Pre-Algebra Notes SOL 8.7 (11-4 11-5) Surface Area Mrs. Grieser

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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 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)

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

know the area of a triangle (2

1bh), so we add twice the area of the

bases. We have 3 rectangular faces, so we add these areas.

S = 2(2

1(4)(3)) + (5)(6) + (4)(6) + (3)(6) = 12 + 30 + 24 + 18 = 84 cm2

Surface Area of a Cylinder

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

Surface area is the sum of the areas of each surface (face) of a figure.

(This is on your formula sheet – yay!)

Pre-Algebra Notes SOL 8.7 (11-4 11-5) Surface Area Mrs. Grieser

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Surface Area of a Pyramid

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(2

1bl) + B (the area of the base)

We can rewrite 4(2

1bl) as (4b)

2

1(l). But 4b is the perimeter of the base (think about it!). So

the formula for surface area of a pyramid = 2

1lp + B, where p is the perimeter of the base, l is

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 = 2

1lp + B =

2

1(8.2)(4)(6) + (6)(6) = 98.4 + 36 = 134.4 m2

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)