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

2. Identical isosceles triangles
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

3. 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 = + +

4. 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

5. The Cone + + = Volume of a 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 =

6. Exercise #1 = (1/3)(3.14)(1)2(2) = 3.14(1)2(2) = 2.09 m3 = 6.28 m3
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=1 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

7. 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)

8. 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

9. 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
h = 172
h2 = 225
h = 15
17ft 15 h
8ft
16ft

10. 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)

11. 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

12. 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 = 670cm cm3
VT = cm3

13. 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

14. 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