so hmmm notice the picture below
a perpendicular line, from the opposite side to the right-angle in a triangle, gives us, three similar triangles, a Large, containing the other two, a Medium, and a Small one
now, using proportions, let's say, "h" is the distance from the mall to home, and "p" the distance to the park
[tex]\bf \cfrac{\textit{medium side}}{\textit{small side}}\qquad \cfrac{p}{6}=\cfrac{8}{p}\qquad \qquad \cfrac{\textit{large side}}{\textit{medium side}}\qquad \cfrac{14}{h}=\cfrac{h}{8}[/tex]
[tex]\bf \cfrac{p}{6}=\cfrac{8}{p}\implies p^2=48\implies p=\sqrt{48}\\\\
-----------------------------\\\\
48\implies 2\cdot 2\cdot 2\cdot 2\cdot 3\implies 2^2\cdot 2^2\cdot 3\implies (2^2)^2\cdot 3\implies 4^2\cdot 3\\\\
-----------------------------\\\\
p=\sqrt{4^2\cdot 3}\implies \boxed{p=4\sqrt{3}}[/tex]
[tex]\bf \cfrac{14}{h}=\cfrac{h}{8}\implies 112=h^2\implies \sqrt{112}=h\\\\
-----------------------------\\\\
112\implies 2\cdot 2\cdot 2\cdot 2\cdot 7\implies 2^2\cdot 2^2\cdot 7\implies (2^2)^2\cdot 7\implies 4^2\cdot 7\\\\
-----------------------------\\\\
\sqrt{4^2\cdot 7}=h\implies \boxed{4\sqrt{7}=h}[/tex]
