The Mickey Mouse Problem – Find the area!

Hello my lovelies, it's Susanna and today I want to show you how to solve this problem. Three identical circles lie inside a rectangle. They touch each other as well as the sides of the rectangle. What is the area of the rectangle? We have these three identical circles in this rectangle here. The circles touch each other as well as the sides of the rectangle here everywhere and we are given the radius of the circle. It is of length one and we have to find the area of this rectangle. I call the area A and to find the area we need the lengths of the rectangle. So I call this side x and I call this side y and the area is x times y. Then this is what we have to find to solve this problem. Let's start with the length of x. We have the radius here of the circle. It's of length one here of this circle as well and they touch each other here. So I can draw the radius from the center to the edge of the circle here as well and have length of one as well. The same here from the center to the edge of the circle. This is also the radius of length one. So now because the circles touch the sides of the rectangle and they touch each other I have the length one, two, three, four for my x then and I can replace this already for my x. I write the four and then I only have to find y to solve this problem. So let's try and find the length of y. What do we know about the length of this vertical line here? Well the radius is part of it and then we have the radius here again of length one. We don't know how long this part here is but from this touching point up to the edge of my rectangle this is the radius again because it is the same as going here from the center up to the edge which is my radius of length one. So I know that this part here is of length one as well but I don't know how long this part here in between is. So I'm going to undo all these steps again. We have to find a different way how to find the length of y. But we haven't used these touching points here yet, right? So usually if you have touching points they usually help you to solve these kinds of problems. So let's connect maybe the centers of our circles with these touching points. If we do that we have some of these touching points so let's connect all of the centers with the attaching points. Like this we have created a triangle here and we even know the lengths of the sides of this triangle because we have radius, radius. So it's one, one so it's two in total. Here the same two in total and here we have one and here the radius of one again. How does this help us now with the length of our y? Well y consisted of this radius here, right? We already found this out and here we had the length of one of the radius as well. We knew these things already but what we haven't known so far is the length of this line here in between. But this actually is the height of my triangle now because this is also perpendicular. So I can find the length of the height of my triangle because I have a right triangle here on the left. I can just use the pythagorean theorem to find the length of h. Okay then let's do this. So for the pythagorean theorem we first have to find the hypotenuse. This is always the side that lies across the right angle. So this is my hypotenuse and the pythagorean theorem then says take one of the sides so your h and square it plus the length of the other side so the one and square it and then you get the hypotenuse squared so two squared. And we now have to solve this equation for h and then we will have solved this problem in total to find the length of y. But first h. Okay let's calculate this a little bit. One squared equals one. Two squared equals four. To solve for h we want to get rid of the one here so we subtract one on both sides of the equation. Then we have h squared here. This cancels out and on the other side we have four minus one which equals three. To solve for h we want to get rid of the square here. We can do that by taking the square root on both sides of the equation. We only have to keep in mind that we always get two solutions. h1 is my first positive solution and h2 is going to be my negative solution. My positive solution is the square root of three and my negative is negative the square root of three. But we don't need the negative solution here because h is a length of a side so we only need the positive solution. But we now know that this part here is of length squared of three so this is what we can see here then. And with this we can find the length of y because we have one here one here which is two in total plus the square root of three here in the middle. So we have two plus the square root of three for our y. With this we go back to the beginning where we needed our y but we have it now. So the area is four times for our y we take this thing here in parentheses so it's two plus the square root of three and if you calculate this we get a result of 14.93 for the area of our rectangle. I hope you enjoyed it. Please give this video a thumb up it helps me a lot and I hope to see you in one of my next videos. Take care!