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1 Standards 8, 10, 11 Classifying Solids PROBLEM 1PROBLEM 2 Classifying Pyramids Surface Area of Pyramids Volume of a Right Pyramid Reviewing Perimeters.

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Presentation on theme: "1 Standards 8, 10, 11 Classifying Solids PROBLEM 1PROBLEM 2 Classifying Pyramids Surface Area of Pyramids Volume of a Right Pyramid Reviewing Perimeters."— Presentation transcript:

1 1 Standards 8, 10, 11 Classifying Solids PROBLEM 1PROBLEM 2 Classifying Pyramids Surface Area of Pyramids Volume of a Right Pyramid Reviewing Perimeters PRESENTATION CREATED BY SIMON PEREZ RHS. All rights reserved END SHOW

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 PYRAMID TRIANGULAR PYRAMID RECTANGULAR PYRAMID PENTAGONAL PYRAMID HEXAGONAL PYRAMID OCTAGONAL Standards 8, 10, 11

5 5 L= l 1 2 x + x + x + x + x + x P= x The perimeter of the BASE is: LATERAL AREA IS: L = l P 1 2 or xlxl 1 2 + x l xlxl 1 2 + x l xlxl 1 2 + x l xlxl 1 2 + x l x + x x + x x + x x + x x + x xlxl 1 2 + l x TOTAL SURFACE AREA: T = P l 1 2 + B P= perimeter of base B= Area of base polygon l = slant height h h= height xlxl 1 2 x l L = P l 1 2 x x Calculating Lateral Area: SURFACE AREA IN PYRAMIDS l Standards 8, 10, 11

6 6 VOLUME OF A PYRAMID: where: B= Area of the base h= height Standards 8, 10, 11 x l h V = Bh 1 3

7 7 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

8 8 Find the lateral area and the surface area and volume of a right pyramid whose slant height is 9 in and whose height is 7 in. Its base is an equilateral triangle whose side is 10 in. Round your answers to the nearest tenth. 10 in h 7 in = l 9 in = L = ( )( ) 1 2 L =(15 in )(9 in) L = P l 1 2 LATERAL AREA: Base perimeter: P = 3( ) 10 in P = 30 in 10 in 30 in 7 in L = 135 in 2 Base Area: 10 5 3 5 5 3 B = ( ) 1 2 = 5 5 3 TOTAL SURFACE AREA: T = P l 1 2 + B = 25 3 in 2 T = 135 in + 43.3 in 2 2 T = 135 in + 25 3 in 2 2 T = 178.3 in 2 VOLUME: V = B h 1 3 1 3 ( )( ) 25 3 in 2 9 in V 101 in 3 10 30° 60°

9 9 Standards 8, 10, 11 Find the lateral area, the surface area and volume of a right pyramid with a height of 26 ft whose base is a regular hexagon with side of 6 ft. Round your answers to the nearest tenth. 6 ft Calculating base area: a 60° 6 30° 60° B = Pa 1 2 3 3 36 B= 1 2 3 3 18= B= 54 3 feet 2 P = 6( ) 6 feet P = 36 feet L = ( )( ) 1 2 L =(18ft )(26.5 ft) L = P l 1 2 LATERAL AREA: 36 ft L = 477 ft 2 26.5ft 3 Perimeter: 26 ft TOTAL SURFACE AREA: T = P l 1 2 + B T 570.5 ft 2 VOLUME: V = B h 1 3 1 3 ( )( ) V 810. ft 3 3 3 we need to find the slant height, using the Pythagorean Theorem: l = 676 + 27 2 l = 703 2 l 26.5 ft 93.5 ft 2 T = 477 ft 2 + B 93.5 feet 2 93.5 ft 2 l h 26 ft = 3 3 l = 26 + ( ) 2 2 2 3 3 3 3 2 2 (9)(3) 27


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