Slide 1
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
Slide 2
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
Exercise #1
Volume of a Cone = (1/3) pr2h
= (1/3)(3.14)(1)2(2)
= 2.09 m3
Slide 3
Find the area of a cone with a radius r=3 m and height h=4 m.
Use the Pythagorean Theorem to find l
l 2 = r2 + h2
l 2= (3)2 + (4)2
l 2= 25
l = 5
Surface Area of a Cone
Surface Area of a Cone
= pr2 + prl
= 3.14(3)2 + 3.14(3)(5)
= 75.36 m2
r = the radius h = the height l = the slant height
Slide 4
Textbook:
P. 421 - 422 # 2a, 3b, 9
P. 439 – 441 # 2abcd, 3, 4c, 5ab, 10abc
Cones – Practice Questions
Slide 5
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
Slide 6
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
+
+
=
Slide 7
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 ( 60m2)(8m)
= 160 m3
= ½ (5)(6)(4)
= 60 m2
Slide 8
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 m2
Slide 9
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)