Math Tech 1 Unit 11

Surface Area

Name ______________

Pd ________

11-8: Surface Area of Prisms

Recall that the prism is a solid figure where the 2 bases are congruent faces in parallel planes. Bases can be triangles, quadrilaterals, or any other polygon. The other faces are all parallelograms and are called lateral faces. The lateral faces intersect at the lateral edges (parallel segments) and the lateral area is found by taking the sum of all the areas of the lateral faces. lateral face Base (4)

(2)

lateral edge

The rectangular prism above is a right prism since the faces are perpendicular to the base. To find the surface area of a box you must add the area of all six sides.

• TOP

BOTTOM

TOP

RIGHT LEFT

FRONT BACK

• BOTTOM • LEFT SIDE • RIGHT SIDE • FRONT • BACK

5 in.

4 in.

3 in.

SA = TOP

BOTTOM

LEFT RIGHT FRONT BACK

Total Surface Area = If the lateral edges are not perpendicular to the bases, it is an oblique prism.

Surface area is usually broken into two categories

#1 Area of the Bases

• There are always two bases and they are always parallel to each other • They also have to be the same size and shape Formula: 2B B = Base Area (Triangle: ½ bh Rectangle: Lw)

#2 Lateral Area

• The lateral area is the area of the vertical sides that make up the prism • They do not have to be parallel to each other • They do not have to be the same size or shape

Formula: Ph P = perimeter of the base, h = height of the prism

Surface Area = Base Area + Lateral Area Two formulas we will use are: L = Ph L (lateral area), P (base perimeter), h (prism height) SA = Ph + 2B SA (total or surface area), L (lateral area), B (base area) EX 1: Find the lateral area of the regular hexagonal prism.

5 cm 12 cm

It is usually easier to find the perimeter of the base (P) and then multiply it by the height of the prism. Thus the formula for lateral area of a prism L = Ph where L is lateral area P is perimeter of the base H is the height of the prism.

EX 2: A solid block of marble will be used for a sculpture. If the block is 4 feet wide, 3 feet long, and 9 feet high, find the surface area of the block.

4 ft 3 ft

9 ft Answer: The surface area would be SA = Ph+2B To find Ph: The base area B would be (rectangular) B = lw So the surface area SA=

EX 3 Find the surface area of the triangular prism. 10 ft 8 ft 3 ft 6 ft Answer: The surface area would be SA = Ph+2B To find Ph: The base area B would be (triangle) B = ½ bh So the surface area SA =

11-8 Classwork Prism Find the lateral area of each prism. LA. = Ph

9.4

Find the surface area of each prism. Round answers to nearest tenth. SA = 2B + Ph

10.3

14.3

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11-8 Prism Find the lateral area of each prism. LA. = Ph

Find the surface area of each prism. Round answers to nearest tenth. SA = 2B + Ph

8.1 10.8

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11-9: Surface Area of Cylinders

Recall that the cylinder is a solid figure where the 2 bases are congruent parallel circles. The axis of a cylinder is a segment with endpoints that are centers of the circular bases. If the axis is perpendicular to the base, it is a right cylinder. Otherwise it is an oblique cylinder.

Axis Base (2)

Altitude

Base (2)

Right cylinder Oblique Cylinder Two formulas we will use are: L = C h L (lateral area), C (circumference of base), h (prism height)

or using the information we know about circumference: L = 2πrh SA= L + 2B SA (total or surface area), L (lateral area), B (base area) and

since base is a circle the total surface area becomes: SA = 2πrh + 2πr2

EX 1: A fruit juice can is cylindrical with aluminum sides and bases. The can is 10 centimeters tall, and the diameter of the can is 3 cm. How many square centimeters of aluminum are used to make the sides of the can (lateral area)?

3 cm.

10 cm.

NOTE : the lateral area is really a rectangle if you drew a net for the cylinder.

Surface area is lateral area plus 2 base areas L = 2πrh

Lateral area: L = 2πrh EX 2: Find the surface area of the cylinder:

14 ft

18 ft

B = πr2 B = Surface area = 2πrh + 2πr2

EX 3: Find the height of the base of a right cylinder if the surface area is 528π square feet and the radius is 12 ft. Use the formula for surface area: T = L + 2B T = 2πrh + 2(πr2)

11-9 Classwork Cylinders Find the surface area of a cylinder with the given dimensions. Round to the nearest tenth. SA = 2π r2 + 2π rh

1. r = 5 in, h = 11 in 2. r = 7 cm, h = 9 cm

3. d = 2 ft, h = 18 ft 4. d = 10 yd, h = 15 yd

Find the surface area of each cylinder shown.

5. 6. 6 ft

9.5 m

7 m 8 ft Find the height of each cylinder:

7. Surface area is 226.2 square centimeters, radius is 4 meters.

8. Surface area is 1520.5 square yards, radius is 10 yards.

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11-9 Cylinders Find the surface area of a cylinder with the given dimensions. Round to the nearest tenth. SA = 2π r2 + 2π rh

1. r = 10 in, h = 12 in 2. r = 8 cm, h = 15 cm

3. r = 5 ft, h = 20 ft 4. d = 20 yd, h = 5 yd

5. d = 8 m, h = 7 m 6. d = 24 mm, h = 20 mm

Find the surface area of each cylinder shown.

7. 8. 5 ft

8.5 m

4 m 7 ft Find the height of each cylinder:

9. Surface area is 603.2 square meters, radius is 6 meters

10. Surface area is 100.5 square inches, radius is 2 inches

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11-8 and 11-9 Homework Prism: SA = PH + 2B Cylinder : SA = 2π rh + 2π r2

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11-10: Surface Area of Pyramids

Pyramids have only one base and have triangles for sides that meet at a point at the top of the pyramid The lateral area is made up of the triangles Pyramids have two heights:

• One is the actual height (how tall it is) (used for finding the volume) • The other is a slant height (height of triangular sides) (used for finding surface area)

Pyramids Base Rectangle: B = LW Triangle: B = 1/2bh B

P perimeter of base l slant height – height of one triangle face B area of base

l

Surface area formula we will use: SA = ½ P l + B

EX 1: A candle store offers pyramidal candles that burn for 20 hours. The square base is 6 cm on a side and the slant height of the candle is 22 cm. Find the surface area of the candle. SA = ½ P l + B P = l = B = SA = Ex 2: Find the surface area of the regular pyramid to the nearest tenth. 6 m

4 m 8 m

8 m

Ex 3: Find the surface area of the regular pyramid to the nearest tenth.

12 cm

15 cm

s = 10.4 cm

B = 280.8 cm2

11-10 Classwork Pyramid Find the surface area of each pyramid. Round to the nearest tenth. SA = ½ P l + B

Base = 21.2 m2

Base = LW

Base = 43 ft2 Base= 26.2 cm2

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Base = 586.9 mm2

side = 13.1 mm Side = 9.6 in

Base = LW l = 12.2 m

Base = 23.4 yd2

l = 9.8 yd

9. The roof of a gazebo is a regular octagonal pyramid. If the base of the pyramid has sides of .5 meters and the slant height of the roof is 1.9 meters, find the area of the roof.

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11-10 Pyramid Find the surface area of each pyramid. Round to the nearest tenth. SA = ½ P l + B

Base = LW

Base = 166.3 in2

Base = LW L = 13.9 Base = 142.8 m2

l = 12.5

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Base = LW Base = 15.6 mm2

l = 10.4

Base = LW l2 = 122 + 92

Base = LW l 2= 202 + 82

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11-11: Surface Area of Cones

Cones have only one base that is a circle. The lateral area is made up of a part of another circle. Cones have two heights:

Slant Height• One is the actual height (how tall it is) (used for finding the volume) • The other is a slant height (used for finding surface area)

Altitude (Height)

SA = π r l +π r2 Where r is radius of base

r

l l is slant height

To find the surface area of a cone add the area of the base to the area of the side by using the formula below L = π r l SA = π r l +π r2

EX 1: A sugar cone has an altitude of 8 inches and a diameter of 2 ½ inches. Find the lateral area of the sugar cone. L = π r l r =

l2 =

1.25

l 8

L = EX 2: Find the surface area of the cone.

SA = π r l +π r2

r = 1.4 cm

L = SA = 3.2 cm

11-11 Classwork CONES Find the surface area of each cone. Round to the nearest tenth. SA = πr l + πr2

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11-11 CONES Find the surface area of each cone. Round to the nearest tenth. SA = πr l + πr2

l2 = 52 + 142 l2 = 252 + 102

l2= 182+72

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r2 = 152 - 122

r2 = 122 - 92

l2= 212+72

r2= 16.42 - 142

r2= 102 - 82

l2= 52+42

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11-10 and 11-11 Homework

1. Identify the solid.

2. Find the surface area of the solid. SA = ½ P l + B 3. Find the surface area of the solid. SA = ½ P l + B where B = 1/2bh

b = 12 h = 10.4

4. Find the surface area of the solid. SA = πr l + πr2

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5. Find the surface area of the solid. SA = πr l + πr2

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6. Find the surface area of the solid. SA = 2πr l + 2πrh

11-12: Surface Area of Spheres

Sphere is the set of all point that are a given distance from one point called the center. There are several special segments and lines that relate to spheres:

Radius DC , DA , DB

Chord FG , AB

Diameter AB

Tangent JH Great circle is formed when a plane intersects a sphere through a diameter (center of the sphere)

The great circle divides the sphere into 2 hemispheres. The formulas we will use: SA = 4 π r2 (sphere) SA = 2 π r2 + π r2 (hemisphere + area of the great circle)

EX 1: a. Find the surface area of a sphere, given a great circle with an area of 907.9 sq cm. SA = 4 π r2

Since the area of the great circle includes the radius of the sphere: A = π r2 so: SA = b. Find the surface area of the hemisphere with a radius of 3.8 in SA = 2 π r2 + π r2 So the hemisphere would be

half the SA + the area of the great circle SA (hemisphere) = SA (great circle) = EX 2: Find the surface area of a ball with circumference of 24 inches to determine how much leather is needed to make the ball. C = 2π r SA = 4 π r2

11-12 Classwork SPHERES In the figure, A is the center of the sphere, and plane T intersects the sphere in circle E. Round to the nearest tenth if necessary.

1. If AE = 5 and DE = 12, find AD. a2 + b2 = c2

2. If the radius (AD) of the sphere is 18 units and the radius (ED) of circle E is 17 units, find AE. a2 + b2 = c2

Find the surface area of each sphere or hemisphere. Round to the nearest tenth. SA = 4π r2

3. A hemisphere with a radius of the great circle 8 yards. 4. A hemisphere with a radius of the great circle 2.5 millimeters.

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5. A sphere with the area of a great circle 28.6 inches.

8. Find the surface area of a sphere with a radius of 8 ft.

9. Find the surface area of a sphere with a diameter of 24 cm.

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11-12 SPHERES

Find the surface area of each sphere given the radius or diameter of the sphere. 11. r = 8 cm 12. r = 2.3 ft 13. r = π in 14. d = 10 in 15. d = 16 yd 16. d = 8.4 cm

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17. Find the surface area of a hemisphere with a radius of 12 cm. 18. Find the surface area of a hemisphere with a diameter of 4π .

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REVIEW WORKSHEET FOR SURFACE AREA NAME___________________________

Name each of the figures A-L.

1. _________________________ 2. _________________________ 3. _________________________ 4. _________________________ 5. _________________________ 6. _________________________ 7. _________________________ 8. _________________________

9. _________________________ 10. _________________________ 11. _________________________ 12. _________________________

For each formula for surface area, identify what each letter represents.

13. rectangular prism SA= 2wh + 2wl + 2lh l = _________ w = __________ h = ___________

14. cylinder SA= 2π rh + 2π r2 r = _________ h = ___________

15. cone SA= π r2 + π r 22 hr + r = __________ h= ___________

or SA=π r2 + π r l l = _________

16. pyramid SA= ½ Pl + B P = ___________ l = _________ B = _______________

17. sphere SA= π d2 or 4π r2 r = ___________ d = ____________

18. what is the value of π ? ________________ - 20 -

Find the total surface area for each of the following figures.

19. 20. 21. 22. 23. 24.

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6 in.

4 ft

25. How many square inches of cardboard are needed to build a box that has of length

7 inches, a width of 10 inches, and a height of 8 inches? (show all work)

26. Use the drawing of the water tank to answer the following questions. (show all work)

a) What three separate geometric figures are used

to make the water tank in the drawing ?

b) What is the overall height of the tank?

c) How much metal would it take to make the top part of the tank?

d) How much metal would it take to make the middle part of the tank?

e) How much metal would it take to make the bottom part of the tank?

f) How much metal would it take to make the entire tank?

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