Download 12.5 Surface Area of Pyramids

 Find lateral areas of regular pyramids.
 Find surface areas of regular pyramids.
PYRAMIDS
All pyramids have the following characteristics.
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All of the faces, except the base, intersect at one point
called the vertex.
The base is always a polygon.
The faces that intersect at the vertex are called lateral
faces and form triangles. The edges of the lateral
faces that have the vertex as an endpoint are called
lateral edges.
The altitude is the segment from the vertex
perpendicular to the base.
Pyramids
 If the base of a pyramid is a regular polygon and
the segment with endpoints that are the center of
the base and the vertex is perpendicular to the base
is called a regular pyramid.
 An altitude is the segment with endpoints that are
at the center of the base and the vertex. All of the
lateral faces are congruent isosceles triangles. The
height of each lateral face is called the slant
height l of the pyramid.
Parts of a Pyramid
Vertex
Lateral Face
Lateral Edge
Altitude
Slant
Height
Base
Altitude
Square
Pyramid
Base
(a regular polygon)
Regular Square
Pyramid
Key Concepts

If a regular pyramid has a lateral area of L
square units, a slant height of l units, and its
base has a perimeter of P units, then L = 1/2P l.

If a regular pyramid has a surface area of T
square units, a slant height of l units, and its
base has a perimeter of P units and an area of B
square units, then T = 1/2P l + B.
Example 1: Use Lateral Area to
Solve Problem

A farmer is building a hexagonal
barn. The base of the roof has
sides of 15 feet, and the slant
height of the roof is 30 feet. The
farmer needs to know exactly
how much wood to use for the
roof. Find the amount of wood
used for the roof.
Example 1: Continued
L = ½P l
= ½(90)(30)
= 1350
Lateral area of a regular pyramid
P = 90, l = 30
Multiply
So, 1350 square feet of wood are used to
cover the roof of the barn.
Example 2:
Surface Area of a Square Pyramid
Find the surface area of the square
pyramid.
To find the surface area, first find the slant
height of the pyramid. The slant height is the
hypotenuse of a right triangle with legs that
are the altitude and a segment with a length
that is one-half the side measure of the base.
24 m
l
18 m
Example 2: Continued
c2 = a2 + b2
l2 = 92 + 242
l = √657
Pythagorean Theorem
a = 9, b =24, c = l
Simplify
Now find the surface area of a regular pyramid. The perimeter of
the base is 4(18) or 72 meters, and the area of the base is 182 or 324
square meters.
T=½Pl+B
Surface area of a regular pyramid
T = ½(72)√657 + 324 P = 72, l = √657, B = 324
T ≈ 1246.8
Use a calculator
Example 3: Surface Area of Pentagonal
Pyramid
Find the surface area of the regular pyramid.
The altitude, slant height, and apothem form a right triangle.
Use Pythagorean Theorem to find the apothem. Let a
represent the length of the apothem.
Example 3: Continued
c2 = a2+b2
(17)2 = a2=152
8=a
Pythagorean Theorem
b =15, c = 17
Simplify
Now find the length of the sides of the base. The central angle of the
pentagon measures 360º/5 or 72º. Let x represent the measure of the
angle formed by a radius and the apothem. Then, x = 72/2 or 36.
opposite
Tan 36 = ( ½ s)/8
tan x=adjacent
8(tan 36) = ½ s
16(tan 36) = s
11.6 ≈ s
Multiply each side by 8
Multiply each side by 2
Use a calculator
Example 3: Continued #2
Next, find the perimeter and area of the base.
P = 5s
≈ 5(11.6) or 58
B = ½Pa
≈ ½(58)(8) or 232
Finally, find the surface area.
T = ½Pl + B
≈ ½(58)(17)+232
≈ 726.5
Surface area of a regular pyramid
P ≈ 58, l =232 ,B ≈232
Simplify
The surface area is approximately 726.5 square inches
Pre-AP Geometry
Pg. 663 # 7-16, 18-23, 27