Reteaching , or 37.5% 360. Geometric Probability. Name Date Class


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1 Name ate lass Reteaching Geometric Probability INV 6 You have calculated probabilities of events that occur when coins are tossed and number cubes are rolled. Now you will learn about geometric probability. Geometric Probability You can find the probability of an event occurring by dividing the number of ways the event can occur by the total number of outcomes. To know the probability of something falling into a part of a whole area or region, you can use geometric probability, which is the ratio of the size of the partial region to that of the whole region. epending on the situation, size can be measured in degrees, length, area, or volume. size of partial region P(partial region) size of whole region small meteor lands in the field to the right. What is the probability that it has landed in the flower garden? 0 IEL 50 m LOWER GREN 100 m area of the flower garden (0 m)(50 m) 1000 m area of the field (100 m)(100 m) 10,000 m P(meteor lands in flower garden) 1000 m % 10,000 m 100 m ompete the steps to find the probability. 1. ind the probability that a dart thrown at the dartboard lands in region 3. The degree measure of region 3 is 135. The dartboard is in the shape of a circle. The number of degrees in a circle is 360. region 3 The probability of a dart hitting region 3 is total degrees ind the probabilities. Region Region º 135º Region , or 37.5% 360. t the local pizza shop, if you drop a coin in the little bowl in the fish tank, you get a free slice of pizza. What is the probability of getting the coin in the fish tank if you have to drop it in while blindfolded? approximately 17.4% 1.5 ft 8 ft 3 ft 0.5 ft 3. ind the probability that a soccer ball kicked on goal goes over the goalie s head % 4 ft 4.5 ft Saxon. ll rights reserved. 131 Saxon Geometry
2 Reteaching continued INV 6 Experimental Probability When you say that the probability of flipping a coin and getting heads is 1, you are talking about the theoretical probability of flipping heads. If we flipped a coin 50 times, the experimental probability of heads would be the actual percentage of flips that resulted in heads. If a coin is tossed 50 times and heads occurs 6 times, what is the experimental probability of heads? P(heads) number of heads 6 heads 0.5 5% number of tosses 50 tosses In general, the experimental probability of an event is the number of times it occurs divided by the total number of trials: experimental probability # of times the event occurred # of trials s the number of trials goes up, an event s experimental probability will get closer and closer to its theoretical probability. ind the experimental probabilities. 4. Kyle rolled a number cube 50 times and got a one 9 times. ind the experimental probability of a one. experimental probability of a one number of ones number of trials % 50 What fraction will the experimental probability get closer to if the cube is tossed more? The experimental probability will get closer to the theoretical probability of 1 because experimental probability approaches theoretical probability 6 as the number of trials increases. 5. Make a spinner like the one to the right. Spin it 50 times and record the number of times it lands on three. ind the experimental probability of landing on three after 10, 5, and 50 spins. What number does it look like the experimental probability is getting closer to? Would it get closer to it if you performed more trials? Why? nswers will vary for the probabilities. The experimental probabilities should be getting closer to 1 8, or 0.15 and will get closer if more trials are performed because the experimental probability should approach the theoretical probability, 0.15, as the number of trials increases Saxon. ll rights reserved. 13 Saxon Geometry
3 Name ate lass Reteaching Quadrilateral Is a Parallelogram 61 You have worked with quadrilaterals. Now you will determine whether a quadrilateral is a parallelogram. Identifying Parallelograms parallelogram is a quadrilateral with two pairs of parallel sides. ll parallelograms have the following properties: Opposite sides are congruent. _ G _ HJ _ GH _ G J J H Opposite angles are congruent. H G J G H J Example: In quadrilateral JKLM, _ JK _ ML. Is JKLM a parallelogram? Step 1: Solve for x. _ JK _ J ML Given 3x 1 x 4 Substitute. x 3 Solve. Step : Substitute x 3 to find JM and KL. JM 4(3) 5 7 KL Since _ JK _ ML and JM _ KL _ M, JKLM is a parallelogram. x + 4 3x + 1 4x 5 x + 5 L K omplete the statements. 1. _ _ ; _ _ ; 63 The figure is a parallelogram since opposite angles and sides are congruent. 63 Use the given information to determine whether the figures are parallelograms. State the reason.. Given: _ _ 3. Given: _ PR _ QS x 18 P (4x + 6) (5x 4) x + 15 Q 6x x (3x + 6) (6x 4) x + 6 yes; x 4; Opposite angles are congruent R 7x 0 no; x 5; Opposite sides are not congruent. S Saxon. ll rights reserved. 133 Saxon Geometry
4 Reteaching continued 61 More on Identifying Parallelograms One pair of opposite sides of a quadrilateral are both parallel and congruent to each other. _ _ The diagonals bisect each other. P PH G P GP PJ J H Example: Use the given information to decide whether JLKM is a parallelogram. Given: _ JK _ LM and _ JK _ J LM Step 1: Is there a relationship between the sides? Opposite sides are both parallel and congruent to each other. M Step : If one pair of corresponding sides of a quadrilateral are both parallel and congruent to each other, the figure is a parallelogram. L K omplete the statements. 4. _ TZ _ ZV ; WZ _ ZU The diagonals of the quadrilateral bisect each other. The figure is a parallelogram. Use the given information to determine whether the figure is a parallelogram. State the reason. 5. _ VZ _ ZX and _ YZ _ ZW 6. _ _ and _ _ V W W T Z 8 V U Z E Y yes; iagonals bisect each other. 7. _ PT _ QT and _ ST _ RT 8. _ _ and _ _ P X Q yes; Opposite sides are both parallel and congruent. T E S R no no Saxon. ll rights reserved. 134 Saxon Geometry
5 Name ate lass Reteaching inding Surface rea and Volume of ylinders 6 You have learned how to find the surface area and the volume of prisms. Now you will find the surface area and the volume of cylinders. To find the surface area of a cylinder, use the formula S r rh to combine the lateral area and the area of the two bases. 9 in ind the area of the cylinder in terms of. S r rh S (9) (9)() S S 558 in in omplete the steps to find the surface area of the cylinder in terms of. 1. S (7 ) (7 )(1 ) S S 66 cm ind the surface area of each cylinder in terms of. 14 cm 1 cm. 5 yd 3. ft 14 yd 18 ft 190 yd 638 ft 4. 8 m 5. 6 mm 1 m 3 mm 00 m 456 mm Saxon. ll rights reserved. 135 Saxon Geometry
6 Reteaching continued 6 oth right cylinders and oblique cylinders each use the same formula, V r h, for volume. ind the volume of the right cylinder in terms of. V r h V (15) (10) 10 mm 15 mm V 50 mm 3 ind the volume of the oblique cylinder in terms of. V r h 8 ft V (8) (5) V 30 ft 3 5 ft omplete the steps to find the volume of each cylinder in terms of. 6. V (5 ) (7 ) 7. V (18 ) (5 ) V 175 in 3 V 8100 yd 3 10 in 18 yd 7 in 5 yd ind the volume of each cylinder in terms of mm 9. 4 cm 9 mm 1 cm mm cm ft cm 0 ft 7 cm ft cm Saxon. ll rights reserved. 136 Saxon Geometry
7 Name ate lass Reteaching Introduction to Vectors 63 You have learned how to find points on a coordinate plane. Now you will determine vectors and their magnitudes. ind r. Step 1: Place M on the origin. y Step : N will be 4 units to the right and 5 units up, so the coordinates of the terminal point N are (4, 5). Step 3: Use the distance formula to find the distance between M(0, 0) and N(4, 5). d ( x x 1 ) ( y y 1 ) M ř N d (4 0) (5 0) d Therefore, r is 41. omplete the steps to find the magnitude of the vector. y N 1. ind p. M ρˇ Place M on the origin. N is now at (3, 4). d (3 0) (4 0) d (3) (4) 5 So, p 5. ind the magnitude of each vector in simplified radical form.. m y 3. d y m d e 5. g y y e g 8 85 Saxon. ll rights reserved. 137 Saxon Geometry
8 Reteaching continued 63 dd vectors v and w. Step 1: Write v in component form. The horizontal change is 7. The vertical change is 1. y The component form of v is 7, 1. Step : Write w in component form. The component form of w is 4, 3. v w Step 3: dd the components: 7 4, , 4. The resultant vector from adding v and w is 11, 4. dd vectors s and t. The component form of s is, 1. The component form of t is, 1. The components of s and t sum of 0 because they are opposite vectors. The resultant vector is 0, 0. t y s omplete the steps to find the sum of the vectors. y 6. dd vectors h and k. The component form of h is 5, 4. The component form of k is 3, 6. dd the components: 5 3, 4 6. The resultant vector is 8, dd vectors a and b. The component form of a is 3, 8. The component form of b is 3, 8. h b y k dd the components: 3 3, 8 8. a The resultant vector is 0, 0. Saxon. ll rights reserved. 138 Saxon Geometry
9 Name ate lass Reteaching ngles Interior to ircles 64 You have learned how to construct a tangent line on a circle. Now you will determine the measure of an angle formed by a chord and a line tangent to a circle. The measure of an angle formed by a chord and a tangent is equal to half the measure of the arc that subtends it. m 1 m E Given: _ YZ and _ HI are tangent lines. E ind the indicated measure. a. m XYZ m XYZ 1 m XY m XYZ 1 (170 ) m XYZ 85 x Y 170 Z b. m GH m GHI 1 m GH 75 1 m GH 150 m GH I G 75 H omplete the steps to find each measure. 1. m 40 m 1 m 1 (40 ) I m 10. m PQR Q P m PRT 1 m PQR m PQR T 135 R 70 m PQR ind the value of x in each figure. 3. x x 66 x 18 x p 133 q Saxon. ll rights reserved. 139 Saxon Geometry
10 Reteaching continued 64 The measure of an angle formed by two chords intersecting in a circle is equal to half the sum of the intersected arcs. m 1 1 (m m ) ind the measure of angle x. x 1 (m m E ) x 1 (60 56) x 58 1 x E omplete the steps to find the measure of angle x. 5. x 1 ( m HJ m Lk ) x 1 ( ) x 146 H L 110 x J K x 1 ( m PS m QR ) x 1 ( 45 73) x 59 ind the measure of angle x for each circle. 7. x x 49.5 W 5 T U x 81 V 9. x x L P S P x x M Q 73 R 6 N Z x 153 x E Y Y G 33 Saxon. ll rights reserved. 140 Saxon Geometry
11 Name ate lass Reteaching istinguishing Types of Parallelograms 65 parallelogram is a rectangle if it has one 90 angle. Example: Is this parallelogram a rectangle? Since the consecutive angles of a parallelogram are supplementary, we know that the measure of the two given expressions must add up to 180. Solve for x. (7x+6) (8x 6) (7x 6) x 180 x 1 Substitute the value of x into one of the expressions to find the measure of the angle. 8(1) 6 90 One of the angle measures of the parallelogram is 90, so the parallelogram is a rectangle. omplete the statements to distinguish the parallelogram. (8x6) 1. Is parallelogram MNOP a rectangle? or a parallelogram to be a rectangle, it must have one 90 angle. 6x x 60 4x 60 x 15 M 6x P (6x + 60) N O If x 15, M 6x , and MNOP is a rectangle.. Is this parallelogram a rectangle? no 3. Is this parallelogram a rectangle? yes (5x + 5) (8x + 18) (6x + 4) (11x  9) 4. If y 10, is this parallelogram a rectangle? 5. If y 3, is this parallelogram a rectangle? W X (5x  y) (3x + y) (4x + 10) (4x  37) Z yes Y no Saxon. ll rights reserved. 141 Saxon Geometry
12 Reteaching continued 65 parallelogram is a rhombus if it has two consecutive sides that are congruent. Example: E If x 14, is parallelogram EGH a rhombus? Since we know x 14, we can find the length of side G: G 4x 16 H G 4(16) 16 G 40 Since E and G have the same length, they are congruent. G Since E and G are consecutive sides of equal length, parallelogram EGH is a rhombus. 40 4x If x, is the parallelogram a rhombus? or a parallelogram to be a rhombus, consecutive sides must have equal length. 4x 1 9 4() 1 9 X U 4x + 1 W V 9 Since UV and VW are consecutive sides that have equal lengths, the parallelogram is a rhombus. 7. If x 19, is this parallelogram 8. If a 3, b 5, c 4, and d 4, a rhombus? is this parallelogram a rhombus? J K Q R 3x + 18 a + b I 75 ft L yes P yes c + d S 9. If x y, is this parallelogram a rhombus? no 10. Is this parallelogram a rhombus? yes W X n 50 x n Y y + 5 Saxon. ll rights reserved. 14 Saxon Geometry
13 Name ate lass Reteaching inding Perimeters and reas of Regular Polygons 66 You have learned how to find perimeter and area of many types of figures. Now you will use formulas to find perimeters and areas of regular polygons. The perimeter of a regular polygon can be found by multiplying the length of one of its sides by the number of sides it has. ind the perimeter of the regular octagon. ecause this is a regular octagon, the length of each side is the same as that of the labeled side, 7 centimeters. n octagon has 8 sides, so we can find the area by multiplying 8 by 7. P ns P (8)(7 cm) P 56 cm The perimeter of the regular octagon is 56 centimeters. 7 cm omplete the steps to find the perimeter of the regular polygon. 1. P (5)(9). P (6)(4.) 9 cm 4. ft P 45 cm P 5. ft ind the perimeter of each regular polygon m mm 1 m 11.5 mm yd in. 5.7 yd 1 in. Saxon. ll rights reserved. 143 Saxon Geometry
14 Reteaching continued 66 The area of a regular polygon is equal to its perimeter times the length of the apothem, divided by ( 1 ap ). Here we are not given the length of the apothem, so we have to find it. When the figure is a regular hexagon, we can form an equilateral triangle with the center of the hexagon as the top vertex. The apothem forms a with the side and hypotenuse. Use the Pythagorean Theorem to find that the length of the apothem is 7 3 millimeters. ind the perimeter: P ns P (6)(14) 84 ind the area: 1 ap 1 (7 3 )(84) 94 3 The area of this regular hexagon is 94 3 mm mm 7 mm a 14 mm omplete the steps to find the area of the regular polygon. 7. apothem 4 3 ft P (6)(8) 1 (4 3 )(48) 19 3 ft 8. apothem 11.5 cm P (7)(11) 1 (11.5 cm)(77 cm) cm a 8 ft 11 cm ind the area of each regular polygon in 1 m 10 in m 6.9 in m yd 4 yd ft 7 ft 9.6 ft Saxon. ll rights reserved. 144 Saxon Geometry
15 Name ate lass Reteaching Introduction to Transformations You have worked with different figures in the coordinate plane. Now you will work with moving those figures around. 67 Transformations n isometry is a transformation that does not change the size or the shape of a figure. There are different types of isometries. translation or a slide is a transformation in which all the points of a figure are moved the same distance in the same direction. ' ' ' ' reflection or a flip is a transformation that flips a figure across a line. E' ' ' E rotation is a transformation that turns a figure around a fixed point, which is called the center of rotation. center of rotation ' Rotation ' Example: Identify the type of transformation. E ' ' E' ' ' ' The figure is rotated about a point. ' omplete the statement to determine the type of transformation. 1. ll the points of the figure are moved the same distance in the L' M' same direction. L N' The transformation is a translation or a slide. M etermine the type of transformation.. reflection 3. rotation N E ' ' E' G G' ' ' Saxon. ll rights reserved. 145 Saxon Geometry '
16 Reteaching continued 67 Performing Transformations or a translation or a slide, move each point of the preimage the indicated distance and direction. ' ' or a reflection or a flip, move each point across the line of reflection so that the point and its image are the same distance from the line. ' ' ' ' ' To rotate a figure about a point, keep point fixed and turn each point on a circular path around, as shown. ' ' Example: Reflect the figure across the line. ' ' E E' ' ' omplete each translation. 4. raw the translation of the figure along the given path. 5. raw the rotation of the figure about point P. R' Q' ' P' P R ' ' Q 6. raw the translation of the figure along the given path. 7. raw the rotation of the figure across the line. ' G' ' I' G H' ' I H ' Saxon. ll rights reserved. 146 Saxon Geometry
17 Name ate lass Reteaching Introduction to Trigonometric unctions 68 You have worked with right triangles. Now you will work with special relationships between the angles and sides in a right triangle. Trigonometric Ratios leg opposite sin leg cos adjacent to hypotenuse hypotenuse Example: Write each trigonometric ratio as a fraction. leg opposite sin 4 hypotenuse 5 cos tan leg adjacent to hypotenuse 3 5 leg opposite 4 leg adjacent to 3 tan hypotenuse 5 leg opposite leg adjacent to 3 4 leg adjacent to leg opposite omplete the statement to determine the given trigonometric ratio. 1. sin K leg opposite K hypotenuse cos K leg adjacent to K hypotenuse 8 17 K 8 J H tan K leg opposite K leg adjacent to K 15 8 Use the diagram from Exercise 1 to write each trigonometric ratio as a fraction.. sin H cos H tan H 8 15 Write each trigonometric ratio as a fraction. 5. sin cos tan sin tan cos Saxon. ll rights reserved. 147 Saxon Geometry
18 Reteaching continued 68 Trigonometric Ratios (continued) You can use a calculator to find the value of trigonometric ratios: cos , or about Example: ind WY. adjacent leg cos W Write a trigonometric ratio that involves WY. hypotenuse cos cm x Substitute the given values. x cos Multiply both sides by x. x 7.5 ivide both sides by os 39 cos 39 x 9.65 cm Simplify. The length of WY is approximately 9.65 centimeters. W 7.5 cm 39 Y X omplete the steps to find E. 11. E sine opposite leg hypotenuse sin 7 18 x x sin m 7 E x 18 sin 7 x m Use your calculator to find each trigonometric ratio. Round to the nearest hundredth. 1. sin cos tan ind each length. Round to the nearest hundredth. 15. H 6.01 in. 16. JK 3.91 mm 17. US 55.3 cm 10 in. 31 G K 34.6 mm 18 L J S 66 T.5 cm U H Saxon. ll rights reserved. 148 Saxon Geometry
19 Name ate lass Reteaching Properties of Trapezoids and Kites 69 You have worked with triangles. Now you will find the lengths of sides of trapezoids and kites. Trapezoid Midsegment Theorem trapezoid is a quadrilateral with one pair of parallel sides. The midsegment of a trapezoid is the segment whose endpoints are the midpoints of the legs. The midsegment of a trapezoid is parallel to each base. M N _ _ MN and _ _ LP The length of the midsegment is onehalf of the sum of L the length of the bases. 1 (MN LP ) Use the formula to find the length of the midsegment _ KL. G KL 1 (GH J ) K KL 1 (18 6) Substitute. KL 1 (44) Simplify. KL Simplify. P is the midsegment of LMNP H L J omplete the steps to find each length. 1. MN. J MN 1 ( ) M MN 1 3 ( 3 17) N XY 1 (J GH ) G X (J 48) MN 1 ( 40) 80 J 48 MN 0 J 3 ind each length. 3. KL 4. TV J Y H K N J M P L 14 S P V T Q 5.6 R Saxon. ll rights reserved. 149 Saxon Geometry
20 Reteaching continued 69 iagonals of Kites kite is a quadrilateral with two pairs of congruent adjacent sides. The diagonals of a kite are perpendicular to each other. _ _ ecause the diagonals of a kite are perpendicular to each other, you can use the Pythagorean theorem to find the length of each side. ind the length of _. Use the Pythagorean theorem to find the length of the side. E E 1 5 Substitute Simplify. 1 E Simplify Solve. So, the Length of _ is 13. omplete the steps to find the lengths of _ JK and _ JM. Round each length to the nearest tenth. K 5. Use the Pythagorean theorem to find the length of each side. JK JN NK JM JN NM JK 6 15 JM 6 8 JK 36 5 JM JK 61 JM 100 J 15 6 N 6 8 L JK 61 JM 100 M JK 16. JM 10 ind each length. Round to the nearest tenth. 6. ind the length of _. 7. ind the length of _ E. 7.4 E E Saxon. ll rights reserved. 150 Saxon Geometry
21 Name ate lass Reteaching inding Surface reas and Volumes of Pyramids 70 You have worked with surface area and volume of prisms. Now you will find the lateral area, the surface area, and the volume of pyramids. Lateral and Surface rea of a Regular Pyramid The lateral area of a regular pyramid with base perimeter P and slant height is L 1 P. The surface area of a regular pyramid with lateral area L and base area is S L or S 1 P. ind the lateral area and the surface area of the regular pyramid. L 1 P Lateral area formula L 1 (3)(1) Substitute. 1 m L 19 m Simplify. S L Surface area formula S 19 6 Substitute. S 56 m Simplify. The lateral area is 19 square meters. The surface area is 56 square meters. l 8 m base height 8 m base omplete the steps to find the lateral area and the surface area of the regular pyramid. Round your answer to the nearest tenth. 1. L 1 P 1 (48)(1) 88 ft 1 s 3 P ft S L ft 1 ft 8 ft 3 ft ind the lateral area and the surface area of the regular pyramid. Round to the nearest tenth.. L 98.0 in 3. L cm S 4.1 in S cm 1.9 in. 13 cm 8.5 in in. 1 cm 1 cm Saxon. ll rights reserved. 151 Saxon Geometry
22 Reteaching continued 70 Volume of a Pyramid The altitude, or height, of a pyramid is the perpendicular segment that measures from the vertex of the pyramid to the base of the pyramid. The volume of a pyramid with base area and h height h is V 1 3 h. ind the volume of the pyramid. Round to the nearest tenth if necessary. V 1 3 h 8 in. V 1 (5)(8) Substitute. 3 V 1 (00) Simplify. 3 V 66.7 in 3 Solve. The volume of the pyramid is 66.7 cubic inches. 5 in. 5 in. omplete the steps to find the volume of the pyramid. Round to the nearest tenth. 4. V 1 3 h 11 cm V cm 6 cm 5. V 1 3 h V 1 (84) (11) 3 V 1 (54.) (16.6) 3 (94) V 1 3 (899.7) V 308 cm 3 V 99.9 in 3 ind the volume of the pyramid. Round to the nearest tenth ft m in. 1.9 in. 8.4 in. 10 ft.6 m 7 ft 7 ft 18.1 m 33.8 m Saxon. ll rights reserved. 15 Saxon Geometry
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