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 Question: Area of trapezoid with different lengths AB-272 feet: AC 690 feet; CD-330 Feet; DA-669Feet Calculate the area in feet and Acres.

Hi,

You didn't say which two sides are the parallel sides. There is an interesting note on the Math Forum site concerning this. I am going to describe how you can find the area if you know which two sides are parallel.

Suppose the vertices of the trapezoid are $P, Q, R$ and $S$ and side lengths are $w, x, y$ and $z$ feet and the sides of length $w$ and $y$ feet are parallel.

I added the line segment $QT$ which is parallel to $RS.$ Hence the trapezoid is partitioned into a parallelogram $QRST$ and a triangle $QTP.$ Hence the area of the trapezoid is the sum of the areas of the parallelogram and the triangle. $|ST| = y$ and $|TP| = w - y$ and the height of the triangle and the height of the parallelogram are equal so if you can find the height of the triangle $QTP$ you can find the area of the trapezoid.

I can show you two ways to find the height of the triangle $QTP.$

First method.

Drop a perpendicular $QU$ from $Q$ to meet $TP$ at $U.$

Let the $|TU| = t$ feet then $|UP| = |TP| - t$ then since $QTU$ and $QUP$ are right triangles, Pythagoras Theorem for these two triangles gives two equations in $h$ and $t.$ Solve for $h.$

Second method.

Find the area of triangle $QTP$ using Heron's Formula and us the fact that the area of this triangle is also $\large \frac12 \normalsize |TP| \times h$ to solve for h.

Once you have the area $A$ of the trapezoid in square feet type A square feet in acres into the Google Search window to convert this area to acres.

I hope this helps,
Penny