New SAT Math Workbook Episode 2 part 4 docx

Geometry 225 www.petersons.com Exercise 3 1. (A) 14 140 10 x x = = The rectangle is 30′ by 40′. This is a 3, 4, 5 right triangle, so the diagonal is 50′. 2. (C) The altitude in an equilateral triangle is always 1 2 3side ⋅ . 3. (D) This is an 8, 15, 17 triangle, making the missing side (3)17, or 51. 4. (A) The diagonal in a square is equal to the side times 2 . Therefore, the side is 6 and the perimeter is 24. 5. (C) Triangle ABC is a 3, 4, 5 triangle with all sides multiplied by 5. Therefore CB = 20. Triangle ACD is an 8, 15, 17 triangle. Therefore CD= 8. CB – CD = DB = 12. Exercise 4 1. (A) Find the midpoint of AB by averaging the x coordinates and averaging the y coordinates. 62 2 26 2 44 ++ ,,       = () 2. (C) O is the midpoint of AB. x xx y yy + + + + 4 2 2440 6 2 1624 === === , , − A is the point (0, –4). 3. (A) d = ()() = === 84 63 4 3 16 9 25 5 22 22 − ++ + - 4. (D) Sketch the triangle and you will see it is a right triangle with legs of 4 and 3. Area = ⋅⋅= ⋅⋅= 1 2 1 2 43 6bh 5. (A) Area of a circle = πr 2 πr 2 = 16π r = 4 The point (4, 4) lies at a distance of ()()40 40 32 22 − + − = units from (0, 0). All the other points lie 4 units from (0, 0). Chapter 13 226 www.petersons.com Exercise 5 1. (A) Angle B = Angle C because of alternate interior angles. Then Angle C = Angle D for the same reason. Therefore, Angle D = 30°. 2. (D) Extend AE to F. ∠A = ∠EFC ∠CEF must equal 100° because there are 180° in a triangle. ∠ AEC is supplementary to ∠CEF. ∠ AEC = 80° 3. (E) ∠ = ∠ ∠∠=° ∠ =° 13 23180 250 + 4. (C) Since ∠BEG and ∠EGD add to 180°, halves of these angles must add to 90°. Triangle EFG contains 180°, leaving 90° for ∠EFG. 5. (C) ∠ = ∠ ∠ = ∠ ∠∠= ∠∠ 13 24 12 34++ But ∠3 + ∠4 = 180°. Therefore, ∠1 + ∠2 = 180° Exercise 6 1. (D) Represent the angles as x, 5x, and 6x. They must add to 180°. 12 180 15 x x = = The angles are 15°, 75°, and 90°. Thus, it is a right triangle. 2. (D) There are 130° left to be split evenly between the base angles (the base angles must be equal). Each one must be 65°. 3. (E) The exterior angle is equal to the sum of the two remote interior angles. 4 100 25 375 x x Ax = = ==°Angle 4. (D) The other base angle is also x. These two base angles add to 2x. The remaining degrees of the triangle, or 180 – 2x, are in the vertex angle. 5. (E) ∠ = ∠ = = = AC xx x x 430210 240 20 − + ∠ A and ∠ C are each 50°, leaving 80° for ∠ B. Geometry 227 www.petersons.com Exercise 7 1. (C) A hexagon has 6 sides. Sum = (n – 2) 180 = 4(180) = 720 2. (D) Opposite sides of a parallelogram are congruent, so AB = CD. xx x AD BC x + 4216 20 614 = = == = − − 3. (B) AB = CD xx x x AB BC CD += − = = === 84 4 12 3 4 12 12 12 If all sides are congruent, it must be a rhombus. Additional properties would be needed to make it a square. 4. (B) A rhombus has 4 sides. Sum = (n – 2) 180 = 2(180) = 360 5. (C) Rectangles and rhombuses are both types of parallelograms but do not share the same special properties. A square is both a rectangle and a rhombus with added properties. Exercise 8 1. (C) Tangent segments drawn to a circle from the same external point are congruent. If CE = 5, then CF = 5, leaving 7 for BF. Therefore BD is also 7. If AE = 2, then AD = 2. BD + DA = BA = 9 2. (D) Angle O is a central angle equal to its arc, 40°. This leaves 140° for the other two angles. Since the triangle is isosceles, because the legs are equal radii, each angle is 70°. 3. (E) The remaining arc is 120°. The inscribed angle x is 1 2 its intercepted arc. 4. (A) 50 1 2 40 100 40 60 °= ° () °= ° °= + + AC AC AC 5. (D) An angle outside the circle is 1 2 the difference of its intercepted arcs. Chapter 13 228 www.petersons.com Exercise 9 1. (D) There are 6 equal squares in the surface area of a cube. Each square will have an area of 96 6 or 16. Each edge is 4. V = e 3 = 4 3 = 64 2. (C) V = πr 2 h = 22 7 · 49 · 10 = 1540 cubic inches Divide by 231 to find gallons. 3. (B) V = πr 2 h = 22 7 · 9 · 14 = 396 cubic inches Divide by 9 to find minutes. 4. (B) V = l · w · h = 10 · 8 · 4 = 320 cubic inches Each small cube = 4 3 = 64 cubic inches. Therefore it will require 5 cubes. 5. (A) Change 16 inches to 1 1 3 feet. V = 6 · 5 · 1 1 3 = 40 cubic feet when full. 5 8 · 40 = 25 Exercise 10 1. (E) If the radius is multiplied by 3, the area is multiplied by 3 2 or 9. 2. (D) If the dimensions are all doubled, the area is multiplied by 2 2 or 4. If the new area is 4 times as great as the original area, is has been increased by 300%. 3. (A) If the area ratio is 9 : 1, the linear ratio is 3 : 1. Therefore, the larger radius is 3 times the smaller radius. 4. (B) Ratio of circumferences is the same as ratio of radii, but the area ratio is the square of this. 5. (C) We must take the cube root of the volume ratio to find the linear ratio. This becomes much easier if you simplify the ratio first. 250 128 125 64 = The linear ratio is then 5 : 4. 5 4 25 5 100 20 = = = x x x Geometry 229 www.petersons.com Retest 1. (C) Area of trapezoid = 1 2 12 hb b+ () Area = ⋅ () = 1 2 310 12 33+ 2. (A) Area of circle = πr 2 = 16π Therefore, r 2 = 16 or r = 4 Circumference of circle = 2πr = 2π (4) = 8π 3. (D) The side of a square is equal to the diagonal times 2 2 . Therefore, the side is 42 and the perimeter is 16 2 . 4. (E) d = () () () = () ( ) = == 74 7 34 916 25 5 2 2 22 − + ++ 3 - 5. (D) ∠CDE must equal 65° because there are 180° in a triangle. Since AB is parallel to CD , ∠x = ∠CDE = 65°. 6. (C) Represent the angles as x, x, and 2x. They must add to 180°. 4 180 45 x x = = Therefore, the largest angle is 2x = 2(45°) = 90°. 7. (B) A pentagon has 5 sides. Sum (n – 2)180 = 3(180) = 540° In a regular pentagon, all the angles are equal. Therefore, each angle = 540 5 108=° . 8. (D) An angle outside the circle is 1 2 the difference of its intercepted arcs. 40 1 2 20 80 20 100 = = = ()x x x − − 9. (D) V = l · w · h = 2 · 6 · 18 = 216 The volume of a cube is equal to the cube of an edge. Ve e e = = = 3 3 216 6 10. (B) If the volume ratio is 8 : 1, the linear ratio is 2 : 1, and the area ratio is the square of this, or 4:1. 231 14 Inequalities DIAGNOSTIC TEST Directions: Work out each problem. Circle the letter that appears before your answer. Answers are at the end of the chapter. 1. If 4x < 6, then (A) x = 1.5 (B) x < 2 3 (C) x > 2 3 (D) x < 3 2 (E) x > 3 2 2. a and b are positive numbers. If a = b and c > d, then (A) a + c < b + d (B) a + c > b + d (C) a – c > b – d (D) ac < bd (E) a + c < b – d 3. Which value of x will make the following expression true? 3 510 4 5 << x (A) 5 (B) 6 (C) 7 (D) 8 (E) 9 4. In triangle ABC, AB = AC and EC < DB. Then (A) DB < AE (B) DB < AD (C) AD > AE (D) AD < AE (E) AD > EC 5. In triangle ABC, ∠1 > ∠2 and ∠2 > ∠3. Then (A) AC < AB (B) AC > BC (C) BC > AC (D) BC < AB (E) ∠3 > ∠1 6. If point C lies between A and B on line segment AB, which of the following is always true? (A) AC = CB (B) AC > CB (C) CB > AC (D) AB < AC + CB (E) AB = CB + AC Chapter 14 232 www.petersons.com 7. If AC is perpendicular to BD, which of the following is always true? I. AC = BC II. AC < AB III. AB > AD (A) I only (B) II and III only (C) II only (D) III only (E) I and II only 8. If x < 0 and y > 0, which of the following is always true? (A) x + y > 0 (B) x + y < 0 (C) y – x < 0 (D) x – y < 0 (E) 2x > y 9. In triangle ABC, BC is extended to D. If ∠A = 50° and ∠ACD = 120°, then (A) BC > AB (B) AC > AB (C) BC > AC (D) AB > AC (E) ∠B < ∠A 10. In right triangle ABC, ∠A < ∠B and ∠B < ∠C. Then (A) ∠A > 45° (B) ∠B = 90° (C) ∠B > 90° (D) ∠C = 90° (E) ∠C > 90° Inequalities 233 www.petersons.com 1. ALGEBRAIC INEQUALITIES Algebraic inequality statements are solved in the same manner as equations. However, do not forget that when- ever you multiply or divide by a negative number, the order of the inequality, that is, the inequality symbol must be reversed. In reading the inequality symbol, remember that it points to the smaller quantity. a < b is read a is less than b. a > b is read a is greater than b. Example: Solve for x: 12 – 4x < 8 Solution: Add –12 to each side. –4x < –4 Divide by –4, remembering to reverse the inequality sign. x > 1 Example: 6x + 5 > 7x + 10 Solution: Collect all the terms containing x on the left side of the equation and all numerical terms on the right. As with equations, remember that if a term comes from one side of the inequality to the other, that term changes sign. –x > 5 Divide (or multiply) by –1. x < –5 Chapter 14 234 www.petersons.com Exercise 1 Work out each problem. Circle the letter that appears before your answer. 1. Solve for x: 8x < 5(2x + 4) (A) x > – 10 (B) x < – 10 (C) x > 10 (D) x < 10 (E) x < 18 2. Solve for x: 6x + 2 – 8x < 14 (A) x = 6 (B) x = –6 (C) x > –6 (D) x < –6 (E) x > 6 3. A number increased by 10 is greater than 50. What numbers satisfy this condition? (A) x > 60 (B) x < 60 (C) x > –40 (D) x < 40 (E) x > 40 4. Solve for x: –.4x < 4 (A) x > –10 (B) x > 10 (C) x < 8 (D) x < –10 (E) x < 36 5. Solve for x: .03n > –.18 (A) n < –.6 (B) n > .6 (C) n > 6 (D) n > –6 (E) n < –6 6. Solve for b: 15b < 10 (A) b < 3 2 (B) b > 3 2 (C) b <− 3 2 (D) b < 2 3 (E) b > 2 3 7. If x 2 < 4, then (A) x > 2 (B) x < 2 (C) x > –2 (D) –2 < x < 2 (E) –2 ≤ x ≤ 2 8. Solve for n: n + 4.3 < 2.7 (A) n > 1.6 (B) n > –1.6 (C) n < 1.6 (D) n < –1.6 (E) n = 1.6 9. If x < 0 and y < 0, which of the following is always true? (A) x + y > 0 (B) xy < 0 (C) x – y > 0 (D) x + y < 0 (E) x = y 10. If x < 0 and y > 0, which of the following will always be greater than 0? (A) x + y (B) x – y (C) x y (D) xy (E) –2x [...]... (E) 4 If AB ⊥ CD and ∠1 > 4, then (A) (B) (C) (D) (E) 5 ∠1 > 2 4 > ∠3 2 > ∠3 2 < ∠3 2 < 4 Which of the following sets of numbers could be the sides of a triangle? (A) 1, 2, 3 (B) 2, 2, 4 (C) 3, 3, 6 (D) 1, 1.5, 2 (E) 5, 6, 12 AC < AB BC < AB ∠C > ∠ABC BC > AC ∠ABC > ∠A www.petersons.com 23 7 23 8 Chapter 14 RETEST Work out each problem Circle the letter that appears before your answer 1 If 2x >... factors of 75 (E) None of the above (D) (E) 6 24 3  1 1 1 x x x x +1 x2 x (B) (C) (D) (E) 7 x2 2 y3 x y3 If the domain of f(x) = − x is the set { 2, –1, 0, 5 2} , then f(x) CANNOT equal (A) |–1 – 2| – |5 – 6| – |–3 + 4| = (A) –5 (B) –3 (C) 1 (D) 3 (E) 5 xy 1 If f(x) = x + 1, then f ( x ) × f  x  =   (A) 1 (B) 2 y2 x2 x 6 y3 y is equivalent to: y6 x 3 x − 2 25 1 − 5 0 5 50 Which of the following equations... Exercise 1 Exercise 2 1 (A) 8 x < 10 x + 20 2 x < 20 x > −10 2 (C) 2 x < 12 x > −6 3 (E) x + 10 > 50 x > 40 4 (A) –.4x < 4 1 (D) Angle A will contain 90°, which is the largest angle of the triangle The sides from largest to smallest will be BC, AB, AC 2 (B) Since ∠SRT = ∠STR, ∠SRT will have to be greater than ∠PTR Therefore, PT > PR in triangle PRT 3 (D) Angle ABC = 60° Since there are 120 ° left for ∠A... (C) (D) (E) 5 x> 2 5 x>− 2 2 x>− 5 5 x< 2 5 x 6, then x>3 x 12 x < 12 x > – 12 If 2 (A) (B) (C) (D) (E) 6 If AB = AC and ∠1 > ∠B, then m, n > 0 If m = n and p < q, then (A) m – p < n – q (B) p – m > q –n (C) m – p > n – q (D) mp > nq (E) m + q < n + p 3 x 5 If ∠3 > 2 and ∠1 = 2, then (A) (B) (C) (D) (E) ∠B > ∠C ∠1 > ∠C BD > AD AB > AD ∠ADC > ∠ADB 7 (A) (B) (C) (D) (E) 4 8 In isosceles... equations defines a function containing the (x,y) pairs (–1,–1) and (– 1 ,0)? 2 (A) y = 3x + 2 (B) y = 2x + 1 (C) y = 6x + 5 (D) y = –4x – 2 (E) y = 4x + 3 24 4 Chapter 15 8 The figure below shows the graph of a linear function on the xy-plane If the x-intercept of line l is 4, what is the slope of l ? (A) (B) (C) (D) (E) 2 3 3 4 5 6 6 5 Not enough information to answer the question is given 9 The figure... parabola in the xyplane Which of the following equations does the graph best represent? (A) y = –x2 + 6x – 9 (B) y = x2 – 2x + 6 (C) y = 2 x2 – 4x + 6 3 (D) y = –x2 + x – 3 (E) y = x2 + 3x + 9 10 Which of the following best describes the 2 relationship between the graph of y = 2 and 2 x the graph of x = 2 in the xy-plane? (A) (B) (C) (D) (E) www.petersons.com y Mirror images symmetrical about the x-axis... shortest two) must be greater than the third side Multiply by 10 to remove decimals 4 x < 40 x > −10 5 (D) 03n > –.18 Multiply by 100 3n > −18 n > −6 6 (D) Divide by 15 10 15 2 Simplify to b < 3 b< 7 (D) x must be less than 2, but can go no lower than 2, as (–3 )2 would be greater than 4 8 (D) n + 4. 3 < 2. 7 Subtract 4. 3 from each side n < –1.6 9 (D) When two negative numbers are added, their sum will... Then (A) (B) (C) (D) (E) AD > DC AD < BD AD > AC BD > DC AB > BD www.petersons.com 23 9 24 0 Chapter 14 SOLUTIONS TO PRACTICE EXERCISES Diagnostic Test 1 2 (D) 4x < 6 6 x< 4 3 Simplify to x < 2 (B) If equal quantities are added to unequal quantities, the sums are unequal in the same order c>d (+ ) a = b a+c > b+d 3 (C) 3 x 4 < < 5 10 5 6 (E) AB = CB + AC 7 (C) In right triangle ACB, the longest side is... Subtract 4. 3 from each side n < –1.6 9 (D) When two negative numbers are added, their sum will be negative 10 (E) The product of two negative numbers is positive www.petersons.com 24 1 24 2 Chapter 14 Retest 2 x > −5 5 x>− 2 1 (B) 2 (C) If unequal quantities are subtracted from equal quantities, the differences are unequal in the opposite order m=n 6 (B) If two sides of a triangle are equal, the angles opposite... Since ∠3 > 2 and ∠1 = 2, ∠3 > ∠1 If two angles of a triangle are unequal, the sides opposite these angles are unequal, with the larger side opposite the larger angle Therefore, AB > BD 4 (D) Since ∠1 > 2 and 2 > ∠3, ∠1 > ∠3 In triangle ACD side AD is larger than side AC, since AD is opposite the larger angle 5 (C) x >6 2 x > 12 www.petersons.com BT = 1 1 ST and SA = SR Since ST = SR, 2 2 BT = SA . coordinates. 62 2 26 2 44 ++ ,,       = () 2. (C) O is the midpoint of AB. x xx y yy + + + + 4 2 244 0 6 2 1 6 24 === === , , − A is the point (0, 4) . 3. (A) d = ()() = === 84 63 4 3 16 9 25 5 22 22 −. first. 25 0 128 125 64 = The linear ratio is then 5 : 4. 5 4 25 5 100 20 = = = x x x Geometry 22 9 www.petersons.com Retest 1. (C) Area of trapezoid = 1 2 12 hb b+ () Area = ⋅ () = 1 2 310 12 33+ 2. . < 3 2 (B) b > 3 2 (C) b <− 3 2 (D) b < 2 3 (E) b > 2 3 7. If x 2 < 4, then (A) x > 2 (B) x < 2 (C) x > 2 (D) 2 < x < 2 (E) 2 ≤ x ≤ 2 8. Solve for n: n + 4. 3