1. If two figures are similar, then their corresponding sides are ____________________ and their corresponding angles are ____________________. 2. What are the three sets of conditions of which any one is sufficient to show that two triangles are similar? a. b. c. 3. The hypotenuses of two similar right triangles are 4 centimeters and 6 centimeters long, respectively. If the area of the larger triangle is 27, what is the area of the smaller one? Concept Review 6: Similar Figures ᐉ 1 ᐉ 2 A BC DE Note: Figure not drawn to scale. 4. In the figure above, ᐍ 1 ⏐⏐ ᐍ 2 , AC = 4, BC = 5, and CE = 6. What is the length of DE? 5. In a 5- ϫ 8-inch rectangular photograph, the image of a tree is 3 inches high. The photograph is then mag- nified until its area is 1,000 square inches. What is the height of the tree image in the larger photograph? 390 McGRAW-HILL’S SAT CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 391 1. The ratio of the areas of two squares is 4:1. If the perimeter of the smaller square is 20, what is the perimeter of the larger square? (A) 5 (B) 10 (C) 20 (D) 40 (E) 80 2. A scale drawing of a rectangular patio mea- sures 5 centimeters by 7 centimeters. If the longer side of the actual patio is 21 feet, what is the area, in square feet, of the actual patio? (A) 72 (B) 315 (C) 356 (D) 441 (E) 617 Note: Figure not drawn to scale. 3. In the figure above, C and D are the centers of the two circles with radii of 3 and 2, respec- tively. If the larger shaded region has an area of 9, what is the area of the smaller shaded region? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8 a° a° 3 2 C D 4. In the figure above, ഞ 1 ⎥⎥ ഞ 2 . If EF = x, and EG = y, then which of the following represents the ratio of CD to BC? (A) (B) (C) (D) (E) 5. A circular cone with a base of radius 5 has been cut as shown in the figure above. What is the height of the smaller cone? (A) (B) (C) (D) (E) 104 5 96 5 96 1 2 96 13 8 13 8 5 5 1+ x y 1− x y y x −1 1+ y x 1− y x BC D EF G A ᐉ 1 ᐉ 2 SAT Practice 6: Similar Figures Note: Figure not drawn to scale. 6. In the figure above, what is the perimeter of the shaded trapezoid? 6 3 10 Note: Figure not drawn to scale. 7. In the figure above, BD ––– is parallel to EG ––– , AD = 6, DG = 4, and ΔAEF has an area of 75. What is the area of ΔABC? (A) 27 (B) 36 (C) 45 (D) 54 (E) 63 A GF E D C B 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 392 McGRAW-HILL’S SAT CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 393 Concept Review 6 1. If two figures are similar, then their correspond- ing sides are proportional and their correspond- ing angles are equal (or congruent). 2. a. two pairs of corresponding angles are equal b. two pairs of corresponding sides are propor- tional and the included angles are equal c. all three pairs of corresponding sides are proportional 3. The ratio of the sides is 4:6 or 2:3. The ratio of the areas is the square of the ratio of sides, which is 4:9. If x is the area of the smaller triangle, then x/27 = 4/9. Solving for x gives x = 12. 4. If ᐍ 1 ⎥⎥ ᐍ 2 , then the two triangles must be similar. Since corresponding sides are proportional, AC/AE = BC/DE. Substituting, this gives 4/10 = 5/DE. Cross-multiply: 4DE = 50 Divide by 4: DE = 12.5 5. A 5- ϫ 8-inch rectangle has an area of 40 square inches. The ratio of areas, then, is 40:1,000, or 1:25. This is the square of the ratio of lengths, so the ratio of lengths must be 1:5. If x is the length of the larger tree image, then 3/x = 1/5. Cross-multiplying gives x = 15, so the tree is 15 inches high in the larger photograph. Answer Key 6: Similar Figures ᐉ 1 ᐉ 2 A BC DE 4 5 6 SAT Practice 6 1. D If the ratio of the areas is 4:1, then the ratio of corresponding lengths is the square root: 2:1. If the perimeter of the smaller square is 20, then the perimeter of the larger one is twice as big. 2. B Find the width of the patio with a proportion: 5/7 = x/21 Cross-multiply: 7x = 105 Divide by 7: x = 15 So the patio is a 15- × 21-foot rectangle, which has an area of 15 × 21 = 315 square feet. 3. A The two regions are similar, because the central angles are the same. The ratio of their correspond- ing lengths is 3:2, so the ratio of their areas is 9:4. Since the larger area is 9, the smaller area must be 4. 4. C If EF has length x and EG has length y, then FG must have length y − x, as shown. Since the two lines are parallel, ΔABC is sim- ilar to ΔAEF and ΔACD is similar to ΔAFG. Therefore AC/CF = BC/x and AC/CF = CD/(y − x). So BC/x = CD/(y − x), and therefore CD/BC = (y − x)/x = y/x − 1. 5. B The height of the larger cone can be found with the Pythagorean theorem to be 12. (It’s the old 5-12-13 right triangle!) Since the two trian- gles are similar, x/12 = 8/13. Multiplying by 12 gives x = 96/13. 6. 28 The two trian- gles are similar be- cause their corresponding angles are equal. Since they are right triangles, the missing sides can be found with the Pythagorean theorem. Your diagram should look like the one above. The perimeter is 3 + 8 + 5 + 12 = 28. 7. A Since the lines are parallel, ΔABC i s similar to ΔAEF and ΔACD is similar to ΔAFG. Therefore, AD/AG = AC/AF = 6/10 = 3/5. The ratio of areas between ΔABC and ΔAEF is the square of the ratio of sides, which is (3/5) 2 = 9/25. Since ΔAEF has an area of 75, (the area of ΔABC)/75 = 9/25. So ΔABC has an area of 27. 8 5 5 12 x A GFE D C B 6 4 6 3 10 8 12 5 B C D EF G A ᐉ 1 ᐉ 2 x y-x y Volume The SAT math section may include a question or two about volumes. Remember two things: • The volume of a container is nothing more than the number of “unit cubes” it can hold. • The only volume formulas you will need are given to you in the “Reference Information” on every math section. Example: How many rectangular bricks measuring 2 inches by 3 inches by 4 inches must be stacked together (without mortar or any other material) to create a solid rectangular box that measures 15 inches by 30 inches by 60 inches? Don’t be too concerned with how the bricks could be stacked to make the box; there are many possible arrangements, but the arrangement doesn’t affect the answer. All you need to know is that it can be done. If so, just looking at the volumes is enough: if you use n bricks, then the box must have a volume that is n times larger than the volume of one brick. Each brick has a volume of 2 × 3 × 4 = 24 cubic inches. The box has a volume of 15 × 30 × 60 = 27,000 square inches. The number of bricks, then, must be 27,000/24 = 1,125. Lesson 7:Volumes and 3-D Geometry 3-D Distances If you are trying to find the length of a line seg- ment in three dimensions, look for a right tri- angle that has that segment as its hypotenuse. Example: The figure at right shows a cube with edges of length 4. If point C is the midpoint of edge BD, what is the length of AC ––– ? Draw segment CE –– to see that AC –– is the hypotenuse of right tri- angle ΔAEC. Leg AE –– has a length of 4, and leg EC –– is the hypotenuse of right triangle ΔEBC, with legs of length 2 and 4. Therefore, so . One possible shortcut for finding lengths in three dimensions is the three-dimensional dis- tance formula: If you think of point A in the cube above as being the origin (0, 0, 0), then point C can be considered to be (4, 4, 2). The distance from A to C, then, is dxx yy zz=− () +− () +− () 21 2 21 2 21 2 AC = () += += =20 4 20 16 36 6 2 2 EC =+=+=24 416 20 22 B DC A 4 B DC A 4 4 2 E 394 McGRAW-HILL’S SAT 40 40 20 16164 36 6 222 − () +− () +− () =++= =. CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 395 1. What is the definition of volume? 2. Write the formula for the volume of a rectangular box. 3. Write the 3-D distance formula. 4. Graph the points A(−2, 3, 1) and B(2, 1, −2) on an x-y-z graph. 5. What is the distance from point A to point B in the figure above? 6. The two containers with rectangular sides in the figure above have the interior dimensions shown. Both con- tainers rest on a flat, horizontal surface. Container A is filled completely with water, and then this water is poured, without spilling, into Container B. When all of the liquid is poured from Container A into Container B, what is the depth of the water in Container B? Concept Review 7: Volumes and 3-D Geometry 12 inches 8 inches 8 inches Container B Container A 4 inches 6 inches 10 inches 1. The length, width, and height of a rectangular box, in centimeters, are a, b, and c, where a, b, and c are all integers. The total surface area of the box, in square centimeters, is s, and the volume of the box, in cubic centimeters, is v. Which of the following must be true? I. v is an integer. II. s is an even integer. III. The greatest distance between any two vertices of the box is . (A) I only (B) I and II only (C) I and III only (D) II and III only (E) I, II, and III 2. The figure above shows a rectangular box in which AB = 6, AC = 5, AD = 8, and F is the midpoint of BE. What is the length of the short- est path from A to F that travels only on the edges of the box and does not pass through ei- ther point B or point C? (A) 27.5 (B) 28.5 (C) 29.5 (D) 30 (E) 30.5 A B C F E D abc 222 ++ 3. A pool-filling service charges $2.00 per cubic meter of water for the first 300 cubic meters and $1.50 per cubic meter of water after that. At this rate, how much would it cost to have the service fill a rectangular pool of uniform depth that is 2 meters deep, 20 meters long, and 15 meters wide? (A) $450 (B) $650 (C) $800 (D) $1,050 (E) $1,200 4. In the figure above, a rectangular box has the dimensions shown. N is a vertex of the box, and M is the midpoint of an edge of the box. What is the length of NM ––– ? (A) (B) (C) (D) (E) 125 108 98 77 63 M N 5 6 8 SAT Practice 7: Volumes and 3-D Geometry 396 McGRAW-HILL’S SAT CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 397 5. A cereal company sells oatmeal in two sizes of cylindrical containers. The smaller container holds 10 ounces of oatmeal. If the larger con- tainer has twice the radius of the smaller con- tainer and 1.5 times the height, how many ounces of oatmeal does the larger container hold? (The volume of a cylinder is given by the formula V = πr 2 h.) (A) 30 (B) 45 (C) 60 (D) 75 (E) 90 6. The figure above shows a rectangular solid with a volume of 72 cubic units. Base ABCD has an area of 12 square units. What is the area of rectangle ACEF? A BC D E F 3 7. The figure above shows a wedge-shaped hold- ing tank that is partially filled with water. If the tank is 1/16 full, what is the depth of the water at the deepest part? (A) 3 (B) 2 (C) 1.5 (D) 1 (E) 0.75 12 16 20 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 Concept Review 7 1. The volume of a solid is the number of “unit cubes” that fit inside of it. 2. V = lwh 3. 4. Your graph should look like this one: dxx yy zz=− () +− () +− () 21 2 21 2 21 2 5. Using the 3-D distance formula, 6. Since the water is poured without spilling, the volume of water must remain the same. Con- tainer A has a volume of 4 × 6 × 10 = 240 cubic inches. Since Container B is larger, the water won’t fill it completely, but will fill it only to a depth of h inches. The volume of the water can then be calculated as 8 × 8 × h = 64h cubic inches. Since the volume must remain the same, 64h = 240, so h = 3.75 inches. d =−− () () +− () +− − () = () +− () +− () 2 2 13 21 423 2 22 222 ==++=16 4 9 29 Answer Key 7:Volumes and 3-D Geometry x y z A 3 1 −2 B 2 1 −2 SAT Practice 7 1. E v = abc, so if a, b, and c are integers, v must be an integer also and statement I is true. The total surface area of the box, s, is 2ab + 2bc + 2ac = 2(ab + bc + ac), which is a multiple of 2 and therefore even. So statement II is true. Statement III is true by the 3-D distance formula. 3. D The volume of the pool is 2 × 20 × 15 = 600 cubic meters. The first 300 cubic meters cost 300 × 2 = $600, and the other 300 cubic meters cost 300 × 1.50 = $450, for a total of $1,050. 4. C Draw segment NP ––– as shown. It is the hy- potenuse of a right triangle, so you can find its length with the Pythagorean theorem: NP =+=+=8 5 64 25 89 22 A B C F E D 8 8 2.5 6 5 M N 5 6 8 P 3 5 8 89 NM –––– is the hypotenuse of right triangle ΔNPM, so . NM = () += +=89 3 89 9 98 2 2 2. C The path shown above is the shortest under the circumstances. The length of the path is 8 + 6 + 5 + 8 + 2.5 = 29.5. 398 McGRAW-HILL’S SAT CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 399 5. C If the volume of the smaller container is V =πr 2 h, then the volume of the larger container is π(2r) 2 (1.5h) = 6πr 2 h = 6v. So the larger container holds six times as much oatmeal as the smaller one. The smaller container holds 10 ounces of oat- meal, so the larger one holds 10 × 6 = 60 ounces. 6. 30 Mark up the diagram as shown. If the base has an area of 12, AB must be 4. If the volume of the box is 72, then the height must be 72/12 = 6. AC must be 5, because it’s the hypotenuse of a 3-4-5 triangle. So the rectangle has an area of 5 × 6 = 30. 7. A If the volume of the water is 1/16 the volume of the tank, the smaller triangle must have an area 1/16 that of the larger triangle. The two are simi- lar, so the ratio of the lengths must be 1/4, because 1/16 = (1/4) 2 . Therefore, the depth of water is 1/4 the depth of the tank: 12/4 = 3. A BC D E F 3 4 5 6 . 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 392 McGRAW-HILL’S SAT CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 393 Concept Review 6 1. If two figures are similar, then their correspond- ing sides are proportional and their correspond- ing angles. be 12. (It’s the old 5-1 2-1 3 right triangle!) Since the two trian- gles are similar, x/12 = 8/13. Multiplying by 12 gives x = 96/13. 6. 28 The two trian- gles are similar be- cause their corresponding. length of NM ––– ? (A) (B) (C) (D) (E) 125 108 98 77 63 M N 5 6 8 SAT Practice 7: Volumes and 3-D Geometry 396 McGRAW-HILL’S SAT CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 397 5. A cereal company