Question

applying triangle classification theorems
to get from his house to the lecture hall at school, lin walked west 651 feet. after class, he walked northeast 910 feet to the gym. finally, he walked 615 feet back to his house from the gym.
what general direction did lin walk from the gym to his house, and what type of triangle did his walking path form?
○ lin walked south, creating a right triangle.
○ lin walked southwest, creating an obtuse triangle.
○ lin walked southeast, creating an acute triangle.
○ lin walked directly east, creating a right triangle.
diagram: triangle with vertices labeled lecture hall, gym, lins house; side lengths: lecture hall to lins house 651 ft, lins house to gym 615 ft, gym to lecture hall 910 ft

Explanation:

Step1: Identify triangle side lengths

Let $a=615$, $b=651$, $c=910$

Step2: Square each side

$a^2 = 615^2 = 378225$
$b^2 = 651^2 = 423801$
$c^2 = 910^2 = 828100$

Step3: Sum smaller squared sides

$a^2 + b^2 = 378225 + 423801 = 802026$

Step4: Compare to largest squared side

Since $802026 < 828100$, so $a^2 + b^2 < c^2$. This means the triangle is obtuse.

Step5: Determine direction

Lin walked west then northeast; the reverse path from gym to house is southwest.

Answer:

Lin walked southwest, creating an obtuse triangle.