Definition 4.1.1. Parallelogram.
A parallelogram is a four sided shape for which opposing pairs of sides are parallel.
Section 4.1 Geometric Reasoning
Here we will use a variety of formulas for geometric properties on basic shapes to analyze objects in context and more complicated shapes. These formulas are provided (largely) without explanation. Our goal is to break down complex problems into simpler problems we can solve using these formulas on basic shapes.
Subsection 4.1.1 Properties
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.
Subsection 4.1.2 2D Shapes
Note this includes rectangles, which are parallelograms with four right angles, and rhombi which are parallelograms with four equal length sides. Notice that a square is a rectangle and a rhombus.
Shape |
Perimeter |
Area |
 |
\(2(a+b)\) | \(h_1 a\) |
Example 4.1.3.
What are the perimeter and area of the parallelogram in Figure 4.1.4?
Solution.
The perimeter is the sum of the sides which in this case is
\begin{equation*}
2(4.05+5.00) = 18.1.
\end{equation*}
The area of a parallelogram, given in Table 4.1.2 is \(h_1 a\text{.}\) For this parallelogram that is
\begin{equation*}
\text{Area} = 3.00 \cdot 5.00 = 15.0.
\end{equation*}
Checkpoint 4.1.5.
In Table 4.1.2 the height is labeled \(h_1\text{.}\) Where is another height that could be used?
Checkpoint 4.1.6.
What does \(h_1\) equal in a rectangle?
Definition 4.1.7. Trapezoid.
A trapezoid is a four sided shape for which one pair of opposing sides are parallel.
Shape |
Perimeter |
Area |
 |
\(a_1+b_1+a_2+b_2\) | \(\frac{h}{2}(a_1+a_2)\) |
Example 4.1.9.
What are the perimeter and area of the trapezoid in Figure 4.1.10?
Solution.
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) \approx 10.1.
\end{equation*}
Definition 4.1.11. Triangle.
A triangle is a three sided shape.
Shape |
Perimeter |
Area |
 |
\(a+b+c\) | \(\frac{1}{2}bh\) |
Note that the value \(h\) in the area formula is called the height of the triangle. It is the length of a line segment perpendicularly down from a vertex to the opposing side (or extension of it). The vertical, dashed line segments in Figure 4.1.14 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 4.1.13.
What are the perimeter and area of the triangles in Figure 4.1.14?
Solution.
The perimeter of the triangle on the left is
\begin{equation*}
5.98+10.9+8.19 \approx 25.1.
\end{equation*}
The perimeter of the triangle on the right is
\begin{equation*}
5.98+2.87+8.19 \approx 17.0.
\end{equation*}
The area of a triangle, given in Table 4.1.12 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 \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 \approx 6.36
\end{equation*}
Checkpoint 4.1.15.
In Table 4.1.12 the height is labeled \(h_1\text{.}\) Where are other possible heights and where are their bases?
Theorem 4.1.16. 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 4.1.17.
Consider the triangle on the right in Figure 4.1.14. Consider the segments of length 4.43, 5.98, and the horizontal dashed segment. 5.98 is the length of the side not adjacent to the right angle (\(c\) in the formula). We can calculate the length of the horizontal, dashed segment using the formula.
\begin{align*}
4.43^2+b^2 \amp = 5.98^2.\\
19.6+b^2 \amp = 35.8.\\
b^2 \amp = 16.2.\\
\sqrt{b^2} \amp = \sqrt{16.2}.\\
b \amp \approx 4.06.
\end{align*}
In Section 7.3 we will develop a version of this statement for triangles without a right angle.
Theorem 4.1.18. 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 4.1.19.
Calculate the area of the triangles in Figure 4.1.14.
Solution.
According to Heron’s formula for the triangle on the left
\begin{align*}
s & = \frac{1}{2}(5.98+8.19+10.90)\\
& \approx 12.54.\\
\text{Area} & = \sqrt{12.54(12.54-5.98)(12.54-8.19)(12.54-10.90)}\\
& = 24.2.
\end{align*}
For the triangle on the right
\begin{align*}
s & = \frac{1}{2}(5.98+8.19+2.87)\\
& \approx 8.52.\\
\text{Area} & = \sqrt{8.52(8.52-5.98)(8.52-8.19)(8.52-2.87)}\\
& \approx 6.35.
\end{align*}
Shape |
Perimeter |
Area |
 |
\(2\pi r\) | \(\pi r^2\) |
| \(\pi d\) | \(\pi \frac{d^2}{4}\) |
Example 4.1.21.
For a circle with radius 7.31 what are the perimeter and area?
The perimeter, given in Table 4.1.20, is \(2\pi r\text{.}\) For radius 7.31 the perimeter is
\begin{equation*}
2\pi (7.31) \approx 45.9.
\end{equation*}
The area, given in Table 4.1.20, is \(\pi r^2\text{.}\) For radius 7.31 the area is
\begin{equation*}
\pi (7.31)^2 \approx 168.
\end{equation*}
Example 4.1.22.
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.
\begin{equation*}
\frac{1}{2}\pi (11.7)+11.7 \approx 30.1.
\end{equation*}
The area is simply half of the usual area.
\begin{equation*}
\frac{1}{2}\pi (11.7)^2 \approx 215.
\end{equation*}
Subsection 4.1.3 Applying Geometry
Our first task in using geometry properties is to break down a problem into the kinds of shapes we already know. Then we can use the properties to calculate results.
Example 4.1.23.
(a)
Find the area of this wall given the dimensions given in feet.
Solution.
First we note that we can describe the wall as a rectangle with a triangle on top of it.
The sides of the rectangle area are 7 ft (height) and 24 ft (width). This means that area is \(7 \text{ ft} \cdot 24 \text{ ft} = 168 \text{ ft}^2\text{.}\)
The top is a triangle with two sides of length 13 and one of length 24. We don’t 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+13+24)\\
& = 25.\\
\text{area} & = \sqrt{25(25-13)(25-13)(25-24)}\\
& = 60.
\end{align*}
The total area then is \(168+60=228\) square feet.
(b)
Find the perimeter of this wall given the dimensions given in feet.
Solution.
There are five (5) edges. There sum is \(7+24+7+13+13=64\) feet.
(c)
What is the (tallest) height of the wall?
Solution.
We need the height of the triangular portion of the wall to find the height at the peak. Height is part of the area formula, and from Heron’s formula we already know the height.
\begin{align*}
60 & = \frac{1}{2}24 h.\\
5 & = h.
\end{align*}
The total height is then \(7+5=12\text{.}\)
Example 4.1.25.
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 4.1.26 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.
(a)
What is the total number of feet of lumber (2x4) needed?
Solution.
There are two boards of length 60.0 inches and three boards of length 27.0 inches. The total length is
\begin{equation*}
2 \cdot 60.0\;\text{in} + 3 \cdot 27.0\;\text{in} = 12\bar{0}\;\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}} = 10.0 \; \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?
Solution.
If a 60.0 inch section is cut from a 96.0 inch board, we have \(96.0\;\text{in}-60.0\;\text{in}=36.0\;\text{in}\) left. This is long enough for one of the 27.0 inch segments but not more. Thus two board will cover all but the last 27.0 inch segment. We need 3, 96.0 inch boards.
(c)
If the boards are painted before they are assembled, what is the total surface area of the boards to be painted?
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*}
60.0\;\text{in} \cdot 3.5\;\text{in} \amp = 210\;\text{in}^2,\\
60.0\;\text{in} \cdot 1.5\;\text{in} \amp = 9\bar{0}\;\text{in}^2,\\
1.5\;\text{in} \cdot 3.5\;\text{in} \amp \approx 5.3\;\text{in}^2.
\end{align*}
For the short segments these are
\begin{align*}
27.0\;\text{in} \cdot 3.5\;\text{in} \amp \approx 95\;\text{in}^2,\\
27.0\;\text{in} \cdot 1.5\;\text{in} \amp \approx 41\;\text{in}^2,\\
1.5\;\text{in} \cdot 3.5\;\text{in} \amp = 5.3\;\text{in}^2.
\end{align*}
Thus the total area is
\begin{align*}
2(2)(210\;\text{in}^2)+2(2)(9\bar{0}\;\text{in}^2)+2(2)(5.3\;\text{in}^2)\\
+3(2)(95\;\text{in}^2)+3(2)(41\;\text{in}^2)+3(2)(5.3\;\text{in}^2) \amp = \\
810+360+21.2+570+246+31.8\;\text{in}^2 \amp = 2069\;\text{in}^2
\end{align*}
This is \(2069 \; \text{in}^2\text{.}\)
(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 \(3.5\;\text{in} \cdot 1.5\;\text{in} \approx 5.3\;\text{in}^2\text{.}\) 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 \(12 \cdot 5.3\;\text{in}^2 \approx 64\;\text{in}^2\text{.}\)
Checkpoint 4.1.27.
What is the total area of the birdhouse shown in Figure 4.1.28?
