12.2 surface area of prisms & cylinders surface... · finding the surface area of a prism •...
TRANSCRIPT
12.2 Surface Area
of Prisms &
Cylinders Geometry
Mr. Peebles
Spring 2013
Please get your whiteboards,
markers, and towels then have a
seat and wait for instructions.
Geometry Bell Ringer
Geometry Bell Ringer
Answer: B
Daily Learning Target
(DLT) • “I can understand, apply, and
remember to find the surface area of
prisms and cylinders.”
Assignment Pages 601-604
(1-9, 13-15, 24-26, 30, 55-58)
Assignment Pages 601-604
(1-9, 13-15, 24-26, 30, 55-58)
1. Vertices: 4 4. Faces = 8 Edges: 6 5. Edges = 12
Faces: 4 6. Edges = 12
2. Vertices: 8 7. Vertices = 8 Edges: 12 8. Vertices = 5
Faces: 6 9. Vertices = 9
3. Vertices: 10 13. Two Edges: 15 Concentric
Faces: 7 Circles
Assignment Pages 601-604
(1-9, 13-15, 24-26, 30, 55-58) 14.Triangle
15.Rectangle
24.Triangle
25.Circle
26.Trapezoid
30.Vertices = 60
55.B
56.G
57.D
58.J
Finding the surface area of a
prism
• A prism is a polyhedron
with two congruent
faces, called bases, that
lie in parallel planes.
The other faces called
lateral faces, are
parallelograms formed
by connecting the
corresponding vertices of
the bases. The
segments connecting
these vertices are lateral
edges.
Finding the surface area of a
prism
• The altitude or height of
a prism is the
perpendicular distance
between its bases. In a
right prism, each lateral
edge is perpendicular to
both bases. Prisms that
have lateral edges that
are not perpendicular to
the bases are oblique
prisms. The length of
the oblique lateral edges
is the slant height of the
prism.
Note
• Prisms are classified by the shape of
their bases. For example, the figures
above show one rectangular prism
and one triangular prism. The surface
area of a polyhedron is the sum of the
areas of its faces. The lateral area of
a polyhedron is the sum of the areas
of its lateral faces.
Ex. 1: Finding the surface
area of a prism
• Find the surface
area of a right
rectangular prism
with a height of 8
inches, a length of
3 inches, and a
width of 5 inches.
Ex. 2: Finding the surface
area of a prism
• Find the surface
area of a box of
cereal with a height
of 15 inches, a
length of 8 inches,
and a width of 4
inches.
Ex. 2: Finding the surface
area of a prism
• Find the surface area of a box of
cereal with a height of 15 inches, a
length of 8 inches, and a width of 4
inches.
Faces Dimensions Area of Faces
Left & Right 15” x 8” 120 in2
Front & Back 15” x 4” 60in2
Top & Bottom 8” x 4” 32in2
SA = 2(120) + 2(60) + 2(32) = 424 in2
Nets • Imagine that you cut some edges of a
right hexagonal prism and unfolded it.
The two-dimensional representation of
all of the faces is called a NET.
Ex. 2: Using Theorem 12.2
Ex. 2: Using Theorem 12.2
Ex. 2: Using Theorem 12.2
Finding the surface area of a cylinder
• A cylinder is a solid with congruent circular bases that lie in parallel planes. The altitude, or height of a cylinder is the perpendicular distance between its bases. The radius of the base is also called the radius of the cylinder. A cylinder is called a right cylinder if the segment joining the centers of the bases is perpendicular to the bases.
Surface area of cylinders
• The lateral area of a cylinder is the area of its
curved surface. The lateral area is equal to the
product of the circumference and the height,
which is 2rh. The entire surface area of a
cylinder is equal to the sum of the lateral area
and the areas of the two bases.
Ex. 3: Finding the Surface Area of a Cylinder
Find the surface area of the right cylinder.
Ex. 3: Finding the Surface Area of a Cylinder
Find the surface area of the right cylinder.
Ex. 4: Finding the height of a cylinder
• Find the height of a cylinder which
has a radius of 6.5 centimeters
and a surface area of 592.19
square centimeters.
Ex. 4: Finding the height of a cylinder
• Find the height of a cylinder which
has a radius of 6.5 centimeters
and a surface area of 592.19
square centimeters.
Assignment Pages 611-614 (1, 2, 5, 6, 8, 9, 11, 13, 22,
23, 38-40)
Exit Quiz – 4 Points
Mario’s company makes unusually
shaped imitation gemstones. One
gemstone had 10 faces and 12 vertices.
How many edges did the gemstone
have?
a. 20 edges b. 23 edges
c. 25 edges d. 22 edges
Formula: F + V = E + 2