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AREA

SURFACE AREA

Three dimensional figures

Chap 12

Geometry

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Surface area of 3-Dimensional figures NAME _____________________ DATE ___________ Period ____ TYPE FIGURE SURFACE AREA

FORMULA

TERMS I NEED TO KNOW

PRISM

CYLINDER

PYRAMID

CONE

SPHERE

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NOTES: 12-2 Surface area and nets Net – pattern for a three-dimensional solid that is flat. Look at the net for the cube on the homework worksheet. If you were to cut it out and fold it along the lines drawn, you would have a cube. There is more than one net for some solids. See if you can figure the nets for the cube on the homework sheet. If you can picture a solid as a flat net, you can find the surface area – that is the sum of all the areas of the faces of the solid. Look at page 645 in book – the triangular prism can be flattened to get the net that has 2 triangles and 3 rectangles. You can use the area formulas you know to find the area of each piece. We use rectangular dot paper to draw a net since each face can be represented by the squares of the rectangular dot paper. When 3-dimensional figures are modeled using isometric dot paper, we are not representing a net, but rather the 3-dimensional view of the figure.

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EX 1: Which net could be folded into a triangular prism? (answer is D)

EX 2: Draw a net for the triangular prism. Find the surface area of the prism.

(answer) Surface area (SA) = 2 triangles + 3 rectangles SA = ½ (4)(1.5) + ½ (4)(1.5) + (4)(6) + (2.5)(6) + (2.5)(6) =3+3+24+15+15 = 60 cm2

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12-2 Surface Area DATE __________

Sketch each solid using isometric dot paper. 1. 2. 3. For each solid, draw a net and find the surface area.

4. SHIPPING: Ronald needs to wrap a package to ship to his aunt. The rectangular package measures 2 inches high, 10 inches long, and 4 inches wide. Draw a net of the package. How much wrapping paper does Ronald need to cover the package?

Rectangular prism 3 units high, 3 units long, and 2 units wide

Triangular prism 3 units high, whose bases are right triangles with legs 2 units and 4 units long.

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5. Draw a net for each solid and find its surface area.

6. Double the dimensions of each figure. Find the surface areas. 7. How does the surface area change when the dimensions are doubled? Explain. 8. Make a conjecture about the surface area of a solid whose dimensions have been tripled. (Check your conjecture by finding the surface area of one of the solids above) 9.

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NOTES: 12-3 prism surface area 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.

The rectangular prism above is a right prism since the faces are perpendicular to the base. If the lateral edges are not perpendicular to the bases, it is an oblique prism. Two formulas we will use are: L = Ph L (lateral area), P (base perimeter), h (prism height) T = L + 2B T (total or surface area), L (lateral area), B (base area)

lateral face (4)

Base (2)

lateral edge

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EX 1: Find the lateral area of the regular hexagonal prism. 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 1b: Now find the surface area of the hexagonal prism We do this by adding the 2 bases’ areas to the lateral area T = L + 2B where T is the total surface area (sometimes referred to as SA) L is the lateral area and B is the base area. The base is a regular hexagon, so just as we found the area before A = ½ asn. a = 2.5 3 , s = 5, n = 6 A = 65 T = 360 + 2 (65) = 490 sq feet

5 cm 12 cm

Since the lateral faces are 6 rectangles with length 12 cm and width 5 cm we can find the lateral area by finding one rectangle (5*12) and then multiply that by 6 since there are 6 rectangles. L = 6*(5*12) = 360 cm2 OR We can find the perimeter of the regular hexagon and multiply it by the length dimension since all rectangles will be equal. P = (5*6) = 30 L = 30 * 12 = 360 xm2

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EX 2: A solid block of marble will be used for a sculpture. If the block is 3 feet wide, 4 feet long, and 9 ½ feet high, find the surface area of the block. Answer: The surface area would be T = L+2B To find L: the perimeter would be 2(3) + 2(4) or 14 feet With height 9 ½, the lateral area would be L = Ph or L = (14)(9.5) = 133 square feet The base area B would be (rectangular) B = lw or (3) (4) = 12 So the surface area T = (133)+ (2)(12) 157 square feet

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12-3 Prism DATE __________ Find the lateral area of each prism.

Find the surface area of each prism. Round answers to nearest tenth.

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12-3 Prism A DATE __________ Find the lateral area of each prism. Round to the nearest tenth.

Find the surface area of each prism. Round to the nearest tenth.

9. Lauren made a rectangular jewelry box in her art class and plans to cover it in red silk. If the jewelry box is 6 ½ inches long, 4 ½ inches wide, and 3 inches high, find the surface area that will be covered.

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NET EXPLORATION: NAME _________________ The net below can be folded into a cube. Do you see it? In your mind, try to figure out how it happens.

There are exactly 11 nets that will form a cube. Which ones do you think will form a cube?

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Answer the following questions. What properties are common to all nets that will form a cube? What type of nets will not work? Why not? Without folding, is there a quick way to determine whether or not a net will fold into a cube? How can you determine if two nets are identical? What sort of properties does your final cube have? How do these compare to the properties of the nets? Challenge problem: (choose one) The ACME box company wants to make these jewelry boxes as efficiently as possible. They can save money by fitting as many nets as possible on one piece of cardboard. If the company uses a piece of cardboard that measures 20cm x 20cm, how many nets (of any type) can you arrange to fit on one piece of cardboard? You may use any of the working net designs you created and you may arrange them in any way on your piece of cardboard. Draw the net for a typical cereal box. Draw a sketch, and then cut it out and fold it. See if you can design a net for other boxes you have seen. Draw a net on an 8.5 x 11 piece of paper that will result in the largest cube possible. Which net will you use? What is its volume?

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NOTES: 12-4 CYLINDER surface area 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.

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 = 2ππππrh T = L + 2B T (total or surface area), L (lateral area), B (base area) and since base is a circle the total surface area becomes: T = 2ππππrh + 2ππππr 2

Axis Base (2)

Altitude

Base (2)

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EX 1: A fruit juice can is cylindrical with aluminum sides and bases. The can is 12 centimeters tall, and the diameter of the can is 6.3 cm. How many square centimeters of aluminum are used to make the sides of the can (lateral area)? Lateral area: L = C h or L = 2πrh = 2π(3.15)(12) = 237.5 cm2

12 cm

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

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EX 2: Find the surface area of the cylinder: Surface area is lateral area plus 2 base areas L = C h or L = 2ππππrh = 2ππππ(14)(18) = 1582.56 B = ππππr 2 B = ππππ(14)2 = 615.44 Surface area = L + 2B = 1582.56 + 2 (615.44) = 2813.44 ft2

14 ft

18 ft

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Example 3: Find the radius of the base of a right cylinder if the surface area is 528ππππ square feet and the height is 10 feet. Use the formula for surface area: T = L + 2B T = 2ππππrh + 2(ππππr 2) 528ππππ = 2ππππr(10) + 2(ππππr 2) 528ππππ = 20ππππr + 2ππππr 2 0 = 2ππππr 2 + 20ππππr - 528ππππ Factor and solve for r 2ππππr 2 + 20ππππr - 528ππππ 2ππππ (r2 + 10r – 264) 2ππππ (r + 22) (r – 12) so r = -22 or r = 12 The radius is 12 feet

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12-4 Cylinders DATE __________ Find the surface area of a cylinder with the given dimensions. Round to the nearest tenth.

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.

Find the radius of the base of each cylinder:

9. Surface area is 603.2 square meters, height is 10 meters

10. Surface area is 100.5 square inches, height is 6 inches

11. Surface area is 226.2 square centimeters, and the height is 5 cm.

12. Surface area is 1520.5 square yards, height is 14.2 yards.

7 ft

5 ft

8.5 m

4 m

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12-4 Cylinder A DATE __________ Find the surface area of a cylinder with the given dimensions. Round to the nearest tenth.

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

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

Find the surface area of each cylinder shown. 5. 6.

Find the radius of the base of each cylinder in 7 –10:

7. Surface area is 603.2 square meters, height is 10 meters

8. Surface area is 100.5 square inches, height is 6 inches

9. Surface area is 226.2 square centimeters, and the height is 5 cm.

10. Surface area is 1520.5 square yards, height is 14.2 yards.

11. Some pencils are cylindrical and others are hexagonal prisms. If the diameter of the cylinder is the same length as the longest diagonal of the hexagon, which has the greater surface area? Explain. Assume that each pencil is 11 inches long and unsharpened.

7 ft

5 ft

8.5 m

4 m

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NOTES: 12-5 Pyramid surface area

Surface area formula we will use: T (or SA) = ½ P l + B

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

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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 = 6*4 = 24 l = 22 B = 6*6 = 36 SA = ½ (24) (22) (36) = 9504 cm2 Ex 2: Find the surface area of the regular pyramid to the nearest tenth.

SA = ½ P l + B P = 8*4 = 32 to find l we need to use Pythagorean theorem with legs 4 and 6 l2 = 42 + 62 l2 = 16+36

l = 52 or 7.2 B = 8*8 = 64 SA = ½ (32)(7.2) + (64) = 179.2 m2

4 m 8 m

8 m

6 m

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

SA = ½ P l + B Leg of triangle formed by slant height 15 and altitude 12 is (152-122 = 9) With a = 9 and angles of the isosceles triangle 60 degrees, we have isosceles triangles on the base such that : So the short leg (half the side length) is 9/3or 5.2. This means the side length is 10.4 cm P = (10.4)*(6) = 62.4 l = 15 B = ½ asn = ½ (9)(10.4)(6) = 280.8 SA = ½ (62.4)(15) + (280.8) = 748.8 cm2

12 cm

15 cm

12 15

9 60

sl

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12-5 Pyramid DATE __________ Find the surface area of each pyramid. Round to the nearest tenth.

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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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12-5 Pyramid II DATE __________ Find the surface area of each pyramid. Round to the nearest tenth.

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NOTES: 12-6 Cones surface area

formulas we will use: L = ππππ r l T or SA = ππππ r l +ππππ 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 = ππππ r l with r = 1.25 (half the diameter) l2 = 82 + 1.252 l2 = 65.5625 l = 8.1 L = ππππ (1.25)(8.1) = ππππ (10.125) = 31.8 in2

r

l

SA = ππππ r l +ππππ r2 Where r is radius of base l is slant height

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1.25

l

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EX 2: Find the surface area of the cone.

SA = ππππ r l +ππππ r2

r = 1.4 L = 3.2 SA = ππππ (1.4 )(3.2) + ππππ (1.4)2 = 4.48 ππππ + 1.96 ππππ = 6.44 ππππ = 20.2 cm2

1.4 cm

3.2 cm

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12-6 CONES DATE __________ Find the surface area of each cone. Round to the nearest tenth.

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12-6 b MIXED DATE __________

1. Name the faces of the solid

2. Identify the solid.

3. How many faces does a dodecahedron have?

4. Draw a net for the solid.

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5. Find the surface area of the solid.

6. Find the lateral area of a triangular prism with a height of 8 centimeters, and

with bases having sides that measure 4 centimeters, 5 centimeters, and 6 centimeters.

7. Find the surface area of the prism.

8. Find the surface area of the solid.

9. A gallon of paint costs $12.99 and covers 400 square feet. How many gallons are needed to paint two coats on the walls and ceiling (not the floor) or a rectangular room that is 30 feet long, 18 feet wide, and 8 feet high? Round to the next whole gallon.

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NOTES: 12-7 Spheres surface area 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

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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: T or SA = 4 ππππ r2

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

Since the area of the great circle includes the radius of the sphere: A = ππππ r2 907.9 = ππππ r2 289 = r2 17 = r so: SA = 4 ππππ 172 = 3631.6 cm2 b. Find the surface area of the hemisphere with a radius of 3.8 in SA = 4 ππππ r2 So the hemisphere would be

half the SA + the area of the great circle SA (hemisphere) = (2 ππππ 3.82 ) + (ππππ 3.82) = 90.7 + 45.4 = 136.1 in2

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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. SA = 4 ππππ r2 With circumference 24 24 = 2 ππππ r 3.82 = r so SA = 4 ππππ 3.822 = 183.3 in2

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12-7 SPHERES DATE __________

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.

2. If AE = 7 and DE = 15, find AD

3. If the radius of the sphere is 18 units and the radius of circle E is 17 units, find AE.

4. If the radius of the sphere is 10 units and the radius of circle E is 9 units, find AE.

5. If M is a point on circle E and AD = 23, find AM.

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Find the surface area of each sphere or hemisphere. Round to the nearest tenth.

8. A hemisphere with a radius of the great circle 8 yards. 9. A hemisphere with a radius of the great circle 2.5 millimeters. 10. A sphere with the area of a great circle 28.6 inches.

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