Algebra Review: Area of a Trapezoid

There are lots of nice and simple and very visual derivations of the area of a trapezoid. Those are the ones I hope my students hang on to and rely on when they are thinking about how to find and calculate its area. (We’ve already talked about my reservations with simply memorizing area formulas, and the problems that can arise from rote memorization).

However, in my geometry classes, I like to present this algebraic derivation of the area formula for a trapezoid. This is about the point in the year where students have forgotten a lot of their algebra skills from last year, and even cycling through some of those skills in classwork and homework only does so much. Playing around with some ideas, including thinking about a trapezoid as a decapitated triangle (because that’s clearly what it is, right?) led me to this derivation. I won’t claim that I’m the first to come up with it, and searching around online I’ve found some similar derivations, but I really like this one because of all the ideas it touches on: similar triangles, somewhat complicated proportional relationships, distribution and factoring of algebraic expressions, manipulating fractions, and more. Even though a few eyes start to glaze over when I first tell students I’m going to lead them through a derivation filled with lots of algebra, they start to get interested when they see that there are places where those algebra skills can really help make sense of a situation.

Here’s the derivation:

Deriving the Area of a Trapezoid (With Lots of Algebra Review)

Suppose we have trapezoid \square BCED with an altitude of length h drawn through points F on \overleftrightarrow{BC} and G on \overleftrightarrow{DE}. Let BC=b_1 and let DE=b_2.

Trapezoid BCDE

Extend the legs until they intersect at point A, creating \triangle ADE and \triangle ABC. Since \overline{BC} \parallel \overline{DE}, we know that \triangle ABC \thicksim \triangle ADE. Also, label the intersection of \overline{DE} and the line perpendicular to \overline{DE} through A as G, and label the intersection of \overline{BC} and \overline{AG} as F. Then FG = h is the altitude of trapezoid \square BCED, and the altitude of  \triangle ABC is AF=h_2.

Since we have similar triangles, we can write the following proportion: \frac{h_2}{h}=\frac{b_1}{b_2 - b_1}. Also, since \alpha \triangle ABC + \alpha \square BCED = \alpha \triangle ADE, we can subtract to find the area of the trapezoid: \alpha \square BCED = \alpha \triangle ADE - \alpha \triangle ABC. The areas of the triangles can be calculated: \alpha \triangle ABC = \frac{1}{2} (b_1)(h_2) and \alpha \triangle ADE = \frac{1}{2} (b_2)(h_2 + h). When we substitute the area calculations, we find that:

\alpha \square BCED =  \frac{1}{2} \big[(b_2)(h_2 + h) - \frac{1}{2} (b_1)(h_2)\big]

At this point, we have the area of the trapezoid in terms of the measurements of its two bases (b1 and b2), the measurement of its height (h), and the measurement of the height of , h2. We can use the proportion we found, \frac{h_2}{h}=\frac{b_1}{b_2 - b_1}, and solve it for h2 to find  h_2=\frac{b_1 h}{b_2 - b_1}. Substituting this in the equation above gets an expression for the area of trapezoid  in terms of the two bases and the height of the trapezoid.

\alpha \square BCED = \frac{b_2 h + b_2 \bigg(\frac{b_1 h}{b_2 - b_1}\bigg) - b_1 \bigg(\frac{b_1 h}{b_2 - b_1}\bigg)}{2}

Factoring out \frac{1}{2} and simplifying the fractions:

\alpha \square BCED = \frac{1}{2} \bigg( b_2 h + \frac{b_1 b_2 h}{b_2 - b_1} - \frac{(b_1)^2 h}{b_2 - b_1} \bigg)

Then multiply b_2 h by \frac {b_2 - b_1}{b_2 - b_1} to make all the terms inside the parentheses have the same denominator:

\alpha \square BCED = \frac{1}{2} \bigg( b_2 h \bigg( \frac {b_2 - b_1}{b_2 - b_1} \bigg ) + \frac{b_1 b_2 h}{b_2 - b_1} - \frac{(b_1)^2 h}{b_2 - b_1} \bigg)

Now distribute and place everything inside the parentheses into one fraction and simplify:

\alpha \square BCED = \frac {1}{2} \bigg( \frac {(b_2)^2h - b_1 b_2 h + b_1 b_2 h - (b_1)^2h}{b_2 - b_1} \bigg)

\alpha \square BCED = \frac {1}{2} \bigg( \frac {(b_2)^2h - (b_1)^2h}{b_2 - b_1} \bigg)

Factoring out the h leaves:

\alpha \square BCED = \frac {h}{2} \bigg( \frac {(b_2)^2 - (b_1)^2}{b_2 - b_1} \bigg)

Then factoring the difference of squares gives:

\alpha \square BCED = \frac {h}{2} \bigg( \frac {(b_2 + b_1)(b_2 - b_1)}{b_2 - b_1} \bigg)

Now divide out the b_2 - b_1 and rearrange to get our familiar area formula:

\alpha \square BCED = \frac {h (b_1 + b_2)}{2}

There are a lot of things I like about this approach, but the engagement of students as I lead them through this proof is a lot of fun. I rarely stand up and lecture or give students almost everything in a proof, but there are times when lecturing is effective and shouldn’t be summarily abandoned. Plus, this also gives me a chance, as we go through each step, to walk around and see which students may need more help with fractions or factoring as we continue through the year, and I can give them some additional help and resources. It’s also an opportunity to talk with the Algebra 2 teacher about the upcoming class and their current skill levels with important concepts.

 

 

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