The perimeters of regular figures Y and Z are 7 yards and 14 yards respectively. If the area of figure Z is 40 square yards, find the area of figure Y.

Answer 1

We know that the perimeter of a regular polygon with $n$ sides, each with length $s$, is given by $P = ns$. Therefore, we can write:

$7 = ns_y$
$14 = ns_z$

where $s_y$ and $s_z$ are the side lengths of figures Y and Z, respectively.

We also know that for a regular polygon with perimeter $P$ and apothem (the distance from the center to the midpoint of a side) $a$, the area $A$ is given by:

$A = \frac{1}{2}Pa$

For figure Z, we know the perimeter is 14 and the area is 40. We can use this to find the apothem:

$40 = \frac{1}{2}(14)a$
$a = \frac{40}{7}$

Now we can use this apothem and the perimeter of figure Y to find its side length:

$7 = ns_y$
$s_y = \frac{7}{n}$

To find $n$, we can use the fact that the apothem of a regular polygon with $n$ sides and side length $s$ is given by:

$a = \frac{s}{2\tan(\frac{\pi}{n})}$

Plugging in our values for figure Z, we have:

$\frac{40}{7} = \frac{s_z}{2\tan(\frac{\pi}{n_z})}$

Solving for $n_z$, we find:

$n_z = \frac{\pi}{\arctan(\frac{4}{7})}$

Using a calculator, we get $n_z \approx 11.44$, so we can round to 11 or 12 sides (since we're dealing with regular polygons, either answer would work).

Let's assume $n_z = 12$ for simplicity, so each side of figure Z has length:

$s_z = \frac{14}{12} = \frac{7}{6}$

Now we can use the formula for the apothem of a regular polygon to find the apothem of figure Y:

$\frac{7}{n_y} = \frac{7}{6} \cdot \tan(\frac{\pi}{n_y})$

Using a calculator or trial and error, we find that $n_y = 6$ gives a valid solution:

$\frac{7}{6} = \frac{s_y}{2\tan(\frac{\pi}{6})}$

$s_y = 2\sqrt{3}$

Finally, we can use the formula for the area of a regular polygon to find the area of figure Y:

$A_y = \frac{1}{2}(7)(\frac{s_y}{2\tan(\frac{\pi}{6})})$

$A_y = \frac{147}{4\sqrt{3}}$ or $\approx 25.4$ square yards (rounded to one decimal place).