Download presentation

Presentation is loading. Please wait.

Published bySavannah Rowberry Modified over 2 years ago

1
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

2
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.

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

4
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

5
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

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

7
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

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

9
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

10
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

11
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

12
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

13
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?

14
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

15
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?

16
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

17
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

18
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).

19
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

20
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 =

21
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)

22
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

23
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

24
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?

Similar presentations

OK

Chapter 8. 8-1 Estimating Perimeter and Area Perimeter – total distance around the figure Area – number of square units a figure encloses.

Chapter 8. 8-1 Estimating Perimeter and Area Perimeter – total distance around the figure Area – number of square units a figure encloses.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on project tiger for class 10 Ppt on porter's five forces apple Animated ppt on magnetism perfume Ppt on polynomials and coordinate geometry practice Ppt on types of energy for kids Ppt on albert einstein at school Ppt on computer hardware and networking Run ppt on html website Topics for ppt on environmental issues Ppt on second law of thermodynamics example