The figures seem to describe an irregular pentagon. Without some angles it isn't possible to determine any areas with side measurements alone. You can see this if you imagine the sides are all connected with hinges at the vertices. Many shapes and their areas are possible because the sides are free to move on the hinges, even though all the sides are connected. Some other properties need to be known. For example, are there any right angles?
The best way to find the area is to divide the figure into shapes, like triangles, for which you can calculate the separate areas. A way to start is to take one side as a baseline. The longest side is 143 (feet, I presume). Use a graph and mark 143 on the x axis. The side AE has coordinates A(0,0) and E(0,143). The two sides, AB and DE, on either side of the longest side can be drawn next: AB=48 and DE=51. But you don't know either of the angles, so set your compasses first to 48 (circle at A) and then 51 (circle at E) and draw circles at the ends of 143.
The remaining two sides are quite long. Choose a point B on the circumference of the circle at A at, say, a position between 12 o'clock and 6 o'clock, viewing the circles as clock faces. On that point draw a circle using a radius equal to the representational length of the next side (119). Now draw a circle with radius equal to the remaining side (129) from a point D on the circumference of the circle at E. The point on the circumference must allow you to draw a circle that intersects the other big circle. Choose the highest intersection point as point C. You can now join up all five points. This is one shape, ABCDE, that fits all your side measurements, and you can certainly draw other shapes that fit. One of them may resemble the area of the lot. AB=48, BC=119, CD=129, DE=51, EA=143.
As an example, I drew the figure so that angles BAE and DEA were right angles. Drop a perpendicular CX where X is on AE, and another perpendicular AY where Y is on DE. CX and BY intersect at right angles at Z.
What we have now are two trapeziums, or trapezoids, ABCX and CDEX, back to back with CX as their common back. Each of these figures consists of a rectangle and a right-angled triangle. We can find the areas of these components and add them together to give us the area of ABCDE.
But first we need to calculate the length of CX. Draw WD parallel to and equal in length to XE, forming the rectangle WDEX. W lies on CX a distance 3 from Z. Here are the figures for which we need the areas: rectangles BZXA and WDEX, and triangles BCZ and CDW.
The length of CX=CW+51=CZ+48. We also need the length of AX=BZ. XE=ZY=143-AX. Let x=AX and y=CX.
Area of BZXA=48x; area of WDEX=51(143-x); area of BCZ=x(y-48)/2; area of CDW=(143-x)(y-51)/2. Total area=48x+7293-51x+xy/2-24x+143y/2-7293/2-xy/2+51x/2=7293/2-3x/2+143y/2 or (1/2)(7293-3x+143y). Now we need to find y (CX).
In triangle BCZ: 119^2=(y-48)^2+x^2; in triangle CDW: 129^2=(y-51)^2+(143-x)^2 (using Pythagoras' theorem).
So, 119^2=y^2-96y+48^2+x^2 and x^2+y^2-96y=119^2-48^2=71*167=11857, so x^2+y^2=11857+96y.
And, 129^2=y^2-102y+51^2+143^2-286x+x^2 and x^2+y^2-286x-102y=129^2-51^2-143^2=-6409.
Substituting for x^2+y^2: 11857+96y-286x-102y=-6409, so 286x+6y=18266; 143x+3y=9133 and y=(9133-143x)/3.
We can now substitute for y in 119^2=(y-48)^2+x^2 to find x:
(9133-143x-144)^2/9+x^2=119^2=(8989-143x)^2/9+x^2.
127449=(8989-143x)^2+9x^2=80802121-2570854x+20449x^2+9x^2;
so 20458x^2-2570854x+80674672=0 is the quadratic that needs to be solved.
x=(2570854+sqrt(2570854^2-4*20458*80674672)/40916=64.952 or 60.713, according to the formula.
143-x=78.048 or 82.287 and y=(9133-143x)/3=-51.73 or 150.347.
We won't follow up the negative value for y in this example so, x=60.713, y=150.347 and 143-x=82.287. [The negative value for y would in fact give us a different shape, with a negative area, because CX would be below AE instead of above it.] C is the point (60.713,150.347).
[A quick way to check these answers is to use the Pythagorean fact that CZ^2+BZ^2=BC^2 and CW^2+WD^2=CD^2. BZ=x and WD=143-x; CZ=y-48 and CW=y-51. BC=119 and CD=129. When the values for x and y are substituted, the facts are confirmed.]
From these values we can find the area=(1/2)(7293-3x+143y)=14305.24 sq ft=0.3284 acre (43560 sq ft=1 acre).
This is one possible answer, but not the only one, because of the variety of shapes fitting the criteria.