Geometry: A Complete Course

Geometry: A Complete Course (with Trigonometry) Module A Instructor's Guide with Detailed Solutions for Progress Tests Written by: Larry E. Collins ERRATA 4/00

Quiz Form B Class Date Score Unit I - The Structure of Geometry Part A - What is Geometry? Lesson - Origin Lesson - Structure. In our study of Algebra, the symbols used to name numbers were examples of the things of mathematics, or the objects around which our study revolves. How many new things did we discuss in Lesson? 4 them. Point Line Plane Space. Tell what part of mathematical speech each of the following is: a) Operation Symbol b) π Number Symbol c) Relation Symbol d) 7 Number Symbol e) + Operation Symbol f) {} Grouping Symbol g) e Number Symbol h) Relation Symbol. the plane shown in two ways Z Q Plane Z; Plane PQR P R 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part A, Lesson, Quiz Form A. Draw the image of the given rectangle after a rotation of 90 o clockwise around the center of rotation Q. B B A A C C D D D A 90 Q Q C B ) Draw a segment from vertex A to the center of rotation point Q ) Measure a 90 o Angle clockwise at point Q. Draw the angle. ) Use a ruler to locate A on the ray of the angle forming angle AQA. The measure of segment QA will equal the measure of segment QA. 4) Repeat steps through 4 for each vertex. Connect the vertices to form rectangle A B C D. 6 006 VideoTextInteractive Geometry: A Complete Course

Quiz Form A Class Date Score Unit I - The Structure of Geometry Part A - What is Geometry? Lesson 5 - More on Groupings. Write D = {,, 5,...} using set-builder notation. D = {x x is an odd natural number} or D = {x x is an odd counting number}. Write B= { x x = y+, y I, y 0} using the roster method. B = {, 4, 7, 0,, 6...} Rewrite the statements in exercises and 4 using set notation. Use the roster method if possible.. The set made up of even counting numbers less than ten is an improper subset of the set made up of even counting numbers less than ten. {, 4, 6, 8} {, 4, 6, 8} 4. The set whose only element is 0 is not a subset of the empty set. {0} { } 006 VideoTextInteractive Geometry: A Complete Course 7

Unit I, Part A, Lesson 5, Quiz Form B Consider these sets for questions 5 through 0. A= {a, b, c, d, e} B = {a, b, c, d, e, f, g} C = {m, i, n, t} D = {d, e} 5. A B = {, a b,, c d,} e These are the elements that are common to both sets 6. A C = {, a b,, c d,, e m,, i n,} t These are the elements in one, or the other, or both of the sets. 7. A D = {,} d e These are the elements that are common to both sets 8. C D = { } These are the elements that are common to both sets 9. Is D {d, e}? Yes, all of the elements in set D are also in the set {d,e} 0. Does A = C? No, sets A and C do not contain exactly the same elements.. Of 68 people surveyed, most often drive to work, 57 usually take the bus to work, and 7 do both equally as often. How many of these surveyed did neither? 5 Use a Venn Diagram to show your work. Drive 6 7 0 Bus U 5 0 006 VideoTextInteractive Geometry: A Complete Course

Quiz Form A Class Date Score Unit I - The Structure of Geometry Part B - The Scope of Our Geometry Lesson 4 - Solids Complete each sentence in exercises through 5 with the appropriate geometric term(s).. The three basic three dimensional shapes in all the world are, Prisms, Pyramids and. Spheres. A prism has Parallelograms for sides.. A pyramid has Triangles for sides. 4. Cones and cylinders have Circles for bases. 5. A sphere is a surface which is everywhere the same from Distance a fixed point. Identify the solids in exercises 6 through as prisms, cylinders, pyramids, cones, or spheres. Note the shape of the base when naming a prism or pyramid and be as specific as possible. 6. 7. 8. Triangular Prism Cylinder (or circular prism) Square Prism 9. 0.. Trapezoidal Pyramid Triangular Pyramid Cone (or circular pyramid) 006 VideoTextInteractive Geometry: A Complete Course 9

Quiz Form B Class Date Score Unit I - The Structure of Geometry Part B - The Scope of Our Geometry Lesson 4 - Solids Complete each sentence in exercises through 5 with the appropriate geometric term(s).. A prism is named after the shape of its. base. A cone has a circular. base. The sides of a prism are parallelograms. 4. The sides of a pyramid are. triangles 5. A cylinder has two bases. circular Identify the solids in exercises 6 through as prisms, cylinders, pyramids, cones, or spheres. Note the shape of the base when naming a prism or pyramid, and be as specific as possible. 6. 7. 8. Rectangular Pyramid Cone (Circular Pyramid) Triangular Prism 9. 0.. Elliptical Cylinder (or Elliptical Prism) Triangular Pyramid Rectangular Pyramid (top view) 005 VideoTextInteractive Geometry: A Complete Course

Quiz Form A Class Date Score Unit I - The Structure of Geometry Part C - Measurement Lesson - Parallelograms. Find the area and perimeter of the given parallelogram. ft. 6.4 8 Area: 5.6 sq. inches Perimeter: 64 inches Area = base height ft = in = 4in a = 4" 6. 4" a = 5. 6 square inches a = 5. 6 inches Perimeter = sum of lengths of the sides = 4" + 8" + 4" + 8" = " + " = 64". Find the area and perimeter of the given parallelogram. 6 60 o Area: 9 square feet Perimeter: 8 feet Area = base height = ' ' = ( )( )( ) = 9 square feet Perimeter = Sum of lengths of the sides = ' + 6' + ' + 6' = 9' + 9' = 8 feet 006 VideoTextInteractive Geometry: A Complete Course 4

Unit I, Part C, Lesson, Quiz Form A. Find the area and perimeter of the given rhombus. 5 cm 7.5 sq. cm. Area:.5 cm Perimeter: 0 cm. Area = Base Height a = 55. a = 7. 5 square cm Perimeter = Sum of lengths of the sides = 5 + 5+ 5+ 5 = 0cm 4. Find the perimeter and the height of this parallelogram. 6 yd Area: 76 sq yds yd h yd Perimeter: 58 yds. height: yds. Area = base height 76 = 6 h 6 76 = 6 6 h 76 6 = h yards = h Perimeter = Sum of lengths of the sides = 6 + + 6 + = 9 + 9 = 58yards 4 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part C, Lesson, Quiz Form A 5. Find the perimeter and the base of the given rhombus. b Area: 48x 6x Perimeter: x units 8x units base: Area = Base Height 48x = b 6x 48x 6x =b 6x 6x 6 8 x x =b 6x 8x units = b Perimeter = Sum of lengths of the sides = 8x+ 8x+ 8x+ 8x = x units 6. Find the area and perimeter of the given parallelogram. 4 mm 5 5 mm 4 8 6 mm 6 mm Area: Perimeter: mm Area = base height = 6 4 4 = 9 7 4 9 7 = 4 = square mm or 6 mm Perimeter = Sum of lengths of the sides = 6 +55 8 +6 +55 8 = 9 + 45 8 + 9 + 45 8 = 9 8 8 + 45 8 + 9 8 8 + 45 8 = 5+5 +5+5 4 = 574 4 mm or mm or 4 mm 006 VideoTextInteractive Geometry: A Complete Course 4

Quiz Form B Class Date Score Unit I - The Structure of Geometry Part C - Measurement Lesson - Parallelograms. Find the height of the given parallelogram. Area: 84 square inches 7 inches height: ft Area = base height 84 = h 84 = h 7 = h 7 = h. Find the area and perimeter of the given parallelogram. 8 45 o 4 Area: 48 sq. inches Perimeter: 40 inches Area = base height a = " 4 " a = ( ) ( 4) ( ) a = 48 square inches Perimeter = Sum of lengths of the sides = + 8+ + 8 = 0 + 0 = 40 inches 006 VideoTextInteractive Geometry: A Complete Course 45

Unit I, Part C, Lesson, Quiz Form B. Find the area and perimeter of the given rhombus. 4m 6m 4 sq. meters Area: Perimeter: 44 meters Area = base height = 6 4 = 4 square meters Perimeter = Sum of lengths of the sides = 6+ 6+ 6+ 6 = 44 meters 4. Find the perimeter of the given parallelogram. 44cm 48cm Area: 508 sq cm Perimeter: 0 cm Area = base height 508 = b 44 508 44 = b 44 44 44 57 = b 44 57cm = b Perimeter = Sum of lengths of the sides = 57 + 48 + 57 + 48 = 0cm 46 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part C, Lesson, Quiz Form A. Area: 0 sq. units 5 ( 8 + 9 + 6) units Perimeter: 6 Area = base height = 8 5 8 = 5 = 4 5 = 4 5 = 0 square units Pythagorean Theorem a + b = c + 5 = x 4+ 5= x 9 = x 9 = x 6 + 5 = y 6 + 5 = y 6 = y 6 = y Perimeter = Sum of Lengths of the Sides ( ) Perimeter = 8 + 9 + 6 units 4. Find the area of a triangle with base (x + ) units and height (4x - ) units. Area = base height = ( x + ) x 4 ( ) ( )( ) x x = + 4 x 4x+ 4x+ x ( ) + ( ) = 8x + x+ 4x+ 6 = 8x + 8x 6 = 4 ( x + 4x ) = ( ) = 4x + 4x square units ( ) = 4x + 4x square units Area: 50 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part C, Lesson, Quiz Form A 5. Find the area and perimeter of the given triangle. 6 4 yds 9 yds 5 yds Area = base height a = 6 4 5 7 6 a = 4 7 6 a = 4 94 a = 4 a = 9 a = 8 square yards Area: 8 sq. yards yards Perimeter: Perimeter = Sum of lengths of the sides = 6 + 5 + 9 4 7 6 = + + 9 4 7 6 4 9 = + + 4 4 8 64 08 = + + 8+ 64 + 08 = 5 = or yards 6. Find the area and perimeter of the shaded square in the given figure. 4 5 Area: 4 sq. units 5 4 5 4 4 5 Perimeter: 4 4 units Area of Larger Square: Pythagorean Theorem helps us find c. Area = ( 5+ 4) ( 5+ 4) a + b = c = 9 9 4 + 5 = c = 8 square units 6 + 5 = c 4 = c 4 = c c) Area of Shaded Square d) Perimeter is the sum of the lengths of the sides. Area = 4 4 Perimeter = 4 + 4 + 4 + 4 = 4 4 = ( + + + ) 4 = 68 = 4 square units = 4 4units 006 VideoTextInteractive Geometry: A Complete Course 5

Quiz Form B Class Date Score Unit I - The Structure of Geometry Part C - Measurement Lesson - Triangles Find the area and perimeter of the given triangles in exercises through. Note: You may first have to use the Pythagorean Theorem (a + b = c ) to find some missing parts.. Area: 90 sq. inches 6 5 Perimeter: 49 inches Area = base height Perimeter = Sum of Lengths of the Sides = + 6 + = 5 = + 8 5 = = 49 inches 65 = = 65 = 90 square inches sq. feet. Area: 9 4 Area = base height a = 9 ( 4 + 4 )feet Perimeter: Pythagorean Theorem a + b = c Two Missing Pieces: a = 9 a = 7 or square feet + 4 = x 9+ 6= x 5 = x + 5 = y 9+ 5= y 4 = y 5 = x 5 = x 4 = y Perimeter = Sum of Lengths of the Sides ( ) ( 4 4 ) = 9+ 5+ 4 = + feet feet 006 VideoTextInteractive Geometry: A Complete Course 5

Unit I, Part C, Lesson, Quiz Form B. Area: 7 sq. cm 6 cm 9 cm Perimeter: ( 5 + ) cm Area = base height Pythagorean Theorem a + b = c = 6 9 6 + 9 = c 69 = 6 + 8 = c 9 = 7 = c 7 = c = 7 square cm = c = c Perimeter = Sum of Lengths of the Sides = 6+ 9+ = ( 5 + ) cm 4. Find the area of a triangle with base (x - 4) units and height (x - ) units ( ) x 4x + 4 square units Area: Area = base height a = ( x ) x 4 a = ( x 8 x+ 8 ) ( ) ( ) a = x 4x + 4 square units 54 006 VideoTextInteractive Geometry: A Complete Course

Quiz Form A Class Date Score Unit I - The Structure of Geometry Part C - Measurement Lesson 4 - Trapezoids Find the area and perimeter of each of the given trapezoids in exercises through. Assume all measures are in inches.. Area: 5 sq. inches 7 6. 6 0 6.7 Perimeter: 0 inches Area = height sum of the bases Perimeter = sum of lengths of the sides = 0+ 67. + 7+ 6. = 6 ( 0+ 7) = 0 inches 67 = = 7 = 5 square inches. Area: 40 sq. inches 7 inches Perimeter: 9 4 Area = height sum of the bases = 4 ( + 9) 4 = 0 = 0 = 40 square inches Perimeter = sum of lengths of the sides = 4 + + + 9 = 7 inches 006 VideoTextInteractive Geometry: A Complete Course 57

Unit I, Part C, Lesson 4, Quiz Form A. Area: 98 sq. inches 8 8 7 9 Perimeter: 45 inches 0 Area = height sum of the bases = 7 ( 0 + 8) 7 = 8 7 = 4 = 98 square inches Perimeter = sum of lengths of the sides = 8+ 0+ 9+ 8 = 45 inches 4. The area of a trapezoid is 00 square centimeters. The sum of the lengths of the bases is 50 centimeters. Find the height. height: 4 cm Area = height sum of the bases 00 = h ( 50) = 00 h 50 00 = h 50 00 = h 50 50 50 50 4 50 = h 50 50 4cm = h 58 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part C, Lesson 4, Quiz Form A 5. The area of a trapezoid is 40m. The height is m. One base is 0m. Find the length b of the other base. Area = height sum of the bases other base: 50 meters 40 = + b ( 0 ) 40 = ( 0 + b ) 6 40 = ( 0 + b ) 40= 6 0+ 6 b 40 = 0 + 6 b 40 0 = 0 0 + 6b 00 = 0 + 6b 00 = 6 6 6 b 650 6 = b 6 6 50m = b 6. Application: The area of a trapezoid is 66 square units. The length of its longer base is 4 units longer than the length of its shorter base, and its height is 7 units longer than the length of its shorter base. Find the length of each base and the height of the trapezoid. (draw a diagram and label the necessary parts) x Area = height sum of the bases 66 = ( x+ 7) ( x) x +7 + ( x + 4) 66 = ( x+ 7) ( x+ 4) x + 4 66= ( x + 7) ( x + 4) length of short base: 4 units = ( x+ 7) ( x+ 4) length of long base: 8 units = x x+ 7 x+ x 4 + 7 4 height: units = x + 8x+ 8 = x + 8x+ 8 0 = x + 8x 04 0 = x + 9x 5 ( ) ( )( + ) 0 = x 4 x 0 = ( false) 0 = x 4 0+ 4= x 4+ 4 4 = x + 0 4 = x 0 = x + 0 = x + = x + 0 = x ( can' t be negative) x = 4 ( base) x + 7= ( height) x + 4 = 8( base) 006 VideoTextInteractive Geometry: A Complete Course 59

Unit I, Part C, Lesson 5, Quiz Form A. Area: 6 square cm Perimeter: 6 cm Perimeter = Sum of Lengths of the sides Area = measure of a side Apothem = Number of sides number of sides length of each side = n s = s a n = or = 6 cm = a Perimeter = ( ) 6 = 6 = 6 square cm 4. Area: 54 square cm 6 Perimeter: 6 cm Perimeter = Sum of Lengths of the sides Area = measure of a side Apothem = Number of sides the number of sides length of each side = n s a = s a n = 6 6 a = a Perimeter = 6 cm a = ( ) 6 = 8 a a = 54 square cm 66 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part C, Lesson 5, Quiz Form A 5. Area: 440 square cm Perimeter: 80 cm 6 Perimeter = Sum of Lengths of the sides = Number of sides length of each side = n s = 56 = 80 cm Area = measure of a side Apothem number of sides = s a n or = a Perimeter = 80 40 = = 440 square cm 6. Find the area and perimeter of a regular decagon with apothem 6.8cm and side length 4.4cm. Area: 49.6 square cm Perimeter: 44 cm Perimeter = Sum of Lengths of the sides = Number of sides length of each side = n s = 0 ( 4. 4) = 44 cm Area = measure of a side Apothem number of sides = s a n = a Perimeter = ( 6.8) 44 (. 4) 44 = = 49. 6 square cm 006 VideoTextInteractive Geometry: A Complete Course 67

Unit I, Part C, Lesson 5, Quiz Form A 7. Application: Find the indicated measures in the given regular hexagon. a = 6 units r = r a Area = 6 square units Perimeter = Sum of Lengths of the sides = Number of sides length of each side = n s = 6 = 7 units Perimeter = 7 units Hint #: Complete the inside of the hexagon with line segments drawn from the center to all 6 vertices. The small triangles appear to be (and actually are) equilateral. Therefore, r = units Hint #: To find a, we can reason this way. Each triangle is an equilateral triangle and therefore an isosceles triangle. The Apothem, a, is an altitude in the triangle and will bisect the base of. So, our hexagon could look like this: Observe what the apothem does to the side of length. Pythagorean Theorem. a + b = c a a a + 6 = + 6 = 44 + 6 + 6 = 44 + 6 a = 08 6 6 a = 08 a = 6 a = 6 units Area = measure of a side Apothem number of sides = s a n = a Perimeter = 6 7 = 7 = 6 square units 68 006 VideoTextInteractive Geometry: A Complete Course

Quiz Form B Class Date Score Unit I - The Structure of Geometry Part C - Measurement Lesson 5 - Regular Polygons Find the perimeter and area of each regular polygon in exercises through 6. Assume all dimensions are in centimeters.. Area: 600 square cm 0 0 Perimeter: 0 cm Perimeter = Sum of Lengths of the sides Area = measure of a side Apothem = Number of sides number of sides length of each side = n s = s a n = 6 0 or = 0 cm = a Perimeter = 0 0 5 0 = = 600 square cm. Area: 4 square cm 8 Perimeter: 7cm Perimeter = Sum of Lengths of the sides Area = measure of a side Apothem = Number of sides number of sides length of each side = n s = s a n = 48 = a Perimeter = 7 cm Note: a = 8/ or 9 = 9 7 9 6 = = 4 square cm 006 VideoTextInteractive Geometry: A Complete Course 69

Unit I, Part C, Lesson 5, Quiz Form B 64 square cm. Area: 6 Perimeter: 48 cm 8 Perimeter = Sum of Lengths of the sides = Number of sides length of each side = n s = 6 = 48 cm Area = measure of a side Apothem number of sides = s a n = a Perimeter 8 = 48 8 8 = = 64 square cm 4. Area: 086 square cm 8. 5 Perimeter: 0 cm Perimeter = Sum of Lengths of the sides Area = measure of a side Apothem = Number of sides number of sides length of each side = n s Area = s a n = 85 Area = a Perimeter = 0 cm Area = ( 8. ) 0 8. 60 = ( ) = ( 8. ) 60 = 086 square cm 70 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part C, Lesson 5, Quiz Form B 5. Area: 7.5 square cm 0 Perimeter: 50 cm 6.9 Perimeter = Sum of Lengths of the sides Area = measure of a side Apothem = Number of sides number of sides length of each side = n s = s a n = 50 = 50 cm = a Perimeter = ( 6.9) 50 69 5 = (. ) = ( 69. ) 5 = 75. square cm 6. A regular dodecagon has a side of length in. and an approximate area of 44.78 in. Find the length of the apothem to the nearest tenth of an inch. apothem:.7 inches Area = measure of a side Apothem number of sides 44. 78 = s a n 44. 78 = a 44. 78 = a 44. 78 = a 44. 78 = a 7. = a. 7 inches = a 006 VideoTextInteractive Geometry: A Complete Course 7

Unit I, Part C, Lesson 6, Quiz Form A In exercises 7 and 8 find the approximation, correct to the nearest hundredth, of the circumference of a circle with the given radius. Use.4 to approximate π. 7 7. radius = cm Circumference 4.66 cm. 8. radius = 8.6 in. Circumference 54.04 in. Circumference = πr ( ).4 7 cm 7 ( 6.84) cm 4.66 cm Circumference = πr (.4) 8. 6 in ( 6.84 )( 8.6 in) 54.04 in ( ) 9. Find the area of the shaded region. Use.4 for π, and approximate the answer to the nearest tenth. Area 5 50. units Area of shaded region = area of large circle - area of small circle. Area = πr πr (. 4)( 5)( 5) (. 4)( )( ) (. 4) 5 (. 4) 9 (. 4) ( 5 9) (. 4)( 6) 50. 7 50. 7 76 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part C, Lesson 6, Quiz Form B In exercises 7 and 8 find the approximation, correct to the nearest hundredth, of the circumference of a circle with the given radius. Use.4 for π. 7. r =. km Circumference =. 0 km 8. r = cm. Circumference = 8.80 cm 7 5 Circumference = πr = ( )( = ( 684. )(. ) 4.. ) =. 964 =. 0 km Circumference = πr e ( 4) 7. 5 e (. 4)(. 4) e ( 684. )( 4. ) e 8. 7976 e 880. cm 9. Find the area of the shaded region. Use.4 for π. Approximate the answer to the nearest tenth Area = 8.6 square units 5 5 Area of shaded region = Area of circle Area of square = πr s = π 55 ( 5 )( 5 ) = π 5 5 ( 5 )( 5 ) = π 5 ( 5 5 ) ( ) = 5π 5 = 5π 50 5(. 4) 50 78. 55 50 8. 55 square units 80 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part C, Lesson 7, Quiz Form A 4. Lateral Area = 55in 4 in 4 in Total Area = 6.9in 5 in Volume =.in. in in Lateral Area of a Prism (L.A.) = Perimeter of the Base Multiplied by the height of the prism L.A.= P h L.A.= ( 4 + 4 + ) 5 L.A.= 5 L.A.= 55 in Total Area of a Prism (T.A.) = The sum of the lateral area plus the area of the bases T.A.= L.A.+ b h+ b h Area = 55 + 4.+ 4. Area = 55 +.+. Area = 55 + 4.46 + 4.46 Area = 6.9 in Volume of Prism (V) = Area of one base multiplied by the height of the prism = b h 5 = 4. 5 =. 5 = 4.46 5 =. in 84 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part C, Lesson 7, Quiz Form B. Lateral Area = 44 in 4 9 5 4 Total Area = (44 + 6 ) in Volume = 7 in Lateral Area of a Prism (L.A.) = Perimeter of the Base Multiplied by the height of the prism. Perimeter = Sum of the lengths of the sides. Volume of a Prism (V) = Area of the Base x height Volume = B h = 8 in 9 in = 7 in or 7 cubic inches Perimeter of Base = + 4 + 5 + 4= 6 in. L.A.= P h = 6 9 = 44 in Total Area of a Prism (T.A.)= The sum of the L.A. and the area of the bases Area of one base = h (b +b ) = ( 5) = 8 sq. in. or 8 in Area of the two bases = 8 in + 8 in = 6 in Total area = 44 +6 in = (44 +6 )in 88 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part C, Lesson 7, Quiz Form B 4. Lateral Area = (48 + 4 π )cm Total Area = π + ( 48) cm 8cm Volume = 6 π cm 6cm Lateral Area of a Cylinder (L.A.) = Circumference of the Base multiplied by height Diameter = Radius d = r Circumference = (*full cylinder) Since this figure has a semi-circle for a base, it is of a cylinder. The Lateral Area will be of the Lateral Area of the full cylinder plus the Area of the rectangular side created by cutting the full cylinder in half. π d L.A.= π d h = π 6 cm 8 cm = π 48 cm = 48 π cm Lateral Area of full cylinder = 48π cm = 4 π cm Area of Rectangle = l w = 8 cm 6 cm =48 cm Lateral Area of cylinder = 4π cm + 48 cm = (48 + 4 π ) cm Total Area of a Cylinder (T.A.)= The sum of the L.A. and the area of the bases Area of one base = π radius radius = π r = π cm cm = π 9 cm * = 9 π cm = r 6 cm = r 6 cm = r cm = r Area of one base of cylinder = Area of one base of 9 cm = cylinder = 9 9π cm = π cm Area of two bases = 9 π cm = 9 π cm = 9π cm Total Area = 4 π cm + 48 cm + 9πcm = (4 π+ 48 + 9 π) cm = (4 π+ 9π + 48) cm = (π + 48) cm Volume of a Cylinder (V) = Area of the Base x height Volume of a full cylinder = 9 π cm i 8 cm = 7π cm Volume of a half cylinder = i 7 π cm = 6 π cm or 6 π cubic centimeters 90 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part C, Lesson 8, Quiz Form A Find the lateral area, total area, and volume of the right circular cone shown below.. Lateral Area = 89.π sq mm Total Area = 9.0 π mm.7mm 6.mm Volume = 68.0 π mm Lateral Area of a Cone (L.A.) = times the circumference of the base times the slant of the height. C = d d = r Use Pythagorean Theorem to find slant height. L.A.= π c l l = π d = π r l π 6. mm 4.77 mm = = π 89. mm = 89.π sq mm Total Area of a Cone (T.A.) = the sum of the Base Area and the Lateral Area. T.A.= B.A. + L.A. = (Area of a circle) + L.A. = π r + 89. sq. mm = π (6. mm) + 89. π mm = π 9.69 mm + 89. π mm = (9.69 π+ 89. π ) mm = 9.0 π mm Volume of a Cone (V) = times the Area of the Base times the height of the cone. Use Pythagorean Theorem to find h. V = B h = (Area of a circle) h = π r h = π (6. mm) (6.mm) (.7mm) = π 504.06 = 504.06 π 68.0 π mm 94 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part C, Lesson 8, Quiz Form A Find the lateral area, total area, and volume of the right pyramid shown below. 4. Lateral Area = 40 ft 8 0 84 ft Total Area = Volume = 84 ft Lateral Area of Pyramid (L.A.) = Use Pythagorean Theorem to find square. L.A.= times the perimeter of the Total Area of Pyramid (T.A.) = the sum of the Base Area and Base multiplied by the slant height. the Lateral Area. the length of each side of the P l = ( + + +) 0 = 48 ft 0 ft T.A. = B.A. + L.A. = 44 ft + 40 ft = 84 ft = 4 0 = 40 ft ft Volume of a Pyramid (V) = times the Area of the Base times the height of the pyramid. Use Pythagorean Theorem to find h. V = B h = 44 ft 8 ft = 48 ft 8 ft = 84 ft ft 96 006 VideoTextInteractive Geometry: A Complete Course

Quiz Form B Class Date Score Unit I - The Structure of Geometry Part C - Measurement Lesson 8 - Pyramids Find the lateral area, total area, and volume of the right square pyramid shown below.. Lateral Area = 6 7 ft Total Area = 6( + 7 ) ft 6 Volume = 44 ft Lateral Area of Pyramid (L.A.) = times the perimeter of the Total Area of Pyramid (T.A.) = the sum of the Base Area and Base multiplied by the slant height. the Lateral Area. T.A. = B.A. + L.A. Use Pythagorean Theorem to find slant height. = 6 ft + 6 7 ft L.A.= P l = (6 +6 +6 +6) 5 = (4 ft) 5 ft = 6( + 7 ) ft = 5 = 5 ft = 9 7 ft = 7ft =6 7 ft ft Volume of a Pyramid (V) = times the Area of the Base times the height of the pyramid. Use Pythagorean Theorem to find h. V = B h = 6 ft ft = = 44 ft ft 006 VideoTextInteractive Geometry: A Complete Course 99

Unit I, Part C, Lesson 8, Quiz Form B Find the lateral area, total area, and volume of the right circular cone shown below. 89π 5 mm 4. Lateral Area = 6 8 mm 4 mm 4 ( ) 89π + 5 Total Area = 6 49π mm Volume = 96 mm Lateral Area of a Cone (L.A.) = times the circumference of the base times the slant height of the cone. C = d d = r Use Pythagorean Theorem to find the slant height. L.A.= c = π d = π l l π r l = 7 7 5 π mm mm 4 4 π 7 7 5 = mm 4 4 π 89 5 = mm 6 = 89 π 5 mm 6 Total Area of a Cone (T.A.) = the sum of the Base Area and the Lateral Area. T.A.= B.A. + L.A. = (Area of a circle) + L.A. = r + 89 5 π π mm 6 7 = π 4 mm 7 89π 5 mm + mm 4 6 π 7 = 7 89π 5 mm + mm 4 4 6 = 89 π 89π 5 mm + mm 6 6 89π + 89π 5 = mm 6 = 89 π(+ 5) mm 6 Volume of a Cone (V) = times the Area of the Base times the height of the cone. V = B h = (Area of a circle) h = π r h 0 006 VideoTextInteractive Geometry: A Complete Course = 7 π mm 7 4 4 mm 7 mm π 7 7 7 = 4 4 mm 49π = mm 96

Unit I, Part C, Lesson 9, Quiz Form A. If the volume of a sphere is π cubic units, find the radius and the surface area. radius = 9 units Surface Area = π sq. units Volume of sphere = 4 π r π = 4 πr π = 4 πr 6 π = 4πr 6 π = 4π 4π 4π r 4π 9 = r 4π 9 = r 9 units = r Surface Area of a sphere = 4 π r = 4 π 9 9 = 4 π 9 = 4 π 8 = 4 π 7 =4 π = π sq. units 006 VideoTextInteractive Geometry: A Complete Course 07

Unit I, Part C, Lesson 9, Quiz Form A 9 5. The volume of a sphere is π cubic meters. Find the radius and surface area. 6 radius = 4 meter 9 π square meters Surface Area = 4 Volume of sphere = 4 π r 9 = 4 π π r 6 48 9 π 48 4 π = π 6 r π 6 9 π 6 = 6 4 π π π r 7 = 64r 7 64 = 64 64 r 7 64 = r 7 64 = r 7 = r 64 4 meter = r Surface Area of a sphere = 4 π r = 4 π 4 4 = 4 π 4 4 π 9 = 4 = 9 π square meters 4 006 VideoTextInteractive Geometry: A Complete Course 09

Quiz Form B Class Date Score Unit I - The Structure of Geometry Part C - Measurement Lesson 9 - Spheres Find the surface area and volume of the sphere illustrated below. Round all answers to the nearest tenth. Use.4 as an approximation for π.. Surface Area = 4.0 sq inches Volume = 5. cubic inches 5 in Surface Area of a sphere = 4 π r Total Area = 4 π r r = 4.4 5 in 5 in = 4.4 5 5 in = 4 in or 4.0 sq inches Volume of sphere = 4 π r = 4 π 5 in 5 in 5 in = 4.4 5 5 5 in = 570 in = 5. cubic inches 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part C, Lesson 9, Quiz Form B. The volume of a sphere is 6 π mm. Find the radius and the surface area. radius = mm Surface Area = 6 π sq mm Volume of sphere = 4 π r 6 π = 4 πr 6 π π = 4 4π 4π r 4π 9 = r 4π 7 =r 7 = r mm = r Surface Area of a sphere = 4 π r = 4 π mm mm = 4 π mm = 6 π sq mm 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part C, Lesson 9, Quiz Form B 5. The radius of a sphere is 8 inches. Find the surface area and volume. Surface Area = π 5 square in 4096 π cubic in Volume = Surface Area of a sphere = 4 π r Total Area = 4 π 8 in 8 in = 4 π 8 8 in =4 π 64 in = 5 π square in Volume of sphere = 4 = 4 π r π 8 in 8 in 8 in = 4 π 8 8 8 in = 4 π 8 8 8 in = 4 π 5 in = 40 96π cubic in 006 VideoTextInteractive Geometry: A Complete Course 5

Unit I, Part D, Lesson, Quiz Form A For each group listed in exercises 4 through 6, read the accompanying scenario illustrating a good use of inductive reasoning. Then write a scenario of your own. Check with another person for its validity. 4. Football Players The opponent in the Cougar s next game throws a pass on first down 8 out of 0 times according to statistics from the first five games. The Cougars expect they will need to be prepared to use their pass defense the majority of the time on first down. Answers will vary. 5. Employees Employees of the Discount Mart Variety Store have a meeting every Friday morning one hour before the store opens. Their supervisor has been ten to fifteen minutes late to the meeting for the last 7 weeks. There has been a noticeable increase in the number of employees who are late since the meeting never seems to start on time anyway. Answers will vary. 6. Police Officers The intersection of 5th Street and Cumberland Avenue has been the scene of nine accidents in the last four weeks. Over the last three months the number of speeding citations issued on Cumberland Avenue has increased by 5% over the previous three month period. The Police Department has requested that a study of the daily traffic patterns be conducted to determine a remedy for the dangerous situation at this intersection. Answers will vary. 8 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part D, Lesson, Quiz Form B For each group listed, in exercises 4 through 6, read the accompanying scenario illustrating a good use of inductive reasoning. Then write a scenario of your own. Check with another person for its validity 4. Manufacturers Each year, auto makers introduce new colors for their cars. One way auto makers choose colors for new cars is to find out which colors sold well in the past. Trends which are observed over a long period of time, say five years, help automakers to decide what color of automobile to produce, in an attempt to sell more cars. Answers will vary. 5. Thieves The owner of a small photography shop is observed leaving his business to go to the bank at approximately the same time every day. A thief would use such an observation to plan a confrontation and possible robbery. For the businessman, his responsibility is to avoid following the same routine everyday. Answers will vary. 6. Explorers Throughout history, there have been many explorers from Columbus to Lewis and Clark to Space Shuttle Astronauts. Early explorers had only limited knowledge about conditions they would encounter. More modern explorers can use technology to help them prepare for their travels. However, at all levels, explorers were required to make observations and record patterns of activity which might effect their success. For example, how did Columbus acquaint himself with the prevailing winds needed to push him across the ocean? How did Lewis and Clark know when the best time was to move? When is the best window for launching a space shuttle? Planning had to be based on observations such as weather patterns and moon phases, and using that information to make assumptions to be acted upon, to successfully complete the mission. Answers will vary. 0 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part D, Lesson, Quiz Form A 4. Use inductive reasoning to find the next two terms of the sequence given below. Describe how you found these terms. A, B, D, G, K,, P V Assign numbers to 6 to the letters of the alphabet and numbers to 7 to the position of the letter in the pattern A, B, C, G, K, P, V. The sum of the position of the letter in the alphabet and the positions of the letter in the pattern gives the numerical position in the alphabet of the next letter in the pattern. Example: A > + = B > + =4 D > 4 4+ =7 G > 7 4 7+4 = K > 5 +5 =6 P > 6 6 6+6 = V > ^ Position of letter in alphabet + Position of letter in pattern = Position of next letter in the alphabet. Look for a pattern and predict the next two numbers in each sequence in exercises 5 through 8. Write a sentence describing how you found these numbers. 5., 0, 00, 000,, 0,000 00,000 Each term is multiplied by 0 to get the next term. 6. 80, 60, 540, 70,, 900 080 Each term is found by adding 80 to the previous term. 5 6 7.,,,, 6 6 or 6 8., 0, 0, 00, 50,, 500 50 Each new term is found by adding 6 to the previous term. Alternate solution: Each fraction divided by the next fraction in the 8 sequence, gives the following fraction. This would give anwers of 4 and 9. The pattern is formed by multiplying the first term by 0 to get the second term. Then divide the second term by to get third term. Repeat this process multiply by 0, divide by. 006 VideoTextInteractive Geometry: A Complete Course

Quiz Form B Class Date Score Unit I - The Structure of Geometry Part D - Inductive Reasoning Lesson - Applications in Mathematics. In each of the past 8 months, Joe has received the rent payment for his rental house on the fourth or fifth day of the month. He makes a conjecture that he will receive next month s rent on the fourth or fifth of the month. Is this a good example of inductive reasoning? Yes Explain your answer. Yes, this is a good example of inductive reasoning. His tenants probably get paid on the first of each month and do not mail any bill payments until their pay check is securely in the bank. Their pattern has been established and will probably continue.. Do you think there is a connection between inductive reasoning and the stock market investments made by investors in our country? Yes Explain your answer. Yes, there is a connection. We hear news reports mentioning economic indicators which are factors occurring within our economic system that have been related to good times for growth or bad times for growth. A stock broker will be aware of these indicators in deciding when and in what investments to place money.. Anthony noted that 5 = + and = +. He concluded that every prime number may be expressed as the sum of the squares of two positive integers. Was he correct? No You might want to test some cases. = + = + Yes = + = + or + = + 4 No 5 = + = + 4 = 5 Yes 7 = + = + 4 or + = + 9 No = + = + = 4 + 9 = 7 = + = + 9 or + = 4+ 9 4 = + 6 = 7 Yes Yes 9 = + 4 = 4 + 6 = 0 No = + 4 = 9 + 6 = 5 No 9 = + 5 = 4 + 5 = 9 Yes No Explain your answer In the first 0 prime numbers, there are 5 which cannot be expressed as the sum of squares of two positive integers. 006 VideoTextInteractive Geometry: A Complete Course 5

Unit I, Part D, Lesson, Quiz Form B 4. Write a general formula for the sum of any number (n) of consecutive odd integers by examining the following cases and using inductive reasoning. integers number of integers sum of integers, 4,,5 9,,5,7 4 6,,5,7,9 5 5 Let n = the number of integers added. n = = 4 n = = 9 n = 4 4 = 6 n=5 5 = 5 For n consecutive odd integers, the sum will be Sum can be found by squaring n or Sum of n odd integers is n. Look for a pattern and predict the next two numbers in each sequence in exercises 5 through 8. Write a sentence describing how you found these numbers. 5. 0, 0,,, 46, 60,, 75 9 6.,, 4, 7,, 8,, 9 47 Start at 0. The pattern of numbers is found by adding consecutive integers to each number beginning with 0. 0, 0+0=0, 0+=, += The pattern is formed by adding two adjacent numbers to get the next number, assuming that you start with and. 7.,, 48, 9, 768,, 07,88 8., 9,, 0, 5 6,,,, 6 Each new number is found by multiplying the previous number by 4. Every number in an odd position in the pattern is found by adding 6 to the previous number in an odd position. Every number in an even position in the pattern is found by adding to the previous number in an even position. 6 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part E, Lesson, Quiz Form A. General Statement - No one living in Oklahoma has a house on the beach. Specific Statement - Jeremy has a beachfront house. Conclusion - We might want to conclude that Jeremy does not live in Oklahoma. However, this is not a valid conclusion. The condition of the general statement is that someone must live in Oklahoma. The specific statement does not satisfy that condition. It states the Jeremy has a beachfront house. Okay, but where does he have his house? For the reasoning to be valid, the specific statement must state whether or not Jeremy lives in Oklahoma. 4. General Statement - The new fluoride toothpaste, Dent-Sure, prevents cavities. Specific Statement - Carl had no cavities at his dental check-up today. Conclusion - We cannot arrive at a conclusion here. The condition of the general statement is that someone uses Dent-Sure toothpaste. The specific statement does not satisfy this condition. It states Carl had no cavities. We don t know what. he does to prevent them.. 8 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part F, Lesson, Quiz Form A Suppose p stands for Scientists are not uneducated (a true statement) and q stands for Geology involves the study of the earth (a true statement). Write, in words, each of the statements in exercises through 6. Then decide the truth of each compound statement.. ~( p q) It is false that scientists are not uneducated and geology involves the study of the earth. This is a false statement. ) True and True > True ) Negated True > False. p q It is not the case that scientists are not uneducated, or it is not the case that geology involves the study of the earth OR scientists are uneducated or geology does not involve the study of the earth. This is a false statement. ) False or False > False 4. ~( p q) It is false that scientists are not uneducated or geology involves the study of the earth. This is a false statement. ) True or True > True ) Negated True > False 5. ( p q) Scientists are not uneducated or geology involves the study of the earth. This is a true statement. ) True or True > True 6. p q It is false that scientists are not uneducated and it is false that geology involves the study of the earth. Or, we could write it another way. Scientists are uneducated and geology does not involve the study of the earth. This is a false statement. ) True and True > True ) ~True and ~True > False 006 VideoTextInteractive Geometry: A Complete Course 7

Unit I, Part F, Lesson, Quiz Form B 7. Suppose p stands for Triangle ABC is right isosceles and q stands for Triangle ABC is a right triangle. Use these two statements to form a conjunction, a disjunction, and a negation of p. Conjunction: Disjunction: Negation of p: ABC is right isosceles and ABC is a right triangle. ABC is right isosceles or ABC is a right triangle. ABC is not right isosceles. Given two statements as indicated in exercises 8 through, indicate whether the conjunction, disjunction, and negation of p are true or false. 8. both p and q are true. 9. p is true and q is false. Conjunction: True and True True Conjunction: True and False False Disjunction: True or True True Disjunction: True or False True Negation of q: not True False Negation of q:: not False True 0. both p and q are false.. p is false and q is true. Conjunction: False and False False Conjunction: Disjunction: False or False False Disjunction: Negation of q: not False True Negation of q: False and True False or True not True False True False 40 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part F, Lesson, Quiz Form B Suppose p stands for Algebra is a branch of mathematics (a true statement) and q stands for Geometry is not worthless (a true statement). Write, in words, each of the statements in exercises through 6. Then decide the truth of each compound statement.. p q Algebra is a branch of mathematics or geometry is not worthless. This is a true statement. ) True or True > True. ~( p q) 4. 5. It is not the case that algebra is a branch of mathematics or geometry is not worthless. This is a false statement. ) True or True > True ) Negated True > False ~ p ~ q It is not the case that algebra is a branch of mathematics and it is not the case that geometry is not worthless. We could say this another way by saying algebra is not a branch of mathematics and geometry is worthless. This is a false statement. ) ~True and ~True ~( p q) False and False > False It is not the case that algebra is a branch of mathematics and geometry is not worthless. This is a false statement. ) True and True > True ) Negated True > False 6. ~ p ~ q It is not the case that algebra is a branch of mathematics or it is not the case that geometry is not worthless. This is a false statement. ) ~True or ~True False or False > False 006 VideoTextInteractive Geometry: A Complete Course 4

Unit I, Part F, Lesson, Quiz Form A Consider each of the statements in Exercises 6-0 to be true. State the converse of each, and tell whether the converse is always, sometimes, or never true. 6. If a polygon is a hexagon, then it has exactly six sides. Converse: If a polygon has exactly six sides, then it is a hexagon. (always true) 7. If x =, then x = 9 Converse: If x = 9 then x = (sometimes or partially true) 8. If you are able to finish a triathlon, then you are in good shape. Converse: If you are in good shape, then you are able to finish a triathlon. (sometimes true) 9. If a person is swimming, then that person is wet. Converse: If a person is wet, then that person is swimming. (sometimes true) 0. If two lines have a common point, then the lines are intersecting lines. Converse: If two lines are intersecting, then they have a common point. (always true) When a statement and its converse are both always true, you can combine the two statements into a biconditional using the phrase "if and only if". For exercises through 5, decide which of the statements from exercises 6 through 0 can be written in biconditional form, and if possible, write the biconditional. If not possible, explain why.. (using exercise 6) This can be written as a biconditional. A polygon is a hexagon if and only if it has exactly six sides.. (using exercise 7) This statement and its converse cannot be put together as a biconditional because both are not always true. (the converse is sometimes false). (using exercise 8) This statement and its converse cannot be put together as a biconditional because both are not always true. (the converse is sometimes false) 4. (using exercise 9) This statement and its converse cannot be put together as a biconditional because both are not always true. (the converse is sometimes false) 5. (using exercise 0) This can be written as a biconditional. Two lines have a common point if and only if the lines are intersecting lines. 44 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part F, Lesson, Quiz Form B Consider each of the statements in Exercises 6-0 to be true. State the converse of each, and tell whether the converse is always, sometimes, or never true. 6. If a figure is a pentagon, then it is a polygon. If a figure is a polygon, then the figure is a pentagon. (sometimes true) 7. If a whole number has exactly two whole number factors, then it is a prime number. If a number is a prime whole number, then it has exactly two whole number factors. (always true) 8. If you are an elephant, then you do not know how to fly. If you do not know how to fly, then you are an elephant. (sometimes true) 9. If you are at least years old, then you can legally vote. If you can legally vote, then you are at least years old. (always true) 0. If an angle is acute, then it is smaller than an obtuse angle. If an angle is smaller than an obtuse angle, then the angle is acute. (sometimes true) When a statement and its converse are both always true, you can combine the two statements into a biconditional using the phrase "if and only if". For exercises through 5, decide which of the statements from exercises 6 through 0 can be written in biconditional form, and if possible, write the biconditional. If not possible, explain why.. (using exercise 6) This statement and its converse cannot be combined as a biconditional. The converse is not always true.. (using exercise 7) This statement and its converse can be combined as a biconditional because both the statement and its converse are always true. A whole number has exactly two whole number factors if and only if it is a prime whole number.. (using exercise 8) This statement and its converse cannot be combined as a biconditional because the statements are not both always true. The converse is not always true. 4. (using exercise 9) This statement and its converse can be combined as a biconditional because they are both always true. You can legally vote if and only if you are at least years old. 5. (using exercise 0) This statement and its converse cannot be combined as a biconditional because both statements are not always true. The converse is false if the angle is a right angle. 46 006 VideoTextInteractive Geometry: A Complete Course

Unit I, Part F, Lesson, Quiz Form A 6. A square is a rectangle. Conditional: Contrapositive: If a figure is a square, then the figure is a rectangle. (true) If a figure is not a rectangle, then the figure is not a square. (true) 7. A figure has three segments if it is a triangle. Conditional: Contrapositive: If a figure is a triangle, then the figure has three segments. (true) If a figure does not have three segments, then the figure is not a triangle. (true) 8. x < 0 implies 5x > 6x Conditional: Contrapositive: If x < 0, then 5x > 6x (true) If 5x > 6x, then x < 0 (true) 9. An integer is less than zero given that the integer is a negative integer. Conditional: Contrapositive: If a number is a negative integer, then the number is less than zero. (true) If a number is not less than zero, then the number is not a negative integer. (true) 48 006 VideoTextInteractive Geometry: A Complete Course

Unit I, The Structure of Geometry, Unit Test Form A Continued, Page In exercises 4 through 44, use the given information about a circle to find the missing measures. Use.4 as an approximation for π, and show your work. Give your answers to the nearest tenth, and label your results properly. radius diameter Circumference Area 4. 5 (C-6) 7.5" 47. inches 76.6 sq. in. Diameter = radius 5" = ir i 5" = i i r 5" = r 7.5" = r Circumference = π d = π 5" (.4) (5) inches 47. inches Area = π r 5" 5" = π i (.4) (7.5)(7.5) sq. in. 76.65 sq. in. 76.6 sq. in. radius diameter Circumference Area 4. 8 cm 6 cm 50. cm 0.0 cm (C-6) Diameter = radius = i 8cm = 6 cm Circumference = π d = π 6 cm. 4 6 cm 50. 4 cm 50. cm Area = π r = π i8cm i8cm = π i 64 cm (. 4 )( 64 )cm 00. 96 cm 0. 0 cm 006 VideoTextInteractive Geometry: A Complete Course 69

Quiz Form B Class Date Score Unit I - The Structure of Geometry Part F - Deductive Reasoning Lesson - Negations of Conditionals Write the inverse, converse, and contrapositive of each conditional in exercises through. Then determine which statements are true and which are false.. If a >, then a > a. Inverse: Converse: Contrapositive: If a >, the a > a (false) If a > a, then a > (false) If a > a, then a > (true). If a triangle is an acute triangle, then it has no obtuse angles. Inverse: If a triangle is not an acute triangle, then the triangle has an obtuse angle. (false) Converse: If a triangle has no obtuse angles, then the triangle is an acute triangle. (false) Contrapositive: If a triangle has an obtuse angle, then the triangle is not an acute triangle. (true). If Sean lives in Pittsburgh, then he lives in Pennsylvania. Inverse: Converse: Contrapositive: If Sean does not live in Pittsburgh, then he does not live in Pennsylvania. (false) If Sean lives in Pennsylvania, then he lives in Pittsburgh. (false) If Sean does not live in Pennsylvania, then he does not live in Pittsburgh. (true) In exercises 4 through 9, write the given statement as a conditional in if-then form. Then form the contrapositive and decide the truth value of each statement. 4. All acute angles are congruent. Conditional: Contrapositive: If angles are acute, then angles are congruent. (false) If angles are not congruent, then the angles are not acute. (false) 5. A number is an integer, given that it is greater than zero. Conditional: Contrapositive: If a number is greater than zero, then the number is an integer. (false) If a number is not an integer, then the number is not greater than zero. (false) 006 VideoTextInteractive Geometry: A Complete Course 5

Unit I, Part F, Lesson, Quiz Form B For exercises 0 through, consider exercises 6 through 8 and decide which conditionals, if any, can be written as a biconditional. Then rewrite the biconditional form of the statement. If the conditional cannot be written as a biconditional, say, not possible. A compound statement consisting of a given conditional and, its converse both of which are considered to be true, can be written as a biconditional. 0. Biconditional (if possible): Exercise 6: If x = 4, then x = (true) If x =, then x = 4 (true) x = 4 if and only if x =. Biconditional (if possible): Exercise 7: If a = b or a = b, then a b = 0 (true) If a b = 0, then a = b or a = b (true) a = b or a = b if and only if a b = 0. Biconditional (if possible): Exercise 8: If A, B, and C are on one line, then A, B, and C are collinear. (true) If A, B, and C are collinear, then A, B, and C are on one line. (true) A, B, and C are on one line if and only if A, B, and C are collinear.. (Bonus Problem) Arrange some or all of the given conditionals into an order that allows you to make the given conclusion. Conditionals: If A, then B. If X, then R. If M, then Y. If R, then M. If Y, then A. Conclusion: If X, then B. If X, then R. If R, then M. If M, then Y. If Y, then A. If A, then B. Therefore, if X then B. 006 VideoTextInteractive Geometry: A Complete Course 5

Unit I, The Structure of Geometry, Unit Test Form A Continued, Page (F-). v disjunction (C-9) 4. f sphere (B-) 5. h polygon (A-5) 6. t (B-) 7. y isosceles triangle (B-) 8. u rhombus (C-7) 9. w right prism m) a prism whose lateral faces are at an angle other than 90 O with the base. n) a three dimensional geometric figure created by connecting a polygon to a point not in the plane. o) a system of reasoning, in an orderly fashion, which draws conclusions from specific premises. p) two or more sets which have no members in common. q) a polygon made with eight line segments r) a triangle in which one of the angles is a right angle (90 O ) s) grouping symbol; brackets (B-) 0. x quadrilateral (C-7). i prism (F-4). k fallacy (B-). e geometric figure t) an operation on two or more sets which selects only those elements common to all of the original sets at the same time u) a quadrilateral in which all four sides are of equal measure v) an operation in logic which joins two simple statements using the word or. w) a prism whose lateral faces are at an angle of 90 O with the bases. (A-) 4. s [ ] x) a polygon made with four line segments. (B-) 5. c hexagon (C-6) 6. l radius y) a triangle in which at least two of the three sides are of equal measure z) the process of using a general statement which calls for a conclusion, based on certain conditions, and then applying a specific statement which satisfies those conditions, therefore establishing the validity of the conclusion. 60 006 VideoTextInteractive Geometry: A Complete Course