RECTANGLES, SQUARES, AND PARALLELOGRAMS Rectangles, squares, and parallelograms are types of quadrilaterals—four-sided geometric figures. Here are five characteristics that apply to all rectangles, squares, and parallelograms: The sum of the measures of all four interior angles is 360°. Opposite sides are parallel. Opposite sides are congruent (equal in length). Opposite angles are congruent (the same size, or equal in degree measure). Adjacent angles are supplementary (their measures total 180°). A rectangle is a special type of parallelogram in which all four angles are right angles (90°). A square is a special type of rectangle in which all four sides are congruent (equal in length). For the GMAT, you should know how to determine the perimeter and area of each of these three types of quadrilaterals. Referring to the next three figures, here are the formulas (l 5 length and w 5 width): Rectangle Perimeter 5 2l 1 2w Area 5 l 3 w Square Perimeter 5 4s [s 5 side] Area 5 s 2 Area 5 S 1 2 D (diagonal) 2 Chapter 11: Math Review: Geometry 283 www.petersons.com Parallelogram Perimeter 5 2l 1 2w Area 5 base (b) 3 altitude (a) GMAT questions involving squares come in many varieties. For example, you might need to determine area, given the length of any side or either diagonal, or perimeter. Or, you might need to do just the opposite—find a length or perimeter given the area. For example: The area of a square with a perimeter of 8 is 4. (s 5 8 4 4 5 2; s 2 5 4) The perimeter of a square with an area of 8 is 8 = 2. (s 5 = 8 5 2 = 2;4s 5 4 3 2 = 2) The area of a square with a diagonal of 6 is 18. (A 5 S 1 2 D 6 2 5 S 1 2 D ~36!518) Or, you might need to determine a change in area resulting from a change in perimeter (or vice versa). 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 s5 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%. GMAT 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. 284 PART IV: GMAT Quantitative Section www.petersons.com 8. The length of a rectangle with area 12 is three times the rectangle’s width. What is the perimeter of the rectangle? (A) 10 (B) 12 (C) 14 (D) 16 (E) 20 The correct answer is (D). 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. Or, a question might require you to determine a combined perimeter or area of adjoining rectangles. 9. 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 GMAT question about a non-rectangular parallelogram might focus on angle measures. These questions are easy to answer. In any parallelogram, opposite angles are congruent, and adjacent angles are supplementary. (Their measures total 180°.) So if one of a parallelogram’s angles measures 65°, then the opposite angle must also measure 65°, while the two other angles each measure 115°. Chapter 11: Math Review: Geometry 285 www.petersons.com 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). 10. In the figure above, AB i CD and AD i BC.IfBC 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 hypotenuse to each leg is = 2:1. The hypotenuse AB 5 2. Thus, the altitude (a)ofABCD 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. 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 GMAT, a trapezoid problem might require you to determine the altitude, the area, or both. 286 PART IV: GMAT Quantitative Section TIP A non-rectangular parallelogram in which all four sides are congru- ent (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 11. 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 GMAT trapezoid problem might require you to find the trapezoid’s altitude by the Pythagorean theorem. 12. 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 Chapter 11: Math Review: Geometry 287 www.petersons.com The correct answer is (C). The figure shows a trapezoid. To find its area, first determine its altitude by creating a right triangle: B C A D 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 CIRCLES For the GMAT, 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 circumference. diameter: The greatest distance from one point to another on the circle’s circum- ference (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). You’ll also need to apply the two basic formulas involving circles (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 GMAT, 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 288 PART IV: GMAT Quantitative Section www.petersons.com 13. 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 : 10p52pr 5 5 r Then, apply the area formula, with 5 as the value of r: A 5p~5! 2 5 25p ADVANCED CIRCLE PROBLEMS GMAT circle problems sometimes involve other geometric figures as well, so they’re inherently tougher than average. The most common such “hybrids” involve triangles, squares, and other circles. In the next sections, you’ll learn all you need to know to handle any hybrid problem. 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). O 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. ALERT! An arc of a circle can be defined either as a length (a portion of the circle’s circumference) or as a degree measure. Chapter 11: Math Review: Geometry 289 www.petersons.com 14. A B C D O 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 The correct answer is (C). Since AO and OB are both radii, we have isosceles DAOB thus making m∠BAO 5 70°. From this we can find 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. 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). 290 PART IV: GMAT Quantitative Section 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°, and AB 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 22r 2 . • The area of each crescent-shaped shaded area is 1 4 ~pr 2 22r 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). 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 Chapter 11: Math Review: Geometry 291 www.petersons.com 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 difference between the two areas—the total shaded area—is pr 2 2 3r 2 = 3 2 . 15. The figure above shows a square that is tangent to one circle at four points, and inscribed in another. If the diameter of the large circle is 10, what is the diameter of the smaller circle? (A) 5 = 3 2 (B) 5 (C) 2p (D) 5 = 2 (E) 7.5 The correct answer is (D). The square’s diagonal is equal in length to the large circle’s diameter, which is 10. This diagonal is the hypotenuse of a triangle whose legs are two sides of the square. The triangle is right isosceles, with sides in the ratio 1:1: = 2. The length of each side of the square 5 10 = 2 ,or5 = 2. This length is also the diameter of the small circle. Tangents and Inscribed Circles A circle is tangent to a line (or line segment) if they intersect at one and only one point (called the point of tangency). Here’s the key rule to remember about tangents: A line that is tangent toacircleisalways perpendicular to the line passing through the circle’s center at the point of tangency. 292 PART IV: GMAT Quantitative Section www.petersons.com . 125%. GMAT 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. 284 PART IV: GMAT. sides multiplied by the altitude). B C A D On the GMAT, a trapezoid problem might require you to determine the altitude, the area, or both. 286 PART IV: GMAT Quantitative Section TIP A non-rectangular parallelogram. figure shows an inscribed square. The square is partitioned into four congruent triangles, each with one vertex at the circle’s center (O). 290 PART IV: GMAT Quantitative Section www.petersons.com Look