The problem asks to calculate the area of a regular polygon circumscribed about a circle, given that the difference between the perimeter and the side length of the polygon is 25.

1. Problem Description

The problem asks to calculate the area of a regular polygon circumscribed about a circle, given that the difference between the perimeter and the side length of the polygon is

2

5.

2. Solution Steps

Let $n$ be the number of sides of the regular polygon, $a$ be the side length, and $r$ be the radius of the inscribed circle.

The perimeter of the polygon is $P = na$.

We are given that $P - a = 25$, so $na - a = 25$, which means $a(n-1) = 25$.

The area of the regular polygon is given by $A = \frac{1}{2}Pa = \frac{1}{2}nar$, where $P=na$ is the perimeter and $r$ is the inradius. Alternatively, we can also express the area as $A = nr^2 \tan(\frac{\pi}{n})$.

We know that $a = 2r \tan(\frac{\pi}{n})$. Substituting this into the equation $a(n-1) = 25$, we get

$2r \tan(\frac{\pi}{n})(n-1) = 25$. Thus, $r = \frac{25}{2(n-1)\tan(\frac{\pi}{n})}$.

The area of the regular polygon is $A = nr^2 \tan(\frac{\pi}{n})$. Substituting the expression for $r$, we get:

$A = n \left(\frac{25}{2(n-1)\tan(\frac{\pi}{n})}\right)^2 \tan(\frac{\pi}{n}) = n \frac{25^2}{4(n-1)^2 \tan^2(\frac{\pi}{n})} \tan(\frac{\pi}{n}) = \frac{625n}{4(n-1)^2 \tan(\frac{\pi}{n})}$

If the polygon is a square, then $n=4$.

$a(4-1)=25$, so $3a = 25$ and $a = \frac{25}{3}$.

The radius of the inscribed circle is $r = \frac{a}{2} = \frac{25}{6}$.

The area of the square is $A = a^2 = (\frac{25}{3})^2 = \frac{625}{9}$.

Also, if the polygon is a square ($n=4$), $A = \frac{625(4)}{4(4-1)^2 \tan(\frac{\pi}{4})} = \frac{625}{9 \tan(\frac{\pi}{4})} = \frac{625}{9}$.

If the polygon is an equilateral triangle ($n=3$), then $a(3-1) = 25$, so $2a=25$, and $a=\frac{25}{2}$.

$A = \frac{3a^2\sqrt{3}}{4}$ and also $A = nr^2 tan(\frac{\pi}{n}) = 3r^2 tan(\frac{\pi}{3}) = 3r^2\sqrt{3}$.

$r=\frac{a}{2\tan(\pi/3)} = \frac{a}{2\sqrt{3}} = \frac{25}{4\sqrt{3}}$

$A = 3(\frac{25}{4\sqrt{3}})^2 \sqrt{3} = 3\frac{625}{16(3)} \sqrt{3} = \frac{625\sqrt{3}}{16}$.

From $a(n-1) = 25$ and $a = 2r\tan(\frac{\pi}{n})$, we can derive

$2r(n-1)\tan(\frac{\pi}{n}) = 25$, $r = \frac{25}{2(n-1)\tan(\frac{\pi}{n})}$

The area of the polygon is $A = nr^2\tan(\frac{\pi}{n}) = n (\frac{25}{2(n-1)\tan(\frac{\pi}{n})})^2 \tan(\frac{\pi}{n}) = \frac{625n}{4(n-1)^2 \tan(\frac{\pi}{n})}$.

Consider the case $n \to \infty$, so we approach the circle.

$A = \pi r^2$, the perimeter approaches the circumference $P = 2\pi r$.

$P - a = 25$, then $a \approx \frac{P}{n}$, so $P - \frac{P}{n} = 25$, $2\pi r (1-\frac{1}{n}) = 25$. When $n \to \infty$, $2\pi r = 25$, so $r = \frac{25}{2\pi}$.

The area is $A = \pi r^2 = \pi (\frac{25}{2\pi})^2 = \frac{625}{4\pi} \approx \frac{625}{12.56} = 49.76$.

If we assume $n$ is very large, then $\tan(\frac{\pi}{n}) \approx \frac{\pi}{n}$. Thus $A \approx \frac{625n}{4(n-1)^2 (\frac{\pi}{n})} = \frac{625n^2}{4(n-1)^2\pi} \approx \frac{625}{4\pi}$

Let's analyze the general formula for $A$. As $n$ gets larger, $\tan(\pi/n) \approx \pi/n$. Then

$A = \frac{625n}{4(n-1)^2 \tan(\pi/n)} \approx \frac{625n}{4(n-1)^2 \pi/n} = \frac{625n^2}{4(n-1)^2 \pi}$.

Taking the limit as $n \to \infty$, we have

$\lim_{n \to \infty} A = \frac{625}{4\pi}$.

3. Final Answer

$\frac{625}{4\pi}$