A Nice geometry math question| solve for h

Hi everyone, today we are asked to solve 4H. You can't imagine this. This is a 6th-grade question, but many people can't solve it—just like we, let's say, adults, can't solve it. Many of us. Now let's go ahead and solve it, don't mind me. Now the questions say we should find 4H. From here to here it's 15, from here to here it's 10. In order for you to be able to solve this, you must have knowledge of similar figures. In this particular diagram, you can see similar triangles. What is a similar triangle? For example, a similar triangle has this form. Let's say you have triangles in this form. In this triangle, we now have two triangles. The larger one is ABC. The smaller one is, let's say, CDE. So we have two triangles here. And as long as this side is parallel to this side, then these are similar triangles. Now, since they are similar, for you to say that two figures are similar means that the ratio of their corresponding sides is equal. The ratios of their corresponding sides are equal. So that means that if I decide to take AB divided by BC for the large triangle. So if I say AB over BC, it must be equal to DE, because DE is the one that corresponds to AB. So their ratio must be equal. So AB and DE. Since AB is what I have as my numerator, I should have DE as the denominator for the smaller triangle, and EC or CE as the denominator. That shows that these two triangles are similar. Therefore, the ratios of their corresponding sides are equal. This is equal to this. The ratio of this should be equal to the ratio of this. The ratio of this to this should be equal to the ratio of this to this. I don't know if you understand what I mean. Now let's go back to the question. According to the question, we can see similar triangles here. Now let us draw the first one out. If I draw the first one out, now this is the first one. This is 15. This is 90 degrees to the right, and I have this. This is H. If we call this place X and this place Y, then this is X and this is Y. Now for the larger triangle, if we say base divided by height, the base over the height. If we talk about base over height, then for the larger triangle, it will be the base, which is X plus Y. So we have X plus Y divided by the height of the larger triangle is 15. Now, since these two triangles are similar, this means that this should be equal to the base of the smaller triangle, which is Y. So we have Y divided by the height of the smaller triangle equals H. So this is what we are going to have. Let's call this equation 1. Now, if we draw out the second one, and you look very carefully, you're going to see the second similar triangle right here. So we draw it out. Here we also have the large one and the small one. Remember that this is X and this is Y. Now, if we also consider base over height, the base of the large one is X plus Y, and the height of the large one is 10. This should be equal to the base of the smaller one, X, divided by the height of the smaller one, H. So let's call this equation 2. Now I'm going to write equation 1 and equation 2 one below the other so that we can understand what I'm about to do. But before that, please click the notification bell so you don't miss our amazing videos. We upload every day. Please tune in at your convenience to watch our latest uploads. Give us a thumbs up if you actually enjoyed this video. Thank you very much. So let me write down the two equations we found. So we found the first equation, which is this. And the second equation states that X plus Y divided by 10 is equal to X over H. This is the second equation. Now, if we add equation 1 to equation 2, that is, equation 1 plus equation 2, We are going to have X plus Y over 15 plus X plus Y over 10 equal to Y over H plus X over H. Now, if you look at this, you notice that both of them have X plus Y in common. Let's factor that out. If we factor it out, we have X plus Y, open bracket. When this is divided by X plus Y, we have a remainder of 15. So we have 1 over 15 remaining. And when this is divided by this, we have 1 over 10 remaining. Now, if you look at this side, they have the same denominator. So this should be over H. Then we have X plus Y. Now let us continue. Let us solve what is inside the parentheses. Let's solve what we have inside the parentheses. So here we have X plus Y in 2. Now the LCM of 15 and 10 is 30. So what should we do to 15 to get 30? We multiply it by 2. So we also multiply the numerator by 2. So we have 1 multiplied by 2 divided by 15 multiplied by 2. Plus 1 multiplied by 3, because here, to get 30 using 10, we have to multiply 10 by 3. So we also multiply the numerator by 3. So we have 1 multiplied by 3 divided by 10 multiplied by 3. Is equal to X plus Y divided by H. So we have X plus Y in 2. 1 multiplied by 2 is 2. And 15 multiplied by 2 is 30. Plus 1 multiplied by 3 is 3. And 10 multiplied by 3 is 30. Is equal to X plus Y divided by H. Alright, have you given us a sumsum yet? If not, please do so. Alright, now let us continue. So from here we have X plus Y in 2. So they have the same denominator. So we have to do it this way. Then we have 2 plus 3 as our numerator. Is equal to X plus Y divided by H. Alright, now here we have X plus Y in 2. 5 divided by 13 equals approximately 0.3846. Is equal to X plus Y divided by H. So we have X plus Y in 2. 5 divided by itself gives us 1. Divide 30 to give us 6. Is equal to X plus Y divided by H. Now, where am I heading? Where am I actually heading? Okay, if you look at this equation closely. You notice that we can factor out X plus Y from the right-hand side as well. So we have X plus Y times one over 6. Equals X plus Y times the quantity 1 over H. Can you see that? Yes. Now, if X plus Y multiplied by 1 over 6. Is equal to X plus Y multiplied by 1 over H. Note that by comparison, this simply implies that 1 over 6 is equal to 1 over H. This implies that H equals 6. And that is the answer. Thank you so much for watching. See you in my next video. Bye.