Question

Kristen lives directly east of the park. The football field is directly south of the park. The library sits on the line formed between Kristen’s home and the football field at the exact point where an altitude to the right triangle formed by her home, the park, and the football field could be drawn. The library is 2 miles from her home. The football field is 8 miles from the library. How far is library from the park? How far is the park from the football field?

Answer 1

Imagine right triangle PHF, where P - park, H - home and F - football field, then PH, PF are legs and HF is hypotenuse . Denote point L to be library. You know that point L lies on the segment FH and FL=8, LH=2. Also you know that PL is an altitude to the hypotenuse.

Use the property of altitude drawn from the vertex of right angle to the hypotenuse (the length of the altitude is geometrical mean between legs' projections onto hypotenuse):

$PL=\sqrt{HL\cdot LF},\\ PL=\sqrt{2\cdot 8}=\sqrt{16} =4$ mi.

This means that the distance between park and libriry is 4 miles.

Consider right triangle PLF ( angle L is right angle and PF - hypotenuse). By the Pythagorean theorem,

$PF^2=PL^2+LF^2,\\ PF^2=4^2+8^2,\\ PF^2=16+64=80,\\ PF=\sqrt{80} =4\sqrt{5}$ mi. The distance between park and football field is $4\sqrt{5}$ miles.

Answer: the distance between park and libriry is 4 miles and the distance between park and football field is $4\sqrt{5}$ miles.