Question

A city has a park shaped as a right angled triangle. The length of the longest side of this park is 80 m. The Mayor of the city wants to construct three paths from the corner point opposite to the longest side such that these three paths divide the longest side into four equal segments. Determine the sum of the squares of the lengths of the three paths.

Accepted Answer

C. Let right-angled vertex be $B$, hypotenuse $AC = 80$. $a^2 + b^2 = 80^2 = 6400$. Points $D, E, F$ divide $AC$ into 4 equal segments of 20 m each. $BE$ is the median to hypotenuse $\implies BE = 40 \implies BE^2 = 1600$. By Apollonius theorem on $	riangle ABE$ and $	riangle EBC$: $BD^2 + BF^2 + 20^2 + 20^2 = \frac{1}{2}(a^2 + b^2 + 40^2 + 40^2) = \frac{1}{2}(6400 + 3200) = 4800$. $BD^2 + BF^2 = 4800 - 800 = 4000$. Sum of squares = $BD^2 + BE^2 + BF^2 = 4000 + 1600 = 5600	ext{ m}^2$.

Right Angled Triangle Park Paths

Explanation

Practise this under exam conditions