Area

The basic shape we will use is the rectangle; the area of a rectangle is \text { (base) } \bullet \text { (height). } If the units for each side of the rectangle are "meters," then the area will have the units \text { ("meters") •("meters") = "square meters" } = \mathrm{m}^{2}. The only other area formulas needed for this section are for triangles, \text { area }=\mathrm{b} \cdot \mathrm{h} / 2, and for circles, \text { area }=\pi \cdot \mathrm{r}^{2}. Three other familiar properties of area are assumed and will be used:

Addition Property: The total area of a region is the sum of the areas of the non–overlapping pieces which comprise the region. (Fig. 1)

Inclusion Property: If region B is on or inside region A, then the area of region B is less than or equal to the area of region A. (Fig. 2)

Location–Independence Property: The area of a region does not depend on its location. (Fig. 3)


Example 1: Determine the area of the region in Fig. 4a

Solution: The region can easily be broken into two rectangles, Fig. 4b, with areas 35 square inches and 3 square inches respectively, so the area of the original region is 38 square inches.


Practice 1: Determine the area of the region in Fig. 5 by cutting it in two ways: (a) into a rectangle and triangle and (b) into two triangles.

We can use the three properties of area to get information about areas that are difficult to calculate exactly. Let A be the region bounded by the graph of f(x)=1 / x, the x–axis, and vertical lines at \mathrm{x}=1 and \mathrm{x}=3. Since the two rectangles in Fig. 6 are inside the region A and do not overlap each, the area of the rectangles, 1 / 2+1 / 3=5 / 6, is less than the area of region A.


Practice 2: Build two rectangles, each with base 1 unit, outside the shaded region in Fig. 6 and use their areas to make a VALID statement about the area of region A.


Practice 3: What can be said about the area of region A in Fig. 6 if we use both inside and outside rectangles with base 1/2 unit?


Example 2: In Fig. 7, there are 32 dark squares, 1 centimeter on a side, and 31 lighter squares of the same size. We can be sure that the area of the leaf is smaller than what number?

Solution: The area of the leaf is smaller than 32+31=63 \mathrm{~cm}^{2}.


Practice 4: We can be sure that the area of the leaf is at least how large?

Functions can be defined in terms of areas. For the constant function \mathrm{f}(t)=2, define \mathrm{A}(x) to be the area of the rectangular region bounded by the graph of f, the t–axis, and the vertical lines at t=1 and t=x (Fig. 8a). \mathrm{A}(2) is the area of the shaded region in Fig. 8b, and \mathrm{A}(2)=2. Similarly, \mathrm{A}(3)=4 and \mathrm{A}(4)=6. In general, \mathrm{A}(x)=(\text { base })(\text { height })= (x-1)(2)=2 x-2  for any x \geq 1. The graph of \mathrm{y}=\mathrm{A}(x) is shown in Fig. 8c, and \mathrm{A}^{\prime}(x)=2 for every value of x>1.


Practice 5: For \mathrm{f}(t)=2, define \mathrm{B}(x) to be the area of the region bounded by the graph of f, the t–axis, and vertical lines at t=0 \text { and } t=x. Fill in the table in Fig. 9 with the values of B. How are the graphs of y=\mathrm{A}(x) \text { and } y=\mathrm{B}(x) related?

Sometimes it is useful to move regions around. The area of a parallelogram is obvious if we move the triangular region from one side of the parallelogram to fill the region on the other side and ending up with a rectangle (Fig. 10). At first glance, it is difficult to estimate the total area of the shaded regions in Fig. 11a . However, if we slide all of them into a single column (Fig. 11b), then it is easy to determine that the shaded area is less than the area of the enclosing \text { rectangle }=(\text { base })(\text { height })=(1)(2)=2.


Practice 6: The total area of the shaded regions in Fig. 12 is less than what number?