55. Choice e is correct. Three of the answer choices can be immediately eliminated because the length of the height cannot be negative. Answer choices b, c, and d all have negative lengths. Call the height h and the base h + 5. The area of a triangle is ᎏ b 2 h ᎏ . Substitute in h and h + 5 for the base and height, and set the area equal to 80 since the are of the given triangle is 80; ᎏ h(h 2 +5) ᎏ = 80 Solve for h. The equation is quadratic, so the quadratic formula will be used; ᎏ h 2 + 2 5h ᎏ = 80 h 2 + 5h = 160 h 2 + 5h − 160 = h Use the quadratic formula to solve. The negative value is eliminated from the answer because it does not make sense. The answer is . 56. Choice i is correct. The easiest way to find the length of one side is to draw the square on the coordi- nate plane and count the spaces. There are 7 spaces between (−2, 3) and (5, 3). Therefore, the length of a side is 7. The distance formula can also be used. First, you must decide which points are consecutive vertices of the square. Let’s use (−2, 3) and (5, 3). The distance formula is then: d = ͙(5 − (− ෆ 2)) 2 + ෆ (3 − 3) ෆ 2 ෆ d = ͙7 2 + 0 ෆ d = ͙49 ෆ d = 7 (−2,3) (5,3) (−2,−4) (5,−4) 7 −5 + ͙665 ෆ ᎏᎏ 2 −5 ± ͙665 ෆ ᎏᎏ 2 −5 ± ͙25 + 6 ෆ 40 ෆ ᎏᎏ 2 −5 ± ͙(5) 2 − ෆ 4(1)(− ෆ 160) ෆ ᎏᎏᎏ 2(1) – ACT MATH TEST PRACTICE– 194 57. Choice e is correct. The slope of the given equation is 3. The slope of a line perpendicular to the line is the opposite, reciprocal of 3. This is − ᎏ 1 3 ᎏ . Arrange each answer choice in the y = mx + b format to quickly find the slope of each choice. a. y = 3x + ᎏ 5 2 ᎏ b. y = 3x − 4 c. y = − ᎏ 2 9 ᎏ x + ᎏ 1 3 ᎏ d. y = −2x + 4 e. y = − ᎏ 1 3 ᎏ x + ᎏ 5 3 ᎏ 58. Choice f is correct. Arrange the equations of the two lines in y = mx + b format. They both become y = ᎏ 2 3 ᎏ x + 4. Therefore, they are the same line. 59. Choice b is correct. To find the midpoint, take the average of the x values and the average of the y values. ( ᎏ −4 2 +3 ᎏ , ᎏ −2 2 +8 ᎏ ) (− ᎏ 1 2 ᎏ , ᎏ 6 2 ᎏ ) (−0.5,3) The midpoint is (−0.5,3). 60. Choice i is correct. Find a common denominator and add the fractions. The common denominator is 15x. ᎏ 3 4 x ᎏ + ᎏ x − 5 1 ᎏ ᎏ 1 2 5 0 x ᎏ + ᎏ 3x(x 15 − 1) ᎏ ᎏ 20 + 3 1 x 5 ( x x − 1) ᎏ ᎏ 20 + 1 3 5 x x 2 − 3x ᎏ ᎏ 3x 2 − 1 3 5 x x +20 ᎏ 61. Choice b is correct. Since the exponent is negative, take the reciprocal of the fraction, then apply the exponent of 3. ( ᎏ 2 1 x 2 ᎏ ) −3 ( ᎏ 2 1 x 2 ᎏ ) 3 (2) 3 (x 2 ) 3 8x 6 – ACT MATH TEST PRACTICE– 195 62. Choice f is correct. Use the substitution method to solve for x and y. The second equation can easily be solved for x in terms of y. 2y − x = 0 2y = x Substitute this value for x in the first equation and solve for y. 4x = 3y + 15 4(2y) = 3y + 15 8y = 3y + 15 5y = 15 y = 3 Next, substitute the value 3 for y in the second equation to find x. 2(3) − x = 0 6 − x = 0 6 = x 63. Choice d is correct. The negative part of the exponent tells you to take the reciprocal of the number. 36 ᎏ − 2 3 ᎏ ( ᎏ 3 1 6 ᎏ ) ᎏ 3 2 ᎏ The denominator of the fractional exponent is the root and the numerator is the power. Therefore, take the square root and raise that answer to the third power. ( Ί ᎏ 3 1 6 ᎏ ๶ ) 3 ( ᎏ 1 6 ᎏ ) 3 ᎏ 2 1 16 ᎏ 64. Choice h is correct. The cube root of a negative number is negative. So, the answer must be negative. (−3) 3 = −27 and (−4) 3 = − 64; −50 falls between −27 and −64. The value of x must be between −3 and −4. 65. Choice e is correct. Every triangle has a total of 180°. 112° are used in the top angle, leaving 68° to be shared equally between the bottom two angles. 68 ÷ 2 = 34°. – ACT MATH TEST PRACTICE– 196 66. Choice f is correct. Recall that all triangles have 180°. Next, using the two angle measures given, find the two bottom angles of the triangle. The bottom left angle is supplementary (adds to 180°) with 120°, therefore, it is 60°. The bottom right angle is a corresponding angle to the 21° angle and, there- fore, is 21°. The three angles in the triangle must add to 180°, so x is 99°. 67. Choice d is correct. Keeping in mind that a tangent line will only intersect the circle in one place, draw the graph of the circle on the coordinate plane to see that the radius must be 4. 68. Choice g is correct. The equation of a circle is in the form (x − h) 2 + (y − k) 2 = r 2 where r is the radius. Since the given equation is already in this form, you can find the radius quickly. r 2 = 36; therefore, r = 6. Use the formula A = πr 2 to find the area of the circle; A = π(6) 2 or 36π. 69. Choice b is correct. Find the measures of angles DBC and BDC by using the supplements given (remember that supplementary angles add to 180°). ∠DBC = 60° and ∠BDC = 70°. The three angles of a triangle must add to 180°. Therefore, ∠BCD = 50°. AB C D E 120° 60° 70° 50° 110° (4,−2) 4 x° 21° 21° 120° 60° – ACT MATH TEST PRACTICE– 197 70. Choice g is correct. A triangle that has sides in the ratio 1:1:͙2 ෆ has angle measures of 45°, 45°, and 90°. The side that measures ͙2 ෆ is opposite the 90° angle and is the hypotenuse. See the diagram below. The smallest angle is one of the 45° angles. To find the sine of 45°, you need to know the side opposite 45°(1) and the hypotenuse (͙2 ෆ ). sin 45 = Rationalize the denominator by multiplying the numerator and denominator by ͙2 ෆ . × = The sine of 45° is . 71. Choice e is correct. −1 ≤ cos x ≤ 1; therefore, −9 ≤ 9cos x ≤ 9. The minimum value is −9. 72. Choice f is correct. A 30-60-90 triangle has side lengths in the ratio 1:͙3 ෆ :2. If the smallest side is 7, the largest side is twice 7, or 14. The hypotenuse is 14. 73. Choice c is correct. Refer to the drawing below to see the dimensions of the pool and the walkway. Notice that the walkway is 10 feet longer and 10 feet wider than the pool (NOT 5 feet) because 5 feet is added on EACH side of the pool. To find the area of the walkway, find the area of the large rectangle (walkway and pool combined), and subtract the area of the pool. Area of the walkway and pool = 34 × 22 = 748 square feet Area of the pool = 12 × 24 = 288 square feet Area of walkway = 748 − 288 = 460 square feet 34 ft. 24 ft. 12 ft. 22 ft. 5 ft. 5 ft. 5 ft. 5 ft. ͙2 ෆ ᎏ 2 ͙2 ෆ ᎏ 2 ͙2 ෆ ᎏ ͙2 ෆ 1 ᎏ 1 ᎏ 45° 45° 1 1 2 √ ¯¯¯ – ACT MATH TEST PRACTICE– 198 74. Choice f is correct. Since Y ෆ W ෆ is an altitude in an equilateral triangle, it bisects the opposite side. X ෆ W ෆ and W ෆ Z ෆ are both 7 inches. See the diagram below. An altitude also makes a right angle and, therefore, the Pythagorean theorem can be used to find the length of the altitude. Refer to triangle WXY. The hypotenuse is 14 inches and one leg is 7 inches. a 2 + b 2 = c 2 7 2 + b 2 = 14 2 49 + b 2 = 196 b 2 = 147 b = ͙147 ෆ b = 7͙3 ෆ The length of the hypotenuse is 7͙3 ෆ . 75. Choice c is correct. The equation is quadratic. Set it equal to zero and factor. 2x 2 − 2x − 12 = 0 2(x 2 − x − 6) = 0 2(x − 3)(x + 2) = 0 Set each factor equal to zero and solve. (2 can be ignored because 2 ≠ 0). x − 3 = 0 x + 2 = 0 x = 3 x = −2 The sum of the solutions is 3 + −2 = 1. 76. Choice j is correct. Use the identity sin 2 A + cos 2 A = 1. sin 2 A + ( ᎏ 1 9 0 ᎏ ) 2 = 1 sin 2 A + ᎏ 1 8 0 1 0 ᎏ = 1 sin 2 A = ᎏ 1 1 0 9 0 ᎏ sin A = Ί ᎏ 1 1 0 9 0 ᎏ ๶ = 77. Choice d is correct. The triangle given is a 45-45-90 triangle so the sides are in the ratio 1:1:͙2 ෆ . Use a proportion to find x. = x = ͙10 ෆ x ᎏ ͙2 ෆ ᎏ 1 ͙19 ෆ ᎏ 10 14 in 14 in Y ZX W 7 in 7 in – ACT MATH TEST PRACTICE– 199 . is 3 + −2 = 1. 76. Choice j is correct. Use the identity sin 2 A + cos 2 A = 1. sin 2 A + ( ᎏ 1 9 0 ᎏ ) 2 = 1 sin 2 A + ᎏ 1 8 0 1 0 ᎏ = 1 sin 2 A = ᎏ 1 1 0 9 0 ᎏ sin A = Ί ᎏ 1 1 0 9 0 ᎏ ๶ =. so the sides are in the ratio 1: 1:͙2 ෆ . Use a proportion to find x. = x = 10 ෆ x ᎏ ͙2 ෆ ᎏ 1 19 ෆ ᎏ 10 14 in 14 in Y ZX W 7 in 7 in – ACT MATH TEST PRACTICE– 19 9 . angles add to 18 0°). ∠DBC = 60° and ∠BDC = 70°. The three angles of a triangle must add to 18 0°. Therefore, ∠BCD = 50°. AB C D E 12 0° 60° 70° 50° 11 0° (4,−2) 4 x° 21 21 12 0° 60° – ACT MATH TEST