Maxwell is walking in a meadow. First, he walks north for $8$ m and then northeast for $5$ m. From there he walks $3$ m south and then southeast for $10$ m. At that point, he turns and walks in a straight line back to his starting point. What is the area (to the nearest square meters) of the region enclosed by the path that Maxwell walked?

Source: NCTM Mathematics Teacher 2008

SOLUTION

Triangle $DAB$ is a right and isosceles triangle with hypotenuse $AB=5$.
$AB=AD\sqrt{2}$
$5=AD\sqrt{2}$
$AD=\dfrac{5}{\sqrt{2}}=\dfrac{5\sqrt{2}}{2}$
Area of triangle $DAB$
$\dfrac{1}{2}AD\cdot DB=\dfrac{1}{2}\cdot\dfrac{5\sqrt{2}}{2}\cdot\dfrac{5\sqrt{2}}{2}$
$=\dfrac{50}{8}$
$=6.25$
Triangle $FCE$ is a right and isosceles triangle with hypotenuse $CE=10$.
$CE=FE\sqrt{2}$
$10=FE\sqrt{2}$
$FE=\dfrac{10}{\sqrt{2}}=\dfrac{10\sqrt{2}}{2}=5\sqrt{2}$
Area of triangle $FCE$
$\dfrac{1}{2}FE\cdot FC=\dfrac{1}{2}(5\sqrt{2})(5\sqrt{2})$
$=\dfrac{50}{2}$
$=25$
Area of rectangle $ADFG$
$FD=FC-DC$
$=FC-(DB-BC)$
$=5\sqrt{2}-\left (\dfrac{5\sqrt{2}}{2}-3\right )$
$=\dfrac{5\sqrt{2}}{2}+3$
$AD\cdot FD=\dfrac{5\sqrt{2}}{2}\left (\dfrac{5\sqrt{2}}{2}+3\right )$
$=12.5+10.61$
$=23.11$
Area of right triangle $GEH$
$GH=AH-AG$
$=8-\left (\dfrac{5\sqrt{2}}{2}+3\right )$
$=5-\dfrac{5\sqrt{2}}{2}$
$GE=GF+FE$
$=\dfrac{5\sqrt{2}}{2}+5\sqrt{2}$
$=\dfrac{15\sqrt{2}}{2}$
$\dfrac{1}{2}GH\cdot GE=\dfrac{1}{2}\left (5-\dfrac{5\sqrt{2}}{2}\right )\dfrac{15\sqrt{2}}{2}$
$=7.77$
$6.25+25+23.11+7.77=62.13$ square meters
Answer: $62.13$ square meters