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h. w. l. Notes Over 10 - 5. Surface Area. l h. l w. h w. h w. l h. l w. Notes Over 10 - 5. Surface Area. r. Area of circle = p r 2. Area of circle = p r 2. - PowerPoint PPT Presentation
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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
Surface AreaFind the surface area of the following figures.
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
Surface AreaFind the surface area of the following figures.
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
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.
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
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.
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
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.
5.
16 mm
12 mm
hwlwlh 2Area
2A 12 12 12 16 12 16
2A
2A 2mm 1056A
144 192 192
528
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.
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
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
Surface AreaFind the surface area of each figure.
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
Surface AreaFind the surface area of each figure.
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
Pg 536 10 – 6 #1 – 6, 7 – 17 Find the surface area of each space figure, to the nearest
square unit.
2.
9 cm
2 4 3.14 9
SA 2 4 r SA
12.56 81
1,017 cm2
Pg 536 10 – 6 #1 – 6, 7 – 17 Find the surface area of each space figure, to the nearest
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
Pg 536 10 – 6 #1 – 6, 7 – 17 Find the surface area of each space figure, to the nearest
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
Pg 536 10 – 6 #1 – 6, 7 – 17 Find the surface area of each space figure, to the nearest
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
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