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Lesson 12.1 Exploring Solids Today, we will learn to… > use vocabulary associated with solids > use Euler’s Formula
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Polyhedron ~ a solid formed by polygons
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Prisms are polyhedra named by their bases name the base
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Prisms have 2 parallel bases, Pyramids have 1 base.
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Cylinders, Cones, & Spheres with curved surfaces are NOT polyhedra
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Parts of Solids Faces Edges Vertex Bases vs Lateral Faces Height
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F + V = E = + = 6814 12
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F + V = E = + = 5510 8
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Euler’s Theorem # Faces + # Vertices = # Edges + 2 F + V = E + 2
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1. In a polyhedron, F = 8 V = 10 E = ? find the number of edges. F + V = E + 2 8 + 10 = E + 2 18 = E + 2 ___ edges16
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2. In a polyhedron, F = 5 V = ? E = 8 find the number of vertices. F + V = E + 2 5 + V = 8 + 2 5 + V = 10 ___ vertices5
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3. In a polyhedron, F = ? V = 6 E = 9 find the number of faces. F + V = E + 2 F + 6 = 9 + 2 F + 6 = 11 ___ faces 5
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A soccer ball is a polyhedron with 32 faces ( 20 hexagons & 12 pentagons ). How many vertices does this polyhedron have? 32 + V = 90 + 2 # Edges = hexagonspentagons 20(6)12(5) ½ (120 + 60) = 90 V = 60 vertices = 120 = 60
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A polyhedron can be convex or concave.
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4.Describe the cross section shown. square
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5.Describe the cross section shown. pentagon
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6.Describe the cross section shown. triangle
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A polyhedron is regular if all of its faces are congruent regular polygons. Platonic Solids
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Lesson 11.1 Angle Measures in Polygons Today, we are going to… > find angle measures in polygons
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# of sides 34568n # triangles sum of all angles measure of each angle in regular polygon 180 ˚ 2346 n-2 360 ˚ 540 ˚ 720 ˚ 1 1080 ˚ (n-2)180 ˚ 60 ˚ 90 ˚ 108 ˚ 120 ˚ 135 ˚ (n-2)180 n
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Theorem 11.1 Polygon Interior Angles Theorem The sum of the measures of the interior angles is ____________ (n - 2)180°
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The measure of one interior angle in a regular polygon is… (n - 2) 180° n Corollary to Theorem 11.1
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1. Find the sum of the measures of the interior angles of a 30-gon. (30 - 2) 180 (28) 180 5040˚
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2. How many sides does a polygon have if the sum of the interior angles is 3240˚? n = 20 sides 180(n-2) = 3240 n – 2 = 18
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3. What is the measure of each interior angle of a regular nonagon? 1260˚ 9 = (9 - 2)180 9 (7)180 9 140
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4. Find x. 180(7 - 2) 180(5) 900˚ How many angles? 7 900° – given angles = x 900 – 783 =117 Sum?
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5. Find x. 180(5 - 2) 180(3) 540˚ How many angles? 5 sum of all angles = 540 43x – 19 = 540 13 Sum? x =
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1 2 3 4 5 Exploring Exterior Angles GSP
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Theorem 11.2 Polygon Exterior Angles Theorem The sum of the measures of one set of exterior angles in any polygon is _________ 360°
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The measure of one exterior angle in any regular polygon is 360° n
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The number of sides in any regular polygon is… 360° ext. # sides =
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One exterior angle and its interior angle are always ________________. supplementary
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6. What is the measure of each exterior angle in a regular decagon? 360˚ 10 36˚
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7.How many sides does a regular polygon have if each exterior angle measures 40˚? 360˚ 40° = 9 sides
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8.How many sides does a regular polygon have if each interior angle measures 165.6˚? Don’t write this down, yet. 180(n – 2) = 165.6 n 180 (n-2) n = 165.6 1 180n – 360 = 165.6 n – 360 = – 14.4 nn =25 sides 180 – 165.6 = 360˚ 14.4 First, find the measure of an exterior angle. = 25 sides 14.4˚
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9. Find x. 3x + 90 + 70 + 80 + 60 =360° x =20 ? ? 90 70
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What’s the measure of each interior angle of a regular pentagon? What’s the measure of each interior angle of a regular hexagon? 180(3) 5 = 108° 180(4) 6 = 120° A soccer ball is made up of 20 hexagons and 12 pentagons.
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Lesson 11.2 Perimeter & Areas of Regular Polygons Today, we are going to… > find the perimeter and area of regular polygons
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10 1. Find the area of this equilateral triangle 5 5 3 60 ˚ 10 (5 3 )(10) 2 25 3 ≈ 43.3
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24 2. Find the area of this equilateral triangle 12 3 (12 3 )(24) 2 144 3 ≈ 249.4
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s s s Equilateral Triangles Area = s2s2 3 2
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A regular polygon’s area can be covered with isosceles triangles.
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side Area = Perimeter = ½ side · apothem side · # sides · # sides side apothem
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Area of Regular Polygons A = ½ (side)(apothem)(# sides) area of each isosceles triangle number of isosceles triangles A = ½ sa n
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Perimeter of Regular Polygons P = (side)(# sides) P = s n
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Find the area of the polygon. 3.a pentagon with an apothem of 0.8 cm and side length of 1.2 cm A = ½ (1.2) (0.8) (5) A = 2.4 cm 2 A = ½ (s) (a) (n)
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Find the area of the polygon. 4.a polygon with perimeter 120 m and apothem 1.7 m A = ½ (1.7) (120) (s)(n) A = 102 m 2 A = ½ (s) (a) (n)
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72°60°45° 360° 5 360° 6 360° 8 36°30°22.5° 360° 2(5) 360° 2(6) 360° 2(8) 5. Find the central angle of the polygon. 6. Find the measure of this angle
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The measure of the angle formed by the apothem and a radius of a polygon is… 360 2(n)
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5 cm 10 cm 7. Find the length of the apothem if the side length is 10 cm. apothem = 5 cm 30° 360° 2(6) 60˚
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A = 10 cm 8. Find the area of the polygon. 259.8 cm 2 ½ (10)(5 )(6) =
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tan A = opposite adjacent cos A = adjacent hypotenuse sin A = opposite hypotenuse
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10 5 a 30° 360° 12 s= a= n= 10 5 3 6 A = ½ san = 259.8
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12 6 a 36° 360° 10 s= a= n= 12 5 A = ½ san 8.3 tan 36° = 6a6a = 249
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12 x a 22.5° 360° 16 s= a= n= 9.2 8 A = ½ san sin 22.5° = x 12 cos 22.5° = a 12 11.1 = 408 12
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A =( )( )= 14. Find the area of the square. 72 units 2 3 2 6 x s =6 2 3 2 6 2 45° a 72 units 2 A = ½ ( )( )(4)= 6 2 3 2
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Worksheet Practice Problems
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Lesson 11.3 Perimeters and Areas of Similar Figures Today, we are going to… > explore the perimeters and areas of similar figures
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1. Find the ratio of their … Sides Perimeters Areas 3 : 2 8 6 12 9 10 15 3 : 2 9 : 436 :54 : 24
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6 8 28 21 12 9 2. Find the ratio of their … Sides Perimeters Areas 3 : 4 9 : 16 60 : 126 : 80 224
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Theorem 11.5 In similar polygons … Sides Perimeters Areas 3 : 2 3 : 2 9 : 4 3 : 4 3 : 4 9 : 16 Do you notice a pattern? a : b a 2 : b 2 a : b
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9 25 3 5 = PV = 6 3. The ratio of the area of Δ PVQ to the area of Δ RVT is 9:25. If the length of RV is 10 and the two triangles are similar, find PV. What is the ratio of their sides? x 10
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4. The ratio of the sides of two similar polygons is 4 : 7. If the area of the smaller polygon is 36 cm 2, find the area of the larger polygon. 4 7 110.25 cm 2 16 49 = What is the ratio of their areas? 36 x
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Lesson 12.7 Similar Solids
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1. Find the ratio of sides. 2:3 Are they similar solids? 4 8 2 6 12 3 3 6 12 2 4 8
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2. Find the ratio of their surface areas. 6 12 3 4 8 2 112 : 252 reduces to4:9 112 units 2 252 units 2
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3. Find the ratio of their volumes. 6 12 3 4 8 2 64 : 216 reduces to8 : 27 64 units 3 216 units 3
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Theorem 12.13 a 2 : b 2 a 3 : b 3 Ratio of sides Ratio of Surface Area Ratio of Volume a : b 2 : 34 : 98 : 27
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4. 1 : 16 1 : 64 Scale Factor SA V 1 : 4 2 : 54 : 258 : 125 9 : 100 3 : 10 27 : 1000
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5. A right cylinder with a surface area of 48π square centimeters and a volume of 45π cubic centimeters is similar to another larger cylinder. Their scale factor is 2:3. Find the surface area and volume of the larger solid. ratio of surface areas? 4:9 4949 48π x = 4x = 9(48 ) x = 108π 339.3
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5. A right cylinder with a surface area of 48π square centimeters and a volume of 45π cubic centimeters is similar to another larger cylinder. Their scale factor is 2:3. Find the surface area and volume of the larger solid. ratio of volumes? 8:27 8 27 45π x = 8x = 27(45 ) x = 151.875π 477.1
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Lesson 12.2, 12.3, 12.6 Surface Area Today, we will learn to… > find the Surface Area of prisms, cylinders, pyramids, cones, and spheres
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Why would we need to find surface area?
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Surface Area of a Right Prism S = 2B + PH Area of the Base Perimeter of the Base Height of the Solid
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S = 749 28 2( ) + ( )( ) = 294cm 2 Shape of the base? square
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S = 2( ) + ( )( ) = 9 16 15 174m 2 Shape of the base? rectangle
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2( )+( )( ) = S = 54 36 3 216 cm 2 Shape of the base? triangle
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Surface Area of a Cylinder S = 2B + PH S = 2π r 2 + 2π r H area of base circumference of base
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S = 2π ( ) 2 + 2π ( )( ) 6 cm 4 cm 4 251.3cm 2 Shape of the base? circle r =4 46
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8 10 8 Slant Height 2 = H 2 + x 2
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Surface Area of a Regular Pyramid S = B + ½ P Area of the Base Perimeter of the Base Slant Height of the Solid
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10.8 64 32 ( ) + ½ ( )( ) = 236.8m 2 S = 2 = 10 2 + 4 2 = 10.8
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Surface Area of a Cone S = B + ½ P area of base ½ the circumference of base S = π r 2 + π r
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17 = 2 = 15 2 + 8 2 628.3in 2 r =8 S = π ( ) 2 + π ( )( ) 8817
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Surface Area of a Sphere S = 4 π r 2
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8 in r = S= 8 4π(64) 804.2in 2
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Each layer of this cake is 3 inches high. One can of frosting will cover 130 square inches of cake, how many cans do we need? 6 in 10 in S = 641.34 in 2 B = ½ san 3 360˚ 12 3 ½ (6)(3 3 )(6) + 36 (3) ½ (6)(5 3 )(10) + 60 (3)+ 5 5 We need 5 cans
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Find the surface area. 72 cm 2 2( )+( )( ) = S = 4 8 8
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Find the surface area. 48 in 2 x 2 = 3 2 + 4 2 x x = 5 3 + 4 + 5 2( )+( )( ) = S = 6 12 3
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Find the surface area. S =2π( ) 2 + 2π( )( ) 967.61 in 2 7 7 15
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Find the surface area. S =( ) + ½( )( ) 144 in 2 4 l l 2 = 3 2 + 4 2 l = 5 64 32 5
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Find the surface area. S = 63.33 m 2 ( 5 / 2 ) 2 ( 3 ) ( ) + ½( )( ) 10.8 15 7 10.8 B?
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Find the surface area. 103.67 in 2 S = π ( ) 2 + π ( )( ) 33 8
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Find the surface area. 282.74 cm 2 x 2 = 5 2 + 12 2 x = 13 S = π ( ) 2 + π ( )( ) 5513
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Find the surface area. S = 4π( ) 2 452.39 cm 2 6
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Lesson 12.4,12.5,12.6 Volume Today, we will learn to… > find the Volume of prisms, cylinders, pyramids, cones, and spheres
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Volume? 5 times 3 is 15 4 layers of 15 60 cubes 5(3)(4) ? Why do we need to find volume?
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Theorem 12.6 Cavaleiri’s Principle If 2 solids have the same height and the same cross-sectional area at every level, then they have the same volume.
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Rectangular Prism V = LWH Volume of a Triangular Prism V = ½ bhH Cube V = s 3
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V = ( ) 3 = 343 cm 3 7
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V =( )( )( )= 135 m 3 359
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V =½ ( )( )( )= 162 cm 3 1293
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Volume of any Prism V = BH Height of the Solid Area of the Base
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Volume of a Cylinder V = BH V = π r 2 H area of base
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V = 4 cm 6 cm (π)( 2 )( )= 301.6 cm 3 46
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Volume of Pyramids and Cones V = B ▪ H 3 Experiment V = π r 2 H 3
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V = 213.3 m 3 ( )( ) 3 64 10
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V = 1005.3in 3 (π)( 2 )( ) 3 8 15
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Volume of a sphere 3 V = 4 π r 3
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5 m r = V = 5 523.6 m 3 3 4 π ( ) 3 5
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Since a hemisphere is ½ of a sphere its volume is ½ the volume of the sphere.
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Find the volume. V = ( )( )( ) 32 cm 3 228
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Find the volume. V = ½ ( )( )( ) =18 in 3 433
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V = (π)( 2 )( ) =2309.07 in 3 Find the volume. 7 15
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V = ( )( ) 3 = 64 in 3 Find the volume. 64 3
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V = π( 2 )( ) 3 69.74 in 3 Find the volume. 8 2 = x 2 + 3 2 x =x = x 3 7.4
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V = 314.16 cm 3 Find the volume. π( 2 )( ) 3 5 12
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V = 4(π)( 3 ) 3 904.78 cm 3 Find the volume. 6
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V = ( )( ) 3 36 6 V = ( )( )( ) 666 288 ft 3 +
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V = 97.9 m 3 π( 2 )( ) 3 2.55 5.1 V = ( )( )( ) 5.1 –
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V = 173.4 in 3 V = (π)( 2 )( ) = 2.5 7.5 π( 2 )( ) 3 2.5 4 +
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Find the volume of the sand. V = π( 2 )( ) 3 3.9 62.12 in 3
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1.Figure ABCDE has interior angle measures of 110˚, 90˚, 125˚, 130˚, and x˚. Find x. 2.A regular polygon has 13 sides. Find the sum of the measures of the interior angles. 3.Find the measures of one interior angle of a regular 22-gon. 4.What is the sum of the measures of one set of exterior angles of a 25-gon? 85° 1980° 163.64° 360°
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5.What is the measure of each exterior angle of a regular octagon? 6.The measure of each exterior angle of a regular polygon is 36˚. How many sides does the polygon have? 7.Find the number of sides in a regular polygon if its interior angles are each 162°. 45° 10 sides 20 sides
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10 5 a 30° 360° 12 s= a= n= 10 5 3 6 A = ½ san
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6.A polygon has interior angle measures of 120˚, 80˚, 135˚, 120˚, 100˚, and x˚. Find x. 7.A regular polygon has 15 sides. Find the sum of the measures of the interior angles. 8.Find the measure of one interior angle of a regular 24-gon. 9.What is the sum of the measures of one set of exterior angles of a 50-gon? 165° 2340° 165° 360°
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10.What is the measure of each exterior angle of a regular 40-gon? 11.The measure of each exterior angle of a regular polygon is 7.2˚. How many sides does the polygon have? 12.Find the number of sides in a regular polygon if its interior angles are each 174°. 9°9° 50 sides 60 sides
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12 6 a 30° 360° 12 s= a= n= 12 6 3 6 A = ½ san = 374
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12 6 a 36° 360° 10 s= a= n= 12 5 A = ½ san 8.3 tan 36° = 6a6a = 249
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12 x a 22.5° 360° 16 s= a= n= 9.2 8 A = ½ san sin 22.5° = x 12 cos 22.5° = a 12 11.1 = 408.5 12
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