Since the question only asked for an approximation to the nearest tenth of an acre, I bypassed all f the trig and approximated the shape as a right triangle. 1/2*b*h with the arc length approximating h.  The appoximation yields 0.29 acres.

My problem is that I sometimes get lost in the weeds and forget to use obvious simplifications.

Just for fun, you can make it accurate using Heron's formula for the small triangle area.. With a bigger triangle it will make the difference.

In fact the problem is not really a trigonometric issue since, with Heron's formula, you don't even need to use sine functions.

Don't keep us in suspense -- show us your workings :-)

OK, the sector area is S=0.5*R²*teta=0.5*L²/teta (L is arc length, teta in rad) and R=L/teta. From this you get the 2 missing sides of the small triangle. Given the 3 sides length (a,b,c), and applying Heron's formula S'=sqrt(p*(p-a)*(p-b)*(p-c)) where p is half perimeter of the triangle p=(a+b+c)/2, you get the answer S-S'.

Alternatively, with the radius of the circle approximated to 216.65 ft.

The arc is approximately 1/12 of the circle.

Area = π * r**2 = 147458. Area / 12 = 12288 sq ft = 0.28 acres.

Removing the small triangle still leaves a number closest to 0.3.

Where would an engineer be without the ability to make "engineering approximations"? LOL

For an arc this small it can reasonably be approximated even further by using the length of the arc as the triangle height and one of the sides as the length, for example: (116 * 216 / 2) / 43560 -> 0.29, though this would probably not pass muster in the real estate world.

I have a bad habit of trying to be precise though.  Given that all three sides of the small triangle to be excluded (triangle BCE) are known, its area can easily be computed by dropping a perpendicular to one of the sides (for example from B to (a newly designated point F on) CE), and using simple trigonometry to find the length of the perpendicular (for example BF), then the area of triangle BCE is half the product of the length of the perpendicular and the length of the side to which it was dropped (for example Area of Triangle BCE = BF * CE / 2).

Restating the full solution then yields the following:

AB = 216.27;
BC = 4.44;
CD = 217.03;
DA = 116.21 (arc);
Theta = 29 35.5' = 29.5917 degrees

Fraction of Circle = Theta / 360 = 0.0822
Circumference = DA / (Fraction of Circle) = 1413.7470
Radius = Circumference / (2 * pi) = 225.0048

BE = Radius - AB = 8.7348;
CE = Radius - CD = 7.9748
BF = BE * sin(Theta) = 4.3134
Area of BCE = BF * CE / 2 = 17.1993

Area of Slice = pi * Radius ^ 2) * (Fraction of Circle) = 13073.9029
Area of Parcel = (Area of Slice) - (Area of BCE) = 13056.7036
Acres of Parcel = (Area of Parcel) / 43560 = 0.2997

In the above, I carried four decimal places through the calculations, then rounded to four significant places (which also happens to be four decimal places) at the end.

I also carried out a more thorough treatment of the errors by considering the uncertainty of the input values which resulted in a range of acreage values from 0.2995 to 0.3000.

I did it the same basic way. A parallelogram 116 x 217, close enough to a rectangle the same size (created out of two triangles).

217 * 116 = 25172 for the rectangle. Divide by 2 to get the triangle = 12586. 43560 sq-ft / acre: 12586 / 43560 ~= 0.29, so the answer is 0.3 acres.

Took about 2 minutes, max.

Good question! Could use it on a math quiz.

It's funny -- I used to know this stuff like the back of my hand -- but it's been a while since I used it -- which explains my "deer in the headlights" look when my wife asked me the question LOL

The problem with knowing something/place like the back of your hand is: if the hand falls off, or otherwise becomes detached - you are lost.

(Happened to me ... once :^)

Mr Bandit -- is this you -- a.k.a. "OneArmedBandit" from the EETimes / Embedded.com sites?

My teachers always emphasized the importance of reading through the whole problem before starting to work on it. What's being requested is an estimate to the nearest 0.1 acre. Given that, one can simply approximate the lot as a section of a circle of radius 217'. Figure the area of the whole circle, multiply by the ratio of the lot's angle (~29.5 degrees) to a whole circle (360 degrees), convert to acres, and end up with 0.278 acres. Having that answer, it's obvious that the small triangular piece can simply be ignored.

Your approach is too difficult. There is a simpler way. For such distances, an arc can be considered as a straight line. So it needs only to get the trapeze area. The trapeze is the shape ABCD where  BC and AD are its bases. I think all know the needed formula: S = (BC + AD)/2*h. Where h is the height of the triangle of AED. It can be got very simple, for example as h = sqrt(AE^2 - (AD/2)^2. Using all of the numbers you will get the result 12614,38' or 0,28958631772 acr

I know, I know -- I couldn't help myself. And the problem is, being a bit obsessive compulsive, even if I'd started with the approximation, I'd have ended up doing the detailed version to see how close the approximation was LOL

Dear author, I have many pending assignments to do so I am leaving a hasty comment without reading other comments so apologies for that 

(the formulation is based on the fact that we need only approximations to 1/10th of an acre):

Area = integral of (1/2) * (AD/Theta)^2 * dTheta from 0 to 0.5164721 radians - integral of (1/2) * (BC/Theta)^2 * dTheta from 0 to 0.5164721 radians = 13054.965 ft^2 = 0.299 ~ 0.3 acres (the algebra can be simplified)

The equation is derived using calculus and the formula of an infinitesimal triangle, (1/2) * R * ds, where ds = altitude of infinitesimal triangle = differential arc length R*dTheta

Cheers for sharing a practical problem!

Thanks for posting your comment -- the fun thing for me is seeing how everyone comes at this from a different angle (no pun intended LOL)

I was disappointed that the triangle was too small to be considered and its area is so small that it can be neglected because the required area is much bigger.

However, here is how I solved it using approximations.

Area of triangle.  

The triangle can be approximated as right angle triangle, the larger side be along AB as AB is shorter in length than CD. The angle is approximately 30 degrees, sin 30 = 0.5, so the upper side should be around 8.8 feet. The lower side = 8.8 x cos 30 = 7.6 approximately. Area of triangle = 7.6 x 4.4 /2 = 16 square feet (approx.)

Now length AE (radius) = 216 + 8.8 = 225 (approximately)

Luckily the angle is about half radian, which makes calculations much easier. I don't have to bother about minutes and seconds :)

Area of circle = pi r^2  for 2 pi radians. So area corresponding to half radian will be 

225 x 225 /4 = 12656 square feet. 

subtract area of small triangle  which is 16 sq. feet

I did not do this subtraction as it is too small quantity 

one acre = 43560 sq. feet, so 

12656 sq feet = 0.2905  which is close to 0.3 !

There is no need to even bother with the small triangle when the answers are so far apart.  It is only a handful of square feet!

I'm not saying you are not right, but not accounting for it would be a constant "niggle" at the back of my brain LOL

The 4.44 measurement appears to be incorrect.

I get 4.33 based on the unambiguous radius and the segment data provided.

Lovely Realtor has made a measurement mistake, it appears.   Not horrible, and not big enough to torpedo your multiple choice quest for an answer, but wrong enough.   (At the very least, it should function as a math check, verifying the radius and it fails to do that.). Did you intentionally do this or am I off base?

I used the radius, the calculated missing segments and the given angle as SAS and law of cosines to get all sides and angles identified.  Side marked 4.44 came out as 4.33.   

I guess I did imply that the 4.44 line is at 90 degrees to a radius line from the center of the circle bisecting the angle of this "slice" -- but to be honest I'm not convinced this is the case -- the way I drew my diagrams was based on the description my wife gave me over the phone -- and when I got to look at the actual drawing it wasn't any more precise -- so I suspect that the 4.44 line was slightly askew (it's a pretty weird lot :-)  

This was a fun problem, but as Donna pointed out, you have extra information.  I would suggest your genius readers solve it exactly(ish) assuming that theta is wrong, since as you stated, it was not originally provided, and not needed, as Donna indicated.

I have not solved it myself yet, but will enjoy working on it, and especially will enjoy seeing the creative methods your readers use!

Thanks for your comment -- it's funny how these little problems can talk a bit of thinking through if it's not the sort of thing you do every day :-)