The area of a circular ring

To find the solution, you build a right triangle using the given segment and the radii of the two circles: Now, labeling with $R$ and $r$ the big and small radius and using Pythagora's Theorem, you have $R^2-r^2=25$ , and since the formula for the area of a circular ring is $\pi(R^2-r^2)$ , the answer is $25\pi$ .

The question did not state what is the size of the inner white circle, therefore I can assume that it's radius is 0. Now the question become very simple, the red line become the radius of the purple circle and the answer is 25π

Relevant wiki: Circles - Area

Let $R$ be the radius of the big circle and $X$ be the radius of the small circle. Consider my diagram. Apply pythagorean theorem in the right triangle. We have

$R^2=X^2+5^2$ $\implies$ $R^2-X^2=25$

The area of the blue region is equal to the area of the big circle minus the area of the small circle, we have

$A=\pi (R^2-X^2)$ but $R^2-X^2=25$ , substituting, we get $A=\pi (25)=25\pi$

The desired answer is $\large{\color{#D61F06}\boxed{25}}$ .

This might be of interest: https://en.wikipedia.org/wiki/Visual_calculus

From the diagram above $R^2 - r^2 = 25 \implies$ the desired area $A = (R^2 - r^2)\pi = 25\pi$ .

The area is pi (R^2-r^2) where R and r are the large and small radii repectively. The power of the outer endpoint of the segment with respect to the inner circle is R^2-r^2, but by POP that is also 5^2. Thus the area is 25 pi

EVERYONE HAS THE SAME SOLUTION. I will now show you a much more unique way of solving this problem. We can think about the shaded area as the region swept out by the segment of length 5. So, draw a triangle from the center of the circle to the point where the segment of length 5 is tangent to the inner circle. Now think about rotating that triangle by N degrees. Now, you may notice that as N gets closer and closer to 0, we see a circle traced out, with radius 5. The area of this circle is pi*5^2=25pi.

Use a right triangle with size x and 5. The radius of the great circle is the hypotenuse of that right triangle. The side x is the radius of the inner circle. Now find the solution of the equation: 25pi + pi x^2 - pi x^2 Well that's easy, you get 25 pi

(Can't use latex on my phone, my solution is not the most beautiful)

As a pythagoras feziomatry

Everyone has a proper explanation. I just thought you have this side and the other side. 5×5= 25 therefore 25π

Assume that R is the large circle radius and r is the small circle radius.

Through ‎Pythagorean theorem:

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

The area is the difference between the large and small circle areas:

$A=\pi\times R^2-\pi\times r^2$

Replacing $R^2$ :

$A=\pi\times(r^2+25)-\pi\times r^2$

$A=\pi\times r^2 + 25\pi -\pi\times r^2$

$A=25\pi$

Aout=PiR2 Ain=Pir2 R2=r2+25 Aou-Ain= pi( R2-r2) = Pi25

Mamikom's proof of pythagorean theorem uses exactly the same approach (i.e. area of annulus, when length of tangent to the inner circle is given.) to show that if the length of the tangent to the inner circle is $r$ then area of annulus = ${\pi}r^2$ . In this case $r = 5$ , hence $Area = \boxed{25\pi}$

Integrate the areas of the isosceles triangles with differential angle between legs of length 5 from 0 to 2 pi. A=1/2 5^2 dTheta. Integrate to get 1/2 5^2 2 pi=25*pi.

Imagine the inner circle radius to be x. Therefore area of the ring is Π(5+x)^2-Πx^2. Taking Π common, it becomes Π(2x+5)*5. Which means that when the given options are divided by 5Π, the quotient left behind is of the form 2x+5, that is the quotient left behind is odd. Of the given options only 25Π satisfies the conditions.

though, not the true way but make the inner circle equals to be a point, the tangent becomes radius and our answer