A Solution: The area of a trapezoid is height (base 1 + base 2) = ( 6) (8 + 18) = ( 6) ( 26) = 78

Transcription

1 10.0 ompute areas of polygons, including rectangles, scalene triangles, equilateral triangles, rhombi, parallelograms, and trapezoids. (cont) Eample 2 Find the area of Trapezoid 8 Solution: The area of a trapezoid is height (base 1 + base 2) 6 18 = ( 6) (8 + 18) = ( 6) ( 26) = 78 Eample 3 Find the area of the triangle shown. Solution: The area of a triangle is b h The base is 8, but we do not have the height. raw an altitude. The altitude cuts the base into two equal parts. Now find the height using Pythagoras (height) 2 = h 2 = 64 h 2 = 48 h = = So, rea of triangle = 8 Homework 1. Find the area of the trapezoid shown. = (0,2) y (3,2) (4,0) 2. Find the area of the trapezoid shown (cont) 18 Geometry 201

2 10.0 ompute areas of polygons, including rectangles, scalene triangles, equilateral triangles, rhombi, parallelograms, and trapezoids. (cont) 3. The square shown has side lengths of 10. Four triangles are removed from the corners as shown. What is the area of the shaded region? (Hint: Find the area of the square and subtract the areas of the 4 triangles.) Rhombus has side lengths of 10. E = 6, E = 8 Find the area of. 6 8 E Geometry 202

3 11.0 etermine how changes in dimensions affect the perimeter, area, and volume of common geometric figures and solids. Rule If two polygons are similar, and the sides are in the ratio of a: b, then Perimeter Ratio = side ratio a : b rea Ratio = (side ratio) 2 a 2 : b 2 Volume Ratio = (side ratio) 3 a 3 : b 3 Eample 1 The circumferences of the bases of two similar cylinders are in the ratio of 3: 5. Find the ratio of their lateral areas. Solution: reas are in the ratio of a 2 : b 2 So, 3 2 : 5 2 = 9 : 25 Eample 2 If the sides of a regular square pyramid are tripled, what happens to the volume? Solution: ssume the sides of the original pyramid have a length of 1 If the sides are tripled, then the new side length would be (3 1) or 3. The side ratio would then be 1: 3 The formula says that the Volumes would be in the ratio of a 3 : b 3 So, 1 3 : 3 3 = 1 : 27. The volume would be multiplied by 27 Eample 3 Two cartons and their dimensions are shown. How many more cubic centimeters can the larger one hold? Solution: Find the volume of each and then subtract. Larger arton rea Smaller arton rea (ase rea) (Height) (ase rea) (Height) L W H L W H = 19 more cubic centimeters in the larger carton. (cont) Geometry 203

4 11.0 etermine how changes in dimensions affect the perimeter, area, and volume of common geometric figures and solids. (cont) Homework 1. If the sides of two similar triangles are in the ratio 4:9, what is the ratio of their perimeters? 2. If the sides of a cube are multiplied by 4, the volume of the cube is multiplied by how much? 3. Find the ratio of the lateral area of a cylinder with radius r to its volume. Hint: Use the cylinder formulas found in Worksheet 9.0. Substitute the formulas into the ratio and then cancel common units. 4. The dimensions of one are shown. one s dimensions are double those of one. one s dimensions are double those of one. How many times greater is the volume of one when compared to the volume of one? one one one In the diagram E = 5 and E = 3. What is the ratio of the areas of E to? E Geometry 204

5 12.0 Find and use measures of sides and of interior and eterior angles of triangles and polygons to classify figures and solve problems. Rule Triangles can be classified in two ways. y their sides or by their angles. Y THEIR SIES Scalene no congruent sides. Isosceles two congruent sides Equilateral three congruent sides Y THEIR NGLES cute- all angles measure less than 90º Right one angle measures eactly 90º Obtuse one angle measures more than 90º Equiangular all angles measure the same (60º) Rule n eterior angle of a triangle is formed by a side of the triangle and an etension of another side. The formula to find the measure of an eterior angle is Eterior ngle Measure Eterior = sum of the two non-adjacent interior angles of the triangle In the triangle to the right, 4 is an eterior angle. It s measure is equal to the sum of the 2 interior angles to which it is not adjacent (not net to). 2 m 4 = m 2 + m Eample 1 In EF, find the measure of the eterior angle at F. Solution: Eterior = sum of two non-adjacent interior (8 + 15) = (3 + 20) + (4 + 5) (8 + 15) = (7 + 25) = 10 m F = (8 + 15) = 8 (10) + 15 = 95º F (8+15) (3+20) (4+5) E Geometry 205

6 12.0 Find and use measures of sides and of interior and eterior angles of triangles and polygons to classify figures and solve problems. (cont) Rule Polygon is a many sided figure. It has the same number of angles as sides. Listed below are some of the more common polygons whose names you should know. Triangle 3 sides Quadrilateral 4 sides Pentagon 5 sides Heagon 6 sides Heptagon 7 sides Octagon 8 sides Nonagon 9 sides ecagon 10 sides odecagon 12 sides Rule The formula we use to find the sum of all the interior angles of a polygon with n number of sides is Sum of ll Interior ngles = 180(n 2) Rule n eterior angle of a polygon is formed by etending one side of the polygon. In the diagram, angle 1 is an eterior angle. F It was formed by etending side E. 1 Rule E No matter what type of polygon, the sum of the eterior angles (one angle at each verte) is LWYS equal to 360º. Sum of Eterior ngles = 360º Note: While it is possible to draw two eterior angles at each verte, the sum of the eterior angles uses only ONE eterior angle at each verte = 360º Rule polygon is called a regular polygon when all of its sides are of the same length and all of its angles are of the same measure.. Rule The formula we use to find the measure of each interior angle of a regular polygon with n sides is Each interior measure of a regular polygon = Geometry 206

7 12.0 Find and use measures of sides and of interior and eterior angles of triangles and polygons to classify figures and solve problems. (cont) Rule The formula we use to find the measure of each eterior angle of a regular polygon with n sides is Each eterior measure of a regular polygon = Eample 2 Find the number of degrees in each interior angle of a regular pentagon. Solution: We know that the number of sides for a pentagon is 5. We use the formula and use 5 for n = = = 108º Eample 3 Each interior angle of a regular polygon measures 135º. How many sides does the polygon have? Solution: Use the formula: interior angle measure = Substitute 135 = ross multiply 135n = 180(n 2) 135n = 180n n = 360 n = 8 The polygon has 8 sides. Homework 1. Find the measure of each interior angle of a regular heagon. 2. The measure of each eterior angle of a regular polygon is 45º. How many sides does the polygon have? 3. What type of polygon has the sum of the interior angles equal to 360º and the sum of the eterior angles equal to 360º? 4. In, = 48º, = 24º, what type of triangle is? (Multiple choice) a. acute b. right c. obtuse 5. The verte angle of an isosceles triangle measure 8 times the measure of a base angle. Find the measure of a base angle. 6. In EF, = 37º and F = 56º. Find the measure of an eterior angle at E. 7. Find the sum of the measures of all the interior angles in a 9-sided polygon. 8. In heagon EF,, E + F = E F Geometry 207

8 13.0 Prove relationships between angles in polygons by using properties of complementary, supplementary, vertical, and eterior angles. Rule omplimentary angles are two angles with a sum of 90º Supplementary angles are two angles with a sum of 180º Vertical angles are congruent 1 2 Eample 1 Find the measure of in the figure Solution: Vertical angles are congruent so the missing angle in the pentagon has measure. In a pentagon the total number of degrees is (5-2) (180º) or 540º. So, 100º + 150º + 60º + 120º + = 540º 430º + = 540º 120 = 110º Eample 2 In the figure 1 and 3 are compliments. m 1 = 63º Find the measure of º Solution: Since 1 and 3 are compliments, they sum to 90º = 90 63º + 3 = 90º 3 = 27º y vertical angles, in the top triangle 27º + 10º + 2 = 180º 37º + 2 = 180º 2 = 143º (cont) 27º 2 63º 2 10º Geometry 208

9 13.0 Prove relationships between angles in polygons by using properties of complementary, supplementary, vertical, and eterior angles. (cont) Homework 1. bisects EF. EF and F are vertical angles with F = (2 10) and EF = ( + 36) Find F E F F 2. In the figure, and intersect at G. G = 65º and 5 = 55º, find GF G E In the figure, if, then = E In the figure, find the value of In the figure 1 and 2 are complements. Find m 2. 40º E is a regular pentagon. Find the measure of. E 72º Geometry 209

10 14.0 Prove the Pythagorean theorem. Over 2,500 years ago, a Greek mathematician named Pythagoras developed the relationship between the hypotenuse and the legs in a right triangle. This relationship can be stated as c a, b are legs a c is the hypotenuse b There are several different ways of proving the Pythagorean theorem. Here s one way. Using the given a, b, c triangle draw 4 of them and put them together as shown. a b a a c b c Now, let s find the area of the new larger figure. There are two ways you could find it. rea Method 1 The large figure is a square with sides lengths a+b a b The formula for area of a square is side side So, (a+b) (a+b) fter foiling rea = a 2 + 2ab + b 2 b a b a rea Method 2 The area of the large figure can be found by summing the areas of all the interior regions. 1 c 5 b c c 2 a 2 + b 2 = c 2 Four of the regions are triangles. One triangle area is base height = a b a 4 Four triangle areas = 4 ( a b ) 3 b = 2 a b The 5 th region is a square, its area is side side = c c = c 2 Summing all 5 regions together, we get rea = 2ab + c Now, we know that rea = rea So Method 1 s answer and Method 2 s answer must be equal. So, a 2 + 2ab + b 2 = 2ab + c 2 Subtracting the common term from each side a 2 + b 2 = c 2 This is what the Pythagorean Theorem states. b c c a a b a b Geometry 210

11 14.0 Prove the Pythagorean theorem. (cont) Homework In proving the Pythagorean Theorem, many properties or concepts were used. nswer yes or no in each of the following as to whether the item was used. 1. The area of a square equals side 2 2. The inner square and the large outer square are congruent 3. The four triangles are congruent. 4. The area of a triangle equals a b 5. The triangles are 30º, 60º, 90º triangles 6. The area of a large region is equal to the sum of the areas of its internal regions. 7. The total of the areas of the four triangles is equal to the area of the inner square. Geometry 211

12 15.0 Use the Pythagorean theorem to determine distance and find missing lengths of sides of right triangles. Rule In any right triangle, the lengths of the sides are in the relationship where leg 2 + leg 2 = hypotenuse 2 leg hypotenuse This is normally stated as: a 2 + b 2 = c 2 a c leg Eample 1 Find the measure of in. Solution: = () = () = () 2 10 = Eample 2 triangle has sides 6, 7, and 10. Is it a right triangle? Solution: If the triangle is a right triangle, the longest side must be the hypotenuse. Now check to see if the Pythagorean Theorem is true ? ? Since the Pythagorean Theorem is NOT true, this triangle is NOT a right triangle. Eample 3 ramp was constructed to load a truck. If the ramp is 9 feet long and the horizontal distance from the bottom of the ramp to the truck is 7 feet, what is the vertical height of the ramp? Solution: Since the ramp is described as having horizontal and vertical measurements, a right angle is implied. Solve using the Pythagorean Theorem. a 2 + b 2 = c = = 81 2 = 32 = or (cont) Geometry b 7

13 15.0 Use the Pythagorean theorem to determine distance and find missing lengths of sides of right triangles. (cont) Eample 4 To get from point to point you must avoid walking through a pond. To avoid the pond, you must walk 30 meters south and 40 meters east. How many meters would be saved if it were possible to walk through the pond? Solution: When walking south 30 meters and east 40 meters you walk a total of 70 meters. If you walked through the pond the distance could be found using Pythagoras = (through distance) = (through distance) = (through distance) 2 50 = through distance Homework The number of meters saved would be X right triangle has one leg with length 6 and the other leg with a length of 4. What is the length of the hypotenuse? 2. baseball diamond is a square with sides of 90 feet. What is the shortest distance between first base and third base? 3. Two joggers run 8 miles north and then 5 miles west. Home What is the shortest distance, to the nearest tenth of a mile, they must travel to return to their starting point? (Multiple hoice) a. 5.7 b. 8.5 c. 9.4 d rd ase 2 nd ase 1 st ase 4. In a computer catalog, a computer monitor is listed as being 26 inches. This distance is the diagonal distance across the screen. If the screen measures 10 inches in height, what is the actual width of the screen? 5. tent has sides that are both 5 feet long and the bottom of the tent is 6 feet across. What is the height of the tent, at its tallest point? Hint: raw the triangle. The height cuts the base in half. 5 feet 5 feet 5 6 feet Geometry 213 3

14 16.0 Perform basic constructions with a straightedge and compass, such as angle bisectors, perpendicular bisectors and the line parallel to a given line through a point off the line. For constructions, we will construct geometric figures using only a straightedge and a compass opy Line Segment 1. To draw a line segment congruent to, begin by drawing a reference line, l, with endpoint K K l 2. Place the point of the compass on point. Stretch the compass so that the pencil is eactly on. Make a mark. 3. Without changing the span of the compass, place the compass point on K and swing the pencil so that it crosses the reference line. Label this crossing point Y. K Y l Your copy, KY, and the original,, are congruent line segments. (cont) Geometry 214

15 For constructions, we will construct geometric figures using only a straightedge and a compass. 1. To draw an angle congruent to, begin by drawing a ray with endpoint. opy n ngle 2. Place the compass on point and draw an arc across both sides of the angle. Without changing the compass radius, place the compass on point and draw a long arc crossing the ray. Label the three intersection points as shown. 3. Set the compass so that its radius is. Place the compass on point E and draw an arc intersecting the one drawn in the previous step. Label the intersection point F. 4. Use the straightedge to draw ray F. EF Geometry 215

16 For constructions, we will construct geometric figures using only a straightedge and a compass. onstruct the Perpendicular isector Of Line Segment 1. egin with line segment XY. 2. Place the compass at point X. djust the compass radius so that it is more than (1/2)XY. raw two arcs as shown here. 3. Without changing the compass radius, place the compass on point Y. raw two arcs intersecting the previously drawn arcs. Label the intersection points and. 4. Using the straightedge, draw line. Label the intersection point M. Point M is the midpoint of line segment XY. M Segment XF has now been bisected and XM YM Geometry 216

17 For constructions, we will construct geometric figures using only a straightedge and a compass isect n ngle 1. Let point P be the verte of the angle. Place the compass on point P and draw an arc across both sides of the angle. Label the intersection points Q and R. 2. Place the compass on point Q and draw an arc across the interior of the angle. 3. Without changing the radius of the compass, place it on point R and draw an arc intersecting the one drawn in the previous step. Label the intersection point W. 4. Using the straightedge, draw ray PW. This is the bisector of QPR. has now been bisected. Geometry 217

18 For constructions, we will construct geometric figures using only a straightedge and a compass onstruct a line parallel to a given line through a point off the line (We are going to construct a corresponding angle at point P that is congruent to angle Q) 1. egin with point P and line k. 2. raw an arbitrary line through point P, intersecting line k. all the intersection point Q. Now the task is to construct an angle with verte P, congruent to the angle of intersection. 3. enter the compass at point Q and draw an arc intersecting both lines. Without changing the radius of the compass, center it at point P and draw another arc. 4. Set the compass radius to the distance between the two intersection points of the first arc. Now center the compass at the point where the second arc intersects line PQ. Mark the arc intersection point R. 5. Line PR is parallel to line k. PR is the line through P parallel to line k Geometry 218

19 Homework: 1. onstruct a segment whose measure is twice the measure of the segment below. 2. Segment and are given. Use them to construct + W X Y Z 3. onstruct the perpendicular bisector of LM. L M 4. onstruct a line through point X perpendicular to line m X m 5. onstruct an angle congruent to 6. onstruct a 90 angle. E F 7. escribe the procedure that could be used to construct a 45 angle. 8. This drawing shows how to. isect segment.. Find the perpendicular bisector to. opy segment. onstruct an angle 9. This drawing shows how to. opy an angle. opy a segment. isect an angle. Find the perpendicular to a given line (cont) E F Geometry 219

20 10. This drawing shows how to. onstruct an angle bisector. onstruct a perpendicular bisector of a segment. opy a segment. opy an angle. 11. This drawing shows how to. isect a segment. isect an angle. opy an angle. onstruct an altitude. 12. This drawing shows how to. isect an angle. onstruct the perpendicular bisector of an angle. opy a segment. onstruct a line parallel to a given line 13. To construct the angle bisector of, point Y would be constructed by opening your compass.. more than half of. less than half of. equal to Y. equal to 14. Which procedure should be followed to construct a 90 angle?. onstruct 3 angles of 30. onstruct 2 parallel lines.. onstruct the perpendicular bisector of a line. Find the midpoint of a line segment. Geometry 220

21 17.0 Prove theorems by using coordinate geometry, including the midpoint of a line segment, the distance formula, and various forms of equations of lines and circles. Midpoint Formula istance Formula Slope Formula (,y) = d = m = Parallel Lines same slope Perpendicular Lines negative reciprocal slopes (to get a negative reciprocal of a number, flip the number over and change the sign) Slope Intercept Form of a Line Equation of a ircle with enter at (h,k) and radius r y = m + b ( h) 2 + (y k) 2 = r 2 Eample 1 What is the equation of a circle whose diameter is 24 and whose center is at the origin? Solution: If the diameter is 24, the radius is 12. The center is at (0,0) Plugging into the formula for a circle ( 0) 2 + (y 0) 2 = y 2 = 144 Eample 2 The point (5,4) lies on a circle. What is the length of the radius of this circle if the center is located at (3,2)? Solution: Using the distance formula: = = = (cont) Geometry 221

22 17.0 Prove theorems by using coordinate geometry, including the midpoint of a line segment, the distance formula, and various forms of equations of lines and circles. (cont) Eample 3 has coordinates (0, 0), (2,3), (4, 0). Prove is isosceles. Solution: If a triangle is isosceles it has 2 sides congruent. Find the length of all 3 sides of and if it is isosceles, 2 sides will be the same length. Using (0, 0), (2,3) find = = Using (0, 0), (4, 0) find = = 4 Using (2,3), (4, 0) find = = Since two sides have lengths Eample 4 is a square with the given coordinates. Find the coordinates of point., the triangle is isosceles. Solution: The -coordinate of point gives the right/left movement. The y-coordinate of point gives the up/down movement. Point is n units right and n units up. The coordinates are (n,n) Homework 1. The coordinates of rectangle are (0,2), (4,8), (7, 6) and (3, 0). Show that the diagonals are equal in length. 2. The coordinates of a quadrilateral are (1,1), (2,5), (5,7), and (7,5). Which statement would prove that is a trapezoid? (multiple choice) a. Find the distance of all 4 sides and show 2 sides are congruent b. Find the slopes of all 4 sides and show 2 sides are parallel and the other 2 sides are not parallel. c. Find the lengths of the 2 diagonals and show they are congruent d. Show the diagonals bisect each other. 3. State the coordinates of the center of this circle. (multiple choice) ( + 3) 2 + (y 5) 2 = 16 a. ( 3, 5) b. ( 3, 5) c. (3, 5) d. (3, 5) (cont) Geometry 222

23 17.0 Prove theorems by using coordinate geometry, including the midpoint of a line segment, the distance formula, and various forms of equations of lines and circles. (cont) 4. is shown. Which statement would prove that is a right triangle? (multiple choice) a. Show distance from to = distance from to b. Show distance from to = distance from to c. Show (slope of ) = (slope of ) d. Show (slope of ) (slope of ) = 1 5. Give the coordinates of point P in parallelogram LMNP. L (b,d) P M (a,0) N (c,0) 6. Give the coordinates of the point of intersection of the diagonals. (n,n) Geometry 223

24 18.0 Know the definitions of the basic trigonometric functions defined by the angles of a right triangle. Know and use elementary relationships between them. Rule In a right triangle there are three relationships called the sine (sin), cosine (cos), and tangent (tan). SOH H TO sin = cos = tan = Eample 1 In, find cos Solution: y the formula, cos = = The hypotenuse is not given in the figure. We must find it using Pythagoras = hyp = hyp = hyp 2 17 = hyp 8 15 So, cos = = Eample 2 If cos =, find sin Solution: Since cos =, we know that =, so the side adjacent to is 3 and hypotenuse is 5 Find the third missing side, leg 2 = 5 2 sin = = 9 + leg 2 = 25 leg 2 = 16 leg = (cont) Geometry 224

25 18.0 Know the definitions of the basic trigonometric functions defined by the angles of a right triangle. Know and use elementary relationships between them. (cont) Eample 3 etermine if the following statement is true or false. sin = Solution: raw a right triangle and label one of the angles. (angle cannot be the right angle) Since no numbers were given let s use the common 3,4,5. (Remember the largest is the hypotenuse) Now, using SOH H TO, get the fraction values. sin = cos = tan = Now plug the fractions into the original problem and see if the two sides are the same. sin = = = hange divide to multiply by the reciprocal = = which is FLSE (cont) Geometry 225

26 18.0 Know the definitions of the basic trigonometric functions defined by the angles of a right triangle. Know and use elementary relationships between them. (cont) Eample 4 M is a right angle in LMN, where m a. (MN) sin 15º = NL b. (ML) sin 15º = MN c. (NL) sin 15º = MN d. (LN) sin 15º = ML Solution: First sketch the triangle. L = 15º. Select the one true statement below. N Since all the choices involve sin 15º, we know sin = M 15 L From the triangle, put the letters of the side opposite sin L = and the letters of the hypotenuse Now, take each multiple choice and get sin 15º by itself and find the one that is hoice a (MN) sin 15º = NL sin 15º = not a match hoice b (ML) sin 15º = MN sin 15º = not a match hoice c (NL) sin 15º = MN sin 15º = is a match. So choice c (cont) Geometry 226

27 18.0 Know the definitions of the basic trigonometric functions defined by the angles of a right triangle. Know and use elementary relationships between them. (cont) Eample 5 In, cos = 0.9, find 3 possible side lengths for side. Solution: cos = We need to write 0.9 as a fraction, so 0.9 = cos = = Putting the numbers on the triangle One possible side length for is second possibility we could write as. Then the triangle looks like nother possible side length for is third possibility we could write as. Then the triangle looks like third possible side length for is Homework Using JKL find the requested values in problems sin J 2. tan 2 L 3. sin L cos L Using MK, decide if each statement is true or false in problems 4-6. J 10 L K 6 4. tan = K 5. sin 2 M + cos 2 M = 1 6. (M) cos = K M 7. Using MK find the numerical fraction value of sin M if tan M = K 8. In UVW, tanw = 0.4 What is the length of? (multiple choice) a. 10 b. 8 c. 20 d. 1 U M V 20 W Geometry 227

28 19.0 Use trigonometric functions to solve for an unknown length of a side of a right triangle, given an angle and a length of a side. RULE In a right triangle there are three relationships called the sine (sin), cosine (cos), and tangent (tan). SOH H TO sin = cos = tan = Eample 1 Find the length of side in right triangle Solution: In relation to the given angle of 40º the two sides are opposite the angle and adjacent to the angle The trig relationship for opposite and adjacent is tan. tan = Substituting the numbers we write tan 40º = Using a calculator we can find the tan 40º to be the decimal approimately 0.84 We then substitute the decimal value in for tan 40º and get = Now cross multiply and solve for 0.84 = = 16.8 (cont) Geometry 228

29 19.0 Use trigonometric functions to solve for an unknown length of a side of a right triangle, given an angle and a length of a side. (cont) Eample 2 S puts up a 72 foot pole. To help support the pole, a wire is attached on the side. The wire forms a 65º angle with the ground as shown. How long is the wire? sin 65º 0.91 cos 65º 0.42 tan 65º Solution: In relation to the angle the two sides are opposite and the hypotenuse. The trig function with these two is the sin. sin = Substituting the numbers we write sin 65º = You don t need a calculator because the sin 65º is given. (The cos 65º and the tan 65º are also given but they are not needed.) Substituting in the decimal value for sin 65º you get 0.91 = Now, cross multiply = 72 Solve for = The wire is 79.1 feet long. Eample 3 If you were finding the length of side in LMN, which equation would you use? (multiple choice) a. = 17 sin 37º b. = 17 cos 37º c. = 17 tan 37º d. = L = 79.1 N M Solution: In relation to the given angle of 37º the two sides are adjacent and hypotenuse. The trig relationship for adjacent and hypotenuse is cos. cos = Substituting the numbers we write cos 37º = Now solve for by cross multiplying cos 37º = This is choice b. 17 cos 37º = Geometry 229

30 19.0 Use trigonometric functions to solve for an unknown length of a side of a right triangle, given an angle and a length of a side. (cont) Homework 1. Find the length of side in right triangle PQR. P Q 2. Tell which equation you would use to find the length of side in right triangle V R a. = 7 sin 25º b. = 7 cos 25º 25 c. = 7 tan 25º d. = W 7 U 3. jet climbs at a steady 20º. When it has traveled 2 km through the air, what would its altitude be? sin 20º = 0.34 cos 20º = 0.94 tan 20º = foot ladder is leaning against a building. The angle the ladder makes with the ground is 70º. How far up the building is the ladder? Geometry 230

31 20.0 Use angle and side relationships in problems with special right triangles, such as 30º, 60º, and 90º triangles and 45º, 45º, and 90º triangles. Rule There are 2 special right triangles. One is a 45º, 45º, 90º triangle. The other is a 30º, 60º, 90º triangle. The lengths of the sides can be obtained using special formulas. 45º,45º,90º Triangle 30º, 60º, 90º Triangle 45 a 45 a 30 2a 60 a Note: lways put a on the sides that are opposite the 45º angles. Note: lways put a on the side opposite the 30º Put 2a always opposite the right angle Put a always opposite the 60º angle Eample 1 Find the measures of sides k and m in the given triangle. k m Solution: You ll notice that one side is given. Find that side in the formula picture for the 30º, 60º, 90º triangle. So the value of a is 7. In the formula triangle the hypotenuse is 2a or 2(7) = 14. This would be the value of k. 30 2a 60 a In the formula triangle the long leg is a or This is the value of m (cont) Geometry 231

32 20.0 Use angle and side relationships in problems with special right triangles, such as 30º, 60º, and 90º triangles and 45º, 45º, and 90º triangles. (cont) Eample 2 Find the length of side in the given triangle Solution: The one side that is given is the hypotenuse. y the formula we see that the hypotenuse is normally 2a. So 2a = 28 a = 14 The side we want is opposite the 60º angle. y the formula that side is normally a and since a is 14 we get an answer of 14 Eample 3 In the given figure is a square with side length 3. Find the perimeter of EF. E F Solution: Fill in the remaining angles and congruent sides. In the two small triangles, the hypotenuse would be a and the a value is 3, so we get Now, add up all eterior segments EF + F + E ( + ) + (3 + 3) + (3 + 3) E F = + 12 (cont) Geometry 232

33 20.0 Use angle and side relationships in problems with special right triangles, such as 30º, 60º, and 90º triangles and 45º, 45º, and 90º triangles. (cont) Homework 1. Find the values of and y in the triangle shown. 60 y is an equilateral triangle with side lengths 8. ltitude is shown. Find the length of The diagonal of a square has length 6, find the length of a side of the square The lower base angles in this isosceles trapezoid each measure 45º. The length of the shorter base is 10 inches and the altitude is 8 inches. Find the length of the longer base The lengths of the two adjacent sides of parallelogram are 8 and 14 inches. If the degree measure of the included angle is 60º, what is the length of the height of the parallelogram? foot telephone pole has wires on each side supporting the pole. Find the length of each wire, represented in the diagram by m and k. m 9 k Geometry 233

34 21.0 Prove and solve problems regarding relationships among chords, secants, tangents, inscribed angles, and inscribed and circumscribed polygons of circles. Formulas entral ngle central angle of a circle is an angle formed by two intersecting radii with the verte at the center of the circle. entral ngle = Intercepted rc O is a central angle Its intercepted arc is the minor arc from to O = O Inscribed ngle n inscribed angle is an angle with its verte on the circle and whose sides contain chords of the circle Inscribed ngle = Intercepted rc is an inscribed angle Its intercepted arc is the minor arc from to. = Tangent hord ngle n angle formed by an intersecting tangent and chord has its verte on the circle. Tangent hord ngle = Intercepted rc is a tangent chord angle Its intercepted arc is the minor arc from to. = Geometry 234

35 ngle Formed Inside of a ircle by Two Intersecting hords When two chords intersect inside a circle, the measure of each angle is related to one-half the sum of the measures of the intercepted arcs. c d ngle Formed Inside by Two hords = Sum of Intercepted rcs = ( c + d ) ngle Formed Outside of a ircle When chords and/or tangents intersect to form an angle outside a circle, the measure of the angle is related to one-half the difference of the intercepted arcs. ngle Formed Outside = ifference of Intercepted rcs Two Tangents Two Secants d c d c = (d c) = (d c) a Tangent and a Secant d c = (d c) Geometry 235

36 Line Segments reated by Intersecting hords When two chords intersect inside a circle, the product of the segment of one chord equals the product of the segments of the other chord. r s = u w r u w s Line Segments reated by Two Secants rawn to a ircle From an Eternal Point When two secants are drawn to a circle from an eternal point, the product of one secant segment and its eternal segment equals the product of the other secant segment and its eternal segment. Whole segment Outside piece = Whole segment Outside piece a b = y a b y Geometry 236

37 Line Segments reated by a Secant and a Tangent rawn to a ircle From an Eternal Point When a secant and a tangent are drawn to a circle from an eternal point, the product of the secant segment and its eternal segment equals the square of the tangent segment. (a) (b) = 2 a b Eample 1 Find the length of segment in the diagram Solution: the formula states that the product of the pieces of one segment equals the product of the pieces of the other segment. So, 4 = = 24 = Eample 2 Find the measure of angle given the arc lengths shown. Solution: The formula says m = ( 168 y) To find the measure of the arc y we add all arcs to 360º. y = 360 y = 360 y = 74 y Now, substitute into the original equation: m = (168 74) 118 m = (168 74) m = ( 94) m = 47 Geometry 237

38 Eample 3 Find the measure of the arc length labeled Solution: STEP 1 STEP 2 STEP 3 First, find the angles in the The third angle in the triangle The arc length is triangle. The 80º arc is cut to make 180º would be 68º double the68º angle, so in half and yields an angle 136º. of 40º Eample 4 Find the measure of if m O = 31º and is a diameter of circle O. 31 O Solution: STEP 1 STEP 2 Step 3 would be twice the 31º angle. Since is a diameter m is half the So, = 62 measures 180º and measure of since measures 62º So, m = = = 118 (118) = 59º 31 O O O 62 Geometry 238

39 Eample 5 Find the area of the shaded region in circle F. = 5, = 6 5 F Solution: STEP 1 STEP 2 F is a radius of the circle and its measure is 5. The third side of the would Therefore the diameter measures 10. measure 8 by Pythagoras m = 90º since it intercepts the diameter () = 10 2 and m = 180º and the angle is half the arc F F 6 6 STEP 3 To find the shaded area, find the area of the circle and subtract the area of the triangle. F = = 24 Geometry 239

40 E Eample 6 Find the measure of 85 Solution: F G STEP 1 STEP 2 In a quadrilateral, opposite angles The formula says the arc is twice and angle. are supplementary, So, = 2 95 = 190º so m F = = 95 E 190 E F F G G Homework 1. Find the length of the segment labeled. 2. Find the measure of angle O Find the measure of angle 4. square with side length 12 is circumscribed about a circle. Find the area of the shaded region is an isosceles with 6. Find the length of segment Find m Geometry 240

41 22.0 Know the effect of rigid motions on figures in the coordinate plane and space, including rotations, translations, and reflections. Rule Translations are SLIES The original object and its translation have the same shape and size, and they face the same direction. The word translate means carried across. Eample 1 Translate 5 units to the right and one unit down. Solution In math, the translation of an object is called its image. If the original object was labeled with letters, such as, the image may be labeled with the letters Take point and slide it 5 units right and 1 down and label it Take point and slide it 5 units right and 1 down and label it Take point and slide it 5 units right and 1 down and label it onnect points,, and and you have ( translation moves an object without changing its size or shape and without turning it or flipping it.) Eample 2 Line segment has endpoints (5,1) and (3, 4). Give the coordinates of if is translated (,y) ( 3, y + 2) Solution: Take the original points and subtract 3 to the coordinate and add 2 to the y coordinate. becomes (5 3, 1 + 2) and becomes ( 3 3, 4 + 2) (2,3) and (0, 2). Geometry 241

42 Rule Reflections are FLIPS n object and its reflection have the same shape and size, ut the figures face in opposite directions. In a mirror, for eample, right and left are switched. Eample 3 Reflect polygon about line m. m Solution: m The line (where a mirror may be placed) is called the line of reflection. reflection can be thought of as a flipping of an object over the line of reflection. Eample 4 Line segment has endpoints (1,3 ) and (5, 2). Give the coordinates of if is reflected over the -ais. Solution: You flip line over the ais as shown. Geometry 242

43 Rule Rotations are TURNS. n object and its rotation are the same shape and size, but the figures may be turned in different directions. n eample of a rotation is when a person who is standing up lies down or stands on their head. rotation of 90º turns the object a quarter turn right or left. 90º rotation clockwise rotation of 90º clockwise turns the object a quarter turn right. 90º rotation counterclockwise rotation of 90º clockwise turns the object a quarter turn left. 180º rotation rotation of 180º turns the object a half turn. Eample 5 Rotate the triangle 90º clockwise. Solution: First put a line at the top that we can use as a reference and can rotate or turn. L I N E Now, turn the entire figure turn right. L I N E Geometry 243

44 Eample 6 Rotate the triangle 180º Solution: First put a line at the top that we can use as a reference and can rotate or turn. L I N E Now, turn the entire figure ½ turn. L I N E Eample 7 Rotate the triangle 90º counterclockwise. Solution: First put a line at the top that we can use as a reference and can rotate or turn. Now, turn the entire figure ¼ turn left L I N E L I N E Geometry 244

45 Homework In problems 1-3 identify whether each figure pair represents a translation, reflection, or rotation If line is rotated 180º about the origin, what are the coordinates of? 5. Find the coordinates of M if point M (7, 5) is translated 2 units in the negative direction and 6 units in the positive y direction. 6. a. Give the coordinates of point if trapezoid is reflected over the y-ais. b. Give the coordinates of point if trapezoid is reflected over the -ais. c. Give the coordinates of point if trapezoid is rotated 90º to the left. Geometry 245