 Let us see how we can find out the area of a circular annulus, which looks something like this. Annulus is kind of a ring with certain width and since it is a ring, it is formed by subtracting the area of one smaller circle from one bigger circle. Note that here, if the whole circle was grey at the start and then if we took out the green circle area from it, we will be left with the annulus. Now note that the centers of both these circles are coinciding. That means the center of both the circles is the same. Now how do we proceed on finding out the area of these two circles? Now I will draw the radius of the smaller circle again since it was not showing. So this is the radius of the smaller circle in blue and in the orange, we see the radius of the larger circle. Let us just use some variables but before that, can I just write a quick equation here mentioning area of annulus, area of annulus to be equal to area of a bigger circle minus area of a area of the smaller circle. Now for the bigger circle, let us say this orange radius, let us say this orange radius is capital R and let us use 8 centimeter for it and if the blue radius is a 5 centimeter and let us say this is small r, how can we calculate the area? So area of the annulus is equal to area of the bigger circle which will be equal to pi times the larger radius which is capital R square minus area of the green circle which is a smaller circle with radius small r which is equal to 5 centimeter. So pi times small r square. Now we will rearrange this equation to look nice. We see that pi appears in both the terms. So we want to write pi outside and inside the bracket, we will write capital R square minus the small r square. Usually what people mistake this for is this is how they write it, pi r minus small r square. So there is no bracket square remember, the square is individually distributed on both the radii here. Now let us calculate the answer for this question. So if capital R is 8 centimeters and small r is 5 centimeters, how will the answer look like? So we have pi times 8 square minus 5 square and the value for this is pi times the square of 8 is 64 minus square of 5 is 25. That would mean this is pi times 39 or better way to write this is as 39 pi. Remember if you did pi times capital R minus small r bracket square, your answer would be 8 minus 5 bracket square which will be 9 pi and this will be a wrong answer. So don't put bracket square here. Always remember to have a square of the larger one minus square of the smaller one. Now an interesting thing comes to my mind. Also note that because this is square minus square, there is another way to solve this. So we have 8 square minus 5 square. Just remember the formula a square minus b square which is a plus b times a minus b and so if we apply that formula, we will be able to write this as pi times 8 plus 5 times 8 minus 5. So this is another way in which you can solve this. So this will be pi times 13 times 3 and you can simply write 39 pi. Now note that what if we were only given with the width of the annulus, will we be able to solve this? If we just know the width of the annulus, in this case this will be 3 and if we just know the width that is basically the subtraction of the 2 radii, so that means if we just knew capital R minus small r, we won't be able to find out the area of the annulus because we also need to know capital R plus small r in order for this formula to work. So remember in order to be able to find out the area of the annulus, you will need to know the radius of the larger circle as well as the smaller circle.