New SAT Math Workbook

B in the general equation y mx b the slope m is given

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Unformatted text preview: xpressed as 1 the fraction 3 . 12. The correct answer is 3/16. Given that the ratio of the large circle’s area to the small circle’s area is 2:1, the small circle must comprise 50% of the total area of the large circle. The shaded 3 areas comprise 8 the area of the small circle, and so the probability of randomly selecting a point in one of these three regions is 3× 1 = 3 . 8 2 16 triangle, you can easily apply either the sine or the cosine function to determine the length of the hypotenuse. Applying the function cos60° = 1 2 , set the value of this function equal to 1x x adjacent hypotenuse , then solve for x: 2 = 10 ; 2x 10 = 10 ; x = 5. 3. (B) The question describes the following 30°60°-90° triangle: Since the length of one leg is known, you can easily apply the tangent function to determine the length of the other leg (x). Applying the 3 , set the value of this 3 x opposite function equal to 6 adjacent , then solve for 3x x: = ; 3x = 6 3 ; x = 2 3 . 3 6 function tan30° = www.petersons.com 306 Chapter 16 4. (D) The area of a triangle = height. Since the figure shows a 30°-60°-90° triangle with base 3, you can easily apply the tangent function to determine the height (the vertical leg). Applying the function tan60° = 3 , let x equal the triangle’s height, set the x opposite value of this function equal to 3 adjacent , 3x then solve for x: = ; x = 3 3 . Now you 1 3 1 × base × 2 Exercise 2 1. The correct answer is 1. AB is tangent to PO ; therefore, AB ⊥ PO . Since the pentagon is regular (all sides are congruent), P bisects AB . Given that the perimeter of the pentagon is 10, the length of each side is 2, and hence AP = 1. (D) Since AC is tangent to the circle, m∠OBC = 90°. ∠BCO is supplementary to the 140° angle shown; thus, m∠BCO = 40° and, accordingly, m∠BOE = 50°. Since ∠BOE and ∠DOE are supplementary, m∠DOE = 130°. (This angle measure defines the measure of minor arc DE.) (B) Since AB is tangent to circle O at C, you can draw a radius o...
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