Walk in a Meadow

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
Area of enclosed meadow
6.25+25+23.11+7.77=62.13 square meters

Answer: 62.13 square meters

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , , . Bookmark the permalink.

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