Sat - MC Grawhill part 42 pdf

Circle Basics Okay, we all know a circle when we see one, but it often helps to know the mathematical definition of a circle. • A circle is all of the points in a plane that are a certain distance r from the center. • The radius is the distance from the center to any point on the circle. Radius means ray in Latin; a radius comes from the center of the circle like a ray of light from the sun. • The diameter is twice the radius: d = 2r. Dia- means through in Latin, so the diameter is a segment that goes all the way through the circle. The Circumference and Area It’s easy to confuse the circumference formula with the area formula, because both formulas contain the same symbols arranged differently: circumference = 2πr and area =πr 2 . There are two simple ways to avoid that mistake: • Remember that the formulas for circumfer- ence and area are given in the reference in- formation at the beginning of every math section. • Remember that area is always measured in square units, so the area formula is the one with the “square:” area =πr 2 . Tangents Lesson 8: Circles diameter radius tangent radius A tangent is a line that touches (or intersects) the cir- cle at only one point. Think of a plate balancing on its side on a table: the table is like a tangent line to the plate. A tangent line is always perpendicular to the radius drawn to the point of tangency. Just think of a bicycle tire (the circle) on the road (the tangent): notice that the center of the wheel must be “directly above” where the tire touches the road, so the radius and tangent must be perpendicular. M P R l M P R l 7 5 400 McGRAW-HILL’S SAT Example: In the diagram above, point M is 7 units away from the center of circle P. If line l is tangent to the circle and MR = 5, what is the area of the circle? First, connect the dots. Draw MP and PR to make a triangle. Since PR is a radius and MR is a tangent, they are perpendicular. Since you know two sides of a right triangle, you can use the Pythagorean theorem to find the third side: 5 2 + (PR) 2 = 7 2 Simplify: 25 + (PR) 2 = 49 Subtract 25: (PR) 2 = 24 (PR) 2 is the radius squared. Since the area of the circle is πr 2 , it is 24π. CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 401 1. What is the formula for the circumference of a circle? 2. What is the formula for the area of a circle? 3. What is a tangent line? 4. What is the relationship between a tangent to a circle and the radius to the point of tangency? 5. In the figure above, AB –– is a tangent to circle C, AB = 8, and AD = 6. What is the circumference of circle C? 6. In the figure above, P and N are the centers of the circles and are 6 centimeters apart. What is the area of the shaded region? Concept Review 8: Circles C A D B 8 6 r r P N 1. Two circles, A and B, lie in the same plane. If the center of circle B lies on circle A, then in how many points could circle A and circle B intersect? I. 0 II. 1 III. 2 (A) I only (B) III only (C) I and III only (D) II and III only (E) I, II, and III 2. What is the area, in square centimeters, of a circle with a circumference of 16π centimeters? (A) 8π (B) 16π (C) 32π (D) 64π (E) 256π 3. Point B lies 10 units from point A, which is the center of the circle of radius 6. If a tangent line is drawn from B to the circle, what is the dis- tance from B to the point of tangency? Note: Figure not drawn to scale. 4. In the figure above, AB ––– and AD ––– are tangents to circle C. What is the value of m? Note: Figure not drawn to scale. 5. In the figure above, circle A intersects circle B in exactly one point, is tangent to both cir- cles, circle A has a radius of 2, and circle B has a radius of 8. What is the length of CD ––– ? CD sruu C A B D C A D B m° 3m° SAT Practice 8: Circles 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 402 McGRAW-HILL’S SAT CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 403 Concept Review 8 1. circumference = 2πr 2. area =πr 2 3. A tangent line is a line that intersects a circle at only one point. 4. Any tangent to a circle is perpendicular to the radius drawn to the point of tangency. Answer Key 8: Circles 6 2 + x 2 = 10 2 Simplify: 36 + x 2 = 100 Subtract 36: x 2 = 64 Take the square root: x = 8 4. 45 Since AB ––– and AD ––– are tangents to the circle, they are perpendicular to their respective radii, as shown. The sum of the angles in a quadrilateral is 360°, so m + 3m + 90 + 90 = 360 Simplify: 4m + 180 = 360 Subtract 180: 4m = 180 Divide by 4: m = 45 C A D B m° 3m° SAT Practice 8 5. Draw BC ––– to make a right triangle, and call the length of the radius r. Then you can use the Pythagorean theorem to find r: 8 2 + r 2 = (r + 6) 2 FOIL: 64 + r 2 = r 2 + 12r + 36 Subtract r 2 : 64 = 12r + 36 Subtract 36: 28 = 12r Divide by 12: 7/3 = r The circumference is 2πr, which is 2π(7/3) = 14π/3. C A D B 8 6 r r 6. Draw the segments shown here. Since PN is a ra- dius of both circles, the radii of both circles have the same length. Notice that PN, PR, RN, PT, and NT are all radii, so they are all the same length; thus, ΔPNT and ΔPRN are equilateral triangles and their angles are all 60°. Now you can find the area of the left half of the shaded region. This is the area of the sector minus the area of ΔRNT. Since ∠RNT is 120°, the sector is 120/360, or 1 ⁄3, of the circle. The circle has area 36π, so the sec- tor has area 12π. ΔRNT consists of two 30°-60°-90° triangles, with sides as marked, so its area is . Therefore, half of the origi- nal shaded region is , and the whole is . 24 18 3 π − 12 9 3 π − 12 6 3 3 9 3/ () () () = P N R T S N R T 6 6 3√3 3√3 3 BA B A BA 1. E The figure above demonstrates all three possibilities. 2. D The circumference = 2πr = 16π. Dividing by 2π gives r = 8. Area =πr 2 =π(8) 2 = 64π. 3. 8 Draw a figure as shown, including the tangent segment and the radius extended to the point of tangency. You can find x with the Pythagorean theorem: A B 6 10 x 5. 8 Draw the segments shown. Choose point E to make rectangle ACDE and right triangle AEB. No- tice that CD = AE, because opposite sides of a rec- tangle are equal. You can find AE with the Pythagorean theorem: (AE) 2 + 6 2 = 10 2 Simplify: (AE) 2 + 36 = 100 Subtract 36: (AE) 2 = 64 Take the square root: AE = 8 Since CD = AE, CD = 8. C A B D 2 2 6 2 8 E 404 McGRAW-HILL’S SAT 405 ESSENTIAL ALGEBRA 2 SKILLS 1. Sequences 2. Functions 3. Transformations 4. Variation 5. Data Analysis 6. Negative and Fractional Exponents CHAPTER 11 ✓ Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use. 406 McGRAW-HILL’S SAT Analyzing Sequences A sequence is just a list of numbers, each of which is called a term. An SAT math question might ask you to use a sequential pattern to solve a problem, such as “How many odd numbers are in the first 100 terms of the sequence 1, 2, 3, 1, 2, 3, . . . ?” An SAT sequence question usually gives you the first few terms of a sequence or the rule for generating the sequence, and then asks you ei- ther to find a specific term in the sequence (as in “What is the 59th term of this sequence?”) or to analyze a subset of the sequence (as in “What is the sum of the first 36 terms of this se- quence?”). To tackle sequence problems: 1. Use the pattern or rule to write out the first six to eight terms of the sequence. 2. Try to identify the pattern in the sequence. Notice in particular when the sequence be- gins to repeat itself, if it does. 3. Use this pattern, together with whole-num- ber division (Chapter 7, Lesson 7), if it’s helpful, to solve the problem. Example: Ϫ1, 2, Ϫ2, . . . The first three terms of a sequence are shown above. Each term after the second term is found by dividing the preceding term by the term before that. For exam- ple, the third term is found by dividing the second term, 2, by the first term, Ϫ1. What is the value of the 218th term of this sequence? Don’t panic. You won’t have to write out 218 terms! Just write out the first eight or so until you notice that the sequence begins to repeat. The fourth term is Ϫ2 ÷ 2 = Ϫ1, the fifth term is Ϫ1 ÷ Ϫ2 = 1/2, and so on. This gives the sequence Ϫ1, 2, Ϫ2, Ϫ1, 1 / 2 , Ϫ 1 / 2 , Ϫ1, 2, . . . . Notice that the first two terms of the sequence, Ϫ1 and 2, have come back again! This means that the first six terms in the sequence, the underlined ones, will just repeat over and over again. Therefore, in the first 218 terms, this six-term pattern will repeat times, or 36 with a remainder of 2. So, the 218th term will be the same as the second term in the sequence, which is 2. Example: Ϫ1, 1, 0, Ϫ1, 1, 0, . . . If the sequence above repeats as shown, what is the sum of the first 43 terms of this sequence? Since the sequence clearly repeats every three terms, then in 43 terms this pattern will repeat 43 ÷ 3 = 14 (with remainder 1) times. Each full repetition of the pattern Ϫ1, 1, 0 has a sum of 0, so the first 14 rep- etitions have a sum of 0. This accounts for the sum of the first 14 ϫ 3 = 42 terms. But you can’t forget the “remainder” term! Since that 43rd term is Ϫ1, the sum of the first 43 terms is Ϫ1. You won’t need to use the formulas for “arith- metic sequences” or “geometric sequences” that you may have learned in algebra class. Instead, SAT “sequence” questions simply require that you figure out the pattern in the sequence. 218 6 36 2 3 ÷= Lesson 1: Sequences There’s a lot of detail to learn and understand to do well on the SAT. For more tools and resources that will help, visit our Online Practice Plus at www.MHPracticePlus.com/SATmath. Concept Review 1: Sequences 1. What is a sequence? 2. If the pattern of a sequence repeats every six terms, how do you determine the 115th term of the sequence? 3. If the pattern of a number sequence repeats every four terms, how do you find the sum of the first 32 terms of the sequence? 4. If a number sequence repeats every five terms, how do you determine how many of the first 36 terms are negative? 5. What is the 30th term of the following sequence? , 1, 3, 9, . . . 6. The first term in a sequence is 4, and each subsequent term is eight more than twice the preceding term. What is the value of the sixth term? 7. The word SCORE is written 200 times in a row on a piece of paper. How many of the first 143 letters are vowels? 8. The third term of a sequence is x. If each term in the sequence except the first is found by subtracting 3 from the previous term and dividing that difference by 2, what is the first term of the sequence in terms of x? 9. A 60-digit number is created by writing all the positive integers in succession beginning with 1. What is the 44th digit of the number? 1 9 , 1 3 CHAPTER 11 / ESSENTIAL ALGEBRA 2 SKILLS 407 408 McGRAW-HILL’S SAT SAT Practice 1: Sequences 1. The first term in a sequence is x. Each subse- quent term is 3 less than twice the preceding term. What is the fifth term in the sequence? (A) 8x Ϫ 21 (B) 8x Ϫ 15 (C) 16x Ϫ 39 (D) 16x Ϫ 45 (E) 32x Ϫ 93 1 / 8 , 1 / 4 , 1 / 2 , . . . 2. In the sequence above, each term after the first is equal to the previous term times a con- stant. What is the value of the 13th term? (A) 2 7 (B) 2 8 (C) 2 9 (D) 2 10 (E) 2 11 3. The first term in a sequence is 400. Every sub- sequent term is 20 less than half of the imme- diately preceding term. What is the fourth term in the sequence? 4. In the number 0.148285, the digits 148285 re- peat indefinitely. How many of the first 500 digits after the decimal point are odd? (A) 83 (B) 166 (C) 167 (D) 168 (E) 332 5, 6, 5, 6, 6, 5, 6, 6, 6, 5, 6, 6, 6, 6, 5, . . . 5. In the sequence above, the first 5 is followed by one 6, the second 5 is followed by two 6s, and so on. If the sequence continues in this manner, how many 6s are there between the 44th and 47th appearances of the number 5? (A) 91 (B) 135 (C) 138 (D) 182 (E) 230 6. The first term in a sequence is Ϫ5, and each subsequent term is 6 more than the immedi- ately preceding term. What is the value of the 104th term? (A) 607 (B) 613 (C) 618 (D) 619 (E) 625 7. What is the units digit of 3 36 ? (A) 0 (B) 1 (C) 3 (D) 7 (E) 9 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 210, 70, . . . 8. After the first term in the sequence above, each odd-numbered term can be found by multiply- ing the preceding term by three, and each even-numbered term can be found by multi- plying the previous term by 1 ⁄3. What is the value of the 24th term? 9. The first two terms of a sequence are 640 and 160. Each term after the first is equal to one- fourth of the previous term. What is the value of the sixth term? 6, 4, . . . 10. After the first two terms in the sequence above, each odd-numbered term can be found by divid- ing the previous term by 2. For example, the third term is equal to 4 ÷ 2 = 2. Each even- numbered term can be found by adding 8 to the previous term. For example, the fourth term is equal to 2 + 8 = 10. How many terms are there before the first noninteger term? (A) 3 (B) 4 (C) 5 (D) 6 (E) 7 Ϫ2, 4, Ϫ8, . . . 11. The first three terms of a sequence are given above. If each subsequent term is the product of the preceding two terms, how many of the first 90 terms are negative? (A) 16 (B) 30 (C) 45 (D) 60 (E) 66 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 CHAPTER 11 / ESSENTIAL ALGEBRA 2 SKILLS 409 . 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use. 406 McGRAW-HILL’S SAT Analyzing Sequences A sequence is just a list of numbers, each of which is called a term. An SAT math. The circle has area 36π, so the sec- tor has area 12π. ΔRNT consists of two 30 -6 0 -9 0° triangles, with sides as marked, so its area is . Therefore, half of the origi- nal shaded region is , and the. number? 1 9 , 1 3 CHAPTER 11 / ESSENTIAL ALGEBRA 2 SKILLS 407 408 McGRAW-HILL’S SAT SAT Practice 1: Sequences 1. The first term in a sequence is x. Each subse- quent term is 3 less than twice the preceding term.