Transcript Springboard, Page 264, #1

Springboard, Page 272, #1
This problem has an infinite
number of answers.
Below is just one example,
but the premise is the
same, no matter which
numbers you use.
•Choose possible numbers for two of the dimensions and substitute them into the surface area
equation.
•Start with small numbers because the surface area isn’t a large number. Below, we start with a
length (l) of 2 feet and height (h) of 3 feet.
•Solve the problem like a linear equation.
96 = 2(2)w + 2(2)(3) + 2(3)w
96 = 4w + 12 + 6w
96 = 10w + 12
-12
- 12
84 = 10w – Divide both sides by 10
8.4 = w
A possible set of dimensions for this rectangular prism is 2 x 3 x 8.4.
Springboard, Page 272, #2
Remember, surface area means you must
find the area of each surface on the
shape and add them together. Even if
you don’t have a formula for the specific
shape, the idea remains the same.
Area of trapezoids = ½ x 4(6+12) = 2(18) = 36
Area of rectangle = 12 x 6 = 72
Area of rectangles on sides = 5 x 6 = 30
Area of square = 6 x 6 = 36
Add them all up.
36 + 36 + 72 + 30 + 30 + 36 = 240 mm2
Springboard, Page 272, #3
Find the area of each face and add them together.
Area of each triangular base = ½(6)(4) = 12
Area of two side rectangular panels = 5 x 8 = 40
Area of bottom rectangular panel = 6 x 8 = 48
Add them all together.
SA = 12 + 12 + 40 + 40 + 48 = 152 ft.2
Springboard, Page 272, #4
P = Perimeter of base (Add all the sides of
the square base) = 14
ℓ = slant height = 6
B = Area of base (Multiply the base and
height of the square base) = 12.25
SA = ½ Pℓ + B
SA = ½(14)(6) + 12.25
SA = 42 + 12.25
SA = 54.25 ft2
Springboard, Page 272, #5
r = radius, which starts at the center of the circle
and stretches to any point on the circle. This
problem gives you the diameter, which stretches all
the way across the circle. Divide this in half to get
the radius. r = 10 in.
SA = 2πrh + 2πr2
SA = 2π(10)(18) + 2 π(10)2
SA = 360π + 200π
SA = 560π or 1758.4 in2
Springboard, Page 272, #6
There are two different ways to solve this
problem, given the information. We can
use the formula for the area of a circle and
use the 4/5 information given, or we can
use the formula for surface area of a cone.
Area of a circle = πr2
Surface Area of a cone = ½ (2 πr) ℓ + B
Option 1: Find the area of both circles.
A = π42 (Full Circle)
A = 16π
A = 16 π or 50.24 cm2
A = 16 π + 20 π
A = 36 π cm2
Area of a circle = πr2
A = π52 (Partial circle)
A = 25 π
The directions tell us that this sector is 4/5 of
the area of a circle, so multiply the area by 4/5.
A = 25 π x 4/5 = 20 π or 62.8 cm2
Add these two totals together.
or
or
A = 50.24 + 62.8
A = 113.04 cm2
Springboard, Page 272, #6
There are two different ways to solve this
problem, given the information. We can
use the formula for the area of a circle and
use the 4/5 information given, or we can
use the formula for surface area of a cone.
Area of a circle = πr2
Surface Area of a cone = ½ (2 πr) ℓ + B
Surface Area of a cone = ½ (2 πr) ℓ + B
R = 4cm
ℓ = 5 cm
B = πr2
SA = ½ (2 x 3.14 x 4) 5 + π42
SA = ½ (25.12) 5 + 16 π
SA = ½ (125.6) + 50.24
SA = 62.8 + 50.24
SA = 113.04 cm2
Springboard, Page 272, #7
As with other problems we
have seen in our homework,
we must look at this as two
different shapes. We will find
the surface area of each, then
add them together.
Let’s start with the base, which is a cube. All faces
of a cube have the same dimensions, so they’ll all
have the same area. If you find the area of one
side, you’ll know the area for each side.
A = bh
A=4x4
A = 16 in.2
There are five faces (sides) of this cube that
contribute to the surface area. The top is not on
the outside and doesn’t need to be counted.
SA = 16 x 5 = 80 in2
And now the pyramid
SA = ½ Pℓ – In this case, we don’t need B because
the base of the pyramid isn’t contributing to
the surface area of the pyramid. It’s on the
inside of the shape.
SA = ½ Pℓ, P = 4 + 4 + 4 + 4 = 16, ℓ = 3
SA = ½(16)(3) = 24 in.2
Now that we have the surface area of the cube and
pyramid, we can add them together.
SA = 80 in2 + 24 in2 = 104 in2
Springboard, Page 264, #4
What is the volume of a cone with a radius of 12
inches and a height of 11.2 inches?
V = ⅓(πr2)h
V = ⅓(π122)11.2
V = ⅓(π144)11.2
V = 537.6 π
V = 1,688.064 in.3
Springboard, Page 264, #5
The area of the base of a pyramid is 85 square
centimeters. If its volumes is 255 cubic centimeters,
find the height of the pyramid.
V = ⅓Bh
255 = ⅓(85)h – Multiply both sides by 3 to get rid of ⅓.
765 = 85h – Divide both sides by 85 to get h all by itself.
9=h
The height is 9 cm.
Springboard, Page 264, #6
Take this problem one step at a
time. We don’t have a formula
to find the volume of the shape
like it is, but we can take it as
two separate prisms and find the
volume of each.
The bottom of shape is a rectangular prism.
V = bwh
V = 22 x 15 x 12
V = 3,960 cm3
The top of shape is a triangular prism.
V = ½(bh)h
V = ½(22 x 10) x 15
V = 1,650 cm3
After you have found the area of both the rectangular prism and the
triangular prism, add the two volumes together.
3,960 + 1,650 = 5,610 cm3
Springboard, Page 264, #1
Determine the volume of the figure sketched on the dot paper.
6
4
3
Count the cubes to determine the dimensions of the prism.
Volume of a rectangular prism = bwh
b (base) = 3, w (width) = 4, h (height) = 6
Plug these dimensions into the formula.
V=3•4•6
V = 72 cubic units or 72 units3
Springboard, Page 264, #2
Find the volume of the cylinder.
Volume of a cylinder = πr2h
π (pi) = 3.14, r (radius) = 4 cm, h (height) = 12 cm
Plug these dimensions into the formula.
V = 3.14 • 42 • 12
V = 3.14 • 16 • 12
V = 602.88 cm3
Springboard, Page 264, #3
Find the volume of the triangular prism.
Volume of a triangular prism = (½bh)h
b (base of triangle) = 8 in., h (height of triangle) = 4 in, h (height
of prism) = 10 in.
Plug these dimensions into the formula.
V = (½ • 8 • 4) • 10
V = 160 cubic inches or 160 in.3