Finding the useable area of a triangular lot.

Question

I have purchase a lot to build a cabin. the lot is triangle shaped. There are streets on 2 sides and a neighboring cabin on the other. The lot dimensions are 103.43' at the neighbors property line, 93.53' along the left side and 99.10' along the right side (facing from the neighbors lot). My setbacks are 15' from my neighbors lot and 20' from each street. Can anyone help? Thanks in advance!! David

Answer 1

A construction to scale using Cinderella shows the area to be approximately 531.88. The edge $AB$ in the following image is the one to your neighbors.

If this approximation wasn't good enough, you'd have to start computations. Using the law of cosines you could compute the angles of your triangles. From the intercept theorem you could conclude the distance between the points where the line $DE$ intersects the outer triangle: that length relates to the length $AB$ like the original height of the plot relates to the height minus 15'. That original height can in turn be computed from one angle and the adjacent edge. Now you still need the distances between the points $D$ and $E$ and the corresponding endpoints on the outer triangle. You can obtain these by dividing 20' by the sine of the angle at that corner. I'm pretty sure that this is correct, but giving a sound proof would take a bit of time which I don't have just now. Once you have the length $DE$, you can obtain the other two inner lengths as well, since the inner and outer triangle are similar. You can therefore compute the area from these lengths using Heron's formula.

Computed with perfect precision (treating your decimal numbers as exact), the area computed as outlined above will be

$$A = \frac{33217}{200}\cdot\sqrt{\frac{7496394733369083}{5868607457510}} - \frac{108081}{20} \approx 531.88454926$$

This was done using sage and algebraic numbers:

AB = QQbar(10343/100)
AC = QQbar(9353/100)
BC = QQbar(9910/100) 
cosA = (AB^2 + AC^2 - BC^2)/(2*AB*AC) #   0.49746779652302
sinA = sqrt(1 - cosA^2)               #   0.86748244444630
cosB = (AB^2 + BC^2 - AC^2)/(2*AB*BC) #   0.57418604431081
sinB = sqrt(1 - cosB^2)               #   0.81872485397641
h1 = sinA*AC                          #  81.1356330290626
h2 = h1 - 15                          #  66.1356330290626
DE = h2/h1*AB - 20/sinA - 20/sinB     #  36.82486242666910
EF = BC*DE/AB                         #  35.28322407892205
DF = AC*DE/AB                         #  33.30010038447608
s = (DE+EF+DF)/2                      #  52.70409344503361
T = sqrt(s*(s-DE)*(s-EF)*(s-DF))      # 531.8845492639687
T0, T1, T2 = T.minpoly().coefficients() 
(-T1+sqrt(T1^2-4*T0*T2))/(2*T2)