**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!*