Surface Area of Cylinders and Triangular Prisms
Surface Area of Cylinder:
The following website has an interactive visual for the surface area formula:
http://www.mathopenref.com/cylinderarea.html
The formula for the surface area of a cylinder is 2∏r2 + 2∏rh
A cylinder is created from 2 circles (the bases) and a
rectangle. The first part of the formula (2∏r2 ) gives the
area of the two circles. The area of the rectangle is base x
height. The height of the rectangle is the same as the height
of the cylinder. The length of the base is the same as the
circumference of the circle. In the formula, the 2∏r
represents the circumference of the circle, which is also the
length of the base. This is multiplied by the height to give
the area of the rectangle.
Example 1:
(See figure to right)
S.A. = 2∏r2 + 2∏rh
S.A. = 2 (3.14) (3)2 + 2(3.14) (3) (7)
S.A. = 2 (3.14) (9) + 131.88
S.A. = 56.52 + 131.88
S.A. = 188.08 cm2
Example 2:
S.A. = 2∏r2 + 2∏rh
S.A. = 2 (3.14) (2)2 + 2(3.14) (2) (10)
S.A. = 2 (3.14) (4) + 125.6
S.A. = 25.12 + 125.6
S.A. = 150.72 cm2
Surface Area of Triangular Prism:
The following website has an interactive visual for the surface area formula of triangular prisms:
http://www.shodor.org/interactivate/activities/SurfaceAreaAndVolume/
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To find the surface area of a triangular prism, using a net is helpful. Draw one of the triangular
bases with the measurement of each side written down and then draw rectangles from each site.
The length of each rectangle is the length of the triangular prism.
Find the area of each piece, remembering that there are two triangles.
Add the areas together.
Example 1:
8 cm
12 cm
10 cm
12 cm
6 cm
12 cm
The area of the triangle is 6 x 8 which is 24 cm2.
2
Since there are two triangles multiply by 2:
The area of the three rectangles are: 8 cm x 12 cm =
6 cm x 12 cm =
10 cm x 12 cm =
Add these to get the surface area:
48 cm2
96 cm2
72 cm2
120 cm2
336 cm2
Surface Area Practice
Name: ____________________________
Solve each problem below!
1. Find the surface area of the cylinder.
2. How much canvas would be needed for the
tent below.
Surface Area = __________units2
Surface Area = ___________ m2
3. A candy bar is in the shape of a triangular
prism with a height of 6 inches. Each side of the
triangle is 1 inch with a height of 0.8 in. How
much paper will be needed for the wrapping?
4. A steel barrel is a cylinder with a radius of 2
feet and a height of 5 feet. How many square
feet of steel were needed to make the barrel?
Surface Area = ___________ in2
Surface Area = ___________ m2
Surface Area Practice Answer Key
Name: ____________________________
Solve each problem below!
1. Find the surface area of the cylinder.
S.A. = 2∏r2 + 2∏rh
S.A. = 2 (3.14) (3)2 + 2(3.14) (3) (6)
S.A. = 2 (3.14) (9) + 113.04
S.A. = 56.52 + 113.04
S.A. = 169.56 cm2
2. How much canvas would be needed for the
tent below.
Area of 2 triangle bases:
(1/2)(2)(1.5) = 1.5 x 2 = 3
Area of rectangle faces:
2x3=6
1.8 x 3 = 5.4
1.8 x 3 = 5.4
S.A. = 3m + 6m + 5.4m + 5.4m = 19.8m2
Surface Area = 169.56 units2
Surface Area = 19.8 m2
3. A candy bar is in the shape of a triangular
prism with a height of 6 inches. Each side of the
triangle is 1 inch with a height of 0.8 in. How
much paper will be needed for the wrapping?
4. A steel barrel is a cylinder with a radius of 2
feet and a height of 5 feet. How many square
feet of steel were needed to make the barrel?
2 ft
Area of 2 triangle bases:
(1/2)(1)(0.8) = 0.4 x 2 = 0.8
Area of rectangle faces:
1x6=6
1x6=6
1x6=6
S.A. = 0.8 in + 6 in + 6 in + 6 in = 18.8 in2
Surface Area =
18.8
in2
5 ft
S.A. = 2∏r2 + 2∏rh
S.A. = 2 (3.14) (2)2 + 2(3.14) (2) (5)
S.A. = 2 (3.14) (4) + 62.8
S.A. = 25.12 + 62.8
S.A. = 87.92 cm2
Surface Area = 87.92 m2