Answer:
[tex]6 \sqrt{2}[/tex] ,[tex]6 \sqrt{11}[/tex]
Step-by-step explanation:
Refer the attached figure according to question.
Katrina lives directly east of the park.
So, Distance between Park and Katrina's home = AB
The football field is directly south of the park.
So, Distance between football field and park = AD
Now we are given that he library sits on the line formed between Katrina’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 4 miles from her home i.e. BC = 4 miles
The football field is 18 miles from the library.i.e. CD = 18 miles
So, Distance between football field and home is BD=CD+BC= 18+4=22 miles
Using the property of similar triangles
[tex]AD^2 = BD \times CD[/tex]
So, [tex]AD^2 = 22 \times 18[/tex]
[tex]AD^2 =396[/tex]
[tex]AD =\sqrt{396}[/tex]
[tex]AD =6 \sqrt{11}[/tex]
b) Thus the park [tex]6 \sqrt{11}[/tex] from the football field.
Now to Find Distance between library and park i.e. AC
[tex]Hypotenuse^2=Perpendicular^2+base^2[/tex]
[tex]AD^2=AC^2+CD^2[/tex]
[tex]396=AC^2+18^2[/tex]
[tex]396=AC^2+324[/tex]
[tex]396-324=AC^2[/tex]
[tex]72=AC^2[/tex]
[tex]\sqrt{72}=AC[/tex]
[tex]6 \sqrt{2}=AC[/tex]
a)Thus the library is [tex]6 \sqrt{2}[/tex] from the Park.
hence Option C is correct.
