Largest rectangle inscribed in a given triangle:
For largest rectangle BDNE inscribed in the triangle ABC
BE = 0.5 BC and
BD = 0.5 BA (see figure below)
For mathematical proof, see Problem 32 of Maxima and Minima.
The same proportion is true if the rectangle is given and the triangle is unknown, which is the case of this problem.
Smallest triangle that can circumscribe a given rectangle:
Triangle ABC is circumscribing the rectangle BDNE, for ABC to be smallest
BE = 0.5 BC and
BD = 0.5 BA
It will follow that for minimum area of the triangle ABC, the Narra tree must be at the midpoint of the dividing fence AC.
By Pythagorean Theorem
From triangle CEN
$\left( L/2 \right)^2 = x^2 + y^2$
From triangle BEN
$\left( L/2 \right)^2 = 1009^2$
$L = 2018 ~ \text{m}$ ← Answer = [ D ]
For the area:
Cosine Law for triangle BFN
$1009^2 = 1898^2 + 2261^2 - 2(1898)(2261) \cos \theta$
$\cos = \dfrac{1898^2 + 2261^2 - 1009^2}{2(1898)(2261)}$
$\theta = 26.27^\circ$
From triangle DFN
$x = 1898 \sin \theta$
$x = 1898 \sin 26.27^\circ$
$x = 840 ~ \text{m}$
$2261 - y = 1898 \cos \theta$
$y = 2261 - 1898 \cos 26.27^\circ$
$y = 559 ~ \text{m}$
Area of triangle ABC
$A = 4\left( \frac{1}{2}xy \right) = 4 \times \frac{1}{2}(840)(559)$
$A = 939,120 ~ \text{m}^2$ ← Check!
