Here’s a little tidbit that was mentioned in Chapter 3, but bears repeating. In some of the formulas presented in this chapter (and throughout this book), there are defined constants. In the four equations immediately above, you will see constants of 2, 180, π, and 360. When you want to decide on the number of justifiable significant figures in the outcome of a calculation, the values of con- stants of this sort can be considered “exact.” They are accurate to as many significant figures as you need. In the case of the constant 2, for example, you can call it 2.00000 . . . , with any number of zeros you want. In the case of π, you can extend its decimal rendition to as many digits as necessary. The same is true for obvious whole-number quantities, such as the number of sides in a polygon. When any two “exact values” are added to each other, subtracted from each
r θ
Fig. 8-10. Dimensions of a circular sector.
other, multiplied by each other, or divided by each other, or when an “exact value” is taken to a whole-number power, the result is another “exact value” for purposes of calculation.
PROBLEM 8-9
Suppose a regular decagon (10-sided polygon) is inscribed within a circle whose radius is 53.25 cm. What is the perimeter of this decagon in meters?
SOLUTION 8-9
Use the formula for the perimeter of a regular polygon inscribed within a circle, setting n=10 (exactly!) and r=53.25. The perimeter in cen- timeters is found using this formula:
B=2nrsin (180 /n)º
=2×10×53.25×sin (180 /10)º
=1065×sin 18º
=1065×0.309017
=329.1 cm
Because 1 m =100 cm, this is equal to 3.291 m.
PROBLEM 8-10
What is the interior area of the decagon described in Problem 8-9, in square meters?
SOLUTION 8-10
Use the formula for the interior area of a regular polygon inscribed within a circle, setting n=10 and r=53.25. The interior area in square centimeters (cm2) is found using this formula:
A=(nr2/ 2) sin (360 /n)º =10×(53.252/ 2)×sin (360 /10)º =14,178×sin 36º =14,178×0.58779 =8333.7 cm2 =8.334×103cm2
Quick Practice
Here are some practice problems that cover the material presented in this chap- ter. Solutions follow the problems.
PROBLEMS
1. Find the interior area, in square meters, of an equilateral triangle meas- uring 2.000 m on each side.
2. Find the interior area, in square centimeters, of a right triangle having sides measuring 300 mm, 400 mm, and 500 mm.
3. Suppose a square has a perimeter in meters that is exactly equal to its interior area in square meters. What is the length of each side in meters? (Do not consider the trivial case, where the length of each side is 0.) 4. If the length of the diagonal of a square is increased by a factor of 5, by
what factor does its perimeter increase? By what factor does its interior area increase?
5. Suppose a trapezoid has two parallel sides measuring 4.57 m and 6.03 m. These parallel sides are 1.00 m apart. What is the interior area of the trapezoid in square meters?
SOLUTIONS
1. Use the second formula for the interior area of a triangle given earlier in this chapter. In an equilateral triangle, all three angles measure 60º. Set s1=2.000,s2=2.000, and q=60º (this angular measure is exact). Then:
A=(s1s2sinq) / 2
=2.000×2.000×(sin 60º) / 2
=4.000×0.8660254 / 2
=1.732 m2
2. First, convert all lengths to centimeters, because we want to find the area in square centimeters. Consider the base length to be 30.0 cm. Then the height is 40.0 cm, because the angle between the 30.0 cm and 40.0 cm
sides is a right angle. Using the first formula for the interior area of a tri- angle given earlier in this chapter, set s1=30.0 and h=40.0. Then:
A=s1h/ 2
=30.0×40.0 / 2
=600 cm2
3. Let sbe the length of each side of this square. Then consider the formu- las for perimeter, B, and area, A, of a square, as follows:
B=4s A=s2
Because the perimeter and the area are represented by the same number (even though the units differ in terms of dimension), we can set B =A, obtaining this single-variable equation:
4s=s2
We are told that s≠0. Therefore, it is all right to divide each side of the above equation by s. This yields the solution directly as 4 = s, which means that the square we seek has sides that are each exactly 4 m long. The perimeter of such a square is exactly 16 m, and the interior area is exactly 16 m2.
4. If the length of the diagonal of a square increases by a factor of 5, then the length of each side also increases by a factor of 5. In effect, we mag- nify the square by a linear factor of 5, because all sides of a square are of equal length. That means the perimeter of the square increases by a fac- tor of 5. However, the interior area of the square increases by a factor of 52, or 25. This is true regardless of the initial size of the square.
5. We do not have to know the lengths of the sides connecting the ends of the parallel sides of this trapezoid in order to determine its interior area. Use the formula for the interior area of a trapezoid, letting s1=4.57,s3= 6.03, and h=1.00, as follows: A=(s1h+s3h) / 2 =(4.57×1.00+6.03×1.00) / 2 =(4.57+6.03) / 2 =10.60 / 2 =5.30 m2