Definition 4.1.2. Parallelogram.
A parallelogram is a four sided shape for which opposing pairs of sides are parallel.
Section 4.1 Geometric Reasoning 2D
This section addresses the following topics.
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Read and use mathematical models in a technical document
This section covers the following mathematical concepts.
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Identify shapes and apply their properties (skill)
This section presents geometric properties, illustrates identifying shapes in applications, and illustrates breaking down complex shapes into simple ones.
Subsection 4.1.1 Formulae
This section defines the two properties of interest and provides the formulae for some common shapes. Memorizing all of the formulae is not likely useful: in a job you will be able to look them up. However, anything you use a lot (e.g., triangles) is worth memorizing.
Two of the properties of shapes we will consider are perimeter and area. The perimeter of a shape is a measure of the size of its border (edges). The area of a shape is a measure of what it takes to fill the shape. In the figure below showing the outline of Kapi’olani Regional Park the perimeter is the distance someone would travel walking around the edge of the park. The area determines how much grass seed it would take to spread over the park.
Rounding.
How we round on geometric problems will depend as always on the application. For example a carpenter will round to units that can be measured with a tape measure like 1/16 of an inch. Contextless examples in this section will be rounded using significant digits. When using significant digits treat all numbers in the formulae as exact numbers.
Parallelograms include rectangles, which are parallelograms with four right angles, and rhombi, which are parallelograms with four equal length sides. A square is a rectangle and a rhombus.
For the formulas we need a segment that is perpendicular to one pair of parallel sides. This is called the height. One of these parallel sides is called the base.
The perimeter formula is the sum of the four sides. Because the sides come in two pairs which each have the same length we end up with \(a+b+a+b = a+a+b+b = 2a+2b = 2(a+b)\text{.}\)
Example 4.1.4.
What are the perimeter and area of this parallelogram?
The perimeter is the sum of the side lengths which in this case is
\begin{equation*}
2(4.05+5.00) = 18.1\text{.}
\end{equation*}
The area of a parallelogram, given in Table 4.1.3 is \(h_1 a\text{.}\) For this parallelogram that is
\begin{equation*}
\text{Area} = 3.00 \cdot 5.00 = 15.0\text{.}
\end{equation*}
Example 4.1.5.
What are the perimeter and area of this parallelogram?
The perimeter is the sum of the side lengths which in this case is
\begin{equation*}
2(18.1+28.2) = 92.8.
\end{equation*}
The area of a parallelogram, given in Table 4.1.3 is \(h_1 a\text{.}\) Because this is a rectangle a side length can be used as the height.
\begin{equation*}
\text{Area} = 18.1 \cdot 28.2 = 510.42 \approx 510\text{.}
\end{equation*}
Checkpoint 4.1.6.
Definition 4.1.7. Trapezoid.
A trapezoid is a four sided shape for which one pair of opposing sides are parallel.
Like a parallelogram a trapezoid has a height. This is a segment perpendicular to the parallel sides.
Because none of the side lengths need be the same, the perimeter “formula” is just the sum of the four sides. The height must be between the two, parallel sides. Any line between the other two sides is not perpendicular to both and the formula does not work.
Example 4.1.9.
What are the perimeter and area of this trapezoid?
The perimeter of this trapezoid is the sum of the four side lengths
\begin{equation*}
4.05+1.76+3.04+5.00 = 13.85\text{.}
\end{equation*}
The area of a trapezoid, given in Table 4.1.8 is \(\frac{h}{2}(a+b)\text{.}\) For this trapezoid that is
\begin{equation*}
\text{Area} = \frac{3.00}{2}(1.76+5.00) = 10.14 \approx 10.1\text{.}
\end{equation*}
Note the following is also a trapezoid and the formulae still apply.
Example 4.1.10.
What are the perimeter and area of this trapezoid?
The perimeter of this trapezoid is the sum of the four side lengths
\begin{equation*}
10.00+10.00+8.54+13.00 = 41.54\text{.}
\end{equation*}
The area of a trapezoid, given in Table 4.1.8 is \(\frac{h}{2}(a+b)\text{.}\) For this trapezoid that is
\begin{equation*}
\text{Area} = \frac{8.00}{2}(10.00+13.00) = 92.0.
\end{equation*}
Checkpoint 4.1.11.
Definition 4.1.12. Triangle.
A triangle is a three sided shape.
Because the three sides can all be different, the perimeter formula is just the sum of the lengths of each side. The height is a segment from a vertex down (perpendicularly) to the opposite side which is called the base. Any vertex/side pair can be used. Be aware that the height may not intersect the opposite side; consider Figure 4.1.15. The vertical, dashed line segments are heights for those two triangles. The one on the left is from the top vertex down to the bottom side. The one on the right is from the top vertex down to the extension (to the left) of the bottom side.
Example 4.1.14.
What are the perimeter and area of the triangles in Figure 4.1.15?
The perimeter of the triangle on the left is
\begin{equation*}
5.98+10.9+8.19 = 25.07 \approx 25.1.
\end{equation*}
The perimeter of the triangle on the right is
\begin{equation*}
5.98+2.87+8.19 = 17.04 \approx 17.0.
\end{equation*}
The area of a triangle, given in Table 4.1.13 is \(\frac{1}{2}b h\text{.}\) For the triangle on the left that is
\begin{equation*}
\text{Area} = \frac{1}{2}10.90 \cdot 4.43 = 24.\underline{1}435 \approx 24.1.
\end{equation*}
For the triangle on the right the area is
\begin{equation*}
\text{Area} = \frac{1}{2}2.87 \cdot 4.43 = 6.3\underline{5}705 \approx 6.36
\end{equation*}
Checkpoint 4.1.16.
Checkpoint 4.1.17.
Theorem 4.1.18. Pythagorean Theorem.
For a triangle containing a right angle
\begin{equation*}
a^2+b^2=c^2
\end{equation*}
where \(a\) and \(b\) are the lengths of the sides adjacent to the right angle and \(c\) is the third side.
Example 4.1.19.
Consider the triangle in Figure 4.1.15.(b). Consider the segments of length 4.43, 5.98, and the horizontal dashed segment. If we want to know the length of the dashed segment, we can use this theorem. 5.98 is the length of the side not adjacent to (opposite from) the right angle (\(c\) in the formula).
\begin{align*}
4.43^2+b^2 \amp = 5.98^2.\\
19.\underline{6}249+b^2 \amp = 35.\underline{7}604.\\
b^2 \amp = 16.\underline{1}355.\\
\sqrt{b^2} \amp = \sqrt{16.\underline{1}355}.\\
b \amp \approx 4.0\underline{1}690179\\
\amp \approx 4.02.
\end{align*}
Example 4.1.20.
Another use of this theorem is determining if a triangle is a right triangle. Notice that the triangle in Figure 4.1.15.(a) appears to have a right angle at the top (opposite the side of length 10.90). If this is a right triangle, then \(a^2+b^2\) will equal \(c^2\text{.}\)
\begin{align*}
a^2+b^2 & =\\
5.98^2 + 8.19^2 & = \\
35.7604 + 67.0761 & = 102.8365.\\
c^2 & = 102.8365.\\
c & = \sqrt{102.8365}\\
& \approx 10.14083330\\
& \approx 10.1\text{.}
\end{align*}
Because \(10.1 \ne 10.9\) this is not a right triangle, but it appears to be close.
Checkpoint 4.1.21.
In Section 7.3 we will develop a version of this statement for triangles without a right angle.
The formula for area of a triangle above requires that we be able to calculate the height. In Section 7.1 we will learn to calculate this height if we know an angle. This next formula enables us to calculate the area of a triangle without knowing any angles or the height.
Theorem 4.1.22. Heron’s Formula.
The area of a triangle can be calculated using the three sides.
\begin{equation*}
A = \sqrt{s(s-a)(s-b)(s-c)}
\end{equation*}
where \(s=\frac{1}{2}(a+b+c)\text{.}\)
Example 4.1.23.
Calculate the area of a triangle with side lengths 4, 5, and 7.
According to Heron’s formula
\begin{align*}
s & = \frac{1}{2}(4+5+7)\\
& = 8.\\
\text{Area} & = \sqrt{8(8-4)(8-5)(8-7)}\\
& = \sqrt{96}\\
& \approx 9.79795897\\
& \approx 9.8\text{.}
\end{align*}
Rounding to one decimal place here is an arbitrary choice for this example.
The next example illustrates tracking significant digits while using Heron’s formula to calculate area.
Example 4.1.24.
Calculate the area of the triangles in Figure 4.1.15.
According to Heron’s formula for the triangle in Figure 4.1.15.(a)
\begin{align*}
s & = \frac{1}{2}(5.98+8.19+10.90)\\
\end{align*}
Each term is precise to the 100ths (2 places)
\begin{align*}
& = 12.5\underline{3}5. \text{ Addition maintains to the 100ths (4 sigfigs)}\\
\text{Area} & \approx \sqrt{12.5\underline{3}5(12.5\underline{3}5-5.98)(12.5\underline{3}5-8.19)(12.5\underline{3}5-10.90)}\\
\end{align*}
Terms precise to 100ths, 3 sigfigs
\begin{align*}
& \approx \sqrt{12.5\underline{3}5(6.5\underline{5}5)(4.3\underline{4}5)(1.6\underline{3}5)} \text{ Product maintains 3 sigfigs}\\
& \approx \sqrt{58\underline{3}.7199977} \text{ Root maintains 3 sigfigs}\\
& \approx 24.\underline{1}6\\
& \approx 24.2.
\end{align*}
For the triangle in Figure 4.1.15.(b)
\begin{align*}
s & = \frac{1}{2}(5.98+8.19+2.87) \text{ Each term precise to 100ths}\\
& \approx 8.52. \text{ Sum maintains to the 100ths}\\
\text{Area} & \approx \sqrt{8.52(8.52-5.98)(8.52-8.19)(8.52-2.87)}\\
\end{align*}
Sums maintain to the 100ths
\begin{align*}
& \approx \sqrt{8.52(2.54)(0.33)(5.65)} \text{ Smallest factor has 2 sigfigs}\\
& \approx \sqrt{4\underline{0}.34927160} \text{ Root maintains 2 sigfigs}\\
& \approx 6.\underline{3}5210765\\
& \approx 6.4.
\end{align*}
Notice that the results from Heron’s formula are slightly different from the results using the \(\frac{1}{2}bh\) formula. This reminds us that different calculations can result in different rounding. If the difference does not impact the application we do not care. If it does have a negative impact, then we must measure more precisely.
Checkpoint 4.1.25.
When performing calculations with \(\pi\) we will need to select an appropriate approximation. Because it is possible to obtain an approximation with arbitrary precision (as many decimal places as we want), we will select the approximation so that rounding is not affected.
Example 4.1.27.
For a circle with radius 7.31 what are the perimeter and area?
The perimeter, given in Table 4.1.26, is \(2\pi r\text{.}\) For radius 7.31
\begin{align*}
\text{perimeter} & = 2\pi (7.31)\\
& \approx 2(3.14159)(7.31)\\
& = 45.\underline{9}300458.\\
& \approx 45.9.
\end{align*}
We need to approximate \(\pi\) to at least as many significant digits as the others numbers to avoid reducing precision. We could use more.
The area, given in Table 4.1.26, is \(\pi r^2\text{.}\) For radius 7.31
\begin{align*}
\text{area} & = \pi (7.31)^2\\
& = (3.14159)(7.31)^2\\
& \approx 16\underline{7}.8743174.\\
& \approx 168.
\end{align*}
Example 4.1.28.
What are the perimeter and area of a semi-circle with diameter 11.7?
The perimeter includes half the usual perimeter plus the length of the diameter. We use the diameter version of the perimeter formula: \(\pi d\text{.}\)
\begin{align*}
\text{perimeter} & = \frac{1}{2}\pi (11.7)+11.7\\
& = \frac{1}{2}(3.14159)(11.7)+11.7\\
& = 18.\underline{3}783015 + 11.7\\
& = 30.\underline{0}783015\\
& \approx 30.1\text{.}
\end{align*}
The area is simply half of the usual area.
\begin{align*}
\text{area} & = \frac{1}{2}\pi (11.7)^2\\
& = \frac{1}{2}(3.14159) (11.7)^2\\
& \approx 21\underline{5}.0261276.\\
& \approx 215.
\end{align*}
Checkpoint 4.1.29.
Subsection 4.1.2 Applying Geometry
When we encounter geometric questions, they will often involve shapes consisting of more than one basic shape such as the park boundry above. Our first task then is to break down these shapes into pieces that are basic shapes (ones we already know). Then we can use the geometric properties to calculate results.
Note that often we can break down a complex shape in more than one way. Which way we choose will depend on what we are trying to calculate and what information we have available.
Example 4.1.30.
(a)
Calculate the area of this wall given the dimensions given in feet.
First we notice that we can describe the wall as a rectangle with a triangle on top of it. The side of the rectangle not drawn will also have length 24.0. This is also the base side of the triangle.
The sides of the rectangular area are 7.0 ft (height) and 24.0 ft (width). This means that area is \(7.0 \text{ ft} \times 24.0 \text{ ft} = 1\bar{6}8 \text{ ft}^2\text{.}\)
The top is a triangle with two sides of length 13.0 and one of length 24.0. We do not know the height of the triangle so it will be easier to use Heron’s formula for area.
\begin{align*}
s & = \frac{1}{2}(13.0+13.0+24.0)\\
& = 25.0.\\
\text{area} & = \sqrt{25.0(25.0-13.0)(25.0-13.0)(25.0-24.0)}\\
& = \sqrt{25.0(12.0)(12.0)(1.0)}\\
& = \sqrt{3600}\\
& = 6\bar{0}.
\end{align*}
The total area then is \(1\bar{6}8+6\bar{0} = 228 \approx 230\) square feet.
(b)
Calculate the perimeter of this wall given the dimensions given in feet.
There are five (5) edges. There sum is \(7.0+24.0+7.0+13.0+13.0=64.0\) feet.
(c)
What is the (tallest) height of the wall?
The height comes from the center of the wall (peak of the triangle). The height of the wall is the height of the rectangle (7.0 ft) plus the height of the triangle. We must calculate the height of the triangle. At this time we know the area and two area formulae. Height is part of the first area formula, and from Heron’s formula we already know the area. Putting these two together gives us the following.
\begin{align*}
6\bar{0} & = \frac{1}{2}24.0 h.\\
6\bar{0} & = 12.0 h.\\
\frac{6\bar{0}}{12.0} & = h.\\
5.0 & = h.
\end{align*}
The total height is then \(7.0+5.0=12.0\text{.}\)
Example 4.1.31.
Calculate the square footage of the combined living room and hallway.
This layout is rectangular. The living room is a rectangle with height 12 ft 10 in and width 15 ft 7 in. The hallway adds an additional rectangle of height 3 ft 1 in and width 11 ft 1 in.
To calculate square footage we need to convert everything to the same unit. To avoid rounding issues we will convert to inches then back to feet in the end.
The living room height is \(12 \text{ ft} \cdot \frac{12 \text{ in}}{1 \text{ ft}} + 10 \text{ in} = 154 \text{ in}\text{.}\) The width is \(15 \text{ ft} \cdot \frac{12 \text{ in}}{1 \text{ ft}} + 7 \text{ in} = 187 \text{ in}\text{.}\) The area is \((154 \text{ in})(187 \text{ in}) = 28798 \text{ in}^2\text{.}\)
The hallway height is \(3 \text{ ft} \cdot \frac{12 \text{ in}}{1 \text{ ft}} + 1 \text{ in} = 37 \text{ in}\text{.}\) The width is \(11 \text{ ft} \cdot \frac{12 \text{ in}}{1 \text{ ft}} + 1 \text{ in} = 133 \text{ in}\text{.}\) The area is \((37 \text{ in})(133 \text{ in}) = 4921 \text{ in}^2\text{.}\)
The total floor surface is \(28798 \text{ in}^2+4921 \text{ in}^2=33312 \text{ in}^2\text{.}\) We can now convert that back to feet.
\begin{equation*}
33719 \text{ in}^2 \cdot \left( \frac{1 \text{ ft}}{12 \text{ in}} \right)^2 = \frac{33719}{144} \text{ ft}^2 \approx 234.1597 \text{ ft}^2 \approx 234 \text{ ft}^2.
\end{equation*}
We are using significant digits rounding. All measurements were accurate to one inch. If we converted to decimal feet we would end up with measurements like \(3 \text{ ft } 1 \text{in} \approx 3.0833 \text{ ft}\text{.}\) The repeated decimal would have to be rounded whereas the inch measurement did not need to be rounded. Avoiding rounding avoids introducing error.
Example 4.1.32.
Calculate the area to be painted on this wall with a circular window.
The wall consists of a rectangle with a triangle on top. The rectangle has width 12 ft and height 5 ft for an area of \((5 \text{ ft})(12 \text{ ft})=60 \text{ ft}^2\text{.}\)
We have the base of the triangle (12 ft), but we do not have the height of the triangle. However, we know that the total wall height is 11 ft, and the part of the height from the rectangal is 5 ft. This means that the rest is from the triangle which has height \((11 \text{ ft})-(5 \text{ ft})=6 \text{ ft}\text{.}\) Thus the area is \(\frac{1}{2}(12 \text{ ft})(6 \text{ ft}) = 36 \text{ ft}^2\text{.}\)
The area of the circle is \(\pi\left( \frac{6 \text{ ft}}{2} \right)^2 \approx 28.27433389 \text{ ft}^2 \approx 28 \text{ ft}^2\text{.}\)
The total wall area is therefore
\begin{equation*}
60 \text{ ft}^2 + 36 \text{ ft}^2 - 28.27433389 \text{ ft}^2 \approx 67.72566611 \text{ ft}^2 \approx 68 \text{ ft}^2\text{.}
\end{equation*}
It is important to round up to have enough paint.
The following example requires geometric thinking, but does not require any formula. Especially important is deciding on the type of rounding which, in this case, is determined by physical constraints.
Example 4.1.33.
Katie is building a large scale abacus for a park. Her plan is to build it from treated 2x4 lumber. Her plan is shown in Figure 4.1.34 Note the depth of each piece of wood is 3.5". If you are wondering why a 2x4 is 1.5 in x 3.5 in, note that the nominal size (2x4 in this case) is based on the initial cut. The lumber shrinks as it cures and again when it is planed smooth.
Because we must have enough wood, we will round up all approximations. Measurements are accurate to an 8th of an inch.
(a)
What is the total number of feet of lumber (2x4) needed?
Solution.
There are two boards of length 60 inches and three boards of length 27 inches. The total length is
\begin{equation*}
2 (60\text{ in}) + 3(27.0\text{ in}) = 120\text{ in}\text{.}
\end{equation*}
We convert this to feet using a ratio.
\begin{equation*}
120 \text{ in} \cdot \frac{1 \text{ ft}}{12 \; \text{in}} = 10 \text{ ft}
\end{equation*}
(b)
If a standard 2x4 is 96.0 inches long, what is the smallest number of boards Katie can purchase to have enough lumber?
(c)
If the boards are painted before they are assembled, what is the total surface area of the boards to be painted? Paint cans are rated for the number of square feet they can cover. We will want the solution in units of square feet.
Solution.
Each board has six surface. Each surface size appears twice (e.g., top and bottom). For the long segments these areas are
\begin{align*}
60\text{ in} \cdot 3.5\text{ in} \amp = 210\text{ in}^2,\\
60\text{ in} \cdot 1.5\text{ in} \amp = 90\text{ in}^2,\\
1.5\text{ in} \cdot 3.5\text{ in} \amp = 5.25\text{ in}^2.
\end{align*}
For the short segments these are
\begin{align*}
27\text{ in} \cdot 3.5\text{ in} \amp = 94.5\text{ in}^2,\\
27\text{ in} \cdot 1.5\text{ in} \amp = 40.5\text{ in}^2,\\
1.5\text{ in} \cdot 3.5\text{ in} \amp = 5.25\text{ in}^2.
\end{align*}
Thus the total area is
\begin{align*}
2(2)(210\text{ in}^2)+2(2)(90\text{ in}^2)+2(2)(5.25\text{ in}^2)\\
+3(2)(94.5\text{ in}^2)+3(2)(40.5\text{ in}^2)+3(2)(5.25\text{ in}^2) \amp = \\
810+360+21+567+246+31.5\text{ in}^2 \amp = 2035.5\text{ in}^2
\end{align*}
Finally we convert this to square feet using the method of Example 1.1.23.
\begin{align*}
2035.5\text{ in}^2 \cdot \left(\frac{1 \text{ ft}}{12 \text{ in}}\right)^2 & = \\
\frac{2035.5\text{ in}^2}{12^2 \text{ in}^2} \text{ ft}^2 & = \\
\frac{2035.5}{144} \text{ ft}^2 & \approx 14.13541667 \text{ ft}^2.\\
& \approx 15\text{.}
\end{align*}
We round up because we want to be safe and because we likely cannot order an amount to cover exactly 14.1 square feet.
(d)
What is the area that is hidden, that is, cannot be seen after assembling?
Solution.
This would be where the three short boards touch the long boards. There are six places where this happens which are all the same shape \(3.5\text{ in} \cdot 1.5\text{ in} = 5.25\text{ in}^2\text{.}\) The covered surface is on both the short boards and the matching spot on the long boards, so there are 12 of these surfaces for \(12 \cdot 5.25\text{ in}^2 \approx 63\text{ in}^2\text{.}\)
The top and bottom (longest) sides are 60.0 inches long. The left, right, and divider sides are 27.0 inches long. All lumber is 1.5 inches wide. Depth of the wood is not shown in the diagram. The divider piece is 15.0" from the left edge.
The following example also illustrates rounding according to physical constraints.
Example 4.1.35.
We are purchasing carpet squares to carpet a rectangular room that is 15.5 ft by 19.0 ft. These carpet squares are 18 in by 18 in. A box contains 10 squares. How many boxes of carpet squares do we need purchase? This image shows four carpet squares together.
We need to know how many square across and down the room. For the number of squares across we can divide the width of the room (15.5 ft) by the width of a carpet square (18 in). This requires matching units, so \(15.5 \text{ ft} \cdot \frac{12 \text{ in}}{\text{ft}} = 186 \text{ in}\) Thus the number of squares across is \(\frac{186}{18} = 10.333\text{.}\) This means the final square will be a third of a square in width. While it might seem efficient to reuse the rest of that carpet square in the rows above, we cannot because the pattern would not continue correctly. Thus we simply round up to 11 squares across.
Next we calculate the number of squares down. \(19.0 \text{ ft} \cdot \frac{12 \text{ in}}{\text{ft}} = 228 \text{ in}\text{.}\) Thus the number of squares down is \(\frac{228}{18} = 12.666\text{.}\) Again to have enough squares we must round up to 13 squares down.
Thus the total number of carpet squares needed is \(11 \cdot 13 = 143\text{.}\) Because each box has 10 squares each, we need \(\frac{143}{10} = 14.3\) boxes. To have enough we need 15 boxes. In practice we would likely buy at least one extra box in case there are problems.
Example 4.1.36.
At first the following might seem to be a reasonable way to solve this problem. First, calculate the total square inches of floor. Using the conversion to inches from above \(186 \cdot 228 = 42408\) square inches of floor.
Next we can divide by the number of square inches each carpet squares covers to count the number of squares. Each square covers \(18^2 = 324\) square inches. Thus we calculate that we need \(\frac{42408}{324} = 130.888\) carpet squares.
This is substantially fewer than the 143 we calculated previously. The reason is this calculation assumes we can use each of the partial squares somewhere else in the room. This is true when the pattern does not need to match, but not in this case.
Now you may understand why such repeating patterns are avoided in most carpets.
Checkpoint 4.1.37.
Subsection 4.1.3 Quantitative Literacy
We use many methods to help us remember formulae and to use the correct one. We can use dimensional analysis (watching units) to help us use the correct formula in geometry.
Example 4.1.38.
A circle has radius \(r=5.3\) in. To calculate the area of the circle we will use either the formula \(2\pi r\) or \(\pi r^2\text{.}\) One is area, and the other is circumference (perimeter). If we recall that area has square units then we notice \(\pi r^2\) will end up with \((5.3 \text{ in})^2\) which has squared units. Thus \(\pi r^2\) is the area formula. \(2\pi r\) must be circumference.
Next we consider if there is any obvious relationship between the circumference and area of a circle. Specifically we ask if one is always bigger than the other.
Example 4.1.39.
Is the area of a circle is always a larger number than the circumference of a circle? This might seem reasonable because the circumference is just the edge of the circle and area is all of it. Guesses are an important part of mathematics. After we guess, we check our conjecture. Consider the following.
\begin{align*}
r \amp = 1.\\
2\pi(1) \amp \approx 6.2832.\\
\pi(1)^2 \amp \approx 3.1416\\
2\pi(2) \amp \approx 12.5664.\\
\pi(2)^2 \amp \approx 12.5664\\
2\pi(3) \amp \approx 18.8496.\\
\pi(3)^2 \amp \approx 28.2743
\end{align*}
Immediately we can see that the circumference is neither always less or always more than the area. If \(r=1\text{,}\) circumference is a larger number than the area. For \(r=3\) the relationship is reversed.
Generally, there is no relationship between perimeter and area of shapes.
Exercises 4.1.4 Exercises
Practice Formulae.
Use the appropriate formulae to calculate requested values.
Compound Shapes.
Use one or more formulas to calculate the areas shown.
Geometry in Applications.
Identify shapes and use appropriate formulae to calculate the results requested.
Quantitative Literacy.
Use you knowledge of geometry to determine what makes sense.
