Mathe RÄTSEL Geometrie – Kannst du es lösen?

Hello everyone, today I have brought you this math puzzle that I found online in the mirror. Let's take a look at it together. A square is given with the specified lengths, in which two sides are cut in a right angle. Here is a sketch. We can see here quite nicely that the square is mapped. We get a few lengths here on the sides and in the square two sides are cut in a right angle. Calculate the missing part, which is marked in red here. We are supposed to find this length. I'll call it x, because that's what we're looking for. And then let's see how we can find it. The information that this is a square we keep in mind that all sides must be the same length. We get, for example, that this piece here is 4 meters long and that is repeated on the other side. So if we draw a vertical line down here and enter the right angle here, then we know that this piece here is also 4 meters long. Then we would have divided this side into three individual parts. At least two lengths are known of these three. This length here is not known. I'll just call it a. But this is a side of this triangle that has now been created. So I'll mark it so that we have a triangle here, whose one side we don't know here. Maybe we don't know it yet. But we can do the same with the 7 meters. If we draw a vertical line down here and enter the right angle here, then we know that this piece here is 7 meters long on the other side. And then we also divided this square into three parts, but we only know one of them. Because here we already have a side of a triangle whose length we don't know. I'll just call it b. But at least we have a lying triangle here that I mark orange. And these two triangles look damn similar. It looks like this was just turned 90 degrees. Here is also the right angle. That would of course be very practical if we could show that. Because then this piece here is just as long as this piece. And then we don't need a and b anymore. Then a and a would be enough. The less unknowns we have, the better. But how can we justify that they are really the same length? If it were really just turned, then these two triangles are congruent to each other. If two triangles are congruent, then there are different sets of congruence. We can see what is identical with these two triangles. The right angle is the same. And this one side here is just as long as this side here. That's the length of the square sides. So from up here to down here, that's the length of the square. And just as from left to right, we have the length of the square. So this side is just as long as this one. So if we collect that now, that we have a side that is congruent in both triangles and a limiting angle on this side that is congruent in both triangles, then a set of congruence that is called WSW fits. If we could now show that the other limiting angle on this side is equally large in both triangles, then the WSW set of congruence says, then your triangles are congruent to each other. That would be very good. So now we just have to show that this angle, that is, the one that limits on this side here on the left, this angle, which I call alpha, that it is equally large in both triangles. So that it also appears here in the green, here at this point, the alpha here again. And then we have to see how we can do that. Maybe we'll take a look at the orange triangle first. Then we have the 90 degree angle, alpha, and the third angle I call beta. That means, whenever we are in a triangle now, in which a 90 degree angle occurs and alpha, then we know that the third angle is beta large. Because we now have a few more triangles that we can make here. For example, here this little triangle, in which alpha also occurs, that is, here, what is so small in here, in which a right angle also occurs here at this point. Here opposite, as a diagonal angle, this one is just as big as this one. And then we have alpha in the small triangle, a right angle. And then we automatically know, well, then the other angle has to be beta large. If you can still read that somehow. And then we have a new right-angled triangle here, for example, that looks up here, this green part. Because we have a right angle here. Then we have beta down here. And then we know, if a right angle and beta occur in a triangle, then the last third angle has to be alpha large. And that's our chosen angle. We have shown, yes, thank God, it's alpha large. Our triangles are congruent to each other. That means, this side is just as long as this one here. We can then, if this is a, we don't have to call this b, but this is also a long. And with that, I tidied it up a bit, we have divided our square into three parts, each of which has two unknowns. Okay, x and a. But let's write down the equation, because this square has to be just as long as this square. So 4 plus a plus 6, if we write that down, has to be just as long as 7 plus a plus x. And maybe you've already noticed that the a just appears on both sides. If we calculate minus a, that just comes out. And only x is in our equation. And we can convert that to x. 4 plus 6 is 10 on the left side. Here's only 7 plus x. And so that x is left alone, let's put 7 on the other side. With minus 7, 10, minus 7 is 3. And that's our x. And with that, we have solved the problem. I hope you enjoyed it. Feel free to write how you did. I'm sure there are more ways to solve it than the ones I showed you here. Otherwise, have a wonderful day and I'll see you in the next video. Take care!