Lesson 11.1 I can identify and name 3-dimensional figures I can identify and name parts of 3-D figures (faces, bases, edges, vertices) I can identify the shape of cross sections of 3-D figures and find the area of the cross sections Lesson 11.1

solid flat surfaces polygons

2 parallel congruent 2 parallel congruent 1 base point vertex shape base(s)

Rectangular prism Pentagonal pyramid

Triangular prism Trapezoidal prism

Square (or rectangular) pyramid Pentagonal prism Square (or rectangular) pyramid

∆ABC , ∆DEF ABED, BEFC, ADFC AB BC CA DE EF FD BE CF AD A, B, C, D, E, F

parallel bases circles circular base vertex

plane rectangle circle

rectangle rectangle

4 4 4 Isosceles triangle square A = ½ bh A = bh A = ½ 18(12) A = 4(4) A = 108 cm2 A = 16 un2

triangle rectangle A = ½ bh A = bh A = ½ 9(6) A = 10(6) A = 27 un2 A = 60 in2

ASSIGNMENT 11-1 worksheet

I can draw 2-d models for 3-d figures I can surface area using nets Lesson 11-2

2 pattern Top Front Left 1 6 2 Bottom Right Back

SA = square + 4 triangles 10 = 64 + 4 (40) = 224 un2 8 8

7 5 35 5 15 3 21 15 SA = 6 rectangles 5 35 = 142 un2 3 21 7

≈ 477.52 cm2 SA = rectangle + 2 circles 15(8π) + 2π(42) 120π + 32π = 152π 4 8π ≈ 477.52 cm2 15

Investigation #1 - Prisms Use the solids to answer the questions for each shape. Draw pictures! Shape of Lateral Faces Number of Bases Base & Number of Sides Number of Edges Number of Vertices Ex Octagonal Prism 8 2 24 16 1 Square Prism 4 12 Rectangular Prism 3 Pentagonal 5 15 10 Triangular 9 6

Curved surface when flattened Investigation #3 – Solids Involving Circles Use the solids and the nets to answer the questions for each shape. Draw pictures! Number of curved surfaces Shape of Curved surface when flattened Bases Base Number of curved edges Number of apexes (like a vertex) 8 Cylinder 1 2 none 9 Cone 10 Sphere 11 Hemisphere

Investigation #2 - Pyramids Use the solids to answer the questions for each shape. Draw pictures! Shape of Lateral Faces Number of Bases Base & Number of Sides Number of Edges Number of Vertices Ex Pentagonal Pyramid 5 1 10 6 Square Pyramid 4 8 Octagonal 16 9 7 Triangular Pyramid 3

11.1 Identify parts of 3d figures Summarize the differences between prisms and pyramids Prisms Pyramids Lateral Faces Rectangles Triangles Bases Two matching polygons One polygon Edges Base edges times 3 Base edges times 2 Vertices Base vertices times 2 Base vertices +1 Name ____________ prism __________ pyramid When the base is a circle it’s called a… Cylinder Cone (name of base) (name of base)

Warm Up 1. Draw and label a net for the following solid. Then find the surface area.

3 6 2 3 2 3 8 16 8 24 16 8 8 24 8 3 2 3 2 2 6 2 3 SA = 16+16+24+24+6+6 = 92 un2

ASSIGNMENT 11-2 worksheet

I can find the volume of prisms I can find the volumes of cylinders Lesson 11-3

V = Bh B = area of base V = πr2h

V = πr2h = π(82) (17.5) = 1120π mm3 ≈ 3518.58 mm3

V = πr2h = π(82) (30) = 1920π cm3 ≈ 6031.86 cm3

V = πr2h = π(62) (11.5) = 414π in3 ≈ 1300.62 in3

24 ft V = πr2h = π(3.52) (24) = 294π ft3 ≈ 923.63 ft3

V = Bh = 4.5 (6) B = 1/2h(b1+b2) = 27 cm3 = 1/2(1.5)(2 + 4) = 4.5

V = Bh = 93.6 (12) B = 1/2Pa = 1123.2 in3 = 1/2(36)(5.2) = 93.6

V = Bh = 150 (12) B = bh = 1800 ft3 = 10(15) = 150

V = Bh = 27 (10) B = 1/2bh = 270 un3 = 1/2(9)(6) = 27

V = Bh 735 = 42h 42 42 17.5 cm = h

V = Bh 196 = 16B 16 16 16 = B 4 in by 4 in 16 ? ?

ASSIGNMENT 11-3 worksheet

I can find the volume of pyramids I can find the volume of cones Lesson 11-4

segment vertex base height lateral face segment vertex perpendicular base

V = 1/3 Bh B = area of base V = 1/3 πr2h

V = 1/3 πr2h = 1/3 π(42) (7) = 37.3π in3 ≈ 117.29 in3

V = 1/3 πr2h 24 = 1/3 π(102) (24) = 800π ft3 ≈ 2513.27 ft3

V = 1/3 πr2h 8 = 1/3 π(62) (8) = 96π cm3 ≈ 301.59 cm3

= 424.3π yd3 ≈ 1333.08 yd3 V = 1/3 πr2h = 1/3 π(102) (12.73) r = 10

V = 1/3 Bh = 1/3 (64) (10) B = bh = 21.33 ft3 = 8(8) = 64

8 B = 1/2bh V = 1/3 Bh = 1/2(6)(8) = 1/3 (24) (15) = 24 = 120 ft3

4 3 V = 1/3 Bh = 1/3 (48) (3) B = bh = 48 yd3 = 8(6) = 48

5 V = 1/3 Bh 10 10 = 1/3 (100) (12) B = bh = 400 cm3 = 10(10) = 100

V = 1/3 πr2h 96π = 1/3 πr2 (8) π π 96 = 1/3 r2(8) 288 = r2(8) 36 = r2 π π (3) (3) 96 = 1/3 r2(8) 288 = r2(8) 8 8 36 = r2 6 = r

V = 1/3 πr2h 2500π = 1/3 π 52 (h) π π 2500 = 1/3 (25)(h) 7500 = 25h π π (3) (3) 2500 = 1/3 (25)(h) 7500 = 25h 25 25 300 = h

ASSIGNMENT 11-4 worksheet

I can find surface area of spheres I can find volume of spheres Lesson 11-5

SA = 4πr2 V = 4πr3 3 V = 2πr3 3

V = = SA = 4πr2 = 4π(6)2 = 144π cm2 ≈ 452.39 cm2 4πr3 3 4π(6)3 3

V = SA = 4π(4)2 = 64π cm2 ≈ 201.06 cm2 4π(4)3 3 = 85.3π cm3

V = 2πr3 3 = 2π(16)3 3 = 2730.7π m3 ≈ 8578.64 m3

V = 2π(10)3 3 = 666.6π in3 ≈ 2094.44 in 3

C = πd SA = 4π(9.23)2 58 = πd ≈ 1070.57 m2 18.46 = d 9.23 = r

9.23 = r V = 4π(9.23)3 3 ≈ 3293.77 cm3

V = 4πr3 3 288π = 4πr3 3 SA = 4π(6)2 = 144π cm2 216 = r3 6 = r

SA = 4πr2 324π = 4πr2 V = 4π(9)3 3 81 = r2 9 = r = 972π in3

ASSIGNMENT 11-5 worksheet

I can find the volume of composite figures Lesson 11-6

cone + cylinder cone: cylinder: Total : 57.9π ≈ 181.9 cm3 V = 1/3 πr2h 4 cone: cylinder: V = 1/3 πr2h V = πr2h = 1/3 π(3)24 = π(3)25.1 = 12π = 45.9π Total : 57.9π ≈ 181.9 cm3

triangular prism V = Bh = 120 (70) = 8400 rectangular prism V = Bh Total = 29,400 ft3 = 2100 (10) = 21000

ASSIGNMENT 11-6 worksheet

I can identify properties of similar solids I can find volume and surface area of similar solids Lesson 11-7

shape size a : b a2 : b2 a3 : b3

congruent similar 1 : 3

neither similar 2 : 1

1 : 2 8 : 5 12 : 22 or 1 : 4 82 : 52 or 64 : 25 13 : 23 or 1 : 8 83 : 53 or 512 :125

12 : 15 4 : 5 4 5 10 x 12.5 m = 42 : 52 or 16 : 25 16 25 280 x 437.5 m2 = 43 : 53 or 64 :125 64 125 400 x 781.25 m3 =

8 : 6 4 : 3 4 3 11 x 8.25 cm = 16 9 x 325 42 : 32 or 16 : 9 577.78 cm2 = 43 : 33 or 64 :27 64 27 1345 x 567.42 cm3 =

ASSIGNMENT 11-6 worksheet