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**Surface Area of Pyramids Volume of a Right Pyramid**

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

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

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

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**SURFACE AREA IN PYRAMIDS**

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

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**V = B h Standards 8, 10, 11 VOLUME OF A PYRAMID: l h x 1 3 where:**

B= Area of the base h= height

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

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

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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 = Pl 1 2 h 26 ft = l P = 6( ) 6 feet L = ( )( ) 1 2 P = 36 feet 36 ft 26.5ft B = Pa 1 2 L =(18ft )(26.5 ft) L = ft 2 B= 1 2 36 3 TOTAL SURFACE AREA: 3 18 = T = Pl 1 2 + B 6 ft 3 B= 54 3 feet 2 93.5 ft 2 T = ft + Calculating base area: B feet 2 T ft 2 we need to find the slant height, using the Pythagorean Theorem: 60° VOLUME: 60° V = B h 1 3 l = ( ) 2 3 a l = 2 3 2 V = 1 3 ( )( ) 30° 93.5 ft 2 26 ft 3 60° l = 703 2 (9)(3) 6 V ft 3 27 3 l ft

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