Geometric Solids and Their Properties
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
=
+
+
=
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
=
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
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
= r
2
+ h
2
l
2
= (3)
2
+ (4)
2
l
2
= 25
l = 5
Surface Area
of a Cone
Surface Area of a Cone
=
r
2
+
rl
= 3.14(3)
2
+ 3.14(3)(5)
= 75.36 m
2
r = the radius h = the height
l
= the slant height
Textbook:
P. 421 - 422
# 2a, 3b, 9
P. 439 – 441
# 2abcd, 3, 4c, 5ab, 10abc
Cones – Practice
Questions
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
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
+
+
=
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
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
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
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
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
This powerpoint was kindly donated to
www.worldofteaching.com
http://www.worldofteaching.com
is home to over a
thousand powerpoints submitted by teachers. This is a
completely free site and requires no registration. Please
visit and I hope it will help in your teaching.
Explore the concepts of cones and pyramids, including their volumes and surface areas. Learn how to calculate the volume of a cone or pyramid, find the surface area of their bases and lateral sides, and solve practice questions to enhance your understanding of these geometric figures.
Download Presentation
Please find below an Image/Link to download the presentation.
The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author. Download presentation by click this link. If you encounter any issues during the download, it is possible that the publisher has removed the file from their server.
E N D
Presentation Transcript
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 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 = r2h Volume of a Cone = (1/3) r2h = (1/3)(3.14)(1)2(2) = 3.14(1)2(2) = 2.09 m3 = 6.28 m3
Surface Area of a Cone Find the area of a cone with a radius r=3 m and height h=4 m. r = the radius h = the height l = the slant height Use the Pythagorean Theorem to find l l 2 = r2 + h2 Surface Area of a Cone = r2 + rl = 3.14(3)2 + 3.14(3)(5) l 2= (3)2 + (4)2 = 75.36 m2 l 2= 25 l = 5
Cones Practice Questions Textbook: P. 421 - 422 # 2a, 3b, 9 P. 439 441 # 2abcd, 3, 4c, 5ab, 10abc
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
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 = + +
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 = Pa a = (5)(6)(4) = 60 m2 s
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 = Pa h l a = (5)(6)(4) = 60 m2 s
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 = base (height) = (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
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
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
This powerpoint was kindly donated to www.worldofteaching.com http://www.worldofteaching.com is home to over a thousand powerpoints submitted by teachers. This is a completely free site and requires no registration. Please visit and I hope it will help in your teaching.
