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The Cone

The Cone. A Cone is a three dimensional solid with a circular base and a curved surface that gradually narrows to a vertex. +. +. =. Volume of a Cone =. Exercise #1.

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The Cone

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  1. The Cone A Coneis a three dimensional solid with a circular base and a curved surface that gradually narrows to a vertex. + + = Volume of a Cone =

  2. Exercise #1 Find the volume of a cylinder with a radius r=1 m and height h=2 m. Find the volume of a cone with a radius r=1 m and height h=2 m Volume of a Cylinder = base x height = pr2h = 3.14(1)2(2) = 6.28 m3 Volume of a Cone = (1/3) pr2h = (1/3)(3.14)(1)2(2) = 2.09 m3

  3. 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

  4. Ex.3: Find the volume of the following right cone w/ a diameter of 6m. Circle V = ⅓Bh = (⅓)r2h = (⅓)(3)2(11) = (⅓)99 = 33 = 103.7m3 11m 3m

  5. 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

  6. 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

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

  8. Pyramids A Pyramid is a three dimensional figure with a regular polygon as its base and lateral faces are identical isosceles triangles meeting at a point. Identical isosceles triangles base = quadrilateral base = heptagon base = pentagon

  9. Volume of Pyramids Volume of a Pyramid: V = (1/3) Area of the base x height V = (1/3) Ah Volume of a Pyramid = 1/3 x Volume of a Prism = + +

  10. Exercise #2 Find the volume of the pyramid. height h = 8 m apothem a = 4 m side s = 6 m Volume = 1/3 (area of base) (height) = 1/3 ( 60m2)(8m) = 160 m3 h Area of base = ½ Pa a = ½ (5)(6)(4) = 60 m2 s

  11. Area of Pyramids Find the surface area of the pyramid. height h = 8 m apothem a = 4 m side s = 6 m Surface Area = area of base + 5 (area of one lateral face) What shape is the base? Area of a pentagon h = ½ Pa = ½ (5)(6)(4) = 60 m2 l a s

  12. Area of Pyramids What shape are the lateral sides? Find the surface area of the pyramid. height h = 8 m apothem a = 4 m side s = 6 m Area of a triangle = ½ base (height) = ½ (6)(8.9) = 26.7 m2 Attention! the height of the triangle is the slant height ”l ” h l l 2 = h2 + a2 = 82 + 42 = 80 m2 l = 8.9 m a s

  13. Area of Pyramids Surface Area of the Pyramid = 60 m2 + 5(26.7) m2 = 60 m2 + 133.5 m2 = 193.5 m2 Find the surface area of the pyramid. height h = 8 m apothem a = 4 m side s = 6 m h l a s

  14. Cones – Practice Questions Textbook: P. 421 - 422 # 1, 2, 3, 8 P. 439 – 441 # 1, 2, 3, 4 Challenging Questions: P. 421 - 422 # 6, 9

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