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Transcript
TAKS Tutorial Geometry Objectives 6 – 8 Part 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 Segment-Angle Relationships Arclength Parallel & Perpendicular Lines Dimension Changes Similarity & Scale Factor Views of 3D figures 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. x 60o 2x 45o 30o x 3 x 2 x 45o x On the chart, the angles and the lengths of the sides opposite them are listed in order of size. 60 x o 2x 30o x 3 The side across from the 30o angle (the smallest angle) is the shortest side, the leg x. The side across from the 60o angle (the midsized angle) is the mid-sized side, the leg x 3 . The side across from the 90o angle (the largest angle) is the longest side, the hypotenuse, 2x. On the chart, the angles and the lengths of the sides opposite them are listed in order of size. 45 x 2 o x 45o x The sides across from each of the 45o 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. The side across from the 90o angle (the x 2 largest angle) is the longest side, the hypotenuse, . Notice that the hose is perpendicular to one side of the garden. 11 ft 11 ft 11 ft You have two different ways that you can find the length of the Now you have a right hose. triangle to work with. Option 1: 30o-60o-90o Relationship The side of the garden forms the hypotenuse of the right triangle. 11 ft 11 ft 5.5 ft In an equilateral triangle, each angle measures 60o. The hose divides its angle in half, making it 30o. And, of course, the right angle is 90o. 11 ft 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 Option 1: 30o-60o-90o Relationship 11 ft 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 11 ft Option 2: Pythagorean Theorem a (5.5) (11) 2 a (11) (5.5) 2 2 2 2 2 a 11 5.5 2 2 11 ft 11 ft a b 5.5 ft The hose is an altitude the Calculatorfortime equilateral triangle. It cuts the 11 ft perpendicular side in half, making the c short side of the right triangle 11/2 ft or 5.5 ft If you don’t remember the Pythagorean Theorem, it is on the formula chart: 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 30o60o-90o triangle. 2003 Angle of depression—Mr. Ryan looking down from the horizon You should know that the measures of the depression and of elevation are the same. 30o 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. 2003 Since the height of the plane is across from the 30o angle, 2400 ft is the shortest leg of the right triangle. Recall from the math chart, the shortest leg is x. ? 30o 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! 2006 Because the figure is a cube, all of its sides have the same length. 45o s 45o 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 45o-45o-90o 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”. 45o s 2 s 45o s We now know both sides of the shaded rectangle. A lw A ( s )( s 2 ) As 2 2 Since the radii of circles must be the same length (or you wouldn’t have a circle!), both of these triangles are isosceles. 45o 45o Because the radii are perpendicular, we have 45o-45o-90o triangles. To find the length of a leg in a 45o-45o-90o triangle, we divide the hypotenuse by the square root of 2. 2 2 2 2 3.5355 2 45o 45o 2 2 The missing lengths that make up segment LT are also radii. 3.5355 3.5355 5 2 3.5355 2006 If you bring your m3 +own, m8 you + m9 = 180o may use map m3 + m6 pencils or + m9 = 180o This fact is not highlighters an onanswer the test. choice. There So arenow lots of You know the 3 angles of a triangle o look at your m1 + m6 + m11 = 180 angles here. o. add up to be 180 symbols/colors. Use the same Start replacing color to denote these angles m6 is in each which angles answer choice, with they areones congruent. so start by are equal to until (No map replacing the 8 you find the right pencils? Use a with 6. match. symbol.) 2006x = 2 y=3 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 Solveeach each equation: equation: Solve y ++23==4y and add 7 to each sideeach side 2x 3x-+71 subtract subtracty 2x and subtract 1 from 29==x3y divide by 3 so 3 = y x=2 y=3 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 so the diagonal MQ is 14 and 3x + 1 = 3(2) + 1 = 7 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. 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. C 2 r 2(3.14)(5.5) C 34.54 inches 45 1 .125 360 8 Remember that the handle is NOT going all the way around so we just want part of the circumference. The handle is going around 45o of the 360o of the complete turn. 2006 .125(34.54 inches) = 4.3175 inches All the way around the circle is 34.54 inches, but the door handle only moves 45o of the circle which is 4.3175 inches 2004 C d 3.14(12) C 37.68 inches An Remember, arc52 is a .fractional we only want part of thethe part of 144444 360 where circumference the plate ofthe a circle. peas are located. First, we need to goes find the circumference of thethe circle The entire circle around 360o. Subtract other two made if the door handle turned all the way around. sections of the plate to find the central angle of the peas o. section: 360of–the 203plate – 105 The diameter is = 1252 inches. The circumference formula is on = the mathinches chart ifofyou We just want .14444(37.68) 5.442 the don’t plate edge. remember it. 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. 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 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 90o. 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. 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. Down 4, right 5, and I Parallel pass the line. X X Positive slope Positive slope Down 4, right 5, and I hit the line.This is it! X lines must have the same slope. Look for a line with a slope of -4/5. X X X See why the slopes must be opposite reciprocals to be perpendicular? These angles are definitely not right angles! Do check this answer! You have one equation in slopeintercept form already. You know the slope and yintercept of the other line. Graph both on the calculator. same x-int 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 yintercept is changed! 2003 All of these lines go through (0, 4). Yes X X X This slope is negative. We need a positive slope. Perpendicular lines have opposite reciprocals for slopes. We need positive 2/1. 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. 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. 2003 Problem part 1 continued on next slide on previous slide 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. It is easiest to count vertically and horizontally, so let’s count what would be the slope oflikethe sides. Looks this isdiagonal the similar triangle—all the sides have the same ratio. Triangle F is half of the original triangle. 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. We really should check all of the Triangle G goes down 3, right 2 triangles. Do not jump while to conclusions too 4, the original goes down 2. These corresponding fast. right sides are not in the ratio 1:1. 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. 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. One last answer choice to check! Triangle J doesn’t even have the same shape as the original triangle. 3 right, 1 down does not give us a side that is half the size of the original triangle. The horizontal sides are in the ratio 2:4 which is 1:2 Definitely, F is the correct answer choice. 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. 2004 For choice F: 2 3 4 is not true. 8 6 12 1 1 1 4 2 3 2004 For choice G: 4 2 8 is not true. 8 6 12 1 1 2 2 3 3 2004 For choice H: 2 1 6 is not true. 8 6 12 1 1 1 4 6 2 2004 That leaves J: But check it first! 4 3 6 IS true. 8 6 12 1 1 1 2 2 2 2006 A = ½bh 10 = ½(b)(4) 10 = 2b 5=b 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. 2006 A = ½bh 10 = ½(b)(4) 10 = 2b 5=b Now that we know that the base of the smaller triangle is 5, we can compare bases and find the scale factor. 8.75 1.75 5 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 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 Start with a visual—draw a picture of what you see. If you look at the formula: V = Bh = r2h for the original If the height doubles V = Bh = r2(2h) = 2r2h means that the volume is doubled. As you can see, if only the height doubles, then the Volume would double. 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 Looking at perimeter… Looking at the lengths of the sides: s 2s 3s Ratio: 2 to 1 2 times as large Ratio: 3 to 1 3 times as large Looking at the perimeter of the base: 4 sides that are s units long P = 4s 4 sides that are 2s units long P = 4(2s) = 2(4s) 2 times longer than the original original 4 sides that are 3s units long original P = 4(3s) = 3(4s) 3 times longer than the original 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. Looking at Area… Looking at the lengths of the sides: s 2s Ratio: 2 to 1 2 times as large 3s Ratio: 3 to 1 3 times as large Looking at the surface area: 6 sides that are s2 in area A= 6s2 6 sides that are 2s by 2s in area 6 sides that are 3s by 3s in area A = 6(2s)2 = 6(4s2) = 4(6s2) A = 6(3s)2 = 6(9s2) = 9(6s2) 4 times more surface area than the original Ratio: 4 to 1 4 times as large 9 times more surface area than the original Ratio: 9 to 1 9 times as large 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). Looking at Volume… Looking at the lengths of the sides: s 2s 3s Ratio: 2 to 1 2 times as large Ratio: 3 to 1 3 times as large 8 cubes that are s3 in volume 27 cubes that are s3 in volume Looking at volume: 1 cube V= s3 V = (2s)3 = 8s3 8 times more volume than the original Ratio: 8 to 1 8 times as large A = (3s)3 = 27s3 27 times more volume than the original Ratio: 27 to 1 27 times as large 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). Let’s look again at the cubes we drew… s 1-D: same increase 2-D: increase squared 3-D: increase cubed 2s Increase (2)1: 2 times as large Increase (2)2 4 times as large Increase (2)3 8 times as large 3s Increase (3)1 : 3 times as large Increase (3)2 9 times as large Increase (3)3 27 times as large The dimension is the exponent for the increase. The catch: All the sides must increase by the same amount for this to happen! Area is 2-D It’s increase is a factor of 4 Length is 1-D. 4 2 We need to square root the increase of the 2-D to find the increase of the 1-D. 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. 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. 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 X X 2006 X 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. 2006 X X 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. 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. X X X 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. Practice Problems All sides are same length. Divide perimeter by 3. 37/3 = 12.33333 cm is the length of each side. 30o 12.3333 12.3333 10.6809 6.1666 12.3333 60o Altitude divides the perpendicular side in half 12.333333/2 =6.16666666 Longer leg is shorter leg times square root of 3 6.16666 3 10.6809 A = bh/2 = (12.3333)(10.6809)/2 A = 65.86537 5 5 2 7.071 2003 2003 too short too short too short Needs to be longer than 5280 ft + 10,560 ft which means longer than 15,840 ft 2004 2004 1 4 130o 0 90o sum of all 5 angles = 3(180) = 540o 90o 540 – 90 – 90 – 90 – 130 = 140o 90o 2003 2003 360o 6 0 Y = 360/6 = 60o Regular hexagon—6 congruent sides means six congruent intercepted arcs means six congruent central angles. Original V = lwh = 24 dm3 Changed V = (½l)(½w)(½h) = 1/8lwh 1/8(24 dm3) = 3 dm3 8 ft 5 ft 3 ft