Sat - MC Grawhill part 38 ppsx

Concept Review 1: Lines and Angles Questions 1 and 2 refer to the diagram above. 1. List all of the different pairs of angles that are congruent (equal). 2. List all of the different sets of angles that have a sum of 180°. ______________________________________________________________________________________________________ Mark the figure to show the following information: AD ⏐⏐ HN, AI ⏐⏐ BM, and HD ⏐⏐ JL. Then list the angles in the figure that have the given characteristic: A D H B C NM J I L G F E K 1 23 45 6 7 8 9 10 11 12 13 14 15 a b c d e l 1 l 2 r 3. One angle equal to ∠10 4. Two angles supplementary to ∠6 5. Two angles supplementary to ∠9 6. One angle equal to ∠15 7. Three angles supplementary to ∠13 State whether each of the following pairs is supplementary (has a sum of 180°), equal, or neither. 8. ∠13 and ∠79.∠13 and ∠6 10. ∠2 and ∠9 11. ∠14 and ∠9 12. ∠7 and ∠3 13. ∠1 and ∠2 14. ∠6 and ∠7 l 1 and l 2 are lines and r is a ray. 360 McGRAW-HILL’S SAT CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 361 SAT Practice 1: Lines and Angles Note: Figure not drawn to scale. 1. The figure above shows the intersection of three lines. x = (A) 16 (B) 20 (C) 30 (D) 60 (E) 90 2. The figure above shows a parallelogram with one side extended. If z = 40, then y = (A) 40 (B) 60 (C) 80 (D) 110 (E) 120 Note: Figure not drawn to scale. 3. In the figure above, if ഞ 1 ⏐⏐ഞ 2 , then a + b = (A) 130 (B) 270 (C) 280 (D) 290 (E) 310 130° a° b° ᐉ 1 ᐉ 2 y° z° z° z° 4. In the figure above, if l 1 ⏐⏐l 2 , then what is the value of n in terms of m? (A) 355 − 2m (B) 185 − 2m (C) 175 − 2m (D) 95 − 2m (E) 85 − 2m 5. In the figure above, if l 1 ⏐⏐l 2 , then x = (A) 43 (B) 69 (C) 79 (D) 101 (E) 111 6. In the figure above, if and FJ –– bisects ∠HFG, what is the measure of ∠FJH? (A) 14 (B) 38 (C) 40 (D) 56 (E) 76 FG HJ F JH G 76° 43° 36° x° ᐉ 1 ᐉ 2 m° m + 5° n° ᐉ 1 ᐉ 2 x° 2x° 3x° Note: Figure not drawn to scale. 7. In the figure above, if ഞ 1 ⏐⏐ഞ 2 and ഞ 2 ⏐⏐ഞ 3 , then a = (A) 50 (B) 55 (C) 60 (D) 65 (E) 70 130° 80° a° ᐉ 1 ᐉ 2 ᐉ 3 Note: Figure not drawn to scale. 8. In the diagram above, if ഞ 1 ⏐⏐ഞ 2 , then x = (A) 65 (B) 60 (C) 50 (D) 45 (E) 40 50° 40° 30° x° ᐉ 1 ᐉ 2 362 McGRAW-HILL’S SAT There’s a lot of detail to learn and understand to do well on the SAT. For more tools and resources that will help, visit our Online Practice Plus at www.MHPracticePlus.com/SATmath. CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 363 Concept Review 1 1. Only c and e are congruent. 2. a + b + c = 180°, c + d = 180°, d + e = 180°, e + a + b = 180° 3. ∠5 4. ∠7 and ∠13 5. ∠12 and ∠14 6. ∠4 7. ∠11, ∠8, and ∠6 Answer Key 1: Lines and Angles 8. equal 9. supplementary 10. neither 11. supplementary 12. neither 13. supplementary 14. supplementary 4. C Notice that the m° angle has a “corresponding” angle below that has the same measure. (Notice that they form an F.) Then (m + 5) + n + m = 180. Simplify: 2m + 5 + n = 180 Subtract (5 + 2m): n = 175 − 2m 5. C Draw an extra line through the vertex of the angle that is parallel to the other two. Notice that this forms two “Z” pairs. Therefore, x =36 + 43 = 79. 43° 36° ᐉ 1 ᐉ 2 43° 36° m° m+5° n° ᐉ 1 ᐉ 2 m° SAT Practice 1 1. C Draw in the three angles that are “vertical,” and therefore congruent, to the angles that are shown. Then choose any three adjacent angles, and notice that they form a straight angle. There- fore, x + 2x + 3x = 180. So 6x = 180 and x = 30. 2. B The opposite angles in a parallelogram must be equal, and any two “consecutive angles” as you move around the figure must be supplementary. (Notice that consecutive angles form C’s or U’s. If you’re not sure why this theorem is true, sketch a few sample parallelograms and work out the angles.) The angle opposite the y° must also measure y°, and when this is added to the three z° angles, they form a straight angle. Therefore, y + 40 + 40 + 40 = 180 and y = 60. 3. E In the triangle, the angles must have a sum of 180°. (See the next lesson for a simple proof.) Therefore, the other two angles in the triangle must have a sum of 50°. Pick values for these two angles that add up to 50°, and write them in. It doesn’t matter how you do it: 25° and 25°, 20° and 30°, 40° and 10°, as long as they add up to 50°. You can then find the values of a and b by noticing that they form straight angles with the interior angles. So if the interior angles are 25° and 25°, then a and b must both be 155. 6. B There are many relationships here to take advantage of. Notice that ∠HFG and the 76° angle form a “Z,” so ∠HFG = 76°. Remember that “bi- sect” means to divide into two equal parts, so ∠HFJ =∠JFG = 38°. Then notice that ∠JFG and ∠FJH form a “Z,” so ∠FJH = 38°. 7. A Consider the triangle with the a° angle. The other two angles in the triangle can be found from the given angles. Notice the “Z” that contains the two 80° angles and the “U” that contains the 130° and 50° angles. Therefore, a + 80 + 50 = 180, so a = 50. 130° 80° a° ᐉ 1 ᐉ 2 ᐉ 3 50° 80° F JH G 76° 38° 38° 38° 104° 8. E Draw two more parallel lines and work with the Z’s. Your figure should look like the one above. 50° 40° 30° ᐉ 1 ᐉ 2 30° 10° 10° 40° 40° 364 McGRAW-HILL’S SAT CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 365 Angles in Polygons Remembering what you learned about parallel lines in the last lesson, consider this diagram: We drew line ഞ so that it is parallel to the opposite side of the triangle. Do you see the two Z’s? The angles marked a are equal, and so are the angles marked c. We also know that angles that make up a straight line have a sum of 180°, so a ϩ b ϩ c ϭ 180. The angles in- side the triangle are also a, b, and c. Therefore, the sum of angles in a triangle is always 180°. Every polygon with n sides can be divided into n − 2 triangles that share their vertices (corners) with the polygon: Therefore, the sum of the angles in any polygon with n sides is 180(n Ϫ 2)°. Angle-Side Relationships in Triangles A triangle is like an alligator mouth with a stick in it: The wider the mouth, the bigger the stick, right? Therefore, the largest angle of a triangle is al- ways across from the longest side, and vice versa. Likewise, the smallest angle is always across from the shortest side. Example: In the figure below, 72 Ͼ 70, so a Ͼ b. 70° 72° a b a° a° b° c° c° ᐉ An isosceles triangle is a triangle with two equal sides. If two sides in a triangle are equal, then the angles across from those sides are equal, too, and vice versa. The Triangle Inequality Look closely at the figure below. The shortest path from point A to point B is the line segment connect- ing them. Therefore, unless point C is “on the way” from A to B, that is, unless it’s on AB –– , the distance from A to B through C must be longer than the direct route. In other words: The sum of any two sides of a triangle is always greater than the third side. This means that the length of any side of a triangle must be be- tween the sum and the difference of the other two sides. The External Angle Theorem The extended side of a triangle forms an exter- nal angle with the adjacent side. The external angle of a triangle is equal to the sum of the two “remote interior” angles. Notice that this follows from our angle theorems: a + b + x = 180 and c + x = 180; therefore, a + b = c a° b° c° x° A C B 12 10 Lesson 2: Triangles 12 − 10 < AB < 12 + 10 2 < AB < 22 5 sides, 3 triangles = 3(180°) = 540° 7 sides, 5 triangles = 5(180°) = 900° 1. The sum of the measures of the angles in a quadrilateral is __________. 2. The sum of the measures in an octagon is __________. 3. In ΔABC, if the measure of ∠A is 65° and the measure of ∠B is 60°, then which side is longest? __________. 4. The angles in an equilateral triangle must have a measure of __________. 5. Can an isosceles triangle include angles of 35° and 60°? Why or why not? 6. Draw a diagram to illustrate the external angle theorem. 7. If a triangle has sides of lengths 20 and 15, then the third side must be less than __________ but greater than __________. 8. Is it possible for a triangle to have sides of lengths 5, 8, and 14? Why or why not? 9. If an isosceles triangle includes an angle of 25°, the other two angles could have measures of __________ and __________ or __________ and __________. 10. In the figure above, QS –– and RT –– are diameters of the circle and P is not the center. Complete the statement below with >, <, or =. PQ + PR + PS + PT __________ QS + RT (not shown) R T S Q P R T S Q Concept Review 2: Triangles 366 McGRAW-HILL’S SAT CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 367 1. In the figure above, if AB = BD, then x = (A) 25 (B) 30 (C) 35 (D) 50 (E) 65 2. In the figure above, a + b + c + d = a° b° c° d° 40° 50° x° A B C D 3. The three sides of a triangle have lengths of 9, 16, and k. Which of the following could equal k? I. 6 II. 16 III. 25 (A) I only (B) II only (C) I and II only (D) II and III only (E) I, II, and III Note: Figure not drawn to scale. 4. Which of the following statements about a and b in the figure above must be true? I. a = b II. a + b = 90 III. a < 60 (A) I only (B) II only (C) I and II only (D) II and III only (E) I, II, and III 5. In the figure above, if AD = DB = DC, then x + y = (A) 70 (B) 80 (C) 90 (D) 100 (E) 120 B D y° x° 100° A C b° a° SAT Practice 2: 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 6. In the figure above, which of the following expresses a in terms of b and c? (A) 180 − (b + c) (B) 180 − (b − c) (C) 90 − (b + c) (D) 90 − (b − c) (E) b + c Note: Figure not drawn to scale. 7. Which of the following represents the correct ordering of the lengths of the five segments in the figure above? (A) AD > AB > DB > BC > DC (B) AD > DB > AB > DC > BC (C) AD > DB > AB > BC > DC (D) AD > AB > DB > DC > BC (E) AD > DB > DC > AB > BC 30° 35° 60° 65° A B C D a° b° c° 8. A triangle has two sides of lengths 4 centi- meters and 6 centimeters. Its area is n square centimeters, where n is a prime number. What is the greatest possible value of n? (A) 11 (B) 12 (C) 19 (D) 23 (E) 24 368 McGRAW-HILL’S SAT CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 369 Concept Review 2 1. 360° 2. 1,080° 3. Draw a diagram. If the measure of ∠A is 65° and the measure of ∠B is 60°, then the measure of ∠C must be 55°, because the angles must have a sum of 180°. Since ∠A is the largest angle, the side op- posite it, BC, ___ must be the longest side. 4. 60°. Since all the sides are equal, all the angles are, too. 5. No, because an isosceles triangle must have two equal angles, and the sum of all three must be 180°. Since 35 + 35 + 60 ≠ 180, and 35 + 60 + 60 ≠ 180, the triangle is impossible. 6. Your diagram should look something like this: a° b° c° x° a + b = c 7. If a triangle has sides of lengths 20 and 15, then the third side must be less than 35 (their sum) but greater than 5 (their difference). 8. No. The sum of the two shorter sides of a triangle is always greater than the third side, but 5 + 8 is not greater than 14. So the triangle is impossible. 9. 25° and 130° or 77.5° and 77.5° 10. Draw in the line segments PQ, PR, PS, and PT. Notice that this forms two triangles, ΔPQS and ΔPRT. Since any two sides of a triangle must have a sum greater than the third side, PQ + PS > QS, and PR + PT > RT. Therefore, PQ + PR + PS + PT > QS + RT. Answer Key 2: Triangles SAT Practice 2 1. A If AB = BD, then, by the Isosceles Trian- gle theorem, ∠BAD and ∠BDA must be equal. To find their measure, subtract 50° from 180° and divide by 2. This gives 65°. Mark up the diagram with this information. Since the angles in the big triangle have a sum of 180°, 65 + 90 + x = 180, so x = 25. 2. 500 Drawing two diagonals shows that the figure can be divided into three triangles. (Remember that an n-sided figure can be divided into n − 2 triangles.) Therefore, the sum of all the angles is 3 × 180°=540°. Subtracting 40° leaves 500°. 3. B The third side of any triangle must have a length that is between the sum and the differ- ence of the other two sides. Since 16 − 9 = 7 and 16 + 9 = 25, the third side must be between (but not including) 7 and 25. 4. A Since the big triangle is a right triangle, b + x must equal 90. The two small triangles are also right triangles, so a + x is also 90. Therefore, a = b and statement I is true. In one “solution” of this triangle, a and b are 65 and x is 25. (Put the values into the diagram and check that everything “fits.”) This solu- tion proves that statements II and III are not necessarily true. 5. C If AD = DB, then, by the Isosceles Triangle theorem, the angles opposite those sides must be equal. You should mark the other angle with an x also, as shown here. Similarly, if DB = DC, then the angles opposite those sides must be equal also, and they should both be marked y. Now consider the big triangle. Since its angles must have a sum of 180, 2x + 2y = 180. Dividing both sides by 2 gives x + y = 90. (Notice that the fact that ∠ADB mea- sures 100° doesn’t make any difference!) 50° x ° A B C D 65° 65° 25° 130° a° b° c° d° 40° b° a° x° B D y° x° 100° A C x° y° . 50. 130° 80° a° ᐉ 1 ᐉ 2 ᐉ 3 50° 80° F JH G 76° 38 38 38 104° 8. E Draw two more parallel lines and work with the Z’s. Your figure should look like the one above. 50° 40° 30° ᐉ 1 ᐉ 2 30° 10° 10° 40° 40° 364 McGRAW-HILL’S SAT CHAPTER. 65 (B) 60 (C) 50 (D) 45 (E) 40 50° 40° 30° x° ᐉ 1 ᐉ 2 362 McGRAW-HILL’S SAT There’s a lot of detail to learn and understand to do well on the SAT. For more tools and resources that will help, visit. “Z,” so ∠HFG = 76°. Remember that “bi- sect” means to divide into two equal parts, so ∠HFJ =∠JFG = 38 . Then notice that ∠JFG and ∠FJH form a “Z,” so ∠FJH = 38 . 7. A Consider the triangle with