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1 Standards 8, 10, 11 Classifying Solids PROBLEM 1PROBLEM 2 PROBLEM 3 Parts of a Prism Classifying Prisms Surface Area of Prisms Volume of a Right Prism Reviewing Perimeters PROBLEM 4 PROBLEM 5PROBLEM 6 PROBLEM 7 PRESENTATION CREATED BY SIMON PEREZ RHS. All rights reserved END SHOW So,What are Cubic Units Algebra Integration Activity

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2 Standard 8: Students know, derive, and solve problems involving perimeter, circumference, area, volume, lateral area, and surface area of common geometric figures. Estándar 8: Los estudiantes saben, derivan, y resuelven problemas involucrando perímetros, circunferencia, área, volumen, área lateral, y superficie de área de figuras geométricas comunes. Standard 10: Students compute areas of polygons including rectangles, scalene triangles, equilateral triangles, rhombi, parallelograms, and trapezoids. Estándar 10: Los estudiantes calculan áreas de polígonos incluyendo rectángulos, triángulos escalenos, triángulos equiláteros, rombos, paralelogramos, y trapezoides. Standard 11: Students determine how changes in dimensions affect the perimeter, area, and volume of common geomegtric figures and solids. Estándar 11: Los estudiantes determinan cambios en dimensiones que afectan perímetro, área, y volumen de figuras geométricas comunes y sólidos.

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3 PRISM PYRAMID CYLINDER SPHERE CONE Standards 8, 10, 11 SOLIDS

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4 VERTICES: A A, B B, C C, D D, E E, F F, G G, H H, I I, J J. FACES: Quadrilaterals HGBC, GFAB, FJEA, JIDE, IHCD. Pentagons EDCBA, JIHGF. EDGES: CB, BA, AE, ED, DC, CH, BG, AF, EJ, DI, HG, GF, FJ, JI, IH. Standards 8, 10, 11

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5 VERTICES: A A, B B, C C, D D, E E, F F, G G, H H, I I, J J. FACES: Quadrilaterals HGBC, GFAB, FJEA, JIDE, IHCD. Pentagons EDCBA, JIHGF. EDGES: CB, BA, AE, ED,DC, CH, BG, AF,EJ, DI, HG, GF, FJ, JI,IH. Standards 8, 10, 11

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6 PRISM CUBE PRISM TRIANGULAR RECTANGULAR PENTAGONAL HEXAGONAL OCTAGONAL Standards 8, 10, 11

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7 SURFACE AREA OF PRISMS a b c c d d e e f f h a b a f e h base L = ahah + bh + ch + dh+ eh + fh P= a a b + b c + c d + d e + e f + f Perimeter of base: Lateral Area: L= ( )h a + b + c + d + e + f Substituting L= Ph Total Area= Lateral Area + 2 Base Area T = Ph + 2B Where: Standards 8, 10, 11 P=perimeter h=height B= base area T = L + 2B L= lateral area

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8 VOLUME OF A RIGHT PRISM: h h V = where: B= Area of the base h= height Standards 8, 10, 11 B B Bh

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9 REVIEWING PERIMETERS P = L L W + W L + L W + W P = 2L + 2W P = L + L + W P = X X +X X X X P = 8X X +X X X P = 6X X P = 5X X P = Y Y Y +Y X +X P= 2Y + X P =4X X +X X Standards 8, 10, 11

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10 Find the lateral area, the surface area and volume of a right prism with a height of 20in and a perimeter of 18 in. whose base is an equilateral triangle. 20 in h= P= 18 in Lateral Area: L= Ph L= ( )( )18 in 20 in L= 360 in 2 Base Area: 30° 60° P= 3X X 18=3X 3 X= 6 6 3 3 3 then the base area: B = ( 6) 1 2 3 3 = 3 3 3 = 9 3 Total Area= Lateral Area + 2 Base Area T = L + 2B 15.6 in B 2 T = 360 in + 2(15.6 in ) 2 2 T = 360 in + 31.2 in 2 2 T = 391.2 in 2 Surface Area: Volume: V=Bh V= ( )( ) 15.6 in 2 20 in V= 312 in 3 Standards 8, 10, 11

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11 Find the lateral area and the surface area of a regular pentagonal prism whose height is 18 ft and whose perimeter is 30 ft. 18 ft h= P= 30 ft Lateral Area: L= Ph L= ( )( )30 ft 18 ft L= 540 ft 2 Base Area: B = Pa 1 2 1 2 = ( )( ) 30 4 B 60 feet 2 6 6 a 3 360° 5 = 72° 72° 2 =36° Tan 36°= 3 a aTan36°= 3 (a) Tan36° a= 4 a 3 36° P= 5X 30=5X 5 X= 6 4 6 Total Area= Lateral Area + 2 Base Area T = L + 2B T = 540 ft + 2( 60 ft ) 2 2 T = 540 ft + 120 ft 2 2 T = 660 ft 2 Surface Area: Standards 8, 10, 11

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12 Standards 8, 10, 11 Find the lateral area, the surface area and the volume of a right prism with a hexagonal base. The side of the base is 15 cm and the height of the prism is 9 cm. 9 cm h= 15cm P = 6X Perimeter: P = 6( ) 15 cm P = 90 cm Lateral Area: L= Ph L= ( )( )90 cm 9 cm L= 810 cm 2 15 a 30° 60° 7.5 3 30° 60° 7.5 a = 7.5 3 a 13 15 13 B= Pa 1 2 1 2 = ( )( ) 90 13 B 585 cm 2 360° 6 = 60° Base Area: T = L + 2B T = 810 cm + 2( 585cm ) 2 2 T = 810 cm + 1170 cm 22 T =1980 cm 2 Surface Area: Total Area= Lateral Area + 2 Base Area Volume: V=Bh V= ( )( ) 585 cm 2 9 cm V = 5265 cm 3

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13 STANDARDS 1 1 1 2 2 2 3 3 3 4 4 4 1x1x1 = 1 3 = 1 1 CUBED 2x2x2 = 2 3 = 8 2 CUBED 3x3x3 = 3 3 = 27 3 CUBED 4x4x4 = 4 3 =64 4 CUBED What is the volume for these cubes?

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14 Standards 8, 10, 11 Find the lateral surface of a cube whose volume is 2197 cubic feet. L L L B V = Bh B = (L)(L) B= L 2 V = L L 2 V= L 3 2197 feet = L 3 3 3 3 3 3 L=13 feet Perimeter of the base: P = 4( ) 13 feet P = 52 feet Lateral Area: L= Ph L= ( )( )52 ft 13 ft L= 676 ft 2 13 feet Note: Lateral Area = Lateral Surface

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15 x x x x x 1 1 1 1 1 1 x V = (x) V = (x) (1) V = (x) (1) V = (1) = x 3 2 = 1 Lets find the volume for this prisms: Can we use this knowledge to multiply polynomials?

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16 Multiply: (x+3)(x+2)(x+1) (x+2) (x+1) (x+3) (x + 2)(x + 3) (3) x (2) x x + x + (2) (3) + F O I L = x + 5x + 6 2 = x + 3x +2x + 6 2 = x+1x+1 X +6+5x +6x +5x 2 x 2 x 3 +6 +11x x 3 +6x 2 x +5x + 6 2

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17 Multiply: (x+3)(x+2)(x+1) = x + 6x + 11x + 6 3 2 (x + 2)(x + 3) (3) x (2) x x + x + (2) (3) + F O I L = x + 5x + 6 2 = x + 3x +2x + 6 2 = x+1x+1 X +6+5x +6x +5x 2 x 2 x 3 +6 +11x x 3 +6x 2 x +5x + 6 2

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18 Multiply: (x+3)(x+2)(x+1) = x + 6x + 11x + 6 3 2 (x + 2)(x + 3) (3) x (2) x x + x + (2) (3) + F O I L = x + 5x + 6 2 = x + 3x +2x + 6 2 = x+1x+1 X +6+5x +6x +5x 2 x 2 x 3 +6 +11x x 3 +6x 2 x +5x + 6 2 (x+2) (x+1) (x+3) So, a third degree polynomial may be represented GEOMETRICALLY, by the VOLUME OF A RECTANGULAR PRISM, in this case with SIDES (x+3), (x+2) and (x+1).

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19 Multiply: (2x+1)(x+3)(x+4) (x+3) (x+4) (2x+1) (2x + 1)(x + 3) (3) x (1) 2x x +2x + (1) (3) + F O I L = 2x + 7x + 3 2 = 2x + 6x +1x + 3 2 = x+4x+4 X +12+28x + 3x +7x 2 8x 2 2x 3 +12 +31x 2x 3 +15x 2 2x +7x + 3 2

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20 Multiply: (2x+1)(x+3)(x+4) (2x + 1)(x + 3) (3) x (1) 2x x +2x + (1) (3) + F O I L = 2x + 7x + 3 2 = 2x + 6x +1x + 3 2 = x+4x+4 X +12+28x + 3x +7x 2 8x 2 x 3 2x +7x + 3 2 +12 +31x 2x 3 +15x 2 +12 +31x 2x 3 +15x 2 =

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21 Multiply: (2x+1)(x+3)(x+4) (2x + 1)(x + 3) (3) x (1) 2x x +2x + (1) (3) + F O I L = 2x + 7x + 3 2 = 2x + 6x +1x + 3 2 = x+4x+4 X +12+28x + 3x +7x 2 8x 2 x 3 2x +7x + 3 2 +12 +31x 2x 3 +15x 2 +12 +31x 2x 3 +15x 2 = (x+3) (x+4) (2x+1)

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22 Standards 8, 10, 11 SIMILARITY IN SOLIDS 6 4 8 3 32.25 8 6 = = 4 3 3 = 1.3 Both solids are SIMILAR

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23 Standards 8, 10, 11 SIMILARITY IN SOLIDS 6 5 10 3 4 2 6 = 5 3 4 2 These solids are NOT SIMILAR = 1.6 = 1.6 = 2

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24 Standards 8, 10, 11 The ratio of the perimeter of two similar prisms is 1:4. If the lateral area of the larger prism is 120 mm, what is the lateral area of the smaller prism? Lateral Area: L= Ph 1 11 2 22 PRISM 1 PRISM 2 IF THEN L P h 1 11 2 22 = P 1 P 2 = h 1 h 2 = 1 4 THEN L 1 = 11 4 4 120 L 1 = 1 16 (120) = 120 16 = 7.5 mm 2 L 1 PRISM 1 < PRISM 2 AND IF They are similar What can you conclude from the ratio of the areas and the ratio of the perimeters?

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Surface Area of Pyramids Volume of a Right Pyramid

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