GCSE Maths Aiming for Grade 9 2026 Paper 1

Added: 2026-05-12

[00:00:00]Hey guys, a bit of a quicker video this time. We're going to do one of the aiming for grade nine papers for spring 2025. And so the way these work is edex put these together and they're a collection of the hardest questions that came up in that exam series for paper one. So as you can see it's only 33 marks because obviously not every question is a grade nine question. Um and they give us this handy mark scheme over here. So essentially people that got a grade three on average only scored 0.4 marks on this. People that got grade four scored one mark. So the idea is if you can get at least one mark on this paper, you should be able to get a grade four, two marks for a grade five, and so on up to a 23 marks for a grade nine, which is quite a lot, but if I can explain how to do all of these questions, I'll hopefully help you in this exam series or future exam series in order to get those higher grades. So let's just jump straight into it. So here's the first kind of harder questions with A, B, C are three solid spheres. So I'm going to write A, B, and C as three separate headings. It says A has a volume of 64. B has a volume of 125. So 64 and 125. And we're basically going to make these giant ratios, right?

[00:01:06]And then it says the radius of sphere C is 50% of the radius of sphere B. So let's go to radius. And it says when the radius is 50% of sphere B. In other words, the ratio of C to B is well, you can do it as two different ways. You can say this is a half and one. But if I times both sides by two, I get a ratio of 1:2 or 2:1 if you want to do it that way. So what I'm going to do is I'm going to put two one for this. And our goal is to work out essentially the ratio of the areas for sphere A and sphere C. So how do you do something like this? Well, the best thing you need the first thing you need to do is create one ratio for all of these. So, what we can do is we can go from radius up to volume or volume down to radius. It's kind of up to you. I think going down would make the most sense. So, I'm actually going to lower this slightly.

[00:01:59]Now, how do you go from a volume kind of scale factor relationship to a to a length one? Well, all you do is you cube root both sides. So, what we're going to do is we're going to cube root 64 and 125, which gives us four to five.

[00:02:16]Now we have two different ratios but they have a letter in common. They have B in common. So I can combine these to make one ratio for the radius. So how do I do that? Well, if I want to make five and two the same number, I need to terms the first ratio by two and the second ratio by five. And that gives me in total it gives me 8 to 10 and then it gives me 10 to 5. So that means the ratio of the radi is 8 to 10 to 5. And then how do we go from the radius to the area? Well, all you need to do is square the whole thing. So our ratio of the areas is 64 to 100 to 25. Now if we go up just to see the letters again, A, B, and C. So we're going to put A, B, and C. And the actual question says, work out the ratio of the surface area of A to the surface area of C. So we're just going to write the 64 to 25 for four marks. So again, I think in this question, it's mainly about how you lay out the information. So I personally like to use like grids or tables for something like this. A circle with equation x^2 + y^2 is 4. Find an equation of the tangent to the point P.

[00:03:28]Okay, so I like to draw diagrams just to make this a bit clearer. So I'm going to draw my axes here. And remember in GCSE, assuming you're not doing further maths, GCSE, all of our circles are centered on the origin. So we're going to put something like that. And we know that the radius, well, that's the radius squared, right? That's r 2, which means r is equal to the<unk> of 4, which is 2.

[00:03:52]Just so you know, this is a non-cal paper. And we have a point p1. So, and p is bigger than zero. Okay, never mind.

[00:03:59]So if p is bigger than zero, it should be probably this point over here somewhere. Um, it doesn't actually tell me. All right, let's just call it P1.

[00:04:06]That's fine.

[00:04:08]And we need to work out the equation of the tangent.

[00:04:11]Um, let's just say something like that.

[00:04:14]Cool. And give your answer in the form y + roo<unk> a of x= b. You must sure you're working out. Well, when we talk about tangents, you probably know the general kind of equation for it, right?

[00:04:27]But we have y= mx plus c. So what we need to do is we need to work out a gradient and we need a coordinate in order to work out something. Now if you have a look at this, we don't actually have a full coordinate. We have P and 1 instead of X and something else. So how can we work out the coordinate? Well, I do have the equation of the circle.

[00:04:49]And does it make sense that this coordinate, it does lie on the green line, but it also lies on the circle.

[00:04:56]Which means if I sub in P as X and Y is 1, that should be a valid equation. So that's what I'm going to do. I'm going to say P ^2 + 1^ 2 = 4. So that's just 1. So that's P ^2 = 3, which means P is equal to the<unk> of 3. And this is actually why they've told us P is bigger than zero because this should be plus and minus. But obviously it's not going to be minus because they've told me it's a plus. So our equation now our coordinate is <unk>3 and 1. Perfect. So we have one of the two things. Now we need to work on the gradient. So again the two things we need are a coordinate and a gradient. We have the coordinate now roo<unk>3 1. Now we need the gradient. Well to work out the gradient is quite nice. You always do the exact same thing. you use the circle theorem that the tangents and the radius will meet at 90 degrees. They're perpendicular to each other. So if I work out the gradient of the radius, all I have to do is take the negative reciprocal to work out the gradient of the tangent. So the gradient of that pink line, well, it's just going to be change in y over change in x. So it's literally just going to be yx. So 1 / roo<unk>3, which is just 1 over<unk>3 or roo<unk>3 over 3 if you want to simplify. which means the gradient of the green line is just minus <unk>3 over 1. So in other words, it's just minus roo<unk>3.

[00:06:18]Okay, so now we can sub it into our equation of a line. y = - <unk>3x plus c because remember we now know that n is minus roo<unk>3. Now we're going to sub in our coordinate x being <unk>3 and y being 1 in order to find out what y is. So we have 1 = -<unk>3 * <unk>3 + c. So we have 1 equals now roo<unk>3 * roo<unk>3 would be<unk> 9 square<unk> of 9 is just 3. So that's just - 3 + c.

[00:06:49]So c equals 4.

[00:06:52]So this isn't our final answer because they did ask for it in a slightly odd form. And do remember that they you have to give it in this form here. So all they've done is they've moved the ax term to the other side. So our final answer is y + <unk>3 x = 4. And there we go for four marks. Hopefully that's not too bad at the end. I think the tricky part is definitely the fact that there's a missing coordinate. So I actually want to talk about this question because it's it's only have one marker. And to be quite frank, it's not that difficult.

[00:07:23]But again, it's in a grade nine paper because a lot of these students are getting seven, eights, and nines got it wrong, right? which is fair enough. So there's a few ways to think of this. By the way, this is really important if you want to do some kind of science in the future or economics or finance or anything like that because you will get graphs that have different variables on the x and y axis. So it shows the volume of water l in liters in a tank at time t seconds. What does the gradient represent? Well, as you can see, what is this telling me? It's telling me that over time the volume of water is decreasing. So what does the gradient represent? Well, it's telling me volume over time, right? Because remember, gradient is change in y over change in x. So, this is going to be the change in volume over time. In other words, it's how quickly the water is decreasing in a tank. So, how can I help? Well, here's what I'm going to do. I'm going to give you a sentence structure that always works for questions like this. It's the rate or in fact, let's do this. change in the y ais per unit x-axis. So this is generally speaking what you're going to look at. Now if times on the bottom you can say like the rate of change of volume or something like that. But if I wanted to write it for this question I would simply say the change in volume per unit time. Okay. So that sentence structure should help you answer those questions. And again, these are grade nine students. These are students that go on to do maths and further maths that get this part wrong. And again, it does come up in all of the sciences, economics, business, maths in the future. So it's quite important. Uh yeah, this one's a bit disappointing just drawing the graph. Um so again, I'm going to show you how I think of it.

[00:09:18]Now, the only thing you need to remember is if it's s or co that starts at zero.

[00:09:22]That's the only thing you have to remember. So is sin 0 0 or is cos 0 0 and the answer is sin 0 is 0.

[00:09:31]And then all you need to do is you need to know that it goes between 1 and minus one. And every 90° it goes between 0 1 and minus one. So for example after 90° it goes up to one. After another 90° it goes back down to zero.

[00:09:48]90° -1 90° 0. Does that make sense? So every 90° it goes between 01 and minus one. And then you're just drawing a smooth curve. I promise you I can draw this way better in real life as my students will tell you. Um but let's leave that for now. So again, you would just connect the dots, right? Nice smooth curve.

[00:10:11]This second part is where a lot more people lost marks. So it says solve the equation 2 sin x= -1. Now if you do further maths, you actually know how to do this quite nicely, but it is something that is in the regular math spec. It's just not always done. So if I divide both sides by two, it tells me that sin x= minus a half. Now there's a few ways I can teach you this. But just so you're aware, and I have got video on this, the values that you need to know is you need to know 0 30 45 60 and 90° for s cos and tan. Now again, I have a strategy on how to do that where you don't need to memorize it. There also some people that use their like hands and their fingers and stuff. That personally never worked for me, but if it works for you, then I'm really really glad. So, if I just fill in the ones for sign for a second, sine zero we know is zero. Sin 30 is a half and then it keeps going up. So, again, that would be roo<unk>2 over two, roo<unk>3 over two, and then lastly just one.

[00:11:12]Now, the way I'm going to show you this is, and something I want to make really, really clear is this graph is completely symmetrical. So if you have a look if I cut this down the middle right the gap between the gap between all of these values is 90° right so between 0 and one is 90° 1 and 0 90° 0 - 1 90° it's all the same so these widths are the exact same so if I have a look at the values that I've had to memorize 30° is a half so around about here 30° should give me a half now if I go backwards a little bit on the graph. And I think you know what I do need to actually if I go backwards a little bit on the graph like this then you would expect that minus a half is also 30° away from the origin. So for example 90° from the origin gives me 1. 90° from the origin should give me minus one. The gap is the same and so the value should be the same. So the answer to this is actually just x= -30°.

[00:12:23]Now it says solve it between 0 and 360.

[00:12:28]That's where it gets a bit more challenging. So again, we've solved it for when it's less than zero. But how do we get the values between 0 and 360?

[00:12:37]Well, if I draw the line of y= - a half, assuming it auto corrects, we can actually see that it actually intersects at some other values as well. So, can you see that it also intersects over here and over here. So, these are two more values that I need to work out. Now, what did I say to you? Going from 0 to minus a half is 30° and that should be the same across the entire graph because it's symmetrical. So in other words, going from 0 to minus a half should be a gap of 30°. So in other words, this gap here is 30°.

[00:13:21]So 180 + 30 gives me 210. That's my first value. If you don't believe me, type in sin 210 onto your calculator. you should get minus a half because again the gap should be the same. Now likewise over here going from minus a half to 0 should be a gap of 30°.

[00:13:43]Now again 360 is there we want to go back 30° so it should be 330. So your two solutions are just 210 and 330. Just so you know if you've watched my um other videos going through the 2025 papers, it came up in 2025 as well. And this comes up also in A level. So we need to be really careful and you need to know that this is on the spec and you do need to know how to do it. So this is actually another really really interesting probability question and I know exactly why again why students would have made a mistake on this. Um it's because the normal diagram of like a tree diagram doesn't really work. You can do it. It just looks hideous. So it's semi-finals.

[00:14:25]Um, A plays B, C plays D, and then the winner will play each other. And they give us a ton of just different probabilities here, which is great. And it says, work out the probability that A will win the chess tournament. So, how did I lay this out?

[00:14:40]Well, when I was showing this to my year 11s, what I did is I drew it out exactly like they do in an actual tournament, right?

[00:14:50]So, you know how you have something like this, right? You'll have you'll have two people and then they kind of go on and the winner is the one that plays the person in the finals, right? You have something like this, right? So, in this case, who actually plays each other? It looks like A plays against B. Yeah, player A. So, we're going to do A and B here. And then we have C and D on the bottom side. Again, I've drawn this quite big. I didn't draw I wouldn't draw this this big in the exam, but either way, it's still fine.

[00:15:20]Right now, what is the point of me doing this? The point of me doing this is to help visualize what I'm actually doing, right? Like, as in what probabilities am I going to use?

[00:15:32]So, if player A wins the chess tournament, here's one thing that's always going to be true. A has to win in the semi-finals, right? So, we're going to have A wins against B uh twice.

[00:15:47]And then if A wins against B, so we want A to be in the finals. Period. It doesn't matter. What options do we have for the person he plays? Well, that's going to be C or D, right? So, we're going to have two possibilities.

[00:16:02]A wins the semi-final and A wins against C. So, A wins against B and A wins against C. Or which means I'm going to add A wins against B and A wins against D.

[00:16:15]So, let's do this. So for first probability, A wins against B. I think it tells you here it's 0.6.

[00:16:24]Now in order for C to be in the final, C has to win against D. So C wins against D is 0.2.

[00:16:33]And then you want A wins against C, which is 0.5.

[00:16:38]The other possibility, well, again, we have A winning against B because A has to be in the final in order for them to win.

[00:16:44]then you need D winning against C. Now, it doesn't tell us that, but we can use our logic, right? If C has a 20% chance of winning, then D has an 80% chance of winning. And then the probability that A wins against D, which it does tell you is 0.3.

[00:17:00]Hopefully, you agree that those are the only two possibilities, right? And then what we're going to is when we work these out, we're going to add them together. So 0.6* 0.2 is 0.12. Half of that is 0.06 06 cuz it is a non-calculated paper. 0.6 * 0.8. Well, 8 * 6 is 48. Um, and then we need to times by 0.3.

[00:17:24]H I'm trying to think there's an easy way to do that. We can do 48 * 3 and then divide it by 10.

[00:17:31]You know what? Let's just do it. Let's just do it manually. 0.48 * 0.3.

[00:17:36]And we're going to do 48 * 3, which gives me 24.

[00:17:42]Uh, that gives me 12 + 4 is that. And then we need to divide it by,000. So 0.144. Okay, cool. 1 44. And then all we're doing is we're adding those two together. So 0.6 plus 0.144.

[00:17:58]Just remember the six is in the sorry 0.06 is in the second decimal point. So, this is actually going to be 0.204 as our final answer.

[00:18:11]Yep, I can imagine why people got that wrong. It's not too bad. Like, the math isn't that bad once you know what you're doing. But think of ways that you can visualize the working out. So, again, with laws of indices, a lot of students do struggle quite a lot on them. uh because mostly because the way we do indices is we keep it quite basic but they actually have these weird index law equations and stuff and I've given some really really tricky questions to my year 11s um that have come up in past papers and mock papers and stuff like that. So what do we want to do? Well the logic one of my students had and I think this makes complete sense is we're going to find an equation for let's say y and then sub it into here to work out what x is. So find an equation for y and x. Sub it into there. Work out what n is. So how do we do that? Well, let's start with the y.

[00:18:59]How can I write square<unk>2? Well, you can write it as 2 to the^ of a half and that's to the^ of 5. Now what happens when you have a bracket to the power of something? So what happens when you have a power to the power? Well, the answer is you times it. So we're going to do a half* 5, which is 2 ^ 5 / 2. So in other words, we straight away know that y is equal to 5 / 2. Why? Well, think of it this way. If I said 2 ^ x = 2 ^ 3, you'd probably immediately say that x equals 3, right? Because the only way for both of these to be the same is if they're the same power. The same logic here, except it's a fraction, which is a bit uglier, but it's the same logic. Now, let's do the same thing with x. So, we have 2^ x. Now, rearranging this one is a little bit harder. [snorts] We're going to have 2 to the^ of n. And there's nothing we can do with that yet.

[00:19:49]But the cube root of 2 is 2 ^ of a3.

[00:19:53]Now what happens when you divide two things of the same base? You subtract their powers.

[00:20:01]So again, and that equals 2 x. So we have x is equal to n -3r.

[00:20:07]So if we sub that into this equation, I'm going to do this part in green, I suppose. x is n - 3. So we have n - a3 + y which is 5 / 2 = 8. So if we add these two together first, we're going to have a third. We're going to have 5 / 2 minus a3. So we're going to times this one by 3 times this one by two. So we get 15 over 2 minus 15 over two. What am I talking about? 15 / 6 - 1 over uh sorry 2 over 6 which gives me 13 / 6. So we get n + 13 / 6 = 8. And so we're going to move the 13 / 6 to the other side.

[00:20:54]Once again, how do we subtract whole numbers and fractions? We change them into fractions, right? And then again, we're going to times the top and bottom of this one by six in order to work out what that is. 8 * 6 is 48 over 6 - 13 / 6 gives me the answer of 35 / 6, which is a disgusting number, but it also happens to be the correct number. Uh, so that's fine. It doesn't say to leave it as a mixed number, so I'm not going to, although you could if you really, really wanted to.

[00:21:26]I hope you've noticed a pattern with a lot of these grade nine questions is the actual numbers and stuff aren't that bad. It's the problem solving which is quite tricky. So again here I mean we're just talking about volume and surface area of a cuboid which is like grade four but look at how it's packaged right look how it's presented to us. They say the cuboid has a volume of 300 the surface area is 370 and the length is 20. They tell me the width is greater than the height. Okay cool. Work out the height of the cuboid showing all your working out. Um that's it. Here's what I would do. And again I strongly recommend this is we're going to draw out the cuboid. And hopefully you guys will forgive me that this will not be particularly great. It's always the back corner I mess up.

[00:22:10]Eh, it says the length of it is 20. Now, honestly, I've had so many students like tell me like yell at me about length and stuff. It doesn't actually matter which side you label as length. Um, I'm going to say the length is this. And hopefully no one gets upset at me. And then we have the width and the height. So, let's just call that W for width, H for height.

[00:22:31]And that's it.

[00:22:33]So what do we do? Well, here's the problem solving logic I always give to students. And again, this is going to seem really obvious to you. But firstly, they have to give you enough information to complete the question. The question can't be impossible. And secondly, they never give you information you don't need. You will always use every single bit of information in the question. So they've told me the volume for a reason.

[00:22:53]They've told me the surface area for a reason. Let's start with the volume. How do you work out the volume of a cuboid?

[00:22:58]you multiply the sides together, right?

[00:23:01]So, in other words, 20 * W * H should give me 300.

[00:23:09]That's an equation. Okay, I'm going to divide both sides by 20. So, we w= So, crosses 0, divide that by two is 15.

[00:23:17]Okay. Uh, let's see if that actually helps me. By the way, just so you know, that's given me one mark already out of five. 20% done already. Okay. The problem here is there's two unknowns, right? Two unknowns and one equation means I can't solve it because there's an unlimited number of answers for two numbers at times to 15. Okay, but they've given me the surface area. Now, how do you work out the surface area of a cuboid? Well, the surface area of the cuboid is literally as it says, it's the area of the surfaces. So, these are a bunch of rectangles. How many? Well, how many faces does this cuboid have? Six.

[00:23:52]So, it's going to be six rectangles we're adding up. Now, that sounds like a lot, but we're lucky. The special thing about a cuboid is that the opposite sides are actually identical. So, the front face of this cuboid is the same as the back face. So, what I could do is I could just do 2 * the front and then I don't have to work out the back. What is the area of the front? Well, it's W * H.

[00:24:16]Okay, so 2 * WH.

[00:24:19]Wonderful.

[00:24:21]Let's do I don't know. Let's do the right side. The side on the right. Well, the height of that rectangle was h and the width of it is 20. So again, it's going to be 2 * 20 * h. Okay, the top side, that's a little bit trickier. If you want, you can do the bottom side, which is w * 20. Again, there' be two lots of it. So 2 * w * 20 should give me an answer of 370. So let's just simplify this now. So we have 2 w + 40h + 40 w = 370.

[00:25:00]Okay. So we have two equations and two unknowns which means this is a simultaneous equation that we can solve.

[00:25:06]Now the first thing I notice is we have wh here and wh is 15. So this is just 2 * 15 right because wh is 15. So that's just 30.

[00:25:18]And now let's think about actually we're going to move the 30 over to the other side first.

[00:25:30]Now let's think about something. Um solving this normally there's two ways.

[00:25:35]One is elimination, right? Which is where what you do is you make the H's the same or the W's the same and you subtract them. The problem is here is one of them is multiplying and the others aren't. So I can't actually get h or w to have the same coefficient because they're joined together.

[00:25:54]Okay, that's a little bit interesting.

[00:25:56]Well, what's the other way to solve simultaneous equations? I'm think of do when you do quadratic simultaneous equations. But what you do is you rearrange for w or h and then substitute in. Right? You rearrange for x or y and substitute in. Now we want to know the height of this cuboid. So, it makes sense for me to eliminate width because I don't care about the width. So, what I'm going to do is I'm going to rearrange this equation for width, which is 15 over h. And then what we're going to do is we're going to shove that into this equation. Uh, one other thing I'm going to do really quickly is I'm actually going to divide everything by 10 to make my life ever so slightly easier. So, now what do we have? We have 4 H + 4 * W, which is 15 /h = 34.

[00:26:45]So if I expand that now, 4 H + 60 /H = 34.

[00:26:52]Okay, that's interesting. How the hell do I solve that? Well, I said before that we're solving this in the same way we solve quadratic simultaneous equations. Now, I'm willing to bet that you guys thought, "Well, it isn't quadratic."

[00:27:05]Well, the other thing is they tell me this piece of information. Do you remember how I said you always use every piece of information? Well, this implies we're going to get two answers, right?

[00:27:18]Surely, otherwise, how am I meant to tell which one's bigger?

[00:27:22]I'm going to times by h and have a look at what happens. 4 H * H is 4 H 2. 60 / H * H is 60.

[00:27:31]= 34h. It is actually a quadratic. So, let's move this over to one side. 4h ^2 - 34h um + 60 equals z. I'm going to divide by 2 and then we're going to have to try and figure out some interesting numbers.

[00:27:48]So, we want two numbers that times to 60, but we'll add to give -7. And those two numbers are going to be -12 and -5.

[00:27:58]And again the way I I do the brackets is I put them together like this.

[00:28:06]But then our answer is going to be two times too big. So I divide by two. Which bracket can actually divide by two? It's the one on the left. h - 6 2 h - 5 = 0.

[00:28:16]So our two answers are either h= 6 or h = um 5 over2 or 2.5 like that. So [snorts] the question is which one is it? Well, it tells me that the width is greater than the height.

[00:28:30]So, logically, it's going to be the smaller one, right? That's it. But you could also work out what the width is and then do it from there. But, it's going to be the smaller one. Another circle question in the exact same paper.

[00:28:42]So, that's uh fun. So, we have a circle center 0 0 as usual. L is a straight line and these two things intersect at two different points and they give us the coordinates of one of the points and the x coordinate of the other. So, let's let's have a look. Right. So again, draw a diagram. Classic.

[00:29:00]And let's draw a big old circle.

[00:29:04]That's actually not bad. Um I know it auto corrected. And they give us two corners. So the point P. So X is somewhere over here. And it tells me it's a positive gradient, which means it should be going up and to the right. So let's say point Q [snorts] is here where we have that minus 2, something. And point P then is up here somewhere where that's 510.

[00:29:29]And that point there is is K.

[00:29:32]Okay. And we find the value of K. So essentially in my opinion working out the equation of the line is quite important because that will just give us the y intercept and this is the y intercept. So that could be quite useful. Um so yeah fairly interesting.

[00:29:47]Now working out the equation of the circle might also be useful. Now, luckily we do know it's going to be x^2 + y^2= something. And working out the the radius actually isn't that bad because it's just the distance from the center to one of the points. So that would be<unk> 10^ 2 + 5^ 2 uh which is<unk> 125. Now just remember that because it equals r 2 is just going to equal 125. Okay, that's perfect for the first step. Uh and let's see what else we know. Well, again, they have to give us enough information to do the question. And hang on, let me and they never give you information you don't need, right?

[00:30:27]Always remember that kind of golden rule there. So, what else can we do from here? Well, the first thing I notice when I look at this is I could also draw a radius here. Right? Now, these two lengths, the pink line, should be the same distance away. So, what do I know?

[00:30:47]Well, I know the distance from the center to here should be square<unk> of 125, right? So, I'm going to write that as So, we're going to have let's say x squ No, not x squ. Actually, yeah, that does work. x2 + y^2 = 125. And I know that the x coordinate is -2.

[00:31:10]So, we get 4 + y^2 = 125.

[00:31:15]So then y^2 is just 121 which is 11. Now technically should be plus or minus 11.

[00:31:24]Well actually I know exactly what it is already. It's plus or minus 11. But if we think about it why can't it be + 11?

[00:31:31]Well + 11 would mean that it's actually above point p which means the gradient is negative not positive because then it'd be going down. Q would be up here and it' be going down. So it must be minus 11.

[00:31:43]So now what do we have? Well, we have two coordinates. So we can actually work out the equation of the line. y= mx + c.

[00:31:52]m is just the change in y over change in x. That's going to be 10 - - 11 over 5 - - 2. So that' be 21 / 7, which is 3. So we have y is equal to 3x + c. And then we can sub in either one of the coordinates. I'm going to sub in the positive one just because uh I want to.

[00:32:13]So x = 10 y sorry x = 5 y = 10. So we have 10 = 3 * 5 + c. So that means our value for c is -5. And we know c is the same as k, right? Cuz it's the y intercept. So k is just equal to minus 5. That's a pretty tricky question, especially the part down here when when you square root, we should get plus or minus, but we actually take the negative number. That's quite rare. hasn't come up too much. And again, the reasoning for that is because when the the lines go up and to the right, if this corner is 11, then it's not going up into the right, is it? Because 11 is higher than 10. Interesting. Okay. So, here we have ABC and D A B are similar isosles triangles, which means that there is a scale factor between them. Right? So, I'm actually going to draw these out separately. So, we have ABC here.

[00:33:09]And then again, now if you think about DAB, that big triangle doesn't look isoclesles, but I'm willing to bet it's actually been rotated. So it should actually look like this, where this is D, uh, that's A, and that's B.

[00:33:26]We have to work out the ratio of the length of A to B to A to D. Cool. Now, they tell me that the ratio of B C to C D is 4 to 21.

[00:33:38]So four and 21. So what that means is BC is four and B D which is this one over here should be 25. And that's kind of our starting point. We can use that to work out the scale factor between the two uh lines right by just dividing them both. Now from here it's a little bit tricky because we need the length of AB, right? Because we need the ratio of AB to A and this is a little bit tricky. So what we could try and do instead is actually work out the perpendicular height of this whole thing. So if I draw a straight line going down, that is the height of the overall triangle. And if that's true, then that means we have two on either side. And we can use this to try and help us work out what um what AD is or also the perpendicular height because we know that those heights should be the same for both of the triangles. The other thing that we know is because this whole length here B to D is equal to 25 because it's isoclesles A to D should also be 25 right because it's an isosles triangle. So what we can do from here is we can actually work out what AB is by working out the height using um let me draw out two triangles.

[00:34:52]Let me show you.

[00:34:56]And then should we call that point like M for the midpoint.

[00:35:01]So we can use our value here that would be 23 because it's 21 + 2. We can use that to work out the height and then we can use the height to then work out AB by just doing the same thing again. So the height there is going to be square<unk> 25^ 2 - 23^ 2 which is a little bit tricky to do in your head. It does give you<unk> 96 eventually which means I can work out the length of AB. And bear in mind I already know a d it's 25 by doing the same thing again.

[00:35:33]So what I mean by that is [snorts] that side is going to be equal to 2^ 2 +<unk> 96^ squ all square rooted uh which is going to be square root. So that's going to be 4 + 96 square of 100. So it's equal to 10. So if a b is 10 and b d and a d is 25. So the ratio is 10 to 25 which I can divide both sides by five to get 2 to 5 as our final answer. So I think yeah breaking into two isocesles triangles and then using that to work out the height is probably the easiest way to do that.

[00:36:10]That brings us to the end of the paper.

[00:36:11]So hopefully you found that useful.

[00:36:13]Hopefully you can see how the grade 7 89 problem solving works again. And yeah, hopefully you found it useful. There's a link to the group tuition classes in the description and the pin comment.