# Another oldie

• Dec 13th 2008, 09:23 AM
Soroban
Another oldie

A surveyor starts at the center of a circular park with radius 7 miles.

He walks 3 miles directly east and drives a stake into the ground at point $\displaystyle A.$

He walk directly north until he reaches the perimeter of the park.
Then he walks directly west until he is directly north of the center,
. . and drives a stake into the ground at point $\displaystyle B.$

Find the distance $\displaystyle \overline{AB}.$

[The shortest solution, please.]
.
• Dec 13th 2008, 10:10 AM
skeeter
here's the hint:

diagonals of a rectangle are ... ?
• Dec 13th 2008, 01:06 PM
CaptainBlack
Quote:

Originally Posted by Soroban
A surveyor starts at the center of a circular park with radius 7 miles.

He walks 3 miles directly east and drives a stake into the ground at point $\displaystyle A.$

He walk directly north until he reaches the perimeter of the park.
Then he walks directly west until he is directly north of the center,
. . and drives a stake into the ground at point $\displaystyle B.$

Find the distance $\displaystyle \overline{AB}.$

[The shortest solution, please.]
.

You have a rectangle one diagonal of which is a radius, AB is the other diagonal.

CB
• Dec 14th 2008, 07:40 AM
ice_syncer
Use pythagoras theorem, the shortest possible method, no need of trigonometry (Thinking)

CLUE: IT IS A SURD!!!!
• Dec 14th 2008, 11:26 AM
skeeter
Quote:

Originally Posted by ice_syncer
Use pythagoras theorem, the shortest possible method, no need of trigonometry (Thinking)

CLUE: IT IS A SURD!!!!

don't think so ...
• Dec 15th 2008, 02:49 AM
ice_syncer
Quote:

Originally Posted by skeeter
don't think so ...

no, use pythagoras theorem.
• Dec 15th 2008, 03:00 AM
samer_guirguis_2000
let me guess! 7 miles may be?
• Dec 15th 2008, 03:13 AM
Greengoblin
^^The diagonals of a rectangle are equal, so it equals the radius, 7 miles.
• Dec 20th 2008, 09:10 PM
Aryth
I know this is a while afterwards but, the solution is quite simple.

One diagonal is a radius, so the Northern leg of the trip must be:

$\displaystyle 3^2 + b^2 = 7^2$

$\displaystyle b = \sqrt{40}$

So, the parallel side is also $\displaystyle \sqrt{40}$, so therefore the diagonal AB must be:

$\displaystyle 3^2 + (\sqrt{40})^2 = c^2$

$\displaystyle 7 = c$

And there you go.