A man starts at point A, somewhere on cartesian coordinate system. He goes 4 units to the right and then he goes 6 units upwards. Finally he makes an angle of 45° with the x-axis downwards to the left over a distance of 2 units. How far away from A is he?

The answer should be #sqrt(56-20sqrt(2))#, but I don't know how you get there.

1 Answer
CW
Jul 8, 2017

#sqrt(56-20sqrt2)#

Explanation:

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The route that the man travels #(A->B->C->D)# is shown in the figure.

Given #AB=4, BC=6, and CD=2#,

#=> CF=CDcos45=2*(cos45)=2*(sqrt2)/2=sqrt2#

#=> DF=CDsin45=2*(sin45)=2*(sqrt2)/2=sqrt2#

#=> AE=AB-EB=AB-DF=4-sqrt2#
#=> BF=BC-CF=6-sqrt2#

#AD^2=AE^2+DE^2#
#=> AD^2=AE^2+BF^2#
#=> AD^2=(4-sqrt2)^2+(6-sqrt2)^2#
#=> AD^2=(16-8sqrt2+2)+(36-12sqrt2+2)#
#=> AD^2=56-20sqrt2#
#AD=sqrt(56-20sqrt2)#

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