Answer
He has to walk $720ft$ on the boardwalk.
Work Step by Step
At first, we have to assign values and variables to the parts of the trajectory that the man has to walk. It will help us for further calculations (For a better visualization see the image above).
Let's assign $x$ to the part of the boardwalk which the man will have to skip, till the point of the $210ft$ line intersection.
Let's say that the part of the way to walk on sand is $m$ and calculate it using the Pythagoras theorem. In the small right angle triangle, it is hypotenuse which is:
$m=\sqrt{210^2-x^2}=\sqrt{44100-x^2}$
Now, to find the part of the boardwalk that the man has to walk, we need length of the whole boardwalk and it is leg of the big right angle triangle:
Boardwalk = $\sqrt{750^2-210^2}=\sqrt{562500-44100}=\sqrt{518400}=720ft$
So, the part of the boardwalk that the man has to walk will be: $(720-x)ft$
According to the problem, the time man takes to walk on boardwalk plus the time he takes to walk on sand till the umbrella is $4min$ and $45sec$ = $285sec$
$t=\frac{S}{V}$
$t_1=\frac{720-x}{4}$
$t_2=\frac{\sqrt{44100-x^2}}{2}$
$\frac{720-x}{4}+\frac{\sqrt{44100-x^2}}{2} = 285$
$720-x+2\sqrt{44100-x^2}=1140$
$2\sqrt{44100-x^2}=420+x$
$4(44100-x^2)=176400+840x+x^2$
$176400-4x^2=176400+840x+x^2$
$5x^2+840x=0$
$x(x+168)=0$
$x_1=0$ or $x_2=-168$
We simply cross out the negative answer, since a distance cannot be negative.
We have $x=0$, which means that the man has to walk all the way on the boardwalk, till the intersection of the right angle line and then go directly to the umbrella.
So, he has to walk $720ft$ on the boardwalk.
