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# ππ ππ’πππ = 50
# ππ ππ’ππππ ππ’π‘ = 8
Boys : Girls4:5
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Determining Surface Area of Three-
Dimensional Figures Module 5 Lessons 18
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Objective 6.G.A.4
SWBAT represent three-dimensional figures
using nets made up of rectangles and triangles
and use the nets to find the surface area of
these figures IOT solve real world and
mathematical problems.
recognize or discover
something that could happen in reality
relating to math
find an answer to
a polygon with three angles and three sides
a quadrilateral with four right angles and two pairs of opposite equal parallel sides
having three dimensions; length, width, and height
a flat shape which can be folded into a three-dimensional solid
the total area of the faces of a three-dimensional solid
be a symbol for
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6.G.A.4 Word Wallβ’ Dimension - directions that an object
can be measured
β’ Net - a flat shape which can be folded into a three-dimensional solid
β’ Three-dimensional - having three dimensions; length, width, and height
β’ Triangle - a polygon with three angles and three sides
β’ Rectangle - a quadrilateral with four right angles and two pairs of opposite equal parallel sides
β’ Solve - to apply an operation(s) in order to find a value
β’ Surface area - the total area of the faces of a three-dimensional solid
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Rectangular Prism
4ππ Γ 1ππ = 4ππ2
4ππ Γ 1ππ = 4ππ2
4ππ Γ 2ππ = 8ππ2
4ππ Γ 2ππ = 8ππ2
2ππ
Γ1ππ
=2ππ
2
2ππ
Γ1ππ
=2ππ
2
To determine surface area, we found the area of each of the faces and then added those areas
ππ΄ = 2 4ππ Γ 1ππ + 2 4ππ Γ 2ππ + 2(2ππ Γ 1ππ)
Each part of the expression represents an area of one face of the given figure. We were able to write a more compacted form because there are three pairs of two faces that are identical.
ππ΄ = 2 4ππ Γ 1ππ + 2 4ππ Γ 2ππ + 2(2ππ Γ 1ππ)
= 2 4ππ2 + 2 8ππ2 + 2(2ππ2)
= 8ππ2 + 16ππ2 + (4ππ2)
= 28ππ2
WE DO
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4 ππ Γ 2 ππ 4 ππ Γ 2 ππ 4 ππ Γ 1 ππ 4 ππ Γ 1 ππ 2 ππ Γ 1 ππ 2 ππ Γ 1 ππ
8ππ2 2ππ22ππ24ππ24ππ28ππ2
π Γ π€ π€ Γ βπ€ Γ βπ Γ βπ Γ βπ Γ π€
ππ΄ = π Γ π€ + π Γ π€ + π Γ β + π Γ β + π€ Γ β + π€ Γ β
ππ΄ = 2(π Γ π€) + 2(π Γ β) + 2(π€ Γ β)Length
Width
Height
I DO
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15ππ Γ 6ππ 15ππ Γ 6ππ 15ππ Γ 8ππ 15ππ Γ 8ππ 6ππ Γ 8ππ 6ππ Γ 8ππ
90ππ2 48ππ248ππ2120ππ2120ππ290ππ2
π Γ π€ π€ Γ βπ€ Γ βπ Γ βπ Γ βπ Γ π€
Length
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Height
WE DO
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ππ΄ = 2(π Γ π€) + 2(π Γ β) + 2(π€ Γ β)
ππ΄ = 2(20ππ Γ 5ππ) + 2(20ππ Γ 9ππ) + 2(5ππ Γ 9ππ)
ππ΄ = 2(100ππ2) + 2(180ππ2) + 2(45ππ2)
ππ΄ = 200ππ2 + 360ππ2 + 90ππ2
ππ΄ = 650ππ2
Length
Width
HeightI DO
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ππ΄ = 2(π Γ π€) + 2(π Γ β) + 2(π€ Γ β)
ππ΄ = 2(12ππ Γ 2ππ) + 2(12ππ Γ 3ππ) + 2(2ππ Γ 3ππ)
ππ΄ = 2(24ππ2) + 2(36ππ2) + 2(6ππ2)
ππ΄ = 48ππ2 + 72ππ2 + 12ππ2
ππ΄ = 132ππ2
Length
Width
Height
WE DO
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YOU DO
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ππ΄ = 2(π Γ π€) + 2(π Γ β) + 2(π€ Γ β)
ππ΄ = 2(8π Γ 6π) + 2(8π Γ 22π) + 2(6π Γ 22π)
ππ΄ = 2(48π2) + 2(176π2) + 2(132π2)
ππ΄ = 96π2 + 352π2 + 264π2
ππ΄ = 712π2
Length
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ππ΄ = 2(π Γ π€) + 2(π Γ β) + 2(π€ Γ β)
ππ΄ = 2(29ππ‘ Γ 16ππ‘) + 2(29ππ‘ Γ 23ππ‘) + 2(16ππ‘ Γ 23ππ‘)
ππ΄ = 2(464ππ‘2) + 2(667ππ‘2) + 2(368ππ‘2)
ππ΄ = 928ππ‘2 + 1334ππ‘2 + 736ππ‘2
ππ΄ = 2998ππ‘2
LengthWidth
Height
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ππ΄ = 2(π Γ π€) + 2(π Γ β) + 2(π€ Γ β)
ππ΄ = 2(4ππ Γ 1.2ππ) + 2(4ππ Γ 2.8ππ) + 2(1.2ππ Γ 2.8ππ)
ππ΄ = 2(4.8ππ2) + 2(11.2ππ2) + 2(3.36ππ2)
ππ΄ = 9.6ππ2 + 22.4ππ2 + 6.72ππ2
ππ΄ = 38.72ππ2
Length
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