in school, though, why do some earn significantly higher scores than others—perhaps 100 to 200 points higher? Those students are better test takers. Either instinctively or by having been taught, they know and use most of the tactics discussed in this chapter. If you master these strategies, you will be a much better test taker and will earn significantly higher scores, not only on the Math 1 test but also on the PSAT, SAT, and other standardized math tests.

THE TACTICS EXPLAINED IN THIS CHAPTER CAN MAKE THE DIFFERENCE BETWEEN A GOOD SCORE AND A GREAT SCORE.

TACTIC

 1 Use your calculator even when no calculations are necessary

Often, if you get stuck on a calculator inactive question, you can use your calculator to get the right answer.

EXAMPLE 1: If , then  =

(A) a – 2

(B) a

(C) a2

(D)

(E) a

Solution: If you think that this is an algebra question for which a calculator would not be helpful, you would be only partly right. There is an algebraic solution that does not require the use of a calculator. However, if you don’t see how to do it, you can plug in a number for x and then use your calculator.

Let x = 2. Then . So a = 6.25.

Now .

Which answer choice equals 4.25 when a = 6.25? Only choice A, a − 2.

In Example 1, you actually didn’t need your calculator very much. You used it only to square 2.5. Example 2 looks easier because there is only one variable, but it actually requires a greater use of the calculator.

EXAMPLE 2: If x > 0 and , what is the value of ?

(A) 4

(B) 16

(C) 60

(D) 62

(E) 64

Solution (using TACTIC 1): . Now use your calculator to approximate x by guessing and checking. Since x must be a little less than 8.

•       too small

•       just a little too small

•       just a little too big

So 7.8 < x < 7.9.

Now evaluate :

•

•

Only choice D, 62, lies between 60.86 and 62.42.

With a graphing calculator, you can find x by graphing  and tracing along the graph, zooming in if necessary, until the y-coordinate is very close to 8. You could also look at  in a table, using increments of 0.1 or even 0.01.

Mathematical Solution to Examples 1 and 2:

So  is 2 more than .

In Example 1, , so .

In Example 2, , so .

The algebraic solution to the following example doesn’t require the use of a calculator, but many students wouldn’t find that solution when taking the test and would leave the question out. Of course, you would not do that. You would use TACTIC 1.

EXAMPLE 3: If 7x= 2 then 73x=

(A)

(B) 4

(C) 6

(D) 8

(E) 4

Solution (using TACTIC 1): First you need to find (or approximate) x. There are several ways to do that, all of which require a calculator. Here are four methods.

1. Guess and check

 71 = 7 way too big 70.5 = 2.6 still too big 70.4 = 2.17 getting close 70.35 = 1.98 a little too small, but close enough

2. Graph y = 7x and trace until y is very close to 2

3. Look at the table for y = 7x and scroll until you find a y  value very near 2

4. Use logarithms: .

Using x = 0.35, the first value we got, we have 3x = 1.05 and 71.05 = 7.7. Clearly, the correct answer is 8 (especially since we know that x = 3.5 is slightly too small).

Mathematical Solution: Of course, the solution using logarithms is a correct mathematical solution. If you carefully enter  into your calculator, you will get 8. The shortest and nicest solution does not require a calculator at all: 73= (7x)= 23 = 8.

EXAMPLE 4: What is the range of the function f (x) = (x – 2)2 + 2?

(A) All real numbers

(B) All real numbers not equal to 2

(C) All real numbers not equal to –2

(D) All real numbers greater than or equal to 2

(E) All real numbers less than or equal to 2

The easiest, correct mathematical solution is to observe that since the square of a number can never be negative, (x – 2)must be greater than or equal to 0. Therefore, f (x) = (x – 2)+ 2 must be greater than or equal to 2.

If you do not see that, however, and if you have a graphing calculator, you can graph y = (x – 2)+ 2 and see immediately that the graph is a parabola whose minimum value is 2—the turning point is at (2, 2).

TACTIC

 2 Backsolve

Backsolving is the process of working backward from the answers. When you back-solve, you test the five answer choices to determine which one satisfies the conditions in the given problem. This strategy is particularly useful when you have to solve for a variable and you are not sure how to do it. Of course, it can also be used when you do know how to solve for the variable but feel that it would take too long or that you might make a mistake with the mathematics.

Always test choice C first. On the Math 1 test, when the five answer choices for a question are numerical, they are almost always listed in either increasing or decreasing order. (The occasional exceptions occur when the choices involve radicals or π.) When you test a choice, if it is not the correct answer, it is usually clear whether the correct answer is greater or smaller than the choice tested. Therefore, if choice C does not work because it is too small, you can immediately eliminate three choices—C and the two choices that are even smaller (usually A and B). Similarly, if choice C does not work because it is too big, you can immediately eliminate three choices—C and the two choices that are even bigger (usually D and E).

Examples 5 and 6 illustrate the proper use of TACTIC 2.

EXAMPLE 5: For what value of n is 21n= 3 75?

(A) 5

(B) 10

(C) 25

(D) 50

(E) 125

Solution (using TACTICS 1 and 2): Use your calculator to evaluate 3 75 = 4,084,101. Now test the choices, starting with C.

Is 2125 = 4,084,101? No. 2125 = 1.13  1033, which is way too big. Eliminate choice C and choices D and E, which are even bigger, and try something smaller. Whether you now test 5 (choice A) or 10 (choice B) does not matter. However, since your first attempt was ridiculously large, try the smaller value, 5. Is 215 = 4,084,101? Yes, so the answer is A. NOTE: If after eliminating C, D, and E you tried B, you would have found that 2110 = 1.668  1013, which is still much too big and you would have known that the answer is A.

TIP

Did you have to backsolve to answer this question? Of course not. You never have to backsolve. You can always get the correct answer to a question directly if you know the mathematics and if you do not make a mistake. You also did not need to use your calculator.

Mathematical Solution: One of the laws of exponents states that for any numbers ab, and nanb= (ab)n. So 3 75 = (3  7)= 215.

Note that in this case the mathematical solution is much faster. If, however, while taking a test, you come to a question such as Example 5 and you don’t remember the laws of exponents or are unsure about how to apply them, you don’t have to omit the question—you can use TACTIC 2 and be certain that you will get the correct answer.

EXAMPLE 6: Alice, Beth, and Carol divided a cash prize as follows. Alice received  of it, Beth received  of it, and Carol received the remaining \$120. What was the value of the prize?

(A) \$360

(B) \$450

(C) \$540

(D) \$600

(E) \$750

Solution (using TACTIC 2): Backsolve starting with C. If the prize was worth \$540, then Alice received  and Beth received .

So, they received a total of \$396, leaving \$540 – \$396 = \$144 for Carol. Since that is too much (Carol only received \$120), eliminate choices C, D, and E and try B. If the prize was worth \$450, then Alice received  and Beth received .

So, they received a total of \$330, leaving \$450 - \$330 = \$120 for Carol, which is correct.

Mathematical Solution: Let x represent the value of the prize. Here are two ways to proceed.

1. Solve the equation:  using the 6-step method from Chapter 6.

Get rid of the fractions by multiplying each term by 15:

6x + 5x + 1,800 = 15x
Combine like terms:                          11x + 1,800 = 15x
Subtract 11x from each side:                       1,800 = 4x
Divide both sides by 4:                                       x = 450

2. Add  and  to determine that together Alice and Beth received  of the prize, leaving  of the prize for Carol. So

If you are comfortable with either algebraic solution and are confident you can solve the equations correctly, just do it, and save backsolving for a harder problem. If you start to do the algebra and you get stuck, you can always revert to backsolving. Note that unlike the situation in Example 5, in Example 6 the correct mathematical solutions are not much faster than backsolving.

TACTIC

 3 Plug in numbers whenever you have EXTRA variables

To use this tactic, you have to understand what we mean by extra variables. Whenever you have a question involving variables:

• Count the number of variables.

• Count the number of equations.

• Subtract these two numbers. This gives you the number of extra variables.

• For each extra variable, plug in any number you like.

If x + y + z = 10, you have three variables and one equation. Hence you have two extra variables and can plug in any numbers for two of the variables. You could let x and y each equal 2 (in which case z = 6); you could let x = 1 and z = 11 (in which case y = -2); and so on. You could not, however, let x = 1, y = 2, and z = 3—you do not have three extra variables, and, of course, 1 + 2 + 3 is not equal to 10.

If x + y = 10, you have two variables and one equation. Hence you have one extra variable and can plug in any one number you want for x or y but not for both. You cannot let x = 2 and y = 2 since 2 + 2  10. If you let x = 2, then y = 8; if you let x = 10, then y = 0; if you let y = 12, then x = −2.

If 2x + 4 = 10, you have one variable and one equation. So you have no extra variables, and you cannot plug in a number for x. You have to solve for x.

If a question requires you to simplify  you should recognize that you have two variables and no equations. Note that  is not an equation; it is an expression. An equation is a statement that one expression is equal to another expression. Since you have two extra variables, you can let m = 1 and n = 2, in which case .

Of course, since  this is the result you would get if you plugged in any numbers for m and n.

Look at Example 1. Without saying so, TACTIC 3 was used. The given information was . Two variables were given but only one equation. So, we had one extra variable and could have plugged in any number for either x or a. Clearly, it is easier to plug in for x and evaluate a than it would be to plug in a number for a and then have to solve for x. But we didn’t have to replace x by 2; we could have used any number. For example, if we let x = 3:

Although all good test takers use TACTIC 3 when they want to avoid potentially messy algebraic manipulations, TACTIC 3 can also be used on geometry or trigonometry questions that contain variables. The basic idea is to

• replace each extra variable with an easy-to-use number;

• answer the question using those numbers;

• test each of the answer choices with the numbers you picked to determine which choices are equal to the answer you obtained.

If only one choice works, you are done. If two or three choices work, change at least one of your numbers, and test only the choices that have not yet been eliminated.

Now look at a few examples that illustrate the correct use of TACTIC 3.

EXAMPLE 7: If a + a + a + a = b, which of the following is equal to 4b - a?

(A) 0

(B) 3a

(C) 15a

(D) 16a

(E) 10a + 10

Solution (using TACTIC 3): Since you have two variables and one equation, you have one extra variable, so let a = 2. Then

Now replace a by 2 in each of the answer choices and eliminate any choice that does not equal 30.

 • 0  30 Cross out A. • 3(2) = 6  30 Cross out B. • 15(2) = 30 C could be the correct answer, but you still have to test choices D and E. • 16(2) = 32  30 Cross out D. • 10(2) + 10 = 30 E could be the correct answer.

At this point, you know that the correct answer is either C or E. To break the tie, you have to choose another number for a, say 3. When a = 3, b = 3 + 3 + 3 + 3 = 12, and 4b − a = 4(12) − 3 = 45. Now test choices C and E.

 • 15(3) = 45 Choice C still works. • 10(3) + 10 = 40  45 Now choice E does not work. Cross out E.

Mathematical Solution:

EXAMPLE 8: Which of the following is equal to 2 sin 23 + 2cos 23 ?

(A) 1

(B) 2

(C) 3

(D) 6

(E) 2 tan2 3

Solution (using TACTIC 3): First note that since you have one variable, , and no equations,  is an extra variable, and so you can replace it by any number. Pick a really easy-to-use number, say  = 0. Then

Immediately eliminate choices A, C, and D and keep B. Now check whether choice E equals 2 when  = 0: 2 tan2 3(0) = 2(tan 0)= 2(0) = 0  2, so E is not correct. The answer is B.

Mathematical Solution: Let x = 3. Then

EXAMPLE 9: If 2a= 3b, what is the ratio of a to b?

(A) 0.63

(B) 0.67

(C) 1.5

(D) 1.58

(E) 1.66

The correct solution, using logarithms, is as follows:

If you have no idea how to solve the given equation, or if you know that it can be done with logarithms but you do not remember how, use TACTIC 3. Since there are two variables and only one equation, there is one extra variable. Pick a number for either a or b, so let b = 2.

Then 2a= 32 = 9. Immediately, you should see that since 23 = 8, a must be slightly greater than 3 and  must be slightly more than . Certainly, the answer is D or E.

At this point you could guess, but you shouldn’t. Instead, you should now use TACTIC 1 (use your calculator) and TACTIC 2 (backsolve) to be sure.

Since b = 2, if  , then a = (1.58)(2) = 3.16 and if  , then a = (1.66)(2) = 3.32. Finally, 23.16 is much closer to 9 than is 23.32.

Alternatively, you could have graphed y = 2and traced to find where 2= 9; or you could have graphed y = 2and y = 9 and found the point where the two graphs intersect.

TACTIC

 4 Draw diagrams

On some geometry questions, diagrams are provided, sometimes drawn to scale, sometimes not. Frequently, however, a geometry question does not have a diagram. In those cases, you must draw one. The diagram can be a sketch, drawn quickly, but it should be reasonably accurate. Never answer a geometry question without having a diagram, either one provided by the test or one you have drawn.

Sometimes looking at the diagram will help you find the correct solution. Sometimes it will prevent you from making a careless error. Sometimes it will enable you to make an educated guess.

EXAMPLE 10: If the diagonal of a rectangle is twice as long as one of the shorter sides, what is the measure of each angle that the diagonal makes with the longer sides.

(A) 15°

(B) 30°

(C) 45°

(D) 60°

(E) 75°

Solution (using TACTIC 4): The first step is to sketch a rectangle quickly, but don’t be sloppy. Don’t draw a square, and don’t draw a rectangle such as the one below in which the diagonal is 4 or 5 or 6 times as long as a short side.

Draw a rectangle such as this:

From the second sketch, it is clear that x < 45, and the angle is not nearly skinny enough for x to be 15. The answer must be 30°, choice B. In this case, you can be sure you have the right answer. If the answer choices had been

(A) 25°

(B) 30°

(C) 35°

(D) 45°

(E) 60°

you could have eliminated D and E but might have had to guess from among A, B, and C.

Mathematical Solution: Here are two correct solutions.

• If the length of one leg of a right triangle is half the length of the hypotenuse, the triangle is a 30-60-90 triangle, and the measure of the angle opposite the shorter leg is 30°.

• From the drawn above, you can see that

EXAMPLE 11:  is a diameter of a circle whose center is at (1, 1). If A is at (–3, 3), what are the coordinates of B ?

(A) (–5, 1)

(B) (–1, 2)

(C) (5, -1)

(D) (5, 1)

(E) (5, 5)

Solution (using TACTIC 4): Even if you think you know exactly how to do this, first make a quick sketch.

Even if your sketch wasn’t drawn carefully enough, it would be clear that the x-coordinate of B is positive and the y-coordinate is near 0. So you could eliminate choices A, B, and E. If (5, –1) and (5, 1) are too close to tell from your sketch and if you don’t know a correct way to proceed, just guess between C and D. If you drew your diagram carefully (as we did), you could definitely tell that the y-coordinate is negative, and so the answer must be C.

Mathematical Solution: Since the center of the circle is the midpoint of any diameter, (1, 1) is the midpoint of  where A is (–3, 3) and B is (xy). Use the midpoint formula: . So

Even if you know how to do this, you should sketch a diagram. If you make a careless error and get y = 5, for example, your diagram would alert you and prevent you from bubbling in E.

TACTIC

 5 Trust figures that are drawn to scale

On the Math 1 test, some diagrams have the following caption underneath them: “Note: Figure not drawn to scale.” All other diagrams are absolutely accurate, and you may rely upon them in determining your answer.

EXAMPLE 12: In the diagram at the right, the radius of circle O is 4 and diameters  and  are perpendicular. What is the perimeter of BOD?

(A) 6

(B) 6.83

(C) 12

(D) 13.66

(E) 16

Solution (using TACTIC 5): Since the diagram is drawn to scale, you may trust it. The question states that radii  and  are each 4. Looking at the diagram, you can see that chord  is longer than  and is therefore greater than 4. Therefore, the perimeter of BOD is greater than 4 + 4 + 4 = 12. Eliminate choices A, B, and C. You can also see that chord  is much shorter than diameter  and so is less than 8. Therefore, the perimeter of BOD is less than 4 + 4 + 8 = 16. So eliminate choice E. The answer must be D.

If choice E had been 13.5, 13.75, or 14, you would not have known whether the correct answer was D or E, and you would have had to guess.

Mathematical Solution: Since OB = OD = 4, BOD is isosceles. Since  and  are perpendicular, BOD is a right triangle. Therefore, BD = 4 = 5.66, and the perimeter of BOD is 4 + 4 + 5.66 = 13.66.

TIP

TACTIC 5 always allows you to eliminate choices, but you are not always able to eliminate all four incorrect choices.

TACTIC

 6 Redraw figures that are not drawn to scale

Recall that on the Math 1 test, the words “Note: Figure not drawn to scale” appear under some diagrams. When this occurs, you cannot trust anything in the figure to be accurate unless it is specifically stated in the question. When figures have not been drawn to scale, you can make noassumptions. Lines that look perpendicular may not be; an angle that appears to be acute may, in fact, be obtuse; two line segments may have the same length even though one looks twice as long as the other.

Often when you encounter a figure not drawn to scale, it is very easy to fix. You can redraw one or more of the line segments or angles so that the resulting figure will be accurate enough to trust. Of course, the first step in redrawing the figure is recognizing what is wrong with it.

When you take the Math 1 test, if you see a question such as the one in Example 13 below and if you are sure that you know exactly how to answer it, just do so. Don’t be concerned that the figure isn’t drawn to scale. Remember that most tactics should be used only when you are not sure of the correct solution. If, however, you are not sure what to do, quickly try to fix the diagram.

EXAMPLE 13: In ABC below, mA = 15° and AC = 8.

What is the value of x?

Note: Figure not drawn to scale.

(A) 2.07

(B) 2.14

(C) 4

(D) 7.72

(E) 8.23

Solution: In the diagram,  and  appear to be about the same length. If the figure had been drawn to scale you would be pretty confident that the answer is D or E. However, the figure is not drawn to scale. Therefore, you can make no such assumptions.

You are told that the figure is not drawn to scale, and in fact, it isn’t. The measure of ∠A is 15°, but in the diagram it looks to be much more, perhaps 45°. To fix it, create a 45° angle by sketching a diagonal of a square, and then divide that angle into thirds. Now you have an accurate diagram, and  is clearly much less than , nowhere near 8.

In fact, it is clearly less than 4. So the answer must be A or B. Unfortunately, no matter how carefully you draw the new diagram, you cannot distinguish between 2.07 and 2.14. Unless you know how to solve for x, you have to guess between A and B. If choice A had been 1.07 instead of 2.07, you would not have had to guess. From your redrawn diagram, you can tell that  is about four times as long as , not eight times as long, and you would know that the answer has to be B.

Mathematical Solution:

tan 15° =  ⇒ x = (8)tan15° = 2.14

TIP

You must guess whenever you can eliminate choices.

TACTIC

 7 Treat Roman numeral problems as three true-false questions

On the Math 1 test, some questions contain three statements labeled with the Roman numerals I, II, and III, and you must determine which of them are true. The five answer choices are phrases such as “None” or “I and II only,” meaning that none of the three statements is true or that statements I and II are true and statement III is false, respectively. Sometimes what follows each of the three Roman numerals are only phrases or numbers. In such cases, those phrases or numbers are just abbreviations for statements that are either true or false. Do not attempt to analyze all three of them together. Treat each one separately. After determining whether or not it is true, eliminate the appropriate answer choices. Be sure to read those questions carefully. In particular, be aware of whether you are being asked what must be true or what could be true.

Now try using TACTIC 7 on the next two examples.

EXAMPLE 14: ΔABC, m∠c=90°. if m∠A > m∠B, which of the following statements must be true?

I. sin A > cos B

II. cos A > cos B

III. tan A > tan B

(A) II only

(B) III only

(C) I and III only

(D) II and III only

(E) I, II, and III

Solution: First use TACTIC 4 and draw a diagram. Since you are told that m∠A > m∠B, make ∠A much bigger than ∠B. Now test each statement.

I. sin A > cos B Is that true or false?

• sin  and cos .

• So sin A = cos B.

• Statement I is false.

• Eliminate C and E, the two choices that include I.

II. cos A > cos B Is that true or false?

• cos  and cos . Clearly from the diagram a > b, and so .

• Statement II is false.

• Eliminate A and D.

Having crossed out choices A, C, D, and E, you know the answer must be B. You do not have to verify that statement III is true. (Of course it is: tan , which is greater than 1 since a > b and , which is less than 1. So tanA > tan B.)

EXAMPLE 15: If the lengths of two sides of a triangle are 4 and 9, which of the following could be the area of the triangle?

I. 8

II. 18

III. 28

(A) II only

(B) III only

(C) I and II only

(D) II and III only

(E) I, II, and III

Solution: Think of Roman numeral I as the statement, “The area of the triangle could be 8,” (and similarly for Roman numerals II and III). You are free to check the statements in any order.

Start by drawing a right triangle whose legs have lengths of 4 and 9. Then use the formula  to calculate the area: . Therefore, statement II is true. Eliminate choice B, the only choice that does not include II.

At this point, there are several ways to analyze the other choices. Knowing what the area is when C is a right angle, consider what would happen if angle C were acute or obtuse.

If you superimpose PBC, in which ∠C is obtuse, and QBC, in which ∠C is acute, onto ABC, you can see that in each case the height to base  is less than 4. So in each case, the area is less than . The area of the triangle could not be 28. III is false. Eliminate D and E, the two remaining choices that include III. The areas of PBC and QBC are each less than 18, but it may not be clear whether either triangle, or any other triangle with sides 4 and 9, could have an area equal to 8. If you cannot determine whether I is true, guess between A (II only) and C (I and II only).

In fact, the area could be 8, or any other positive number less than 18. For the area to be 8, just let the height be .

A particularly nice way to solve Example 15 is to use the formula for the area of a triangle that relies on trigonometry: , where a and b are the lengths of two of the sides and  is the measure of the angle between them. Since the maximum value of sin is 1, the maximum possible area is . Therefore, III is false. Eliminate choices B, D, and E. Finally, ask, “Could ?”

and since  is in the range of the sine function, the answer is yes. (In fact, , which you should not take the time to evaluate.) I is true. Eliminate A. The answer is C.

Note that the formula  is not part of the Math 1 syllabus, and no question on the test requires you to know it. If, however, you do know it, you are free to use it.

EXAMPLE 16: In the diagram below, is a  diameter and chords and  are parallel. Which of the  following statements must be true?

Note: Figure not drawn to scale.

I. x = y

II. a = b

III. AB = CD

(A) None

(B) I only

(C) II only

(D) III only

(E) I, II, and III

Solution: Here, nothing is wrong with the diagram. You are told that chord  is a diameter, and since it passes through the center, it is correctly drawn. You are told that  and  are parallel, and they are. However, there are many ways you could redraw the diagram consistent with the given conditions.

Remember

Sometimes when you see “Note: Figure not drawn to scale,” there is nothing wrong with the diagram. The diagram just did not have to be drawn the way it was.

For example, you don’t have to draw , and you could draw  closer to . Both of the following diagrams satisfy all the given conditions.

In each of the diagrams shown, x y, and AB CD, so I and III are false. Eliminate choices B, D, and E. Either II is true (in which case the answer is C) or II is false (in which case the answer is A). In both redrawn diagrams, a and b appear to be equal. Unless you can draw a diagram in which a and b are clearly unequal, you should guess that II is true.

In fact, in a circle parallel chords always cut off congruent arcs.

TACTIC

 8 Eliminate absurd choices

There will likely be some questions on the Math 1 test that you do not know how to answer. Before deciding to omit them, look at the answer choices. Very often, two or three choices are absurd. In that case, eliminate the absurd choices and guess among the remaining ones. Occasionally, four of the choices are absurd. In that case, your answer is not a guess, it is a certainty.

What makes an answer choice absurd? Lots of things. For example, you may know that the answer to a question must be positive, but two or three of the choices are negative. You may know that the measure of an angle must be acute, but three of the choices are numbers greater than or equal to 90°. You may know that a ratio must be greater than 1, but two or three of the choices are less than or equal to 1. Even if you know the correct mathematical method for answering a question, sometimes it is faster to start by eliminating answers that are clearly impossible.

EXAMPLE 17: If , where b > 1, which of the following could b be the value of a?

(A)

(B)

(C)

(D)

(E)

Solution: Since you are asked for a possible value of a, you could use TACTIC 2 and backsolve. Before doing that, however, carefully look at the choices. Since b is positive,  is positive, and so a must be greater than 1: eliminate choices A and B. Also, , and so a must be less than 2: eliminate choices D and E. The answer must be C.

You do not then have to prove that the answer is C by solving the equation . You do not get any extra credit for determining that  when b = 2.5.

Smart Strategy

If you have eliminated four answer choices, the remaining choice must be the correct answer. Don’t waste time verifying it.

EXAMPLE 18: In the figure below, a square is inscribed in a circle of radius 4. What is the area of the shaded region?

(A) 4.57

(B) 9.13

(C) 18.26

(D) 25.13

(E) 34.26

Solution (using TACTIC 8): In a question such as this some of the choices are always absurd. The area of a circle of radius 4 is π(4)= 16π, which is approximately 50. Since the diagram is drawn to scale, you can trust it. Clearly, more of the circle is white than is shaded, so the area of the square is more than 25 and the area of the shaded region is less than 25. Eliminate choices D and E. If the area of the shaded region were 9.13, the area of the white square would be greater than 40, which is more than 4 times 9.13. That is surely wrong and 4.57 is an even worse answer.

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