Kathy lives directly east of the park. The football field is directly south of the park. The library sits on the line formed between Kathy'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 9 miles from her home. The football field is 12 miles from the library.
a. How far is the library from the park?
b. How far is the park from the football field?

Look at the picture in the attachment.

Using the Pythagorean theorem, set up a system of three equations:
[tex]x^2+y^2=(12+9)^2 \\ 12^2+z^2=x^2 \\ 9^2+z^2=y^2 \\ \\ x^2+y^2=441 \\ 144+z^2=x^2 \\ 81+z^2=y^2[/tex]

[tex]\hbox{substitute } 144+z^2 \hbox{ for } x^2 \hbox{ and } 81+z^2 \hbox{ for } y^2 \hbox{ in the first equation:} \\ 144+z^2+81+z^2=441 \\ 225+2z^2=441 \\ 2z^2=441-225 \\ 2z^2=216 \\ z^2=\frac{216}{2} \\ z^2=108 \\ z=\sqrt{108} \\ z=\sqrt{36 \times 3} \\ z=6\sqrt{3} \\ z \approx 10.39[/tex]

[tex]81+z^2=y^2 \\ 81+108=y^2 \\ 189=y^2 \\ \sqrt{189}=y \\ \sqrt{9 \times 21}=y \\ y=3\sqrt{21} \\ y \approx 13.75[/tex]

a. The library is approximately 10.39 miles (exactly: 6√3 miles) from the park.
b. The park is approximately 13.75 miles (exactly: 3√21 miles) from the football field.

calculista calculista · Answer 2 · 2018-08-05T20:02:32+02:00

see the attached figure to better understand the problem

Let

z---------> distance from the library to the park in miles

x-------> distance from the park to the to the football field in miles

y-------> distance from the park to Kathy's home in miles

we know that

In the right triangle ABC

Applying the Pythagorean Theorem

[tex] x^{2} +y^{2} =(12+9)^{2} \\ x^{2} +y^{2}=441 [/tex] -----> equation [tex] 1 [/tex]

In the right triangle ABD

Applying the Pythagorean Theorem

[tex] 12^{2} +z^{2} =x^{2} \\ 144 +z^{2}=x^{2} [/tex] -----> equation [tex] 2 [/tex]

In the right triangle BCD

Applying the Pythagorean Theorem

[tex] 9^{2} +z^{2} =y^{2} \\ 81 +z^{2}=y^{2} [/tex] -----> equation [tex] 3 [/tex]

Add equation [tex] 2 [/tex] and equation [tex] 3 [/tex]

[tex] 144 +z^{2}=x^{2} [/tex]

[tex] 81 +z^{2}=y^{2}\\ ------ [/tex]

[tex] 144+81+2z^{2} =x^{2} +y^{2} [/tex] -----> equation [tex] 4 [/tex]

Substitute equation [tex] 1 [/tex] in equation [tex] 4 [/tex]

[tex] 144+81+2z^{2}=441\\ 2z^{2} =441-225\\ 2z^{2}=216\\ z^{2}=108\\ z=\sqrt{108} miles\\ z=10.39 miles [/tex]

Find the value of x

[tex] 144 +z^{2}=x^{2}\\ 144 +\sqrt{108}^{2}=x^{2} \\ x^{2} =144+108\\ x^{2} =252\\ x=\sqrt{252} miles\\ x=15.87 miles [/tex]

therefore

the answer is

Part a) The distance from the library to the park is equal to [tex] 10.39 miles [/tex]

Part b) The distance from the park to the to the football field is equal to [tex] 15.87 miles [/tex]