2. 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 = ++=

3. 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 =  r 2 h = 3.14(1) 2 (2) = 6.28 m 3 Exercise #1 Volume of a Cone = (1/3)  r 2 h = (1/3)(3.14)(1) 2 (2) = 2.09 m 3

4. II. Volume of a Cone 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

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

6. 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

7. Ex.5: Solve for the missing variable. 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

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

9. 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. Pyramids base = quadrilateral base = pentagon base = heptagon Identical isosceles triangles

10. 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 Volume of Pyramids ++ =

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

12. Surface Area = area of base + 5 (area of one lateral face) Area of Pyramids Find the surface area of the pyramid. height h = 8 m apothem a = 4 m side s = 6 m h a s Area of a pentagon l What shape is the base? = ½ Pa = ½ (5)(6)(4) = 60 m 2

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

14. Area of Pyramids Find the surface area of the pyramid. height h = 8 m apothem a = 4 m side s = 6 m h a s l Surface Area of the Pyramid = 60 m 2 + 5(26.7) m 2 = 60 m 2 + 133.5 m 2 = 193.5 m 2

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