6. E Label the two angles that are “vertical” to those marked b and c. No- tice that the angle marked a is an “external” angle. By the External Angle theorem, a = b + c. 7. D Write in the missing angle measures by using the fact that the sum of angles in a triangle is 180°. Then use the fact that the biggest side of a triangle is always across from the biggest angle to order the sides of each trian- gle. 8. A Consider the side of length 4 to be the base, and “attach” the side of length 6. Notice that the triangle has the greatest possible height when the two sides form a right angle. Therefore, the greatest possible area of such a triangle is (1/2)(4)(6) = 12, and the minimum possible area is 0. The greatest prime number less than 12 is 11. 30° 35° 60° 65° A B C D 55° 115° 6 4 370 McGRAW-HILL’S SAT a° b° c° CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 371 The Pythagorean Theorem The Pythagorean theorem says that in any right triangle, the sum of the squares of the two shorter sides is equal to the square of the longest side. If you know two sides of any right trian- gle, the Pythagorean theorem can always be used to find the third side. Example: In the figure below, what is x? Pythagorean theorem: 9 2 + x 2 = 15 2 Simplify: 81 + x 2 = 225 Subtract 81: x 2 = 144 Take the square root: x = 12 You can also use the modified Pythagorean theorem to find whether a triangle is acute or obtuse. If (side 1 ) 2 + (side 2 ) 2 < (longest side) 2 , the triangle is obtuse. (If the stick gets bigger, the alligator’s mouth gets wider!) If (side 1 ) 2 + (side 2 ) 2 > (longest side) 2 , the triangle is acute. (If the stick gets smaller, the alligator’s mouth gets smaller!) Special Right Triangles Certain special right triangles show up frequently on the SAT. If you see that a triangle fits one of these pat- terns, it may save you the trouble of using the Pythagorean Theorem. But be careful: you must know two of three “parts” of the triangle in order to assume the third part. x 15 9 c 2 c b 2 a 2 a b a 2 + b 2 = c 2 3-4-5 triangles More accurately, these can be called 3x-4x-5x triangles because the multi- ples of 3-4-5 also make right triangles. Notice that the sides satisfy the Pythagorean theorem. 5-12-13 triangles Likewise, these can be called 5x-12x-13x triangles because the multi- ples of 5-12-13 also make right triangles. No- tice that the sides satisfy the Pythagorean theorem. 45°-45°-90° triangles These triangles can be thought of as squares cut on the diagonal. This shows why the angles and sides are related the way they are. Notice that the sides satisfy the Pythagorean theorem. x x 2 45° 45° x 5 12 13 2.5 6 6.5 3 4 5 16 12 20 Lesson 3: The Pythagorean Theorem The Distance Formula: d 2 = (x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 so y x (x 2 , y 2 ) (x 1 , y 1 ) O x 2 – x 1 y 2 – y 1 d dxx yy=− () +− () 21 2 21 2 30°-60°-90° triangles These triangles can be thought of as equilateral triangles cut in half. This shows why the angles and sides are re- lated the way they are. Notice that the sides satisfy the Pythagorean theorem. The Distance Formula Say you want to find the distance between two points (x 1 , y 1 ) and (x 2 , y 2 ). Look carefully at this diagram and notice that you can find it with the Pythagorean theorem. Just think of the distance between the points as the hypotenuse of a right triangle, and the Pythagorean theorem becomes—lo and behold— the distance formula! 60° 30° 2x x x 3 372 McGRAW-HILL’S SAT CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 373 Concept Review 3: The Pythagorean Theorem 1. Draw an example of each of the four “special” right triangles. Use the modified Pythagorean theorem and the triangle inequality to find whether a triangle with the given side lengths is acute, obtuse, right, or impossible. 2. 5, 6, 9 3. 2, 12, 12 4. 6, 8, 11 5. 2, 2, 12 6. 3, 4, 7 7. 1.5, 2, 2.5 8. The circle above has its center at P and an area of 16π. If AP = AB, what is the area of ΔABC? ____________ 9. The area of the triangle above is 30. What is the value of h? ______________ 10. What is the height of an equilateral triangle with sides of length cm? ______________ 11. Point P is at (0, 0), point M is at (4, 0), and point N is at (9, 12). ______________ What is the perimeter of ΔMNP? 63 5 h C A B P 1. The length and width of a rectangle are in the ratio of 5:12. If the rectangle has an area of 240 square centimeters, what is the length, in cen- timeters, of its diagonal? (A) 26 (B) 28 (C) 30 (D) 32 (E) 34 2. A spider on a flat horizontal surface walks 10 inches east, then 6 inches south, then 4 inches west, then 2 inches south. At this point, how many inches is the spider from its starting point? (A) 8 (B) 10 (C) 12 (D) 16 (E) 18 3. In the figure above, ABCF is a square and ΔEFD and ΔFCD are equilateral. What is the measure of ∠AEF? (A) 15° (B) 20° (C) 25° (D) 30° (E) 35° A B C D E F 4. In the figure above, an equilateral triangle is drawn with an altitude that is also the diameter of the circle. If the perimeter of the triangle is 36, what is the circumference of the circle? (A) (B) (C) (D) (E) 36π 5. In the figure above, A and D are the centers of the two circles, which intersect at points C and E. CE –– is a diameter of circle D. If AB = CE = 10, what is AD? (A) 5 (B) (C) (D) (E) 10 3 10 2 53 52 C A B E D 12 3π 12 2π 63π 62π 374 McGRAW-HILL’S SAT SAT Practice 3: The Pythagorean Theorem CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 375 6. Point H has coordinates (2, 1), and point J has coordinates (11, 13). If HK ___ is parallel to the x-axis and JK ___ is parallel to the y-axis, what is the perimeter of ΔHJK? 7. A square garden with a diagonal of length meters is surrounded by a walkway 3 meters wide. What is the area, in square meters, of the walkway? 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 2 42 24 2 3 Note: Figure not drawn to scale. 8. In the figure above, what is the value of z? (A) 15 (B) (C) (D) (E) 30 3 30 2 15 3 15 2 x° x° 2x° 2y + 5 3y z 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 3 1. Your diagram should include one each of a 3x-4x-5x, a 5x-12x-13x, a 30°-60°-90°, and a 45°-45°-90° triangle. 2. Obtuse: 5 2 + 6 2 = 61 < 9 2 = 81 3. Acute: 2 2 + 12 2 = 148 > 12 2 = 144 4. Obtuse: 6 2 + 8 2 = 100 < 11 2 = 121 5. Impossible: 2 + 2 isn’t greater than 12 6. Impossible: 3 + 4 isn’t greater than 7 7. Right: 1.5 2 + 2 2 = 6.25 = 2.5 2 8. Since the area of a circle is πr 2 = 16π, r = 4. Put the infor- mation into the diagram. Use the Pythagorean theorem or notice that, since the hypotenuse is twice the shorter side, it is a 30°-60°-90° triangle. Either way, , so the area of the triangle is . bh () = () () =2443283 CB = 43 9. At first, consider the shorter leg as the base. In this case, the other leg is the height. Since the area is (bh)/2 = 30, the other leg must be 12. This is a 5-12-13 triangle, so the hypotenuse is 13. Now consider the hypotenuse as the base. Since 13h/2 = 30, h = 60/13 = 4.615. 10. Your diagram should look like this: The height is . 11. Sketch the diagram. Use the Pythagorean theo- rem or distance for- mula to find the lengths. The perimeter is 4 + 13 + 15 = 32. 33 3 9 ()() = Answer Key 3: The Pythagorean Theorem C A B P 4 4 4 4 3 5 h 12 13 6 3 6 3 3 3 9 P(0, 0) M(4, 0) 4 N(9, 12) 12 5 13 15 y x SAT Practice 3 1. A Draw the rectangle. If the length and width are in the ratio of 5:12, then they can be expressed as 5x and 12x. The area, then, is (5x)(12x) = 60x 2 = 240. So x = 2, and the length and width are 10 and 24. Find the diagonal with the Pythagorean theorem: 10 2 + 24 2 = a 2 , so 100 + 576 = 676 = a 2 and d = 26. (Notice that this is a 5-12-13 triangle times 2!) 2. B Draw a diagram like this. The distance from the starting point to the finishing point is the hypotenuse of a right triangle with legs of 6 and 8. Therefore, the distance is found with Pythagoras: 6 2 + 8 2 = 36 + 64 = 100 = d 2 , so d = 10. (Notice that this is a 3-4-5 triangle times 2!) 3. A Draw in the angle measures. All angles in a square are 90° and all angles in an equilateral tri- angle are 60°. Since all of the angles around point F add up to 360°, ∠EFA = 360 – 60 − 60 − 90 = 150°. Since EF = AF, ΔEFA is isosceles, so ∠AEF = (180 − 150)/2 = 15°. 4. B If the perimeter of the triangle is 36, each side must have a length of 12. Since the altitude forms two 30°-60°-90° triangles, the altitude must have length . This is also the diameter of the circle. The circum- ference of a circle is π times the diameter: . 5. C Draw in AE and AC. Since all radii of a circle are equal, their measures are both 10 as well. Therefore ΔACE is equilateral, and AD divides it into two 30°-60°-90° triangles. You can use the Pythagorean theorem, or just use the 30°-60°-90° relationships to see that . AD = 53 63 π 63 10 6 4 2 6 8 Start Finish A B C D E F 60° 60° 60° 60° 60°60° 90°90° 90° 150° 12 12 66 60°60° 6 3 C A B E D 10 10 10 5 5 376 McGRAW-HILL’S SAT CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 377 6. 36 Sketch a diagram. Point K has coordinates (11, 1). ΔHJK is a right triangle, so it satisfies the Pythagorean theorem. Your diagram should look like this one. The perimeter is 9 + 12 + 15 = 36. 7. 324 Since the garden is a square, the diagonal divides it into 45°-45°-90° triangles. Therefore the sides have a length of 24. The outer edge of the walkway is therefore 24 + 3 + 3 = 30. The area of the walkway is the difference of the areas of the squares: 30 2 − 24 2 = 324. 8. B The sum of the angles is 180°, so x + x + 2x = 4x = 180, and x = 45. Therefore the triangle is a 45°-45°-90° triangle. Since it is isosceles, 3y = 2y + 5, and therefore y = 5. The three sides, then, have lengths of 15, 15, and . 15 2 24 2 3 30 24 H(2, 1) J(11, 13) K(11, 1) x y 12 9 15 Working with Slopes Every line has a slope, which is the distance you move up or down as you move one unit to the right along the line. Imagine walking between any two points on the line. As you move, you go a certain distance “up or down.” This distance is called the rise. You also go a certain distance “left or right.” This distance is called the run. The slope is simply the rise divided by the run. Slope = rise run yy xx = − − 21 21 Plotting Points Some SAT questions may ask you to work with points on the x-y plane (also known as the coordinate plane or the Cartesian plane, after the mathematician and philosopher René Descartes). When plotting points, remember these four basic facts to avoid the common mistakes: • The coordinates of a point are always given alphabetically: the x-coordinate first, then the y-coordinate. • The x-axis is always horizontal and the y-axis is always vertical. • On the x-axis, the positive direction is to the right of the origin (where the x and y axes meet, point (0,0)). • On the y-axis, the positive direction is up from the origin (where the x and y axes meet, point (0,0)). You should be able to tell at a glance whether a slope has a positive, negative, or zero slope. If the line goes up as it goes to the right, it has a positive slope. If it goes down as it goes to the right, it has a negative slope. If it’s horizontal, it has a zero slope. If two lines are parallel, they must have the same slope. y x (3, 5) 3 "right" O 5 "up" y x (x 2 , y 2 ) (x 1 , y 1 ) O run = x 2 − x 1 rise = y 2 − y 1 y x O positive slope negative slope 0 slope Finding Midpoints The midpoint of a line segment is the point that divides the segment into two equal parts. Think of the midpoint as the average of the two endpoints. Midpoint = xxyy 1212 22 ++ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ , y x (x 2 , y 2 ) (x 1 , y 1 ) O ( x 1 +x 2 , y 1 +y 2 ) 2 2 378 McGRAW-HILL’S SAT Lesson 4: Coordinate Geometry CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 379 Concept Review 4: Coordinate Geometry Questions 1–10 refer to the figure below. Horizontal and vertical lines are spaced 1 unit apart. 1. What are the coordinates of point A? __________ 2. What are the coordinates of the midpoint of AB –– ? __________ 3. What is the slope of AB –– ? __________ 4. Draw a line through B that is parallel to the y-axis and label it ഞ 1 . 5. What do all points on ഞ 1 have in common? __________ x y A B 6. Draw a line through B that is parallel to the x-axis and label it l 2 . 7. Draw the line y =−1 and label it ഞ 3 . 8. If point A is reflected over ഞ 2 , what are the coordi- nates of its image? __________ 9. If line segment AB –– is rotated 90° clockwise about point B,what are the coordinates of the image of point A? __________ 10. If point B is the midpoint of line segment AC –– , what are the coordinates of point C? __________ Rectangle ABCD has an area of 108. Note: Figure not drawn to scale. Questions 11–15 pertain to the figure above. 11. k = _____ m = _____ n = _____ p = _____ 12. What is the ratio of AC to the perimeter of ABCD? __________ 13. What is the slope of DB ––– ? __________ 14. At what point do AC –– and DB ––– intersect? __________ 15. If B is the midpoint of segment DF ––– , what are the coordinates of point F? __________ x y A(2, k) B(m, n) C(14, p)D(2, 1) O Questions 16–18 pertain to the figure above. 16. What is the area of the triangle above? __________ 17. If the triangle above were reflected over the line x = 3, what would be the least x-coordinate of any point on the triangle? __________ 18. If the triangle above were reflected over the line y = 1, what would the area of the new triangle be? __________ x y (6, 7) (-1, 3) (-1, -1) . sides satisfy the Pythagorean theorem. 5-1 2-1 3 triangles Likewise, these can be called 5x-12x-13x triangles because the multi- ples of 5-1 2-1 3 also make right triangles. No- tice that the sides satisfy. order to assume the third part. x 15 9 c 2 c b 2 a 2 a b a 2 + b 2 = c 2 3-4 -5 triangles More accurately, these can be called 3x-4x-5x triangles because the multi- ples of 3-4 -5 also make right triangles 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 3 1. Your diagram should include one each of a 3x-4x-5x, a 5x-12x-13x, a 30 -6 0 -9 0°, and a 45 -4 5 -9 0° triangle. 2. Obtuse: 5 2 + 6 2 = 61 < 9 2 = 81 3. Acute: 2 2 +