1. If point A has coordinates (3, 5), point B has coordinates (3, 2), and ABCD is a square, which of the following could be the coordi- nates of point C? (A) (4, 2) (B) (6, 2) (C) (6, 6) (D) (4, 6) (E) (8, 2) 2. If ഞ 1 is a horizontal line passing through (1, 8) and ഞ 2 is a vertical line passing through (−3, 4), then at what point do ഞ 1 and ഞ 2 intersect? (A) (−3, 8) (B) (1, 4) (C) (−1, 6) (D) (−2, 12) (E) (0, 0) 3. The point (−3, 4) is on a circle with its center at the origin. Which of the following points must also be on the circle? I. (0, −5) II. (−4, −3) III. (3, 4) (A) I only (B) II only (C) I and II only (D) II and III only (E) I, II, and III 4. If the point (3, −7) is the center of a circle and the point (8, 5) is on the circle, what is the cir- cumference of the circle? (A) 13π (B) 15π (C) 18π (D) 25π (E) 26π 5. In the figure above, point E is to be drawn so that ΔCDE has the same area as ΔABC. Which of the following could be the coordinates of E? (A) (16, 5) (B) (3, 8) (C) (5, 12) (D) (2, 16) (E) (4, 24) x y 1 1 B CA D 6. What is the area, in square units, of the shaded region in the figure above? (A) 32 (B) 33 (C) 34 (D) 35 (E) 36 Note: Figure not drawn to scale. 7. Points A and B lie on the line y = 6, as shown above. Lines ഞ 1 and ഞ 2 pass through the origin, and ഞ 1 has a slope of 1 ⁄2. If the distance from A to B is 4, what is the slope of line ഞ 2 ? y x y = 6 O ᐉ 1 ᐉ 2 A B x y 1 1 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 6 380 McGRAW-HILL’S SAT SAT Practice 4: Coordinate Geometry CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 381 Answer Key 4: Coordinate Geometry Concept Review 4 1. (−2, 1) 2. (0.5, 2) 3. 2 ⁄5 4. Your line should be vertical (straight up and down) and pass through B. 5. The x-coordinate of all the points is 3, so ഞ 1 can be described by the equation x = 3. 6. ഞ 2 should be horizontal (straight across) and pass through B. 7. ഞ 3 should be a horizontal (straight across) line one unit below the x-axis. 8. (−2, 5) 9. (1, 8) (Notice that the new segment must be the same length as AB, but its slope is the negative reciprocal of AB’s slope, that is, −5/2.) 10. (8, 5) 11. The area of the rectangle is 108, and its length DC is 14 − 2 = 12. So its height is 108/12 = 9. There- fore, k − 1 = 9, and k = 10. m = 14 n = 10 p = 1 12. AC is the hypotenuse of a right triangle with legs of 9 and 12. This is a 3-4-5 triangle times 3—a 9-12-15 triangle—so AC is 15. The perimeter of the rectangle is 9 + 12 + 9 + 12 = 42. So the ratio of the diagonal to the perimeter is 15/42 = 5/14. 13. The slope of DB is 3/4, or .75. 14. AC and DB intersect at the midpoint of each seg- ment. The midpoint of AC is ((14 + 2)/2, (10 + 1)/2) = (8, 5.5). 15. (26, 19) 16. Use the left side of the triangle as the base. This way, the base is vertical and the height is horizontal, so the lengths are easier to find. The base is 4 units and the height is 7 units, so the area is (4 × 7)/2 = 14. 17. The reflection of the triangle over the line x = 3 is shown above. The “leftmost” point has an x -coordinate of 0. 18. No matter how the triangle is reflected, the area remains the same. The area is still 14. x y (6, 7) (-1, 3) (-1, -1) b = 4 x = 3 h = 7 SAT Practice 4 1. B The vertices of a square must always be listed in consecutive order, so point C must follow consecutively after B and can be in either position shown in the figure at right. Therefore, C can be at (0, 2) or (6, 2). 2. A The horizontal line passing through (1, 8) is the line y = 8, and the vertical line passing through (−3, 4) is the line x =−3. So they intersect at (−3, 8). 3. E The distance from (0, 0) to (−3, 4) is 5, which is the radius of the circle. Therefore, any point that is 5 units from the origin is also on the circle. Each of the given points is also 5 units from the origin. 4. E The distance from (3, −7) to (8, 5) is . The circumference is 2πr = 26π. 83 5 7 5 12 1 3 2 2 22 − () +−− () () =+= 5. D ΔABC has a base of 4 and height of 8, so its area is (4 ×8)/2 =16. Since the base of ΔCDE is 2, its height must be 16 if it is to have the same area as ΔABC. The y-coordinate of E, then, must be 16 or −16. 6. E Draw a rectangle around the quadrilateral as in the figure at right. The rectangle has an area of 9 × 8 = 72. If we “remove” the areas of the four right triangles from the corners, the area of the shaded re- gion will remain. 72 − 4 − 12 − 12.5 − 7.5 = 36 7. 3 ⁄8 The y-coordinate of point A is 6, which means the “rise” from O to A is 6. Since the slope of ഞ 1 is 1 ⁄2, the “run” must be 12. The “run” from O to B is 12 + 4 = 16, and the “rise” is 6, so the slope of ഞ 2 is 6 ⁄16 = 3 ⁄8. x y 1 C 2 A C 1 B x y 4 12.5 7.5 12 9 8 y x y = 6 O ᐉ 1 ᐉ 2 A B 4 6 12 The Formulas The only area or perimeter formulas you will need for the SAT will be given at the beginning of each section: Reference Information r w ᐉ h b w ᐉ h h r c a b x 2x x 3 s 2 s s A = p r 2 A = ᐉw C = 2p r A = bh V = p r 2 h V = ᐉwh c 2 = a 2 + b 2 Special right triangles 1 2 45° 45° 60° 30° The arc of a circle measures 360°. Every straight angle measures 180°. The sum of the measures of the angles in a triangle is 180°. Finding the area of an obtuse triangle can be tricky. Keep in mind that any side can be the base—rotate the triangle if it helps. Also re- member that an altitude might be outside the triangle, as in the diagram below. Strange Shapes Don’t panic when you see a strange-looking shape on an SAT. Just notice how the shape re- lates to simple shapes. Example: In the figure below, the shaded region is con- structed of only horizontal and vertical sides. That is, all angles are right angles. What is the perime- ter of the shaded region? Don’t confuse the area formula for a circle (πr 2 ) with the circumference formula (2πr). Just re- member that areas are measured in square units, so the area formula is the one with the radius squared. Using Diagrams If a geometry problem doesn’t include a figure, draw one, because seeing the relationships among the parts is essential to solving geome- try problems! If it does include a figure, mark it up with any information you find! You can use the diagram to estimate angle measures and lengths, unless it is labeled “Note: Figure not drawn to scale,” which means that the figure is drawn inaccurately, or in only one of many different possible ways. In this case, it often helps to redraw the figure. If it’s drawn inaccurately, redraw it accurately, and see whether anything important changes. If it can be drawn in different ways, redraw it so that it is as different as possible from the original, but all of the given information is maintained. 20 15 A B Compare the shaded region to the rectangle, but keep in mind that the question asks about the perimeter, not the area! Even though the area of the shaded region is clearly less than the area of the rectangle, their perimeters must be the same! How do we know? Consider the two different paths from A to B. Notice that all the horizontal segments of the “jagged” path add up in length to 382 McGRAW-HILL’S SAT Lesson 5: Areas and Perimeters height base CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 383 the horizontal part of the “simple” path along the rectangle. The same is true of the vertical parts. So the perimeter is 15 + 20 + 15 + 20 = 70. Example: If the circle with center C has an area of 16π, what is the area of the shaded region? Piece together the strange shape from simple shapes. Notice that the shaded region is simply a quarter of the circle minus the triangle. If the area of the circle is 16π, then a quarter of the circle has an area of 4π. Since πr 2 = 16π, the radius must be 4. Therefore, the base and height of the triangle are both 4, and the area of the triangle is (4×4)/2 = 8. Therefore, the area of the shaded region is 4π−8. C Draw a diagram in the space below for the following situation: The length and width of rectangle ABCD are integers. The area of ABCD is 32. Diagonal AC is extended to point F such that C is the midpoint of AF. 1. If the area of ABCD is 32, what is the area of ΔFDB? 2. What is the ratio of the area of ΔFCB to the area of ΔFAD? Questions 3 and 4 pertain to the diagram below, in which P is the center of the semicircle. 3. What is the area of the figure above? 4. What is the perimeter of the figure above? 5. The figure above consists of a rectangle and a curved path which is made up of 10 semicircles of equal diameter. If the total length of this curved path is 40π, then the area of the rectangle is (A) 40 (B) 80 (C) 96 (D) 192 (E) 384 Concept Review 5: Areas and Perimeters P 5 11 10 384 McGRAW-HILL’S SAT CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 385 1. In the figure above, a circle is inscribed in a square and intersects the square at points A, B, C, and D. If AC = 12, what is the total area of the shaded regions? (A) 18 (B) 36 (C) 18π (D) 24π (E) 72 2. In the figure above, ABCD is a rectangle and P and R are midpoints of their respective sides. What is the area of ΔAPR? (A) 54 (B) 68 (C) 72 (D) 78 (E) 96 AB C D R P 24 8 A B C D 3. A fence encloses three sides of a rectangular garden that is bordered on the other side by a barn, as shown in the figure above. If the total length of the fence is 44 meters and p is the length, in meters, of the fence parallel to the barn wall, then which of the following expresses the area of the garden? (A) 22p − p 2 (B) (C) 22p (D) (E) 44p − p 2 4. What is the maximum number of pieces that can be cut from a paper circle with four straight cuts if none of the pieces are moved or folded between cuts? (A) 7 (B) 8 (C) 9 (D) 10 (E) 11 5. In the figure above, all angles shown are right angles. What is the perimeter of the shaded re- gion? (A) 51 (B) 54 (C) 57 (D) 60 (E) 63 3 10 20 22 2 2 pp− 44 2 2 pp− barn garden p SAT Practice 5: Areas and Perimeters 6. The figure above consists of a circle with radius 2 inscribed in an equilateral triangle in which all three interior angle bisectors are drawn. What is the total area of the shaded regions? (A) (B) (C) (D) 2π+2 (E) π+2 7. In the figure above, C is the center of the circle, AC = 12, and ∠BAD = 60°. What is the perimeter of the shaded region? (A) 12 + 4π (B) (C) (D) (E) 12 3 4+ π 12 3 3+ π 63 3+ π 63 4+ π A B C D 22 3 π − 22 3 π + 23 2 8. In the figure above, PR is the radius of the cir- cle as well as a side of the rectangle. If the cir- cle has an area of 4π and the rectangle has an area of 8, then what is the perimeter of the shaded region? (A) π+8 (B) π+10 (C) π+12 (D) 2π+8 (E) 2π+12 9. The figure above shows a square with sides of 12 centimeters in which a smaller square with sides a centimeters is inscribed. If a is an inte- ger and 2 Յ b Յ 5, then what is one possible value for the area of the shaded square? 12 a b 12 P R 386 McGRAW-HILL’S SAT 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 CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 387 Concept Review 5 Your diagram should look something like the one above, although the rectangle can also have dimen- sions of 1 and 32 or 2 and 16. 1. Think of the area of ΔFDB as the sum of the areas of ΔDCB, ΔFDC, and ΔFCB. Each of these trian- gles has an area that is half of the rectangle, be- cause each one has the same base and height as the rectangle. (If you have a tough time seeing this, think of BC as the base of ΔFCB and CD as the base of ΔFDC. Since these triangles are ob- tuse, their heights are “outside” the triangles.) Therefore, the area of ΔFDB is 16 + 16 + 16 = 48. 2. We just found that ΔFCB has an area of 16. The area of ΔFAD is the area of ΔFCD + the area of ΔADC, which is 16 + 16 = 32. Therefore, the ratio is 1:2. 3. Draw the extra segment as shown and determine its length from the Pythagorean theorem. (It’s a 3-4-5 right triangle times 2!) The area of the semi- circle is 16π/2 = 8π, the area of the rectangle is 40, and the area of the triangle is 24, for a total area of 64 + 8π. Answer Key 5: Areas and Perimeters D C BA 8 4 F 8 4 P 5 1 1 10 5 6 8 4 4 SAT Practice 5 1. E Move the shaded pieces around to see that they make up half of the square. The area of the square is 12 × 12 = 144, so the shaded region has area 144/2 = 72. 2. C Find the area indirectly by subtracting the three right triangles from the rectangle. The rec- tangle has area 8 × 24 = 192, so the triangle has area 192 − 48 − 48 − 24 = 72. A B C D AB C D R P 24 8 12 12 4 4 24 48 48 4. The perimeter of the semicircle is 4π, so the perimeter of the whole figure is 26 + 4π. 5. E Each semicircle has a perimeter of 4π, which means the circumference of a “whole” circle would be 8π and therefore the diameter of each circle is 8. Therefore, the height of the rectangle is 16 and the length is 24. 24 × 16 = 384 3. B The length of the garden is p. The width is half of 44 − p. Therefore, the area is p((44 − p)/2) = (44p − p 2 )/2. 4. E You get the maximum number of pieces (11) by making sure that each cut intersects every other previous cut in a new spot. Your diagram should look something like this, with six points of intersection in the circle. 5. B Draw the extra line shown here to see that the shaded region has a perimeter equal to a 10-by-17 rec- tangle. Therefore its perimeter is 10 +17 +10 +17 =54. 3 10 20 17 6. A Move the pieces together to see that they form a right triangle. Since all of the interior an- gles of an equilateral triangle are 60° and the bisectors divide them in half, the triangle is a 30°-60°-90° triangle, so its base must be , and the area is . 232 2 23 () () () = 23 2 2√3 30° 7. E The two right triangles have two pairs of equal sides (the two radii and the shared hypotenuse), so they must be congruent triangles. Arc BD is 1 ⁄3 of the circle, with circumference 12π. Therefore, you should be able to determine the measures shown in this diagram and see that the perimeter of the shaded region is . 12 3 4+ π A B C D 1230° 30° 60° 60° 6 6 6√3 6√3 8. A If the circle has an area of 4π, its radius is 2. If the rectangle has an area of 8, its length must be 4. The arc portion of the shaded region is 1 ⁄4 of the circle with circumference 4π, so the perimeter is 2 + 2 + 4 +π=8 +π. P R 2 2 4 2 2 π 388 McGRAW-HILL’S SAT 9. 81 or 100 Consider the right triangle in the upper-left corner of the diagram. Notice that it has a hypotenuse of a and legs of length b and 12 − b. The question is asking for the area of the shaded square, which is a 2 . By the Pythagorean theorem, a 2 = b 2 + (12 − b) 2 = 2b 2 − 24b + 144. Since 2 Յ b Յ 5, the maximum possible value of a 2 is 2(2) 2 − 24(2) + 144 = 104, and the minimum pos- sible value of a 2 is 2(5) 2 − 24(5) + 144 = 74. Since a must be an integer, a 2 must be a perfect square, and the only perfect squares between 74 and 104 are 81 and 100. 12 12 - b a b b CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 389 Similar Figures When you think of similar you probably think of “almost the same, but not quite.” In mathe- matics, however, the word similar has a much more specific, technical meaning. Two figures are similar if they are the same shape, but not necessarily the same size. For instance, all cir- cles are similar to each other, and all squares are similar to each other: there is only one “shape” for a circle, and only one “shape” for a square. But there are many different shapes that a rectangle may have, so two rectangles aren’t necessarily similar. If two shapes are similar, then all correspond- ing angles are equal and all corresponding lengths are proportional. Use proportions to find the lengths of un- known sides in similar figures. Example: What is x in the figure at left below? • Two pairs of corre- sponding sides are proportional and the angles between them are equal. • All three pairs of corresponding sides are proportional. Ratios of Areas Consider two squares: one with a side length of 2 and the other with a side length of 3. Clearly, their sides are in the ratio of 2:3. What about their areas? That’s easy: their areas are 2 2 = 4 and 3 2 = 9, so the areas are in a ratio of 4:9. This demonstrates a fact that is true of all similar figures: If corresponding lengths of two similar figures have a ratio of a:b, then the areas of the two fig- ures have a ratio of a 2 :b 2 . Example: A garden that is 30 feet long has an area of 600 square feet. A blueprint of the garden that is drawn to scale depicts the garden as being 3 inches long. What is the area of the blueprint drawing of the garden? It is tempting to want to say 60 square inches be- cause 30:600 = 3:60. But be careful: the ratio of areas is the square of the ratio of lengths! You can draw a diagram, assuming the garden to be a rec- tangle. (The shape of the garden doesn’t matter: it’s convenient to draw the garden as a rectangle, but it doesn’t have to be.) Or you can simply set up the proportion using the formula: Cross-multiply: 900x = 5,400 Divide by 900: x = 6 x 600 3 30 9 900 2 2 == Lesson 6: Similar Figures a° a° b° b° 10 7 8 x The two triangles are similar because all of their corre- sponding angles are equal. (Even though only two pairs of angles are given as equal, we know that the other pair are equal also, because the angles in a trian- gle must add up to 180°.) So we can set up a proportion of corresponding sides: Cross-multiply: 7x = 80 Divide by 7: x = 80/7 = 11.43 Two triangles are similar if any of the follow- ing is true: • Two pairs of corre- sponding angles are equal. (If two pairs are equal, the third pair must be equal, too.) 10 78 = x 30 feet 20 feet 3 inches 2 inches 600 sq. ft 6 sq. in a° a° b° b° a° a° 4 2 6 3 4 2 6 3 3 1.5 . 108/12 = 9. There- fore, k − 1 = 9, and k = 10. m = 14 n = 10 p = 1 12. AC is the hypotenuse of a right triangle with legs of 9 and 12. This is a 3-4 -5 triangle times 3—a 9-1 2-1 5 triangle—so AC. “leftmost” point has an x -coordinate of 0. 18. No matter how the triangle is reflected, the area remains the same. The area is still 14. x y (6, 7) (-1 , 3) (-1 , -1 ) b = 4 x = 3 h = 7 SAT Practice 4 1 length of this curved path is 40 , then the area of the rectangle is (A) 40 (B) 80 (C) 96 (D) 192 (E) 384 Concept Review 5: Areas and Perimeters P 5 11 10 384 McGRAW-HILL’S SAT CHAPTER 10 / ESSENTIAL