A very hard geometry math question that many got wrong.

Yes, we can. Now let's do this together. First of all, permit me to erase this. all the shaded area so that I can put whatever I want inside, okay? Now let's drop a Let's drop a perpendicular to AC. Okay, let's drop a perpendicular to AC, so. If we do that, let's assume this is the perpendicular, all right? So this is 90 degrees. Now let's call this point point M. Okay, now let's call it X. Let's call it X, and I want us to To see that I focus on triangle ADM, a right triangle ADM, so let's draw that out. Triangle ADM. If we were to draw that out, we have this, we have this, we have this. Okay. So this is our M, this is our D, this is our A. So we can know, we see that this is 90 as well. This place is 90 degrees, and this is 45 degrees, so that means this is 45 degrees. Also, angle D is right, so we can see that the base angles of this triangle are equal. that means this place is equal to this place because it is this, which implies that triangle ADM is an isosceles triangle okay, and if it is an isosceles triangle, then this side These two sides are equal. I remember that AM is 2 plus x, so this side is 2 plus x. Therefore, DM It is also 2 plus x, so permit me to write dm equals 2 plus x. All right, that is why I was explaining. All these things, please click the subscription button if this is the first time you have seen it. This lovely channel, give us a thumbs up if you love what we're doing, all right now if You look at this diagram very carefully, you will notice that we have several triangles, but I want us to Consider triangles CDM and DEM. Let's focus on these two triangles. Now you're going to notice something; you're going to notice that angle DMC is equal to angle DME right quote angle DME yes which is equal to 90 degrees angle DMC is equal to angle DME, both being 90 degrees, and prior to that, if we refer to this point Alpha, all right, if we call this place Alpha, and this is what 90 means, whatever that is. staying here, this angle will be equal to 90 minus alpha, right? This point will be 90 minus Alpha, I remember that the entire area of this place is 90, so it simply means that this section is also alpha, right, because alpha plus 90 minus alpha gives us 90 degrees, which is what angle d is In total, so if we should note, we also notice that angle EDM is equal to angle ECD, sorry, angle DCD, MCD, angle EDM is equal to angle MCD, which is equal to alpha. So if you look at this, what does this mean? This implies that triangle CDM is similar to triangle DEM, which is our conclusion, and if they are similar Remember that the law of similarity states that the ratio of the corresponding sides is equal. The corresponding angles of similar triangles are equal; the ratio of their corresponding sides is equal. Therefore, let's look at the ratios of those corresponding sides now. What do we do? if we say mc mc divided by md mc divided by md should be equal to md or dm right? md or dm divided by em, since they are similar, divided by em right now, mc That's what MC is not known, so we write MC. MD is 2 plus x. 2 plus x is equal to MD again. 2 plus x. And em is x, okay? Now let's cross multiply: m times c times x, so x times m times c is equal to 2 plus x squared, right? So it means that MC divided by X divided by X; we remove this, and we have that MC. is equal to 2 plus x, all squared, divided by x, so we'll write here 2 plus x All squared divided by x. Okay, yeah. Now let's continue; there is something we have to notice. All right, now let us compare triangle ABC and triangle ABC. Okay, and triangle DMC as well, triangle CDM as well. So if we do that, you notice that Angle ACB is equal to angle MCD or DCM. The angle is common, which is alpha. Right angle alpha is common. Also, angle BAC is equal to angle DMC. angle DMC right, which is equal to 90 degrees. All right, having said that, this also means that triangle ABC is similar to triangle CDM, therefore the ratio of their corresponding Since the sides are equal, let's examine the corresponding sides and choose their ratios. So if we have MC times MC divided by AC, it should be equal to MD divided by AB. Alright, so the MC is this: let me write it out here—we have 2 plus x all squared. divided by x over ac; ac is from here to here, which means 2 plus x plus this 2 plus x plus (2 + x) squared divided by x, okay, is equal to md times md over ab. Now, md is 2 plus x. So we have 2 plus x divided by ab equals 6 over ab. So let's cross multiply; we have 6. multiplied by 2 plus x squared over x okay is equal to this multiplied by this 2 plus x bracket 2 plus x plus 2 plus x squared divided by x. Okay, now let us divide through. dividing through by 2 plus x, okay, so if we do that, this divided by 2 plus x, one of this will Right, so now we have 6 brackets: 2 plus x divided by x is equal to. Now, we remove this; we are left with this, so we have only the brackets remaining: 2 plus x all squared divided by x okay, if we divide again, dividing through again by 2 plus x, okay, so we are going to have This divided by this, this we remove, we are left with 6 over x is equal to this, we divide. What remains is one plus this divided by this, leaving us with just one of this. So we have 2 plus x over x, so this is 6 over x equals 1 plus this is the same thing as 2 over x. plus x over x, we're almost there. Remember, we are asked to find the area of the shaded area. portion, and we are almost there. Okay, so now we have 6 over x is equal to 1 plus 2 over x. plus 1 because x divided by itself is 1, so we have 6 over x is equal to 1 plus 1, which is 2 plus 2 over x, okay, so we go like this: let this come to this side, 6 over x, take away 2 over x is equal to 2, so this is the same as 6 over 2 divided by耆 is equal to 2, so this sentence 4 over x equals 2. If we cross multiply, we get 2x equals 4. Okay, so now we have 2x. divided by 2 is equal to 4 divided by 2; this removes this, and we have x is equal to 2. Now that we x is equal to 2, x is equal to 2, right? Our answer is here. Yes, when we are asked to find the This side: D, E, C. Let's draw it out. So this is E, this is D, this is C. From here to here. We have 90 degrees; this is m; now this is x; this is 2 plus x squared divided by x. Okay, and this line is 2 plus x. Now, the area of this triangle, which is the shaded portion. The area of the triangle is half the base multiplied by the height, so it will be equal to half. The base is what e c half e c multiplied by h; the height is what d m multiplied by d m. So if we solve for that, we will have half e c e c is x plus 2 plus x squared over x, that is dc. Now multiply by d m d m is 2 plus x, okay, but x is equal to 2, but x is equal to 2. Okay, I'm sorry, I didn't even know the video glitched, and I was just talking, so uh, x is equal to 2 now let's substitute you need to use your board maths or bitmaps carefully here so I Don't make a mistake, okay? So we go like this: the area is half bracket. Anyway, I see x; I'll replace it. So I will have 2 plus 2 plus 2, all squared, divided by 2, in brackets, 2 plus 2. Okay, so area. is equal to half bracket; remember, we solve our brackets first, so inside this bracket we solve First, 2 plus 2 is 4, so we have 4 squared, which is 16, divided by 2. The bracket: 2 plus 2 is 4. Okay, so at At this point, these two can divide this; we have to write, so we have area equal to 2 plus we need. We must divide first before we add, so 16 divided by 2 is 8; remember, this is inside the brackets, okay? We have 2, so a is equal to. We solve inside the brackets before we multiply, so 2 plus 8 is 10, times 2. So the area is 20 square units. Thank you for watching. I'll see you in my next video. Bye.