Notes Over 10 - 5

Surface Area

lw

hl h

l h

l w

l w

h wh w

hwhwlwlwlhlh Area hwlwlh 2Area

Surface Area

Surface Area

r Area of circle = r2

Area of circle = r2

Area of rectangle = l x w where the length is the circumference of the

circle and the width is the height. A = d h

or 2rh. .

h

SA = 2rh + 2r2

Surface AreaFind the surface area of the following figures.

1.

5 cm

12 cm

1 cm

hwlwlh 2Area

2A

2A

2A 2cm 154A

5 1 5 12 1 12

5 60 12

77


5 yd3 yd

2.

SA = 2rh + 2r2

SA = 2( )( ) + 225 3 5

SA = ( )( ) + ( )30 3.14 3.1450

SA = +94.2 157 = 251.2 yd 2


3 ft

4 ft

8 ft5 ft

3. The two right triangles will form

a rectangle

3

4

R1 = 3 ( 8 ) = 24R2 = 5 ( 8 ) = 40

R3 = 4 ( 8 ) = 32

SA = + + + = 108 ft 2 12 24 40 32

Pg 530 10 – 5 #1 – 7, 8 – 19 For the space figure represented by each net, find

the surface area to the nearest square unit.

1.

35 m

7 m7 m

7 m

7 m

7 m

hwlwlh 2Area

2A

2A

2A 2m 1078A

7 7 7 35 7 35

49 245 245

539



2. 12 yd

10 yd

37.7 yd

SA = lw + 2r2

SA = ( )( ) + 2210 37.7 6

SA = ( ) + ( )377 3.1472

SA = + 377 226 = 603 yd 2



3.

30 m

6 m9 m

6 m

9 m

6 m

hwlwlh 2Area

2A

2A

2A 2m 1008A

6 9 6 30 9 30

54 180 270

504

Pg 530 10 – 5 #1 – 7, 8 – 19 Find the surface area for each space figure. If the

answer is not a whole number round to the nearest tenth.

4. 4.8 m

4.6 m

SA = 2rh + 2r2

SA = 2( )( ) + 22

2.4 4.6 2.4SA = ( )( ) +

( )22.08 3.14 3.145.76

SA = + 69.3312 18.0864 = 87.4 m 2



5.

16 mm

12 mm

hwlwlh 2Area

2A 12 12 12 16 12 16

2A

2A 2mm 1056A

144 192 192

528



3 in

4 in

7 in5 in

6. The two right triangles will form

a rectangle

3

4

R1 = 3 ( 7 ) = 21R2 = 5 ( 7 ) = 35

R3 = 4 ( 7 ) = 28

SA = + + + = 96 in 2 12 21 35 28

Pg 530 10 – 5 #1 – 7, 8 – 19 7. The base of a rectangular prism is 3 in. by 5 in., and

the height is 11 in. Draw and label a net for the prism. Find its surface area.

11 in

3 in

5 in

5 m 3 in

hwlwlh 2Area

2A 5 3 5 11 3 11

2A

2A 2in 206A

15 55 33

103

Surface Area

bb

l

lArea of Triangles =

bl

2

14

Area of Base = 2bSA

bl

2

14 2b

Surface Area

Cone

.Area = r

Area = r 2

SA = r r 2

Surface Area

Sphere

SA = 4 2 r

2 2

14 SA

Surface AreaFind the surface area of each figure.

1.

5 m

8 m

5 8 5

80 + 25

= 105 m2

2

2

14 SA bbl

SA


2.

3 m

7 m

2 3.14 3.143 7 3

3.14 3.1421 9

65.94 + 28.26

94.2 m2

SA

2 SA rr


3. 3,963 mi.

2 4 3.14 3,963

12.56 15,705,369

197,259,434.6 mi2

SA 2 4 r SA

Pg 536 10 – 6 #1 – 6, 7 – 17 Find the surface area of each space figure, to the nearest

square unit.

30 cm

20 cm1. SA = r r 2

2 3.14 3.1410 30 10

3.14 3.14300 100

942 + 314

1256 cm2

SA


square unit.

2.

9 cm

2 4 3.14 9

SA 2 4 r SA

12.56 81

1,017 cm2


square unit.

3.5.5 m

4 m

2 2

14 SA

4 5.5 4

44 + 16

= 60 m2

2

2

14 SA bbl

SA


square unit.

4. The base of a cone has radius 3 ft. Its slant height is 8 ft. Find the surface area of the cone.

SA = r r 2

2 3.14 3.143 8 3

3.14 3.1424 9

75.36 + 28.26

104 ft2

SA


square unit.

5. The length of the base of a square pyramid is 5 cm. Its slant height is 8 cm. Find the surface area of the square pyramid.

2 2

14 SA

5 8 5

80 + 25

= 105 cm2

2

2

14 SA bbl

SA

Pg 536 10 – 6 #1 – 6, 7 – 17 6. Which has the greater surface area, a cylinder with

height 2 in. and radius of base 2 in., or a sphere with radius 2 in.? Justify your answer.

2 in

2 in2 in

SA = 2rh + 2r2

SA = 2( )( ) + 222 2 2

SA = ( )( ) + ( )

8 3.14 3.14 8

SA = 25.12 + 25.12 = 50 in2

2 4 3.14 2 SA 2 4 r

SA

12.56 450 in2

Pg 536 10 – 6 #1 – 6, 7 – 17

Pg 530 10 – 5 #1 – 7, 8 – 19

Notes Over 10 - 5

Documents

Transcript of Notes Over 10 - 5