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Construction Geometry. Spheres Surface Area Volume. Sphere. A sphere is a solid comprised of the set of all points in space equidistant from a given point. Sphere. Spheres do not have faces, edges, or vertices. Sphere. “Hemi” = half, a hemisphere is half of a sphere. Surface Area.
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Construction Geometry Spheres Surface Area Volume
Sphere • A sphere is a solid comprised of the set of all points in space equidistant from a given point.
Sphere • Spheres do not have faces, edges, or vertices.
Sphere • “Hemi” = half, a hemisphere is half of a sphere.
Surface Area • The formula for the surface area of a sphere is found on the Math Reference Sheet. • Surface Area = 4πr2
Practice #1 • Determine the surface area of the sphere. • 4πr2 = 4π(3)2 = 36πm2 • SA ≈ 113.1 m2 3m
Application • One great application for a hemisphere is a skate bowl. • Carpenters frame and form for shapes such as these to fit local recreation needs.
Application • The bowls can be made of concrete or tiles can be laid forming the interior wall.
Application 16’
Application • In order to tile or paint the interior of this hemisphere, we need to calculate the surface area to be covered. 16’
Practice #2 • Determine the surface area of the interior for the skate bowl. 16’
Practice #2 • Surface Area of Sphere = 4πr2 • d = 16’, r = 8’ • 4πr2 = 4π(8)2 = 256 πft2 • Hemisphere SA = ½ (Sphere SA) • ½ (256 πft2) • 128 πft2 • SA ≈ 402.1 ft2 16’
Volume • The formula for the volume of a sphere is found on the Math Reference Sheet. • V = πr3
Practice #3 • Determine the volume of the sphere. 9 ft
Practice #3 • V = πr3 • V = π(9 x 9 x 9) • V = (729) π • V = 972 π ft3 • V ≈ 3053.6 ft3 9 ft
Application • Before removing soil for skate park construction, carpenters need to calculate the yardage to be removed.
Application • Accurate volume calculations of excavation materials make formwork quicker and easier.
Practice #4 • How many cubic meters of soil would need to be removed to permit construction of this skate bowl? 12 m
Practice #4 • Sphere: V= πr3 • d = 12 m, r = 6 m • V = π(63) • V = (216)π • V = 288 π • V ≈ 904.8 m3 • V(hemisphere) ≈ ½(904.8) ≈ 452.4 m3 12 m
Practice & Assessment Materials • You are now ready for the practice problems for this lesson. • After completion and review, take the assessment for this lesson.