11 – 5 Volumes of Pyramids & Cones

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Volumes of Pyramids & Cones

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11 – 5 Volumes of Pyramids & Cones Objectives: 1) Find the volume of a right Pyramid. 2) Find the volume of right Cone.

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

Ex.1: Volume of a right Pyramid Find the volume of a square pyramid with base edges of 15cm & a height of 22cm. Square V = (⅓)Bh = (⅓)l•w•h = (⅓)15•15•22 = (⅓)4950 = 1650cm3 22cm 15cm 15cm

Ex.2: Another square pyramid Find the area of a square pyramid w/ base edges 16ft long & a slant height 17ft. V = (⅓)Bh = (⅓)l•w•h = (⅓)16•16•___ = (⅓)3840 = 1280ft3 a2 + b2 = c2 h2 + 82 = 172 h2 = 225 h = 15 17ft 15 h 8ft 16ft

Vc = ⅓Bh II. Volume of a Cone Cone – Is “pointed” like a pyramid, but its base is a circle. h Vc = ⅓Bh r Area of the Base A = r2 Height of the cone, not to be confused with the slant height (l)

Ex.3: Find the volume of the following right cone w/ a diameter of 6in. Circle V = ⅓Bh = (⅓)r2h = (⅓)(3)2(11) = (⅓)99 = 33 = 103.7in3 11in 3in

Ex.4: Volume of a Composite Figure Volume of Cone first! Vc = ⅓Bh = (⅓)r2h = (⅓)(8)2(10) = (⅓)(640) = 213.3 = 670.2cm3 10cm 4cm Volume of Cylinder NEXT! Vc = Bh = r2h = (8)2(4) = 256 = 804.2cm3 8cm VT = Vc + Vc VT = 670cm3 + 804.2cm3 VT = 1474.4cm3

Ex.5: Solve for the missing variable. The following cone has a volume of 110. What is its radius. V = ⅓Bh V = ⅓(r2)h 110 = (⅓)r2(10) 110 = (⅓)r2(10) 11 = (⅓)r2 33 = r2 r = √(33) = 5.7cm 10cm r

Vc = ⅓Bh What have we learned??? Volume of Cones & Pyramids Height is the actual height of the solid not the slant height!!! Volume of Cones & Pyramids Vc = ⅓Bh h Area of Base h r