Area of Annulus

This is an interesting question. Now we know that we can find the area of an annulus with only one measurement. I thought we need to know the internal and external radii.

Back to the question. The area of the annulus $A = \pi (R^{2} - r^{2})$ , where $R$ and $r$ are the external and internal radii of the annulus respectively. Since $A=a\pi$ cm $^{2}$ , then $a=R^{2}-r^{2}$ .

Let the $10$ -cm line meets the external circle at $A$ and $B$ and touch the internal circle at $M$ , and center of the annulus be $O$ . We note that $AO=BO=R$ and $MO=r$ . We also note that $AO^{2}-MO^{2}=AM^{2}$ . And since $AM=BM=5$ cm, we have $R^{2}-r^{2}=5^{2}$ or $a=\boxed{25}$ .

The area between the two circles is

$A=\pi R^2-\pi r^2=\pi(R^2-r^2)$

Using pythagorean theorem, we have

$R^2=r^2+5^2$

$R^2-r^2=25$

Finally,

$A=\pi(25)$

So the desired answer is $25$ .

Let the outer circle have radius R and the inner circle radius r. Then we can form a right triangle using these two radii and one-half of the 10 cm chord. We then have

$R^{2}$ = $r^{2}$ + $5^{2}$ , which can be rewritten as $R^{2}$ - $r^{2}$ = 25.

But the area of the annulus is just $pi$ * [ $R^{2}$ - $r^{2}$ ], so the value of a is $\boxed{25}$ .