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# šš šš¢ššš = 50
# šš šš¢šššš šš¢š” = 8
Boys : Girls4:5
148
Determining Surface Area of Three-
Dimensional Figures Module 5 Lessons 18
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
6.G.A.4 Word Wallā¢ Dimension - directions that an object
can be measured
ā¢ Net - a flat shape which can be folded into a three-dimensional solid
ā¢ Three-dimensional - having three dimensions; length, width, and height
ā¢ Triangle - a polygon with three angles and three sides
ā¢ Rectangle - a quadrilateral with four right angles and two pairs of opposite equal parallel sides
ā¢ Solve - to apply an operation(s) in order to find a value
ā¢ Surface area - the total area of the faces of a three-dimensional solid
Rectangular Prism
4šš Ć 1šš = 4šš2
4šš Ć 1šš = 4šš2
4šš Ć 2šš = 8šš2
4šš Ć 2šš = 8šš2
2šš
Ć1šš
=2šš
2
2šš
Ć1šš
=2šš
2
To determine surface area, we found the area of each of the faces and then added those areas
šš“ = 2 4šš Ć 1šš + 2 4šš Ć 2šš + 2(2šš Ć 1šš)
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.
šš“ = 2 4šš Ć 1šš + 2 4šš Ć 2šš + 2(2šš Ć 1šš)
= 2 4šš2 + 2 8šš2 + 2(2šš2)
= 8šš2 + 16šš2 + (4šš2)
= 28šš2
WE DO
4 šš Ć 2 šš 4 šš Ć 2 šš 4 šš Ć 1 šš 4 šš Ć 1 šš 2 šš Ć 1 šš 2 šš Ć 1 šš
8šš2 2šš22šš24šš24šš28šš2
š Ć š¤ š¤ Ć āš¤ Ć āš Ć āš Ć āš Ć š¤
šš“ = š Ć š¤ + š Ć š¤ + š Ć ā + š Ć ā + š¤ Ć ā + š¤ Ć ā
šš“ = 2(š Ć š¤) + 2(š Ć ā) + 2(š¤ Ć ā)Length
Width
Height
I DO
15šš Ć 6šš 15šš Ć 6šš 15šš Ć 8šš 15šš Ć 8šš 6šš Ć 8šš 6šš Ć 8šš
90šš2 48šš248šš2120šš2120šš290šš2
š Ć š¤ š¤ Ć āš¤ Ć āš Ć āš Ć āš Ć š¤
Length
Width
Height
WE DO
šš“ = 2(š Ć š¤) + 2(š Ć ā) + 2(š¤ Ć ā)
šš“ = 2(20šš Ć 5šš) + 2(20šš Ć 9šš) + 2(5šš Ć 9šš)
šš“ = 2(100šš2) + 2(180šš2) + 2(45šš2)
šš“ = 200šš2 + 360šš2 + 90šš2
šš“ = 650šš2
Length
Width
HeightI DO
šš“ = 2(š Ć š¤) + 2(š Ć ā) + 2(š¤ Ć ā)
šš“ = 2(12šš Ć 2šš) + 2(12šš Ć 3šš) + 2(2šš Ć 3šš)
šš“ = 2(24šš2) + 2(36šš2) + 2(6šš2)
šš“ = 48šš2 + 72šš2 + 12šš2
šš“ = 132šš2
Length
Width
Height
WE DO
YOU DO
šš“ = 2(š Ć š¤) + 2(š Ć ā) + 2(š¤ Ć ā)
šš“ = 2(8š Ć 6š) + 2(8š Ć 22š) + 2(6š Ć 22š)
šš“ = 2(48š2) + 2(176š2) + 2(132š2)
šš“ = 96š2 + 352š2 + 264š2
šš“ = 712š2
Length
Width
Height
šš“ = 2(š Ć š¤) + 2(š Ć ā) + 2(š¤ Ć ā)
šš“ = 2(29šš” Ć 16šš”) + 2(29šš” Ć 23šš”) + 2(16šš” Ć 23šš”)
šš“ = 2(464šš”2) + 2(667šš”2) + 2(368šš”2)
šš“ = 928šš”2 + 1334šš”2 + 736šš”2
šš“ = 2998šš”2
LengthWidth
Height
šš“ = 2(š Ć š¤) + 2(š Ć ā) + 2(š¤ Ć ā)
šš“ = 2(4šš Ć 1.2šš) + 2(4šš Ć 2.8šš) + 2(1.2šš Ć 2.8šš)
šš“ = 2(4.8šš2) + 2(11.2šš2) + 2(3.36šš2)
šš“ = 9.6šš2 + 22.4šš2 + 6.72šš2
šš“ = 38.72šš2
Length
Width
Height