CCGPS Unit 5 Overview
Area and Volume
MCC6.G.1
Find the area of right
triangles, other
triangles, special
quadrilaterals, and
polygons by composing
into rectangles or
decomposing into
triangles and other
shapes; apply these
techniques in the context
of solving real-world and
mathematical problems.
Examples:
• Area of Right Triangles
• Area of Triangles
• Area of Squares
• Area of Kites
• Area of Parallelogram
• Area of Trapezoid
• Real-World Problems with
Area
Vocabulary Words
• Right Triangle
• Square
• Kite
• Parallelogram
• Trapezoid
• Area
A
L
T
I
T
U
D
E
Base
A triangle that has
◦
exactly one 90 angle
4 ft
Substitute the values into the equation.
• b = 3ft
• h = 4ft
We are solving for A.
3 ft
A = ½ (4 ft)×(3 ft)
A = ½ (12 ft2)
A = 6 ft2
8.4 ft
We are solving for A.
A = ½ (8.4 ft)×(6.2 ft)
A = ½ (52.08 ft2)
6.2 ft
A = 26.04 ft2
A
L
T
I
T
U
D
E
A polygon having three
sides.
Base
We are solving for A.
A = ½ (6 ft)×(4 ft)
4 ft
A = ½ (24 ft2)
6 ft
A = 12 ft2
9 ft
We are solving for A.
A = ½(6 ft)×(9 ft) + ½(6 ft)×(9 ft)
6 ft
A = ½(54 ft2) + ½(54 ft2)
A = 27 ft2 + 27 ft2
9 ft
A = 54 ft2
9 ft
We are solving for A.
6 ft
A = (6 ft.)×(9 ft.)
A = 54 ft2
4 ft.
4 ft.
A Rhombus is a four-sided Polygon where all sides have
equal length
(It looks like a someone sat on a square)
4 ft
We are solving for A.
3 ft
4 ft
A = (4 ft)×(3 ft)
A = 12 ft2
15 ft
We are solving for A.
A = (15 ft)×(10 ft)
10 ft
A = 150 ft2
We are solving for A.
A = (10 ft)×(10 ft)
10 ft.
10 ft.
A = 100 ft2
12 in.
Suppose you were asked to find the
area of this kite. Using what we
already know about triangles, how
can we find the area of the kite?
We are solving for A.
36 in. A = ½(24in×12 in) + ½(24 in×36 in)
A = ½(288 in2) + ½(864 in2)
A = 144 in2 + 432 in2
A = 576 in2
24 in.
How can we find the area
of this trapezoid?
12 in.
We are solving for A.
24 in.
12 in.
36 in. 12 in.
12 in.
A=
½(12 in× 24 in)
½(12 in× 24 in)
+ (12 in× 24 in)
A=
½(288 in2)
½(288 in2)
+ (288 in2)
A = 144 in2 + 144 in2 + 288 in2
A = 576 in2
Area = ½ Base x Height
A = ½ (9 feet) x (25 feet)
A = ½ (225 ft2)
A = 112.5 ft2
Mr. and Mrs. Brady purchased this home in Marietta, Georgia. Mrs. Brady is tired of looking at
the brown dirt patch in the median of the road in front of their home. She has asked her
husband to plant something in that area which will give it a more attractive curb appeal. Mr.
Brady knows that the soil will need to be treated with fertilizer in order for it to be able to grow
beautiful flowers. Mr. Brady needs to find the area of the ground to be treated. He has
provided the length of two sides that make a right angle for you to use to determine the area.
One side is twenty five feet. The other side is 9 feet. What is the Area?
Area = Base x Height
A = (2.5 meters) x (50 meters)
A = (125 meters2)
If one lane is 125 m2. and we
have eight lanes than we need to
multiple the area of one lane by
the total number of lanes 8
125 m2 x 8 = 1000 m2
The Beijing National Aquatics Center has request your help. They require a
cover for the Olympic swimming pool that is housed within this beautiful
building. The pool has eight lanes that are each 2.5 meters wide. The length
of each lane is 50 meters long. How long will the cover need to be if you wish
to cover only the surface of the pool?
MCC6.G.2
Find the volume of a right
rectangular prism with
fractional edge lengths by
packing it with unit cubes of
the appropriate fraction edge
lengths, and show that the
volume is the same as it
would be found by
multiplying the edge lengths
of the prism. Apply the
formulas V = lwh and V = Bh
to find the volumes of right
rectangular prisms with
fractional edge lengths in the
context of solving real-world
and mathematical problems.
Examples:
• Use unit cubes to pack a
right rectangular prism
with fractional edge
lengths
• Find the volume of right
rectangular prisms using
the volume formulas
• Real-world problems
Vocabulary Words
• Prism
• Right Rectangular Prism
• Edge
• Base of a Prism
• Volume
• Unit Cubes
Find the volume of a rectangular
prism with fractional edge lengths
You cannot use a whole unit block as
a visual because when you try the
2 units block will extend past the boundary
of the original shape.
To fill a fractional side we will need a
fractional unit cube.
2 units
1.5 units
1 unit
Now we can insert the ½ unit blocks onto the bottom
row of our figure .
6 x ½ = 3 whole unit blocks on bottom row
6 X ½ = 3 whole unit blocks on the top row
3 + 3 = 6 total whole unit blocks
1 unit
1 unit1/2 in.
MCC6.G.4
Represent threedimensional figures
using nets made up of
rectangles and triangles
using nets made up of
rectangles and triangles,
and use the nets to find
the surface area of these
figures. Apply these
techniques in the context
of solving real-world and
mathematical problems.
Examples:
• Nets of triangular prisms
• Nets of rectangular prisms
• Nets of square pyramids
• Nets of rectangular
pyramids
• Surface area of each of the
prisms and pyramids
• Real-world problems
Vocabulary Words
• Nets
• Triangular Prisms
• Pyramid
• Surface Area
6 ft.
By looking at a net of this square
pyramid determine the surface area.
Surface are of the square in the middle:
2
SA
(5ftft.)×(5
ft.)
A ==25
5 ft.
Surface are of one
2 of the triangle segments:
SA = 25 ft
Surface are of one of the triangle segments:
A = 15 ft2
SA
= ½ (5 ft.)×(6 ft.)
Total Surface Area: 1 square & 4 Triangles
2
2
A
SA==25
15ftft2+ 4 (15 ft )
A = 85 ft2
3 ft.
C
A
1 ft.
B
C
A
3 ft.
B
5 ft.
By looking at a net of this
rectangular prism determine the
surface area. Similar rectangles are
By looking at a net of this
labeled.
rectangular prism determine the
Surface area of all six rectangles.
surface area. Similar rectangles are
Rectangle
labeled. A - SA = 3 ft2
2.
Rectangle
A
SA
=
+
3
ft
Surface area of the rectangle C.
B.
A.
Rectangle B - SA = + 15 ft2
SA
= (5
(3 ft.)×(1
ft.)×(5
ft.)= + 15 ft2
Rectangle
B - SA
Rectangle
C - SA = + 5 ft2
2
2
SA = 15
5 ftft
3
Rectangle C - SA = + 5 ft2
Total = SA = 46 ft2