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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.
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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 r2 Area of circle = p r2 Area of rectangle = l x w where the length is the circumference of the circle and the width is the height. A = d p h or 2prh . . Notes Over 10 - 5 Surface Area h SA = 2prh + 2pr2
1. 1 cm 12 cm 5 cm Notes Over 10 - 5 Surface Area Find the surface area of the following figures. 5 1 5 12 1 12 5 60 12 77
2. 3 yd 5 yd Notes Over 10 - 5 Surface Area Find the surface area of the following figures. SA = 2prh + 2pr2 SA = 2p( )( ) + 2p( )2 5 3 5 SA = ( )( ) + ( )( ) 30 3.14 50 3.14 SA = + 94.2 157 = 251.2 yd 2
The two right triangles will form a rectangle 3. 5 ft 8 ft 3 3 ft 4 4 ft Notes Over 10 - 5 Surface Area Find the surface area of the following figures. R2 = 5 ( 8 ) = 40 R1 = 3 ( 8 ) = 24 R3 = 4 ( 8 ) = 32 = 108 ft2 SA = + + + 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. 7 m 1. 7 m 7 m 49 245 245 7 m 539 7 m 35 m 7 7 7 35 7 35
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 + 2pr2 SA = ( )( ) + 2p( )2 10 37.7 6 SA = ( ) + ( )( ) 377 72 3.14 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. 6 m 3. 9 m 6 m 54 180 270 9 m 504 6 m 30 m 6 9 6 30 9 30
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 = 2prh + 2pr2 SA = 2p( )( ) + 2p( )2 2.4 4.6 2.4 SA = ( )( ) + ( )( ) 22.08 3.14 5.76 3.14 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. 144 192 192 16 mm 528 12 mm 12 12 12 16 12 16
Pg 530 10 – 5 #1 – 7, 8 – 19 The two right triangles will form a rectangle 6. 5 in 7 in 3 3 in 4 4 in Find the surface area for each space figure. If the answer is not a whole number round to the nearest tenth. R2 = 5 ( 7 ) = 35 R1 = 3 ( 7 ) = 21 R3 = 4 ( 7 ) = 28 = 96 in2 SA = + + + 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. 15 55 33 103 11 in 3 in 5 m 3 in 5 in 5 3 5 11 3 11
l b b l Notes Over 10 - 6 Surface Area
Cone Notes Over 10 - 6 Surface Area .
Sphere Notes Over 10 - 6 Surface Area
1. 8 m 5 m Notes Over 10 - 6 Surface Area Find the surface area of each figure. 5 8 5 80 + 25 = 105 m2
2. 7 m 3 m Notes Over 10 - 6 Surface Area Find the surface area of each figure. 3.14 3 7 3.14 3 3.14 21 3.14 9 65.94 + 28.26 94.2 m2
3,963 mi. Notes Over 10 - 6 Surface Area Find the surface area of each figure. 3. 3.14 3,963 12.56 15,705,369 197,259,434.6 mi2
Pg 536 10 – 6 #1 – 6, 7 – 17 Find the surface area of each space figure, to the nearest square unit. 1. 20 cm 3.14 10 30 3.14 10 30 cm 3.14 300 3.14 100 942 + 314 1256 cm2
Pg 536 10 – 6 #1 – 6, 7 – 17 Find the surface area of each space figure, to the nearest square unit. 9 cm 2. 3.14 9 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. 5.5 m 3. 4 m 4 5.5 4 44 + 16 = 60 m2
Pg 536 10 – 6 #1 – 6, 7 – 17 Find the surface area of each space figure, to the nearest square unit. • The base of a cone has radius 3 ft. Its slant height is 8 ft. Find the surface area of the cone. 3 8 3.14 3 3.14 3.14 24 3.14 9 75.36 + 28.26 104 ft2
Pg 536 10 – 6 #1 – 6, 7 – 17 Find the surface area of each space figure, to the nearest square unit. • 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. 8 5 5 80 + 25 = 105 cm2
Pg 536 10 – 6 #1 – 6, 7 – 17 • 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. 3.14 2 2 in 2 in 2 in SA = 2prh + 2pr2 SA = 2p( )( ) + 2p( )2 2 2 2 4 12.56 SA = ( )( ) + ( )( ) 8 3.14 8 3.14 50 in2 SA = 25.12 + 25.12 = 50 in 2