1. TAKS Tutorial Geometry Objectives 6 – 8 Part 2

2. The Geometry tested on the Exit Level TAKS test covers High School Geometry. Topics to be covered in today’s tutorial are Special Right Triangles Special Right Triangles Segment-Angle Relationships Segment-Angle Relationships Arclength Arclength Parallel & Perpendicular Lines Parallel & Perpendicular Lines Dimension Changes Dimension Changes Similarity & Scale Factor Similarity & Scale Factor Views of 3D figures Views of 3D figures

3. Special Right Triangles have relationships that are important enough to test. The formula chart now has those relationships listed. Please be sure to consult the chart when you come upon one of those test questions. 60 o 30 o 45 o

4. On the chart, the angles and the lengths of the sides opposite them are listed in order of size. The side across from the 30 o angle (the smallest angle) is the shortest side, the leg x. The side across from the 30 o angle (the smallest angle) is the shortest side, the leg x. The side across from the 60 o angle (the mid- sized angle) is the mid-sized side, the leg. The side across from the 60 o angle (the mid- sized angle) is the mid-sized side, the leg. The side across from the 90 o angle (the largest angle) is the longest side, the hypotenuse, 2x. The side across from the 90 o angle (the largest angle) is the longest side, the hypotenuse, 2x. 60 o 30 o

5. On the chart, the angles and the lengths of the sides opposite them are listed in order of size. The sides across from each of the 45 o angles (the smallest angles) are the shortest sides, the legs x. The sides across from each of the 45 o angles (the smallest angles) are the shortest sides, the legs x. Notice, the angles are the same size (congruent) and so are the sides across from them. Notice, the angles are the same size (congruent) and so are the sides across from them. The side across from the 90 o angle (the largest angle) is the longest side, the hypotenuse,. The side across from the 90 o angle (the largest angle) is the longest side, the hypotenuse,. 45 o

6. 11 ft Notice that the hose is perpendicular to one side of the garden. Now you have a right triangle to work with. You have two different ways that you can find the length of the hose.

7. Option 1: 30 o -60 o -90 o Relationship 11 ft The side of the garden forms the hypotenuse of the right triangle. In an equilateral triangle, each angle measures 60 o. The hose divides its angle in half, making it 30 o. And, of course, the right angle is 90 o. According to the formula on the chart, the hypotenuse measures 2x, double the length of the shortest side, x. We need to divide 11 by 2, and find x = 5.5 ft 5.5 ft

8. Option 1: 30 o -60 o -90 o Relationship 11 ft Again, according to the formula on the chart, the longer leg, which is the hose, measures x times the square root of 3. We need to multiply: 5.5 ft

9. Option 2: Pythagorean Theorem 11 ft The hose is an altitude for the equilateral triangle. It cuts the perpendicular side in half, making the short side of the right triangle 11/2 ft or 5.5 ft 5.5 ft If you don’t remember the Pythagorean Theorem, it is on the formula chart: a c b

10. This next problem cannot be done using the Pythagorean Theorem because only one of the three triangle sides is known. You have no choice but to use the relationships among the sides of a 30 o - 60 o -90 o triangle.

11. 2003 You should know that the measures of the depression and of elevation are the same. Angle of depression—Mr. Ryan looking down from the horizon Angle of elevation— someone looking up at Mr. Ryan from the same location that Mr. Ryan is looking at sees Mr. Ryan at the same angle measurement. 30 o

12. 2003 Since the height of the plane is across from the 30 o angle, 2400 ft is the shortest leg of the right triangle. 30 o Recall from the math chart, the shortest leg is x. We are looking for the horizontal distance, which is the length of the longer leg. That leg measures 2400 times the square root of 3. Get out the calculator! ?

13. 2006 Because the figure is a cube, all of its sides have the same length. s s We now know one side of the shaded rectangle. We cannot find its area until we know the length of the other side. A = lw To find the remaining side of the rectangle, we need to look at the right triangle shown, which happens to be isosceles and 45 o -45 o -90 o 45 o

14. 2006 Using the special triangle relationship, when the leg of a 45-45-90 triangle is x, the hypotenuse is x times the square root of 2. Our “x” just happens to be “s”. s s We now know both sides of the shaded rectangle. 45 o

15. Since the radii of circles must be the same length (or you wouldn’t have a circle!), both of these triangles are isosceles. Because the radii are perpendicular, we have 45 o -45 o -90 o triangles. 45 o

16. To find the length of a leg in a 45 o -45 o -90 o triangle, we divide the hypotenuse by the square root of 2. The missing lengths that make up segment LT are also radii. 45 o 2 2 3.5355 2

17. 2006 If you bring your own, you may use map pencils or highlighters on the test. There are lots of angles here. Use the same color to denote which angles are congruent. (No map pencils? Use a symbol.) You know the 3 angles of a triangle add up to be 180 o. m  3 + m  8 + m  9 = 180 o This fact is not an answer choice. So now look at your symbols/colors. Start replacing these angles with ones they are equal to until you find the right match. m  6 is in each answer choice, so start by replacing the 8 with 6. m  3 + m  6 + m  9 = 180 o m  1 + m  6 + m  11 = 180 o

18. 2006 You are expected to remember that the diagonals of a parallelogram bisect each other. That means: 2x + 3 = 3x + 1 and y + 2 = 4y – 7 Solve each equation: 2x + 3 = 3x + 1 subtract 2x and subtract 1 from each side 2 = x x = 2 Solve each equation: y + 2 = 4y - 7 subtract y and add 7 to each side 9 = 3y divide by 3 so 3 = y y = 3

19. Now substitute those values into the expression on the diagonals. Make sure they are equal and then add them together to find the length of the diagonal. That means: 2x + 3 = 2(2) + 3 = 7 and 3x + 1 = 3(2) + 1 = 7 so the diagonal MQ is 14 x = 2y = 3

20. 2006 An arc is a fractional part of the circumference of a circle. First, we need to find the circumference of the circle made if the door handle turned all the way around.

21. Let’s concentrate on the information they gave us about the door handle. The radius of the circle is 5.5 inches. The circumference formula is on the math chart if you don’t remember it. Remember that the handle is NOT going all the way around so we just want part of the circumference. The handle is going around 45 o of the 360 o of the complete turn.

22. 2006.125(34.54 inches) = 4.3175 inches All the way around the circle is 34.54 inches, but the door handle only moves 45 o of the circle which is 4.3175 inches

23. 2004 An arc is a fractional part of the circumference of a circle. First, we need to find the circumference of the circle made if the door handle turned all the way around. The diameter of the plate is 12 inches. The circumference formula is on the math chart if you don’t remember it. Remember, we only want the part of the plate where the peas are located. The entire circle goes around 360 o. Subtract the other two sections of the plate to find the central angle of the peas section: 360 – 203 – 105 = 52 o. We just want.14444(37.68) = 5.442 inches of the plate edge.

24. In Algebra I, you studied parallel and perpendicular lines. In Geometry, you built on that knowledge. Algebra II expected you to remember that material. Now, TAKS is going to test that knowledge.

25. Parallel lines are going in the same direction. Since slope is the indicator of the line’s direction, the slopes of parallel lines must be the same. Both of these parallel lines have a slope of -2

26. Perpendicular lines must go in opposite directions so their slopes must have opposite signs. However, they must be positioned so that the angles formed at their intersection are 90 o. To ensure that measurement, the slopes must also be reciprocals.-----Opposite Reciprocal slopes! Here, the slopes are -2 and ½. Note, they have opposite signs and 2 and ½ are reciprocals.

27. A vertical line and a horizontal line are perpendicular to each other. The horizontal line has a slope of zero while the vertical line has undefined slope.

28. Parallel lines must have the same slope. Look for a line with a slope of -4/5. X X X Positive slope Down 4, right 5, and I hit the line.This is it! Down 4, right 5, and I pass the line.

29. X X X Parallel lines have the same slope. These slopes have opposite signs. Even though the slopes have opposite signs, the fractions are NOT reciprocals. They distinctly said the y- intercept is changed! Do check this answer! You have one equation in slope- intercept form already. You know the slope and y- intercept of the other line. Graph both on the calculator. See why the slopes must be opposite reciprocals to be perpendicular? These angles are definitely not right angles! same x-int

30. 2003 This slope is negative. We need a positive slope. All of these lines go through (0, 4). X X X Perpendicular lines have opposite reciprocals for slopes. We need positive 2/1. Yes

31. Similar figures are figures with the same shape, although they may be different sizes. To have the same shape, all of the corresponding angles in both figures must have the same degree measure. To have the same shape, all of the corresponding angles in both figures must have the same degree measure. Even though the figures might be different sizes, the ratio of the sides must be the same for all corresponding parts. Even though the figures might be different sizes, the ratio of the sides must be the same for all corresponding parts. Congruent figures are indeed similar. The ratio of their corresponding parts is 1:1. Congruent figures are indeed similar. The ratio of their corresponding parts is 1:1.

32. 2003 Problem part 1 continued on next slide

33. on previous slide

34. We need to compare each answer choice with the original triangle. We’ll start with F. Let’s look at the horizontal side first. The original triangle has that side 4 units long. The horizontal side of F is 2 units long. The ratio of the scale factor is 2:4 or 1:2. That means, all the sides of triangle F must be half the size of the original triangle.

35. It is easiest to count vertically and horizontally, so let’s count what would be the slope of the diagonal sides. Down 4, right 2 for this side. The corresponding side on triangle F should be up 2 left 1 (if a rotation) or up 2 right 1 (if a reflection). That works! Either way Now, to check out the remaining side on the original triangle. Looks like this is the similar triangle—all the sides have the same ratio. Triangle F is half of the original triangle.

36. We really should check all of the triangles. Do not jump to conclusions too fast. Both the original triangle and triangle G have a horizontal side that is 4 units long. That means triangle G needs to be exactly the same size as the original. It definitely is not. Triangle G goes down 3, right 2 while the original goes down 4, right 2. These corresponding sides are not in the ratio 1:1.

37. Now, to look at Triangle H. The horizontal side of triangle H is 6 units long. The ratio of the sides then is 6:4 or 3:2. Triangle H needs to be 1 ½ times as large as the original. 5 up, 3 right gives the ratio 5:3, NOT 3:2. These two triangles are not similar, either.

38. One last answer choice to check! Triangle J doesn’t even have the same shape as the original triangle. The horizontal sides are in the ratio 2:4 which is 1:2 3 right, 1 down does not give us a side that is half the size of the original triangle. Definitely, F is the correct answer choice.

39. 2004 The dimensions are given in terms of length by width by height. The given rectangular solid is 8 units by 6 units by 12 units. You need to look at corresponding ratios; they must all reduce to the same ratio. Remember: in similar figures, corresponding sides must be proportional.

40. 2004 For choice F:

41. 2004 For choice G:

42. 2004 For choice H:

43. 2004 That leaves J: But check it first!

44. 2006 You were told that the triangles ARE similar. That means that the sides and altitude are in proportion. We are going to need to use some algebra to find the base of ΔHKM. We need the base so that we can compare them and find the ratio which is the scale factor used to get ΔRTV. A = ½bh 10 = ½(b)(4) 10 = 2b 5 = b

45. 2006 Now that we know that the base of the smaller triangle is 5, we can compare bases and find the scale factor. A = ½bh 10 = ½(b)(4) 10 = 2b 5 = b We know now that the altitude of the larger triangle is 1.75 times greater than the altitude of the smaller triangle: 4(1.75) = 7 units

46. 2006 Let’s use the Area of a triangle formula again—this time to find the area of the larger triangle RTV. A = ½bh A = ½(8.75)(7) A = 30.625 We know now that the altitude of the larger triangle is 1.75 times greater than the altitude of the smaller triangle: 4(1.75) = 7 units

47. Start with a visual—draw a picture of what you see. As you can see, if only the height doubles, then the Volume would double. If you look at the formula: V = Bh =  r 2 h for the original If the height doubles V = Bh =  r 2 (2h) = 2  r 2 h means that the volume is doubled.

48. To understand the next problem, let’s have a little lesson: Start with a cube… Next, double the length of each side… Now, let’s triple the lengths… s 2s 3s

49. Looking at perimeter… Looking at the lengths of the sides: s 2s3s Ratio: 2 to 1 2 times as large Ratio: 3 to 1 3 times as large Looking at the perimeter of the base: P = 4s P = 4(2s) = 2(4s) P = 4(3s) = 3(4s) 4 sides that are s units long 4 sides that are 2s units long 4 sides that are 3s units long 2 times longer than the original 3 times longer than the original original

50. Measurements that are 1-Dimensional: length, side, width, height, altitude, perimeter, radius, diameter, circumference, slant height all have the same increase or decrease when there is a dimension change. In other words, if the length of the sides of a cube doubles, the perimeter of the base doubles. If the radius of a circle triples, the circumference triples. If the perimeter of a rectangle quadruples, then both the length and the width quadrupled.

51. Looking at Area… Looking at the lengths of the sides: s 2s3s Ratio: 2 to 1 2 times as large Ratio: 3 to 1 3 times as large Looking at the surface area: A = 6s 2 A = 6(2s) 2 = 6(4s 2 ) = 4(6s 2 ) A = 6(3s) 2 = 6(9s 2 ) = 9(6s 2 ) 6 sides that are s 2 in area 6 sides that are 2s by 2s in area 6 sides that are 3s by 3s in area 4 times more surface area than the original 9 times more surface area than the original Ratio: 4 to 1 4 times as large Ratio: 9 to 1 9 times as large

52. Measurements that are 2-Dimensional: any kind of area, all have the square of the increase or decrease when there is a dimension change. In other words, if the length of the sides of a cube doubles, the area of the base quadruples. If the radius of a circle triples, the area of the circle becomes 9 times greater. If the perimeter of a rectangle quadruples, then its area increases by a multiple of 16. On the other hand, if the area of a square increases by 4, the length of the side doubled (square root of 4). If the total area of a pyramid increased by a factor of 9, the perimeter of its base increased by a factor of 3 (square root of 9).

53. Looking at Volume… Looking at the lengths of the sides: s 2s3s Ratio: 2 to 1 2 times as large Ratio: 3 to 1 3 times as large Looking at volume: V = s 3 V = (2s) 3 = 8s 3 A = (3s) 3 = 27s 3 1 cube 8 cubes that are s 3 in volume 27 cubes that are s 3 in volume 8 times more volume than the original 27 times more volume than the original Ratio: 8 to 1 8 times as large Ratio: 27 to 1 27 times as large

54. Measurements that are 3-Dimensional: any kind of volume, all have the cube of the increase or decrease when there is a dimension change. In other words, if the length of the sides of a cube doubles, its volume is increase 8 times. If the radius of a sphere triples, the volume of the sphere becomes 27 times greater. If the perimeter of the base of a rectangular solid quadruples, then its volume increases by a multiple of 64. On the other hand, if the volume of a cube increases by 8, the length of the side doubled (cube root of 8). If the volume of a pyramid increased by 27, the perimeter of its base increased by a factor of 3 (cube root of 27).

55. Let’s look again at the cubes we drew… s 2s3s Increase (2) 1 : 2 times as large Increase (3) 1 : 3 times as large 1-D: same increase 3-D: increase cubed The dimension is the exponent for the increase. 2-D: increase squaredIncrease (2) 2 4 times as large Increase (3) 2 9 times as large Increase (2) 3 8 times as large Increase (3) 3 27 times as large The catch: All the sides must increase by the same amount for this to happen!

56. Area is 2-D It’s increase is a factor of 4 Length is 1-D. We need to square root the increase of the 2-D to find the increase of the 1-D.

57. Diameter is 1-D. It increases 1.5 times Volume is 3-D. Since a diameter change is all that is needed to change the volume, we need to cube the increase… (1.5) 3 = 3.375 The volume increases 3.375 times its original size.

58. Let’s now look at figures with 2-D and 3-D in mind. Looking at the Front view, you should see 4 cubes, 2 cubes, and 1 cube right in the front row. That would eliminate F and J since they each have 3 cubes in the middle of the front row. X X The difference between choices G and H is the middle of the right view. There should be NO cubes in the middle far right. X

59. 2006 Either, imagine taking a scissors and cutting the prism at its edges and laying it flat OR imagine folding the nets into the prism. Answer choice A will not work since the short edge of the triangular base will be too short for the side to which it must attach. X

60. 2006 Either, imagine taking a scissors and cutting the prism at its edges and laying it flat OR imagine folding the nets into the prism. Answer choice B will not work since the short edge of the triangular base will be too short for the side to which it must attach. X X

61. 2006 Either, imagine taking a scissors and cutting the prism at its edges and laying it flat OR imagine folding the nets into the prism. Answer choice D will not work since the long edges of the triangular base will be too long for the sides to which it must attach. X X X

62. Practice Problems

64. All sides are same length. Divide perimeter by 3. 37/3 = 12.33333 cm is the length of each side. 12.3333 60 o 30 o Altitude divides the perpendicular side in half 12.333333/2 =6.16666666 6.1666 10.6809 A = bh/2 = (12.3333)(10.6809)/2 A = 65.86537 Longer leg is shorter leg times square root of 3

66. 5

67. 2003

68. 2003 too short Needs to be longer than 5280 ft + 10,560 ft which means longer than 15,840 ft

69. 2004

70. 2004 sum of all 5 angles = 3(180) = 540 o 90 o 130 o 540 – 90 – 90 – 90 – 130 = 140 o 140

71. 2003

72. 2003 360 o Regular hexagon—6 congruent sides means six congruent intercepted arcs means six congruent central angles. Y = 360/6 = 60 o 60

79. Original V = lwh = 24 dm 3 Changed V = (½l)(½w)(½h) = 1/8lwh 1/8(24 dm 3 ) = 3 dm 3

83. 8 ft 5 ft 3 ft