Standards 8 10 11 Classifying Solids Classifying Pyramids

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

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. 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Standards 8, 10, 11 SOLIDS PYRAMID PRISM CYLINDER CONE PRESENTATION CREATED BY SIMON PEREZ. All rights reserved SPHERE 3

TRIANGULAR RECTANGULAR PENTAGONAL PYRAMID Standards 8, 10, 11 HEXAGONAL OCTAGONAL PYRAMID PRESENTATION CREATED BY SIMON PEREZ. All rights reserved 4

SURFACE AREA IN PYRAMIDS l l h x x x 1 2 xl L= 1 2 l x+x+x+x + xl + 1 2 The perimeter of the BASE is: P= x + x + x + x LATERAL AREA IS: L= 1 2 l. P or x x L= xl x x Calculating Lateral Area: 1 2 x x x + l l l L= 1 2 Pl PRESENTATION CREATED BY SIMON PEREZ. All rights reserved xl + 1 2 xl TOTAL SURFACE AREA: T= 1 2 Pl + B P= perimeter of base B= Area of base polygon l= slant height 5 h= height Standards 8, 10, 11

VOLUME OF A PYRAMID: Standards 8, 10, 11 l h x V= 1 3 Bh where: B= Area of the base h= height 6 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Standards 8, 10, 11 REVIEWING PERIMETERS X L X W W Y Y P= L+W +L +W P =L + W + W X X L X P = Y +Y +X P= 2 Y + X X X P = X +X +X P = 8 X P = 2 L + 2 W X X X P = 6 X P = 5 X P =4 X 7 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Standards 8, 10, 11 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. LATERAL AREA: 9 in =l 7 in =h 10 in P = 3(10 in ) ( 10 ) 5 3 =5 5 3 L= = 25 3 2 1 2 ( 30 in )( 9 in ) in 2 TOTAL SURFACE AREA: T = 1 Pl + B 2 2 Base perimeter: 1 2 L = 1 Pl L =(15 in )(9 in) 10 in B= L = 135 in T = 135 in 2 + 25 3 Base Area: in 2 T = 135 in 2 + 43. 3 in 2 2 10 in T = 178. 3 in 30° 5 3 60° P = 30 in 10 5 VOLUME: 1 V = 3 Bh 1 ( 25 3 V= 3 V 101 in 3 in 2 )( 7 in ) 8

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. LATERAL AREA: Perimeter: L = 1 Pl 2 P = 6( 6 feet ) 1 L = P = 36 feet 2 ( 36 ft )( 26. 5 ft) l B= 26 ft =h 6 ft B= = 3 3 Calculating base area: 3 3 1 2 Pa 36 18 3 3 B= 54 3 feet 2 B 93. 5 feet 2 we need to find the slant height, 60° using the Pythagorean Theorem: 60° l 2= 26 2 + ( 3 3 ) 2 2 l =2 676 + 27 30° 2 a 3 3 2 60° l = 703 (9)(3) 6 27 l 26. 5 ft 3 L =(18 ft )(26. 5 ft) L = 477 ft 2 TOTAL SURFACE AREA: T = 1 Pl + B 2 T = 477 ft 2 + 93. 5 ft 2 T 570. 5 ft 2 VOLUME: 1 V = 3 Bh 2 1 ( )( 26 ft ) 93. 5 ft V= 3 V 810. ft 3 9