Examples

Example for Prims	Example for a Pyramid	Example for a cylinder
Solving for the Surface area . 1 .Find the area and the perimeter of the base   B = 5(10) = 50 in2  P= 10 + 10 +5 +5=30 in   2. Plug it in and solve  S= 2B + Ph   S= 2(50) + 30(6)  S= 100 +  180  S=  280 in 2	Solving for the Surface Area. 1. Find the perimeter and the area of the base . P= 5 +5 + 10 + 10 = 30 in B= 10(5) = 50in.2 2. Plug it in into the formula and solve. S= B + 1/2 x PL   S= 50 + 1/2 x 30 (13) S= 50 +  195 S= 245 in 2	Solving for the Surface Area 1. Plug it into the formula and solve .   S= 2(PI)rh + 2(PI)r2 S= 2( PI) (4)(8) + 2(PI) (4)2 S= (PI) 64  + (PI) 32 S= 96(PI)in square
Solving for the Volume. 1. Find the area of the base .  B= 5(10) = 50 in square 2. Plug it in and solve . V= BH V= 50(6) V= 300 in 3	Solving for the Volume. 1. Find the volume of the base. B = 5(10) = 50 in suquare 2. Plug it in and solve using the formula. V= 1/3 Bh V=1/3 x 50 (12) V= 1/3 x 600 V= 200 in 3	Solving for the Volume. 1. Plug it in and solve .   V= (PI)r2 h V= (PI)( 4) 2(8) V=  (PI) (16)(8) V= 120(PI) in 3

Example of a Cone	Example of a Sphere
Solving for the Surface Area 1. Plug it in and solve usinf the formula .   S= (PI)r 2  + (PI)rL S= (PI)(4)2  + (4)(6) S= 16 (PI) + 24(PI) S= 40(PI)in2	Solving for Surface Area 1.Plug it in and solve.   S= 4(PI) r2 S= 4(PI) 5 2 S= 4(PI) 25 S= 100(PI) in 2
Solving for the volume     1. Plug the corresponding number into the formula and solve .   * use the pythagorean theorem to get the height   V= 1/3 (PI)r 2 h V= 1/3 (PI) (4)  2 (4.47) V= 23. 84 (PI)in3	Solving fot the volume 1. Plug the numbers in and solve. V= (4/3)(PI) r 3 V= (4/3) (PI) (5)3 V= (4/3) (PI) 125 V= (500)(1/3) (PI) V=  166.67 (PI)in 3