Pre-Algebra Notes SOL 8.7 (11-4 11-5) Surface Area
Mrs. Grieser
Name: _______________________________ Block: _______ Date: _____________
Surface Area
We’ve looked at nets, the unwrapping of a solid. The figure at right shows
the net of a rectangular prism. We can see all the faces, or surfaces, of
the solid. If perimeter is a fence, and area is a carpet, and volume is water
in a swimming pool, then surface area is wrapping paper.
Surface area is the sum of the areas of each surface (face) of a figure.
Surface area of a Prism
Let’s add the areas of each surface:
Surface (face)
Area
top and bottom
lw + lw = 2lw
front and back
lh + lh = 2lh
two sides
wh + wh = 2wh
SUM OF AREAS
2lw + 2lh + 2wh or 2(lw + lh + wh) (This is on your formula sheet – yay!)
Example 1: Find the surface area of the rectangular prism shown:
S = 2lw + 2lh + 2wh
S = 2(20)(14) + 2(20)(10) + 2(14)(10)
S = 560 + 400 + 280 = 1240 in2
NOTE: We use square units since we are adding areas!
Example 2: Find the surface area of the triangular prism shown:
Yikes, this solid is not on our formula sheet! But we can still find the
surface area. Our top and bottom (bases) are triangle shaped, and we
1
know the area of a triangle ( bh), so we add twice the area of the
2
bases. We have 3 rectangular faces, so we add these areas.
1
(4)(3)) + (5)(6) + (4)(6) + (3)(6) = 12 + 30 + 24 + 18 = 84 cm2
2
Surface Area of a Cylinder
S = 2(
A net of a cylinder is shown at right. To find surface area,
we need to add the area of the top and bottom (2 circles)
and the rectangle that makes up the body. The length of
the side of the rectangle is the same as the circumference
of the circle (think about it!), so it is 2πr. Therefore the
area of the rectangle is 2πrh. So S = 2πr2 + 2πrh (on our formula sheet – yay!)
Example 3: Find the surface area of the cylinder at right.
S = 2πr2 + 2πrh = 2π12 + 2π(1)(3) = 2π + 6π = 8π ≈ 25.1 m2
1
Pre-Algebra Notes SOL 8.7 (11-4 11-5) Surface Area
Surface Area of a Pyramid
Mrs. Grieser
At right we see a square pyramid and its net. We can see
that we need to add the area of the base with the area of
each triangle attached to the base. We call each triangle a
lateral face, and we call the height of each triangle the
slant height (because it is different from the height of the pyramid itself).
S = 4(
1
bl) + B (the area of the base)
2
1
1
We can rewrite 4( bl) as (4b) (l). But 4b is the perimeter of the base (think about it!). So
2
2
1
the formula for surface area of a pyramid = lp + B, where p is the perimeter of the base, l is
2
the slant height (lateral height), and B is the area of the base. It’s on our formula sheet!
Example 4: Find the surface area of the pyramid at right:
S=
1
1
lp + B = (8.2)(4)(6) + (6)(6) = 98.4 + 36 = 134.4 m2
2
2
Surface Area of a Cone
The base is a circle, so its area is πr2. The area of the body of a
cone is πrl, where l is the slant height (lateral height).
S = πrl + πr2 (on our formula sheet!)
Example 5: Find the surface area of the cone at right:
S = πrl + πr2 = π(10.6)(15) + π(10.6)2
= 159π + 112.36π = 271.36π ≈ 852.5 m2
You try: Find the surface area…
a)
b)
c)
d)
e)
f)
2