Equations are more important to me, because politics is for the present, but an equation is something for eternity. Albert Einstein Today: ◦ Multiple Choice Practice ◦ 12.6 Instruction ◦ Practice
Objectives: Find the volume of a Sphere Find the volume of a Composite Figures Vocabulary:: None
Surface Area Prism S = 2B + Ph Cylinder S = 2B + Ch Pyramid S = B + ½P l Cone S = B + ½C l Sphere S = ? Volume Prism V = Bh Cylinder V = Bh Pyramid V = 1/3Bh Cone V = 1/3Bh Sphere V = ?
Sphere – the locus of points in space that are equal distance from a point. Has no base, but can be cut into circles. Circle at center point called great circle. SA of Sphere = 4 r 2 where r = radius.
Find the surface area of a sphere with radius of 2”. 16 in 2 in 2 2” Find the surface area of the hemisphere of the above figure. Cut in half? 8 in 2 ? 12 in 2 in 2
Find the surface area of the figure. Round answer to nearest tenth. Include units! cm
36 cm 3 cm 3 Find the volume of the sphere if r = 3 cm. r Volume of a Sphere The volume, V, of a sphere is V = 4/3 r 3, where r = radius.
What is the radius to the nearest hundredth of a sphere molded from the cylinder of modeling clay shown? 3.63 cm 2 cm 16 cm
A rectangular pyramid fits exactly on top of a rectangular prism. The prism has length 15 cm, width 5 cm, and height 7 cm, and the pyramid has height 13 cm. Find the volume of the composite figure.
Surface Area Prism S = 2B + Ph Cylinder S = 2B + Ch Pyramid S = B + ½P l Cone S = B + ½C l Sphere S = 4 r 2 Volume Prism V = Bh Cylinder V = Bh Pyramid V = 1/3Bh Cone V = 1/3Bh Sphere V = 4/3 r 3
Equations are more important to me, because politics is for the present, but an equation is something for eternity. Albert Einstein Assignment: Worksheet Project Work Tomorrow Test Tuesday