I’m in a bit of an oddball situation.
I have to measure the square footage of a building that is in the shape of an ellipse that has been truncated on either end of the major axis, so that the resulting chords are parallel to the minor axis. Unfortunately, my only tool measures length, so that I have no way to measure angles.
I have an idea for what I can do if the chords intersect the foci, but if not I am at a complete loss.
How might I find the area? What conditions will make it impossible to calculate it exactly?
Do not assume the shape is symmetrical. The two chords may not be the same length.
If I'm reading right, the truncated ellipse contains the center of the complete ellipse, right?
FIND THE SEMI-MAJOR AXIS
Imagine tying a rope to one of the chords' ends and use it to draw out a circle having radius the length of the chord. Repeat using the opposite end point as the circle's center. These will intersect at one point in the interior and one point in the exterior (if you are using a drawing). Repeat for the other chord. Now draw a line through the two interior points of intersection, you have constructed part of the major axis that also cuts the chords into equal halves.
METHOD FOR FINDING Y-COORDINATES
Using this partial major axis, you can now associate a y coordinate with any point on the ellipse. Suppose you have line L and point P on that line. Pick a point C not on that line as a center of a circle have radius CP. The circle will intersect the line to points, P and a new point S. Draw a line through S and the center of the circle. Call the point where it intersects the circle F. Draw a line through F and P. By Thales' Theorem, this is perpendicular to L. Measure the distance from P through F to the the ellipse and you can assign a y coordinate to where this line intersects the Ellipse. You can now assign an Y coordinate to any point on the truncated ellipse.
FINDING CENTER OF ELLIPSE
You can use the same principle to find the center of the ellipse. Use the above procedure to construct a perpendicular to the end points of the shorter of the two chords. Where these perpendiculars intersect the ellipse will be points on a new chord the same length as the shorter chord. Draw lines through opposite endpoints of the congruent chords. Where they intersect is the center of the ellipse.
FINDING X-COORDINATES Now between the Pythagorean Theorem and the ability to assign a y coordinate to any point on the ellipse, you can assign an x coordinate. Let $l$ be length of rope from center to point. Then $l^2=x^2+y^2$. You know $l$ by measuring. You know $y$ by the process above, so now you can solve for $x$.
FIND LENGTH OF SEMI-MINOR AXIS. Construct a perpendicular through the semi-major axis through the center of the ellipse and measure the distance between the parts that intersect the ellipse. This gives you the minor and semi-minor axes lengths.
FIND EQUATION OF ELLIPSE The equation of the ellipse centered at the origin is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. For any of the endpoints of the original chords, you have $x$ and $y$. You now have one of $a$ and $b$ since you know the semi-minor axis length, so you can solve for the other. At this point, you might not know which is the semi-major or semi-minor axis, but you've narrowed the possibilities down to two options.
FIND AREA
Now that you have a complete equation for the ellipse and you have a way of finding your $x$ coordinates for your limits of integration, you can use calculus to find the area.
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\implies y=\pm b\sqrt{1-x^2/a^2}$
Area=$2\int_{c_1}^{c_2}y= b\sqrt{1-x^2/a^2} dx$