Trapezoid in Middle School Geometry

In today’s middle school geometry curriculum, the trapezoid is an essential shape, typically defined as a quadrilateral with one pair of parallel sides (often called the top base and bottom base). 

For example, if the line segments \(\overline{AB}\) and \(\overline{CD}\) are parallel, then we say “\(ABCD\) is a trapezoid.” In that case, “\(AB\)” is considered the top base and “\(CD\)” the bottom base. According to the Korean educational syllabus, students generally begin formal study of trapezoids - its definition and area formula—in the second year of middle school.

Area of a Trapezoid

The Trapezoid Area Formula

A trapezoid’s area can be calculated by the following formula:

\(\text{Area}(A) = \frac{ (a + b )}{2} \times h\)

Here, (a) is the length of the top base, (b) is the length of the bottom base, and (h) is the perpendicular distance (i.e., height) between those two parallel sides. For instance, if a trapezoid has a top base of 5 cm, a bottom base of 9 cm, and a height of 4 cm, its area is

\(A = \frac{(5 + 9)}{2} \times 4
= \frac{14}{2} \times 4
= 7 \times 4
= 28 \text{ cm}^2\)

One intuitive way to grasp this formula is to think in terms of triangles or parallelograms. If you have two identical trapezoids and rearrange them appropriately, you form a parallelogram, whose area is simply (\text{base} \times \text{height}). Splitting that parallelogram back into two equal parts yields the trapezoid area formula:

\(\frac{(\text{top base} + \text{bottom base})}{2} \times \text{height}\)

Examples and Problem-Solving

1) Simple Integer Example

Let a trapezoid have:

• top base \(a = 5 ,\text{cm}\),
• bottom base \(b = 7 ,\text{cm}\),
• height \(h = 3 ,\text{cm}\).

Its area is:
\(\frac{(5 + 7)}{2}\times 3
= \frac{12}{2}\times 3
= 6\times 3
= 18\text{ cm}^2\)

2) Using Meters in an Example (m)

Suppose a field has a trapezoidal shape with a top base of 12 m, a bottom base of 20 m, and a height of 10 m. Then its area is:

\(\frac{(12 + 20)}{2}\times 10
= \frac{32}{2}\times 10
= 16 \times 10
= 160\text{ m}^2\)

In practical surveying, fields or paddies shaped like trapezoids can be measured easily using this formula.

3) Real-World Problem (Roadway Width Measurement)

Consider a roadway cross-section with a top width of 8 m, a bottom width of 10 m, and a depth (height) of 2 m. Assuming this cross-section is a perfect trapezoid, its cross-sectional area is:

\(\frac{(8 + 10)}{2}\times 2
= \frac{18}{2}\times 2
= 9 \times 2
= 18\text{ m}^2\)

In civil engineering, trapezoidal cross-sections are often used to estimate the required amount of construction material or earthwork.

Connections to Other Polygon Area Formulas

The trapezoid is closely related to parallelograms, rectangles, and rhombi. For example:

• Parallelogram area = \(\text{base} \times \text{height}\).
• Rectangle area = \(\text{length} \times \text{width}\).
• Triangle area = \(\frac{1}{2} \times \text{base} \times \text{height}\).

By combining or reconfiguring two trapezoids, you can form a parallelogram. Conversely, clever cuts and rearrangements can yield triangles. This shows how the trapezoid area formula links to those of other polygons, boosting our understanding of the splitting and recombining of shapes in geometry.

In addition, from a calculus standpoint, the trapezoid area concept underpins the Trapezoidal Rule in numerical integration. This rule approximates the area under a curve by summing the areas of multiple small trapezoids.

Example Problem and Solution

Problem: Suppose a trapezoid \(ABCD\) has parallel sides \(AB\) and \(CD\), where \(AB = 6\text{ cm}\), \(CD = 14\text{ cm}\), and the trapezoid’s height is 5 cm. Let \(E\) be the midpoint of \(AB\), \(F\) the midpoint of \(CD\). We draw segment \(EF\). Find the areas of the smaller trapezoids \(AEFB\) and \(EFCD\) formed by this segment.

Outline of the Solution: