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A man walks due west 800 feet, then due north 900 feet, then due east 1500 feet, and finally due south 100 feet. How far is he from his starting point?

 Aug 3, 2022
 #1
avatar+113 
+2

It helps to draw this, using some drawing app.. GeoGebra online is fine for this sort of thing.

 

We can calculate the length EA by EA = \(\sqrt{EF^2 + FA^2} = \sqrt{700^2 + 800^2} = \sqrt{490,000 + 640,000} = \sqrt{1,390,000} = 1,063.015\)

Now we round this off to whole feet. He is 1,063 feet from where he started.

See https://www.geogebra.org/classic/ujqjzhfh

 Aug 3, 2022
edited by tuffla2022  Aug 3, 2022
edited by tuffla2022  Aug 3, 2022
 #2
avatar+124596 
+1

Welcome aboard, Tuffla!!

 

Here's one  more way

 

Let him start  at (0,0)

 

When he walks 800 feet west he is at  (-800, 0)

When he then  walks 900 ft due north he is at  (-800 , 900)

When he then walks 1500 ft due east he is at  ( -800 + 1500, 900) = (700,900)

When he walks due south 100 ft he is at ( 700 , 900 - 100) =  (700, 800)

 

And his distance from the starting  point (as Tuffla calculated)  is   

 

sqrt [ 700^2 + 800^2 ]  =

 

sqrt [ 100^2  ( 7^2 + 8^2) ]  =

 

100 sqrt [ 113  ]  ft   ≈   1063 ft

 

cool cool cool

 Aug 3, 2022

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