$$
begin{align*}
d&=vt\
t&=dfrac{d}{v}\
&=dfrac{17 text{cm}}{2.51 frac{text{cm}}{text{s}}}\
&=quadboxed{6.77 text{s}}\
end{align*}
$$

$$
begin{align*}
A=r^{2}pi\
end{align*}
$$

Therefore, the area of the circle in question, in meters squared, is

$$
begin{align*}
A&=(12.77 text{m})^{2}pi\
&=quadboxed{512.3 text{m}^{2}}\
end{align*}
$$

$$
begin{align*}
A=dfrac{btimes h}{2}\
end{align*}
$$

where $b$ is base, and $h$ is height.

For a triangular sail boat with a height of $h=4.1 text{m}$ and a base of $b=6.15 text{m}$, the area is

$$
begin{align*}
A&=dfrac{6.15 text{m}times 4.1 text{m}}{2}\
&=quadboxed{13 text{m}^{2}}\
end{align*}
$$