Surface Area of Common Solids

 

Introduction

     In this section, you will learn how to calculate the surface area of common solids to include definition of surface area, prisms, cylinders, pyramids, cones, and spheres. You will be invited to try our quizmasters at the end of each lesson.

 

 


     Within our area section, we had to provide a definition for the meaning of area. That definition rested upon the square -- particularly a unit square. A unit Square can be 1" x 1" or 1 yd x 1 yd or 1 ft x 1 ft or a square by some other unit.

Unit Square
Unit Square

     We saw the area of a figure was nothing more than the sum of all unit squares of a figure. For the surface area of a solid, there is a similar definition, but it applies to the exterior surfaces of the solid. The definition of surface area is the sum of all unit squares that fit on the exterior of a solid.

Surface Area of Prisms

SA = 2(lw + hl + hw)

 

Standard Prism
General Prism

     This is the best figure to begin with when investigating surface area. It is the most simple figure of all the solids. It is also a figure most people have personal experience due to either wrapping or opening gifts.

     All the surfaces of a prism are rectangular. This makes calculating the areas of these surfaces very easy to do. The area of rectangles have been discussed in another section, which is available for review before proceeding, if necessary.

     As the diagram below indicates, there are six surfaces to a rectangular prism. There is a front, back, top, bottom, left, and right to every rectangular prism. The surface are of a prism is nothing more than the sum of all the areas of these rectangles.

Six Surfaces of a Rectangular Prism

     Using the labeling of the general prism diagram above, a formula can be created for dealing with the surface area of prisms. Let's calculate the area of each surface.

Prism Surface Area Formula

Top

: lw

Bottom

: lw

Front

: hl

Back

: hl

Left

: hw

Right

: hw

Total

: lw + lw + hl + hl + hw + hw

 

: 2lw + 2hl + 2hw

 

: 2(lw + hl + hw)

Example 1: Given l = 4 yds, w = 2 yds, and h = 5 yds, the surface area would be
  SA = 2[(4 yds)(2 yds) + (5 yds)(4 yds) + (5 yds)(2 yds)]= 2(8 yds2 + 20 yds2 + 10 yds2) = 2(38 yds2) = 76 yds2.

Example 2: Given l = 6 mm, w = 9 mm, and h = 8 mm, the surface area would be
  SA = 2[(6 mm)(9 mm) + (8 mm)(6 mm) + (8 mm)(9 mm)]= 2(54 mm2 + 48 mm2 + 72 mm2) = 2(174 mm2) = 348 mm2.

Surface Area of Cylinders

SA = 2πr2 + 2πrh

 

General Cylinder
General Cylinder

     A cylinder has a total of three surfaces: a top, bottom, and middle. The top and bottom, which are circles, are easy to visualize.

     The area of a circle is πr2. So, the area of two circles would be πr2 + πr2 = 2πr2.

     The third surface, the lateral surface area, is less easy to visualize for the purposes of calculating its area, especially since it does not appear to be in a shape that fits a known area like a triangle or parallelogram. The surface being referred to is the curved wall of the cylinder.

Lateral Area of a Cylinder

     We can manipulate the curved wall of a cylinder to produce a recognizable shape. Imagine starting with a cylinder, like a typical can, and then cutting the can up its wall. From the bottom to the top of the can, a cut is made. See the dashed segment in step one of the diagram to the right.

     Now, from the newly cut edge, the cylinder wall will be spread open. Almost like opening double doors to a fancy home, the walls spread apart. See step two of the same diagram.

     When the cylinder's wall is completely open, it takes the form a very recognizable, basic shape. It is a rectangle. Fortunately, the area of a rectangle is easy to calculate. But we need to go back to the original cylinder for a moment before we calculate the rectangle's area.      The top of the original cylinder is a circle. We know that the distance around a circle is called its circumference, C = 2πr. The circumference has been marked in red.

     When the cylinder wall is completely open, we see that the circumference of the circle becomes the length of the final rectangle. The dimensions of the rectangle are the circumference, C = 2πr, and the height of the cylinder, h. So, the area of the rectangle is A = l x w = C x h = 2πr x h = 2πrh.

     Finally, if we put all the three surfaces together, we can arrive at the formula for the complete surface area of a cylinder. The top, the bottom, and the walls of the cylinder (the lateral surface area) determine the total surface area. Those values are πr2 + πr2 + 2πrh = 2πr2 + 2πrh. Sometimes this expression is written as 2πr(r + h) in certain geometry texts. [Can you see how 2πr2 + 2πrh = 2πr(r + h) by the distributive property?]

Example 1: If r = 6 ft and h = 4 ft, then the surface area would be
  SA = 2(3.14)(6 ft)2 + 2(3.14)(6 ft)(4 ft) = 2(3.14)(36 ft2) + 2(3.14)(24 ft2) = 226.08 ft2 + 150.72 ft2 = 376.8 ft2.

Example 2: If r = 5 cm and h = 2 cm, then the surface area would be
  SA = 2(3.14)(5 cm)2 + 2(3.14)(5 cm)(2 cm) = 2(3.14)(25 cm2) + 2(3.14)(10 cm2) = 157 cm2 + 62.8 cm2 = 219.8 cm2.

Surface Area of Square Pyramids

SA = s2 + 2sl

 

General Pyramid
General Pyramid

     Pyramids that have a square base have a total of five surfaces. To determine the shapes of those surfaces, we will start with a pyramid from step one below. If we cut along the lateral edges of the pyramid, we can allow the figure to flatten out in step two below. From step two, the individual figures are easily identified as a square and four triangles.

Surfaces of a Pyramid

     We can use the area formulas for a rectangle and a triangle to determine the complete formula for the surface area of the pyramid. The square (or base of the solid) has an area that can be calculated by multiplying its length times its width. Since those dimensions are equal, the area is s x s = s2.

     Now we need to calculate the area of the remaining surfaces. The remaining surfaces happen to form the lateral surface area of the pyramid, which are triangles. The area formula for a triangle is its base times its height divided by two. In the case of one of the triangles above it would be s x l ÷ 2. However, there are four triangles. This would make the total lateral surface area equal to four times the area of one triangle, or 4 x s x l ÷ 2. Upon simplifying the expression, we get 2sl.

     The total surface area of the pyramid is equal to the area of the base plus its lateral surface area or s2 + 2sl. So, given the base length of the pyramid, s, and its slant height, l, the total surface area is not a difficult computation. Yet, when dimensions are provided for a pyramid, one is rarely provided with the slant height, l. Instead, the height of the pyramid is given. This is an important difference, as one is always larger than the other.

Internal Right Triangle of a Pyramid

     Not knowing the slant height makes calculating its value the first goal for finding the lateral area of a pyramid. The diagram to the right will provide insight into finding a relationship for calculating the slant height.

     A right triangle, which rests internally within the pyramid, has been highlighted. The hypotenuse (longest side) is the slant height of the pyramid, l. This is the length we need to know in order to calculate the surface area of the pyramid. The height of the pyramid is a leg of the right triangle. The base of the right triangle is half the length of the base edge of the pyramid, s. At this point, we would use The Pythagorean Theorem to calculate the slant height.

     Pythagoras said leg2 + leg2 = hypotenuse2. In the case of our right triangle, we would get (1/2 s)2 + h2 = l2. Solving for the slant height would then depend on our ability to use the Pythagorean Theorem.

     To review the process, there is an outline of the steps that need to be taken:

  1. Determine the height, h, and base edge, s, of a given pyramid. Make sure their values are in the same unit measure (feet, centimeters, ...).
  2. Use The Pythagorean Theorem to calculate the slant height, l, of the pyramid.
  3. Calculate the area of the base, s2.
  4. Calculate the lateral surface area, 2sl.
  5. Calculate the total surface area by adding the base area to the lateral area, s2 + 2sl.
  6. Report the final surface area value using square units.

     Here are two examples that demonstrate how to calculate the total surface area of a pyramid.

Example 1: Given: s = 8 m and h = 3 m. First we calculate the slant height using The Pythagorean Theorem and half the base length. 32 + 42 = l2 ==> 9 + 16 = l2 ==> 25 = l2, or l = 5 m. The surface area would then require use of the formula
  SA = (8 m)2 + 2(8 m)(5 m) = 64 m2 + 80 m2 = 144 m2.

Example 2: Given: s = 7 in and h = 6 in. First we calculate the slant height using The Pythagorean Theorem and half the base length. (3.5)2 + 62 = l2 ==> 12.25 + 36 = l2 ==> 48.25 = l2, or l = 6.95 in. The surface area would then require use of the formula
  SA = (7 in)2 + 2(7 in)(6.95 in) = 49 in2 + 97.3 in2 = 146.3 in2.

Surface Area of Cones

SA = πr2 + πrl

 

General Cone
General Cone

Example 1: Given r = 4 m and h = 3 m. First we must calculate the slant height using The Pythagorean Theorem. (4 m)2 + (3 m)2 = l2 ==> 16 m2 + 9 m2 = l2 ==> 25 m2 = l2, or l = 5 m. The surface area would then require use of the formula
  SA = (π)(4 m)2 + (π)(4 m)(5 m) = (16 m2)(π) + (20 m2)(π) = 36π m2 (in terms of π) or 113.0 m2.

Example 2: Given r = 7 in and h = 10 in. First we must calculate the slant height using The Pythagorean Theorem. (7 in)2 + (10 in)2 = l2 ==> 49 in2 + 100 in2 = l2 ==> 149 in2 = l2, or l = 12.21 in. The surface area would then require use of the formula
  SA = (π)(7 in)2 + (π)(7 in)(12.21 in) = (49 in2)(π) + (85.47 in2)(π) = 134.47(π) in2 = 422.2 in2.

Surface Area of Spheres

SA = 4πr2

 

General Sphere
General Sphere

Example 1: Given r = 3 ft. The surface area requires the use of the formula
  SA = 4(π)(3 ft)2 = 4(π)(9 ft2) = 36π ft2 (in terms of π) or 113.0 ft2.

Example 2: Given diameter = 20 m or r = 10 m. The surface area requires the use of the formula
  SA = 4(π)(10 m)2 = 4(π)(100 m2) = 400π m2 (in terms of π) or 1256 m2.

 

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