The length of one side of the square - in particular, that of $\displaystyle \overline{SQ}$ - is equal to one fourth its perimeter, so it can be determined that $\displaystyle SQ = 64 \div 4 = 16$. However, no relation among the sides of the hexagon is given - in particular, it is not given that the hexagon is regular - so the length of $\displaystyle \overline{HE}$ cannot be determined. Insufficient information is given.

$\displaystyle P=2l+ 2w$

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

$\displaystyle 42=2l+2(7)$

$\displaystyle 42=2l+14$

Subtract $\displaystyle 14$ from both sides

$\displaystyle 42-14=2l+14-14$

$\displaystyle 28=2l$

Divide $\displaystyle 2$ by both sides

$\displaystyle \frac{28}{2}=\frac{2l}{2}$

$\displaystyle 14=l$

$\displaystyle P=2l+ 2w$

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

$\displaystyle 62=2l+2(8)$

$\displaystyle 62=2l+16$

Subtract $\displaystyle 16$ from both sides

$\displaystyle 62-16=2l+16-16$

$\displaystyle 46=2l$

Divide $\displaystyle 2$ by both sides

$\displaystyle \frac{46}{2}=\frac{2l}{2}$

$\displaystyle 23=l$

$\displaystyle P=2l+ 2w$

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

$\displaystyle 92=2l+2(21)$

$\displaystyle 92=2l+42$

Subtract $\displaystyle 42$ from both sides

$\displaystyle 92-42=2l+42-42$

$\displaystyle 50=2l$

Divide $\displaystyle 2$ by both sides

$\displaystyle \frac{50}{2}=\frac{2l}{2}$

$\displaystyle 25=l$

$\displaystyle P=2l+ 2w$

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

$\displaystyle 59=2l+2(17)$

$\displaystyle 59=2l+34$

Subtract $\displaystyle 34$ from both sides

$\displaystyle 59-34=2l+134-34$

$\displaystyle 25=2l$

Divide $\displaystyle 2$ by both sides

$\displaystyle \frac{25}{2}=\frac{2l}{2}$

$\displaystyle 12.5=l$

$\displaystyle P=2l+ 2w$

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

$\displaystyle 66=2l+2(18)$

$\displaystyle 66=2l+36$

Subtract $\displaystyle 36$ from both sides

$\displaystyle 66-36=2l+36-36$

$\displaystyle 30=2l$

Divide $\displaystyle 2$ by both sides

$\displaystyle \frac{30}{2}=\frac{2l}{2}$

$\displaystyle 15=l$

$\displaystyle P=2l+ 2w$

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

$\displaystyle 60=2l+2(14)$

$\displaystyle 60=2l+28$

Subtract $\displaystyle 28$ from both sides

$\displaystyle 60-28=2l+28-28$

$\displaystyle 32=2l$

Divide $\displaystyle 2$ by both sides

$\displaystyle \frac{32}{2}=\frac{2l}{2}$

$\displaystyle 16=l$

$\displaystyle P=2l+ 2w$

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

$\displaystyle 40=2l+2(6)$

$\displaystyle 40=2l+12$

Subtract $\displaystyle 12$ from both sides

$\displaystyle 40-12=2l+12-12$

$\displaystyle 28=2l$

Divide $\displaystyle 2$ by both sides

$\displaystyle \frac{28}{2}=\frac{2l}{2}$

$\displaystyle 14=l$

$\displaystyle P=2l+ 2w$

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

$\displaystyle 96=2l+2(30)$

$\displaystyle 96=2l+60$

Subtract $\displaystyle 60$ from both sides

$\displaystyle 96-60=2l+60-60$

$\displaystyle 36=2l$

Divide $\displaystyle 2$ by both sides

$\displaystyle \frac{36}{2}=\frac{2l}{2}$

$\displaystyle 18=l$

$\displaystyle P=2l+ 2w$

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

$\displaystyle 90=2l+2(12)$

$\displaystyle 90=2l+24$

Subtract $\displaystyle 24$ from both sides

$\displaystyle 90-24=2l+24-24$

$\displaystyle 66=2l$

Divide $\displaystyle 2$ by both sides

$\displaystyle \frac{66}{2}=\frac{2l}{2}$

$\displaystyle 33=l$