Tài liệu Master the Gre 2010 - Part 30 pdf

7. If a square’s sides are each increased by 50%, by what percent does the square’s area increase? (A) 75% (B) 100% (C) 125% (D) 150% (E) 200% The correct answer is (C). Letting s 5 the length of each side before the increase, area 5 s 2 .If 3 2 s 5 the length of each side after the increase, the new area 5 S 3 2 s D 2 5 9 4 s 2 . The increase from s 2 to 9 4 s 2 is 5 4 , or 125%. GRE questions involving non-square rectangles also come in many possible flavors. For example, a question might ask you to determine area based on perimeter, or vice versa. 8. The length of a rectangle with area 12 is three times the rectangle’s width. What is the perimeter of the rectangle? inches Enter a number in the box. The correct answer is (16). The ratio of length to width is 3:1. The ratio 6:2 is equivalent, and 6 3 2 5 12 (the area). Thus, the perimeter 5 (2)(6) 1 (2)(2) 5 16. A question might involve the properties of a square or rectangle as well as those of another geometric figure, such as a right, isosceles, or equilateral triangle. 9. A rectangular block of wood is to be cut into sections to form either two triangular enclosures or three rectangular enclosures. All wood sections must be equal in length on all sides. Column A Column B The total area of the two proposed triangular enclosures The total area of the three proposed rectangular enclosures (A) The quantity in Column A is greater. (B) The quantity in Column B is greater. (C) The quantities are equal. (D) The relationship cannot be determined from the information given. The correct answer is (A). Since the sections must all be equal in length, the triangles must each be equilateral and the rectangles must each be square. The easiest way to make the comparison is to assign to the length of the original block a number that is the least common multiple of 6 and 12 (the total number of sides among the triangles and squares, respectively). That multiple, of course, is 12. Each triangle side = 2. Applying the area formula for equilateral triangles, the Chapter 11: Math Review: Geometry 273 www.petersons.com area of each triangle is 23 4 3 2 = , and the total area for both triangles is 23 , or about 3.4 (using 1.7 as an approximate value for 3 ). Given a block length of 12, and 12 sides for three squares, the length of each side of each square = 12 12 1= . Accordingly, the area of each square is 1, and the total area of all three squares is 3. Or, a question might require you to determine a combined perimeter or area of adjoining rectangles. 10. In the figure above, all intersecting line segments are perpendicular. What is the area of the shaded region, in square units? (A) 84 (B) 118 (C) 128 (D) 139 (E) 238 The correct answer is (C). The figure provides the perimeters you need to calculate the area. One way to find the area of the shaded region is to consider it as what remains when a rectangular shape is cut out of a larger rectangle. The area of the entire figure without the “cut-out” is 14 3 17 5 238. The “cut-out” rectangle has a length of 11, and its width is equal to 17 2 4 2 3 5 10. Thus, the area of the cut-out is 11 3 10 5 110. Accordingly, the area of the shaded region is 238 2 110 5 128. Another way to solve the problem is to partition the shaded region into three smaller rectangles, as shown in the next figure, and sum up the area of each. A GRE question about a non-rectangular parallelogram might focus on angle mea- sures. These questions are easy to answer. In any parallelogram, opposite angles are congruent, and adjacent angles are supplementary. (Their measures total 180°.) So if PART IV: Quantitative Reasoning274 www.petersons.com one of a parallelogram’s angles measures 65°, then the opposite angle must also measure 65°, while the two other angles each measure 115°. 11. w x z y WXZY is a parallelogram. r = 180 2 s Column A Column B pq (A) The quantity in Column A is greater. (B) The quantity in Column B is greater. (C) The quantities are equal. (D) The relationship cannot be determined from the information given. The correct answer is (D). In the figure, r° and s° are opposite angles, and therefore r = s. Given that r = 180 2 s, r and s must both equal 90. Since the diagonals of WXZY form right angles, WXZY must be a rhombus (a parallelogram in which all four sides are equal in length). However, p = q if and only if WXZY is a square. A more difficult question about a non-rectangular parallelogram might focus on area. To determine the parallelogram’s altitude, you might need to apply the Pythagorean theorem (or one of the side or angle triplets). 12. In the figure above, AB i CD and AD i BC. If BC is 4 units long and CD is 2 units long, what is the area of quadrilateral ABCD? (A) 4 (B) 4 = 2 (C) 6 (D) 8 (E) 6 = 2 The correct answer is (B). Since ABCD is a parallelogram, its area 5 base (4) 3 altitude. To determine altitude (a), draw a vertical line segment connecting point A to BC, which creates a 45°-45°-90° triangle. The ratio of the triangle’s Chapter 11: Math Review: Geometry 275 ALERT! A GRE geometry figure will not necessarily be drawn to scale—unless the figure is accompanied by a note stating that it is. TIP A non-rectangular parallelogram in which all four sides are congruent (called a rhombus) has the following in common with a square: Perimeter 5 4s; Area 5 one half the product of the diagonals. www.petersons.com hypotenuse to each leg is = 2:1. The hypotenuse AB 5 2. Thus, the altitude (a)of ABCD is 2 = 2 ,or = 2. Accordingly, the area of ABCD 5 4 3 = 2, or 4 = 2. Trapezoids A trapezoid is a special type of quadrilateral. The next figure shows a trapezoid. Referring to this figure, all trapezoids share these four properties: Only one pair of opposite sides are parallel (BC i AD). The sum of the measures of all four angles is 360°. Perimeter 5 AB 1 BC 1 CD 1 AD Area 5 BC 1 AD 2 3 altitude (that is, one half the sum of the two parallel sides multiplied by the altitude). B C A D On the GRE, a trapezoid problem might require you to determine the altitude, the area, or both. PART IV: Quantitative Reasoning276 www.petersons.com 13. To cover the floor of an entry hall, a 1' 3 12' strip of carpet is cut into two pieces, shown as the shaded strips in the figure above, and each piece is connected to a third carpet piece as shown. If the 1' strips run parallel to each other, what is the total area of the carpeted floor, in square feet? (A) 46 (B) 48 (C) 52.5 (D) 56 (E) 60 The correct answer is (E). The altitude of the trapezoidal piece is 8. The sum of the two parallel sides of this piece is 12' (the length of the 1' 3 12' strip before it was cut). You can apply the trapezoid formula to determine the area of this piece: A 5 8 3 12 2 5 48 The total area of the two shaded strips is 12 square feet, so the total area of the floor is 60 square feet. A GRE trapezoid problem might require you to find the trapezoid’s altitude by the Pythagorean theorem. Chapter 11: Math Review: Geometry 277 www.petersons.com 14. B C AD 3 5 4 120 ° In the figure above, BC i AD. What is the area of quadrilateral ABCD? (A) 5 = 2 (B) 9 = 3 2 (C) 27 = 3 4 (D) 27 2 (E) 16 The correct answer is (C). The figure shows a trapezoid. To find its area, first determine its altitude by creating a right triangle: BC AD 3 5 120 60 33 2 4 √ ° This right triangle conforms to the 30°-60°-90° Pythagorean angle triplet. Thus, the ratio of the three sides is 1: = 3:2. The hypotenuse is given as 3, so the trapezoid’s altitude is 3 = 3 2 . Now you can calculate the area of the trapezoid: S 1 2 D ~4 1 5! S 3 = 3 2 D 5 S 9 2 DS 3 = 3 2 D 5 27 = 3 4 PART IV: Quantitative Reasoning278 www.petersons.com CIRCLES For the GRE, you’ll need to know the following basic terminology involving circles: Circumference: The distance around the circle (its “perimeter”). Radius: The distance from a circle’s center to any point on the circle’s cir- cumference. Diameter: The greatest distance from one point to another on the circle’s circumference (twice the length of the radius). Chord: A line segment connecting two points on the circle’s circumference (a circle’s longest possible chord is its diameter, passing through the circle’s center). Finding a Circle’s Circumference or Area In solving most GRE circle problems, you apply one or both of the following two formulas (r 5 radius, d 5 diameter): Circumference 5 2pr,orpd Area 5pr 2 Note that the value of p is approximately 3.14, or 22 7 . For the GRE, you won’t need to work with a value for p any more precise. In fact, in most circle problems, the solution is expressed in terms of p rather than numerically. With the two formulas, all you need is one value—area, circumference, diameter, or radius—and you can determine all the others. For example: Given a circle with a diameter of 6: Radius 5 3 Circumference 5 (2)(3)p56p Area 5p(3) 2 5 9p 15. If a circle’s circumference is 10p centimeters long, what is the area of the circle, in square centimeters? (A) 12.5 (B) 5p (C) 22.5 (D) 25p (E) 10p The correct answer is (D). First, determine the circle’s radius. Applying the circumference formula C 5 2pr, solve for r: Chapter 11: Math Review: Geometry 279 TIP If answering a GRE question requires you to use a number value for p, rest assured that a rough approximation—e.g., a little greater than 3—will suffice. www.petersons.com 10p52pr 5 5 r Then, apply the area formula, with 5 as the value of r: A 5p~5! 2 5 25p Arcs and Degree Measures of a Circle An arc is a segment of a circle’s circumference. A minor arc is the shortest arc connecting two points on a circle’s circumference. For example, in the next figure, minor arc AB is the one formed by the 60° angle from the circle’s center (O). A B A circle, by definition, contains a total of 360°. The length of an arc relative to the circle’s circumference is directly proportionate to the arc’s degree measure as a fraction of the circle’s total degree measure of 360°. For example, in the preceding figure, minor arc AB accounts for 60 360 ,or 1 6 , of the circle’s circumference. 16. Circle O, as shown in the figure above, has diameters of DB and AC and has a circumference of 9. What is the length of minor arc BC? (A) 4 (B) 11 3 (C) 7 2 (D) 13 4 (E) 3 PART IV: Quantitative Reasoning280 TIP An arc of a circle can be defined either as a length (a portion of the circle’s circumference) or as a degree measure of the central angle that determines the arc. www.petersons.com The correct answer is (C). Since AO and OB are both radii, DAOB is an isosceles triangle, and therefore m∠BAO 5 70°. It follows that m∠AOB 5 40°. ∠BOC is supplementary to ∠AOB, therefore m∠BOC 5 140°. (Remember: angles from a circle’s center are proportionate to the arcs they create.) Since m∠BOC accounts for 140 360 ,or 7 18 of the circle’s circumference, we have the length of minor arc BC 5 S 7 18 D ~9!5 7 2 . Circles and Inscribed Polygons A polygon is inscribed in a circle if each vertex of the polygon lies on the circle’s circumference. A test question might require you to visualize the possible shapes or proportions of a type of triangle or quadrilateral inscribed in a circle. 17. An equilateral triangle is inscribed in a circle such that each vertex of the triangle lies along the circle’s circumference. Column A Column B The length of any side of the triangle The circle’s diameter (A) The quantity in Column A is greater. (B) The quantity in Column B is greater. (C) The quantities are equal. (D) The relationship cannot be determined from the information given. The correct answer is (B). If any of the triangle’s sides were the length of the circle’s diameter, that side would pass through the circle’s center. In this case, however, it would be impossible to construct an inscribed equilateral triangle: Each of the other two sides would need to extend beyond the circle’s circum- ference. The next figure shows an inscribed square. The square is partitioned into four congruent triangles, each with one vertex at the circle’s center (O). A B O Chapter 11: Math Review: Geometry 281 www.petersons.com Look at any one of the four congruent triangles—for example, DABO. Notice that DABO is a right triangle with the 90° angle at the circle’s center. The length of each of the triangle’s two legs ( AO and BO) equals the circle’s radius (r). Accordingly, DABO is a right isosceles triangle, m∠OAB 5 m∠OBA 5 45°, andAB 5 r = 2. (The ratio of the triangle’s sides is 1:1: = 2.) Since AB is also the side of the square, the area of a square inscribed in a circle is ~r = 2! 2 ,or2r 2 . (The area of DABO is r 2 2 , or one fourth the area of the square.) You can also determine relationships between the inscribed square and the circle: • The ratio of the inscribed square’s area to the circle’s area is 2:p. • The difference between the two areas—the total shaded area—is pr 2 2 2r 2 . • The area of each crescent-shaped shaded area is 1 4 ~pr 2 2 2r 2 !. The next figure shows a circle with an inscribed regular hexagon. (In a regular polygon, all sides are congruent.) The hexagon is partitioned into six congruent triangles, each with one vertex at the circle’s center (O). B A O 60º Look at any one of the six congruent triangles—for example, DABO. Since all six triangles are congruent, m∠AOB 5 60° (one sixth of 360°).You can see that the length of AO and BO each equals the circle’s radius (r). Accordingly, m∠OAB 5 m∠OBA 5 60°, DABO is an equilateral triangle, and length of AB 5 r. Applying the area formula for equilateral triangles: Area of DABO 5 r 2 = 3 4 . The area of the hexagon is 6 times the area of DABO, or 3r 2 = 3 2 . You can also determine relationships between the inscribed hexagon and the circle. For example, the dif- ference between the two areas—the total shaded area—is pr 2 2 3r 2 = 3 2 . PART IV: Quantitative Reasoning282 www.petersons.com . Thus, the area of the cut-out is 11 3 10 5 110. Accordingly, the area of the shaded region is 238 2 110 5 128. Another way to solve the problem is to partition. triangle conforms to the 30 -6 0 -9 0° Pythagorean angle triplet. Thus, the ratio of the three sides is 1: = 3:2. The hypotenuse is given as 3, so the trapezoid’s