1. Volumes of Pyramids & Cones Objectives: 1) Find the volume of a right Pyramid. 2) Find the volume of right Cone.

2. I. Volume of a Pyramid Pyramid – Is a polyhedron in which one face can be any polygon & the other faces are triangles. Pyramid – Is a polyhedron in which one face can be any polygon & the other faces are triangles. h V p = ⅓Bh Area of the Base A = lw A = ½bh Height of the pyramid, not to be confused with the slant height (l)

3. Ex.1: Volume of a right Pyramid Find the volume of a square pyramid with base edges of 15cm & a height of 22cm. Find the volume of a square pyramid with base edges of 15cm & a height of 22cm. 22cm 15cm V = (⅓)Bh = (⅓)lwh = (⅓)151522 = (⅓)4950 = 1650cm 3 Square

4. II. Volume of a Cone Cone – Is “pointed” like a pyramid, but its base is a circle. Cone – Is “pointed” like a pyramid, but its base is a circle. h r V c = ⅓Bh Area of the Base A =  r 2 Height of the cone, not to be confused with the slant height (l)

5. Ex.3: Find the volume of the following right cone w/ a diameter of 6in. 11in V = ⅓Bh = (⅓)  r 2 h = (⅓)  (3) 2 (11) = (⅓)99  = 33  = 103.7in 3 Circle 3in

6. Ex.5: Solve for the missing variable. The following cone has a volume of 110 . What is its radius. The following cone has a volume of 110 . What is its radius. 10cm r V = ⅓Bh V = ⅓(  r 2 )h 110  = (⅓)  r 2 (10) 110 = (⅓)r 2 (10) 11 = (⅓)r 2 33 = r 2 r = √ (33) = 5.7cm

7. Ex.4: Volume of a Composite Figure 8cm 10cm 4cm Volume of Cone first! V c = ⅓Bh = (⅓)  r 2 h = (⅓)(8) 2  (10) = (⅓)(640)  = 213.3  = 670.2cm 3 Volume of Cylinder NEXT! V c = Bh =  r 2 h =  (8) 2 (4) = 256  = 804.2cm 3 V T = V c + V c V T = 670cm 3 + 804.2cm 3 V T = 1474.4cm 3