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UNIT 10.3 SURFACE AREAS OF UNIT 10.3 SURFACE AREAS OF PYRAMIDS AND CONESPYRAMIDS AND CONES
Warm UpFind the missing side length of each right triangle with legs a and b and hypotenuse c.
1. a = 7, b = 24
2. c = 15, a = 9
3. b = 40, c = 41
4. a = 5, b = 5
5. a = 4, c = 8
c = 25
b = 12
a = 9
Learn and apply the formula for the surface area of a pyramid.Learn and apply the formula for the surface area of a cone.
Objectives
vertex of a pyramidregular pyramidslant height of a regular pyramidaltitude of a pyramidvertex of a coneaxis of a coneright coneoblique coneslant height of a right conealtitude of a cone
Vocabulary
The vertex of a pyramid is the point opposite the base of the pyramid. The base of a regular pyramid is a regular polygon, and the lateral faces are congruent isosceles triangles. The slant height of a regular pyramid is the distance from the vertex to the midpoint of an edge of the base. The altitude of a pyramid is the perpendicular segment from the vertex to the plane of the base.
The lateral faces of a regular pyramid can be arranged to cover half of a rectangle with a height equal to the slant height of the pyramid. The width of the rectangle is equal to the base perimeter of the pyramid.
Example 1A: Finding Lateral Area and Surface Area of Pyramids
Find the lateral area and surface area of a regular square pyramid with base edge length 14 cm and slant height 25 cm. Round to the nearest tenth, if necessary.
Lateral area of a regular pyramid
P = 4(14) = 56 cm
Surface area of a regular pyramid
B = 142 = 196 cm2
Example 1B: Finding Lateral Area and Surface Area of Pyramids
Step 1 Find the base perimeter and apothem.
Find the lateral area and surface area of the regular pyramid.
The base perimeter is 6(10) = 60 in.
The apothem is , so the base area is
Example 1B Continued
Step 2 Find the lateral area.
Lateral area of a regular pyramid
Substitute 60 for P and 16 for ℓ.
Find the lateral area and surface area of the regular pyramid.
Example 1B Continued
Step 3 Find the surface area.
Surface area of a regular pyramid
Substitute for B.
Find the lateral area and surface area of the regular pyramid.
Check It Out! Example 1
Find the lateral area and surface area of a regular triangular pyramid with base edge length 6 ft and slant height 10 ft.
Step 1 Find the base perimeter and apothem. The base perimeter is 3(6) = 18 ft.
The apothem is so the base area is
Check It Out! Example 1 Continued
Find the lateral area and surface area of a regular triangular pyramid with base edge length 6 ft and slant height 10 ft.
Step 2 Find the lateral area.
Lateral area of a regular pyramid
Substitute 18 for P and 10 for ℓ.
Step 3 Find the surface area.
Surface area of a regular pyramid
Check It Out! Example 1 Continued
Find the lateral area and surface area of a regular triangular pyramid with base edge length 6 ft and slant height 10 ft.
Substitute for B.
The vertex of a cone is the point opposite the base. The axis of a cone is the segment with endpoints at the vertex and the center of the base. The axis of a right cone is perpendicular to the base. The axis of an oblique cone is not perpendicular to the base.
The slant height of a right cone is the distance from the vertex of a right cone to a point on the edge of the base. The altitude of a cone is a perpendicularsegment from the vertex of the cone to the plane of the base.
Example 2A: Finding Lateral Area and Surface Area of Right Cones
Find the lateral area and surface area of a right cone with radius 9 cm and slant height 5 cm.
L = πrℓ Lateral area of a cone = π(9)(5) = 45π cm2 Substitute 9 for r and 5 for ℓ.
S = πrℓ + πr2 Surface area of a cone= 45π + π(9)2 = 126π cm2 Substitute 5 for ℓ and
9 for r.
Example 2B: Finding Lateral Area and Surface Area of Right Cones
Find the lateral area and surface area of the cone.
Use the Pythagorean Theorem to find ℓ.
L = πrℓ
= π(8)(17)
= 136π in2
Lateral area of a right cone
Substitute 8 for r and 17 for ℓ.
S = πrℓ + πr2 Surface area of a cone
= 136π + π(8)2
= 200π in2
Substitute 8 for r and 17 for ℓ.
Check It Out! Example 2
Find the lateral area and surface area of the right cone.
Use the Pythagorean Theorem to find ℓ.
L = πrℓ
= π(8)(10)
= 80π cm2
Lateral area of a right cone
Substitute 8 for r and 10 for ℓ.
S = πrℓ + πr2 Surface area of a cone
= 80π + π(8)2
= 144π cm2
Substitute 8 for r and 10 for ℓ.
ℓ
Example 3: Exploring Effects of Changing Dimensions
The base edge length and slant height of the regular hexagonal pyramid are both divided by 5. Describe the effect on the surface area.
3 in.
2 in.
Example 3 Continued
original dimensions: base edge length and slant height divided by 5:
S = Pℓ + B 12 S = Pℓ + B
12
Example 3 Continued
original dimensions: base edge length and slant height divided by 5:
3 in.
2 in.
Notice that . If the
base edge length and slant height are divided by 5,
the surface area is divided by 52, or 25.
Check It Out! Example 3
The base edge length and slant height of the regular square pyramid are both multiplied by . Describe the effect on the surface area.
Check It Out! Example 3 Continued
original dimensions: multiplied by two-thirds:
By multiplying the dimensions by two-thirds, the surface area was multiplied by .
8 ft8 ft
10 ft
S = Pℓ + B 12
= 260 cm2
S = Pℓ + B 12
= 585 cm2
Example 4: Finding Surface Area of Composite Three-Dimensional Figures
Find the surface area of the composite figure.
The lateral area of the cone isL = πrl = π(6)(12) = 72π in2.
Left-hand cone:
Right-hand cone:Using the Pythagorean Theorem, l = 10 in. The lateral area of the cone isL = πrl = π(6)(10) = 60π in2.
Example 4 Continued
Composite figure: S = (left cone lateral area) + (right cone lateral area)
Find the surface area of the composite figure.
= 60π in2 + 72π in2 = 132π in2
Check It Out! Example 4
Find the surface area of the composite figure.
Surface Area of Cube without the top side:
S = 4wh + BS = 4(2)(2) + (2)(2) = 20 yd2
Check It Out! Example 4 Continued
Surface Area of Pyramid without base:
Surface Area of Composite:
Surface of Composite = SA of Cube + SA of Pyramid
Example 5: Manufacturing Application
If the pattern shown is used to make a paper cup, what is the diameter of the cup?
The radius of the large circle used to create the pattern is the slant height of the cone.
The area of the pattern is the lateral area of the
cone. The area of the pattern is also of the area
of the large circle, so
Example 5 Continued
Substitute 4 for ℓ, the slant height of the cone and the radius of the large circle.
r = 2 in. Solve for r.
The diameter of the cone is 2(2) = 4 in.
If the pattern shown is used to make a paper cup, what is the diameter of the cup?
Check It Out! Example 5
What if…? If the radius of the large circle were 12 in., what would be the radius of the cone?
The radius of the large circle used to create the pattern is the slant height of the cone.
The area of the pattern is the lateral area of the
cone. The area of the pattern is also of the area of
the large circle, so
Check It Out! Example 5 Continued
Substitute 12 for ℓ, the slant height of the cone and the radius of the large circle.
r = 9 in. Solve for r.
The radius of the cone is 9 in.
What if…? If the radius of the large circle were 12 in., what would be the radius of the cone?
Lesson Quiz: Part I
Find the lateral area and surface area of each figure. Round to the nearest tenth, if necessary.
1. a regular square pyramid with base edge length 9 ft and slant height 12 ft
2. a regular triangular pyramid with base edge length 12 cm and slant height 10 cm
L = 216 ft2; S = 297 ft2
L = 180 cm2; S ≈ 242.4 cm2
4. A right cone has radius 3 and slant height 5. The radius and slant height are both multiplied by . Describe the effect on the surface area.
5. Find the surface area of the composite figure. Give your answer in terms of π.
The surface area is multiplied by .
S = 24π ft2
Lesson Quiz: Part II
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