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ABSCHLUSSPRÜFUNG Realschule Mathe – Geometrie 10. Klasse

May 30, 2024
Hello everyone, today I brought you an assignment from a high school final exam. The task was given in Bavaria in 2021 and is as follows: the sketch shows the rectangle a b c d there we have the correct abc.de with the diagonals ac, that is. from a to c its distance ac then this notation in parentheses exactly represents this line from a to c and bd is also a diagonal, so from b to de we have a section here where we get the diagonal intersections, we can also see here that the diagonals intersect, it is assumed that is the point m and we also obtain the distance ef so from e to f we also have a distance ok, the length of the distance is applied, but that is what this line represents, it must be six meters, so from a to b are six meters, he can do it for us.
abschlusspr fung realschule mathe geometrie 10 klasse
Let's write down the length of the path bc is a penalty meter, so b to c is a penalty meter, so we are told that the point is an element of this path here, so it is here and the path dm is the piece here so we can say that it just suits us on this diagonal, it's basically not here somehow and it's just drawn precisely but it should be on the diagonal just like f should also be on the other diagonal link from c to m in the piece here is the point, there is a new mutation here the ef path is parallel to the cd path so ef that would be the one here, it is supposed to be parallel as you suspect, but tell us explicitly again, ef is parallel to cd hero, so it doesn't just look like that, it's actually supposed to be like that and then there's a name that's a little bit weirder and it means distance, so the distance from ehe to CD, that would be the distance from the point at this distance from up here, which should be two meters, then this piece from there to there should be two meters but at the same time it also tells us that because they are parallel the distance here is Of course, also that two meters everywhere determines information important Why is there something?
abschlusspr fung realschule mathe geometrie 10 klasse

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abschlusspr fung realschule mathe geometrie 10 klasse...

What should we calculate? Calculate the percentage of the content of the area of ​​the triangle. mfg First let's read the content of the area of ​​the rectangle a b c d ok, then we must calculate the content of the area of ​​the triangle. mfg then in effect it would be from mft here this triangle whose area content we need to calculate and also the area content of this rectangle which you can probably see in the rectangle OK, and then we should calculate the percentage, that is, what proportion of the area does it have this triangle in this complete rectangle?
abschlusspr fung realschule mathe geometrie 10 klasse
Okay, let's see at the end, so first of all we definitely need the content of the area. You can start with the contents of the area of ​​our rectangle because we can calculate it quickly, we have both sides. , that would simply be these six meters multiplied by the penalty meters and the hours, we can finish it quickly then we would have 66 square meters and we would have the content of the area of ​​the rectangle, of all things, it becomes more difficult with that the same content of this triangle here because there are still some pieces of information missing that we can now look for and now I've drawn it again on the next page and now we're actually looking for the contents of the area of ​​this triangle.
abschlusspr fung realschule mathe geometrie 10 klasse
First we would need a. formula for the area content of a triangle, the formula is half times the side of the base, multiplied by the corresponding height h, that is, half times the hair, which would be here now, in this case my side of the base, you can take any side of your triangle here, the side is now suitable ef because we also have information about this page about fm is right in the middle here somehow we don't know anything about em here we also don't know anything so now I would use page as the base, which means we would need the length of for our formula This side ef which we don't have yet, what about the height?
The height is the side that is perpendicular to this principle and passes through the opposite corner, that is, let's draw. inside it should look like this, it looks nice vertically that would be our height and we'll see if we can calculate it because it goes from here to the center point, that is, the center point that we know in a rectangle like this is in the middle, so it's more like halve this side here and also halve it. This side means from up here to the center, but it is not a penalty meter but rather half of these eleven meters, that is, 5.5 meters, which gives us that now we know from there to there are 5.5 meters but we also know that this piece Here what we don't want to be tall is two meters, that means two meters plus 3.5 while then it would be this 5.5 so our height has to be 3.5 meters.
I'll make it yellow and then we can do it. We already put it in our formula 1. So we have it good. Now how do we find the length of the side? Well, we guess since they're parallel, it looks a little like a. prayer of rays because we, we, rays, it comes out like this, that's why I have this big upper triangle here on the next page that is called like that from the center up here, the side up here was six meters long, the piece was two meters and it was 35, that was our height. In order to apply the radiation theorem, we first have to see if the requirements are met.
For this to be fulfilled we need a center from which the radiation starts, which is what we have put here, we even have three rays here, in reality we only need two rays for the standard rays, but that's okay, we can select two of them and also we need additional parallels crossing these rays and we have that here too, the side was parallel to this one. , which means we can use the spoke seats, so we take two spokes from those and those from there and then we just do everything here with just the right side of these spokes set here, we still know the length of the side up here because everything was six meters the piece here then half will be three meters because this was in the middle the piece down here we don't know what we want to find out let's call it iks because finally we want to calculate the entire side ef of the piece as if it were the half that is in the way.
Now I've drawn this right side here again, you can see it again very well and now we can use iks to calculate the second ray theorem which applies because it uses these parallels here and it says that this parallel behaves with this short piece in exactly the same way that this parallel behaves with the long piece here, so if we write it, iks behaves so you can do it like this. Break, then write that proportion, so I divided it by This 3.54 to 3.5 meters can only be entered is the same size as the other proportion of this long grass, so from this 3 eleven meters to this long piece, how big is that?
It's 2 + 35, so this 55 meters and now we have an equation. in which iks appears We can easily solve this equation for iks by simply multiplying by these 3.5 meters because then asics is alone and on the right we would have these three meters by the 55 meters and multiplying everything by the 3.5 meters that we can. now type that into the calculators and we'll round it to 1.91, unit that meters and meters are shortened here, but there's still a meter and that's what we have. Of course, we were also hoping that we would get meters here because we were looking for length. of this section here, so it's 1916, but in the end we weren't really interested, but that's just a little bit here, we were looking for the page ef and ef is then double this 1.91, that is 3.82 meters and finally we can plug that into our formula for our area content 1, which we are looking for all the time with 3.82 meters, we have to divide everything with the height that we already had from the beginning, if you calculate it and round up you get 6, 69 and here we would have meters by meters, that is, square meters, which would be the content of the area of ​​our triangle and what was finally sought is that both areas have content found in the triangle and in the rectangle, but finally the percentage was visited from the triangle to the rectangle and what we just had, the ratio or proportion can also be calculated simply by calculating the content of the area of ​​the triangle by the content of the area.
So, from the rectangle you divide these 6.6 square meters, you divide it by the 66 square meters. m there and now you calculate the ratio or the proportion of this triangle to the rectangle and when you calculate that you get approximately, first of all, square meters get shorter from a ratio like this, there is no unit, we get 0.10 14 we're almost done because we were looking the percentage, that is not a percentage yet. We multiply 14 by 100 and we make a calculator or we know that it just shifts two places to the right and then we have 10.14 percent, so the content of the area of ​​the triangle represents 10.14 percent of the content of the area of ​​the rectangle.
That was exactly when we solved the task. I hope you were able to take something with you. If you have any questions, as always, don't hesitate to write them in. comments. Otherwise, have a wonderful day and I'll see you in the next video.

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