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.
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).
To find the area of figure Y, we need to determine its side length first. The perimeter of a regular figure with side length "s" is given by the formula: P = ns, where "n" is the number of sides.
Given that the perimeter of figure Y is 7 yards and figure Z is 14 yards, we can set up the following equations:
7 = nY * sY (equation 1)
14 = nZ * sZ (equation 2)
We also know that the area of figure Z is 40 square yards. The area of a regular figure is given by the formula: A = (s^2 * n) / (4 * tan(π/n))
Substituting the values into the formula, we get:
40 = (sZ^2 * nZ) / (4 * tan(π/nZ))
Let's solve equation 2 for sZ:
14 = nZ * sZ
sZ = 14 / nZ
Now we can substitute sZ into the area formula:
40 = ((14 / nZ)^2 * nZ) / (4 * tan(π/nZ))
We can simplify this equation:
40 = (14^2 * nZ) / (4 * tan(π/nZ))
Now, solve equation 1 for sY:
7 = nY * sY
sY = 7 / nY
Substitute sY into the formula for the perimeter of a regular figure:
7 = nY * sY
7 = nY * (7 / nY)
7 = 7
Since it simplifies to 7 = 7, the number of sides (nY) for figure Y does not affect the perimeter or the area. Therefore, we can conclude that the area of figure Y is also 40 square yards.