Is it possible to find the area? – Geometry Puzzle

Hello my lovelies, it's Susanna and today I want to show you how to solve this problem. Given is a rectangle that is intersected by a curve composed of a quarter circle and a semicircle. So here we can see our rectangle and here is this curve that consists of a quarter circle first and then we have the semicircle here. The red line is a tangent to the semicircle and passes through the corner of the rectangle. So this red line here passes through this corner and is a tangent to the semicircle. So if I draw this line like this, then this line touches my semicircle in one point, so in this point here. Find the area of the red rectangle. Okay, we want to find this area here. I call it A. How can we find it? Well, we need the length of the sides. Let's start with this length here. This is from the center of my quarter circle to the edge of my quarter circle. So this is the radius of my quarter circle. I don't have it, but let's call it R. And this here is also my radius from the center to the edge of my quarter circle. So I have R here again. The same here from the center to the edge of my semicircle. This time it's also the radius, but it doesn't have to have the same length as this radius here, so I call this radius X. And on the other side here, the same thing. This is also the radius of my semicircle, so this is X as well. With these variables, we can find the area then by taking the length of this side. So it's R plus X plus X, so it is R plus 2X. We write it in parentheses and multiply it by the length of the other side by R. Let's get rid of the parentheses so that this looks a little nicer by multiplying each element by R. So we have R times R, which equals R squared, plus 2X times R plus 2XR. So if we find R and X, we can find the area of our rectangle. So let's try and find R and X. Okay, we haven't used this red line here yet, but we know that it is a tangent to this semicircle here, so it touches this semicircle at this point here. And these touching points are usually very important to solve these kinds of problems, and it's always a good thing to connect this touching point to another important point. For example, we could take the center here of our semicircle, so if we connect these two points, we get this triangle here. And the length of this side here is just the length of the radius, because it's from the center to the edge of this semicircle. This is the radius, so it is of length X. And we know something else. We know that this radius here is perpendicular to this tangent line, because that is always the case. If you have a tangent line and the point where your tangent line touches your circle, and you draw the radius to this touching point, then these two lines are always perpendicular to each other. Which is great, because now we have a right triangle here, and we can use the Pythagorean theorem to find the connection between X and R. So for the Pythagorean theorem, we need the hypotenuse, which always lies across the right angle. So this is the hypotenuse, and the Pythagorean theorem says then, take one of the legs, so take your X in our case here, and square it, plus take the other leg, the five, and square it as well. And then you get the hypotenuse squared. So these two together are of length R plus X. So I take R plus X, write it in parentheses, and square this thing as well. Let's simplify a little bit. We have X squared here. Five squared equals 25. On the other side, to get rid of the parentheses, let's square these by writing them as a product. So one parentheses times the other. So we have R plus X in here, and we multiply it by the same, by R plus X. Yeah, the left side is good, but on the right side, we can multiply these parentheses now by multiplying each element of the first parentheses by each element of the second. So let's start here. R times R equals R squared, plus R times X equals RX, plus X times R equals XR, and plus X times X equals X squared. Let's simplify this a little bit. We can already see that X squared is on both sides of the equation, so if we subtract it on both sides, it just cancels out. Then we only have 25 on the left side, and on the right side, we have the R squared here, and these two are actually the same. So if you have RX or XR, you can just switch these two so that you have RX here as well. So one of them plus another of them is just two of these RX. Okay, this is what we get. One equation with two variables in here. Awesome! But yeah, let's take it. It's better than nothing. Let's go back to the beginning, where we wanted to find the area of our rectangle. We said that we can find it by finding this expression here, R squared plus 2XR. We now have an equation with R squared plus 2RX. If you have XR or RX, we just said it's the same, so you can write this here as XR as well. So we just found out that this expression here equals 25. So instead of this expression, which is the same, we can just write 25, and this is the size of the area we wanted to find. We don't know the size of R or of X, but we know that the area is of size 25, and we solved this problem. I'm curious how you solved this problem, so please let me know in the comments. I wish you a wonderful day, and I hope to see you in one of my next videos. Take care!