From my experience a lot of techniques that Andy applies, and the way he applies them in the problems he solves, have genuine practical research and engineering value.
Best part is he shows all his work, even if it’s kinda unnecessary he still shows us the working even down to the most basic multiplication. So anybody can follow along
@@bobtheblob2770 For those who have a CAD package and the ability to use it. What Andy is doing is educating people on the applications of the fundamentals. I hope your computer never fails.
Math is a subset of formal logic, and thus philosophy, using very precise symbols, definitions and minimal axioms. I'm an industrial electrician (sparkmonkey) that translates philosophy on the side. I understand how you feel, and I can offer you this tentative answer: you want to learn something new and true about reality
I watch math channels on RU-vid because I love math and I like to keep my memory fresh and ready to solve problems. But Andy's videos are really exciting. It makes you fall in love with solving math problems even if you don't like math. How Exciting!!
that could have worked as squares of the side lengths give you the area of the square and would have got you straight to 36 without first calculating 6.
You can also solve one of the angles of the blue triangle and then use trig functions to find the side length of the pink triangle, but your method of using pythagoras theorem was more elegant! 👍
I was never much for math, but I do love a logic puzzle. Your channel reminds me that math is just a puzzle and I love that. It's been 20 years since I learned some of this stuff. Some of it I'm not sure I was ever taught. Many of your leaps go over my head, but I'm certain with extended exposure, I will begin to pick up the pattern of your logical deductions.
My memory is spotted so I missed some of the shortcuts related to alternate interior angles and sorta had to do everything from scratch with basic trigonometry and actually work out all the angles. I calculated the hypotenuse of the blue triangle using plain old Pythagorean Theorem. With all the sides and one angle known, I could use sine/cosine for the remaining angles. For the top right angle of the pink triangle, I knew it had to equal 90 minus the top right angle of the blue triangle. With two out of three angles of the pink known (it's a right triangle), we just subtract the sum of the two knowns from 180 to find the last angle. With all angles and one side known of the pink triangle, I opted to use the Law of Sines to find the base of the pink triangle which would also equal the base of the pink square.
Andy, I'm really enjoying these puzzles. Just found them and I've gone through a few; subscribed with alerts. I'm out of practice and you're helping me limber up my math brain, thank you!
I’m learning so much from this. Haven’t been doing too good in geometry lately and these videos have really made me think about it more (in a positive way)
I saw your video with roots and x=69, but I was curious of your content and I saw you have a lot of videos and a website with even college complexity, you made shockingly good impression on me. Even the way you talk and move tells me you’re passionate with it. You’re a great guy. Thank you I’ll be studying differentials with you. Peace!
I used to love math a lot during my high school years, but due to limitation of job here in my country, I further my study in accounting instead. Watching your videos bring lots of good sweet memories. Thank you.
I am absolutely satisfied to find this channel, because I gain two benefits- learning English and learning mathematics, Just thank you very much for creating this channel. 🙏🏻
nice, I did it differently using trigonometry, yours seems neater. I realised that all the right angle triangles made by extending the lines of the squares of the pink square are the same ratio and did the maths. they square of the hypotenuse of the blue triangle was 25 (side is 5) so all that needed to be done was calculate the short side of the small triangle created by extending the pink square's right side down, which turned out to be 1. 5+1=6 is then the side of the pinks quare, so Area is 36.
I did it a little differently: Since the 20-square is exactly 4 times the area of the 5-square, we can cut it up into four squares, each of area 5. The side length of each 5-square is clearly √5. So the side length of the big square is 3√5. The small white triangle is similar to the pink triangle, as Andy showed (right triangles with alternate interior angles). So the ratio of the long leg to the short leg in both triangles must be the same. In the white triangle, the long leg is 2√5 and the short leg is √5. So the ratio is 2:1. The long leg of the pink triangle is s (the side of the pink square), so the short leg must be s/2. The hypotenuse of the pink triangle is the side length of the big square which we already know is 3√5. So, by the pythagorean theorem, s² + (s/2)² = (3√5)² . (3√5)² = 45, so the equation simplifes to 5s²/4 = 45. Solving for s², we get (4/5) x 45 = 36. Finished; no need to find s, itself.
I got this one! Not the numbers, but the principle. I got there through the way you prove that line theorem. Top corner blue (a)+top corner pink(b) = 90, they're both right triangles, so angle sum of 180, 90 for the right angle, so the 90 left goes to a and b
For a first time in a while, I've actually decided to solve the problem myself, and at some points I thought "I might be doing something wrong" (which actually happened at proportion, oops 😬), but it was so much fun to see that you were actually on the right path the whole time
I constantly solve problems with a solution different from how the teacher solves them. This task is no exception. I found that the side of the square (which is the hypotenuse of the triangle shaded in red) and it is equal to √45, the drawing clearly shows that the smaller catheter of the triangle shaded in red is 1/2 of the second Let the small catheter be x, then the second one will be 2x. So we get a triangle with x and 2x catheter whose hypotenuse is equal to √45. According to the Pythagorean theorem, C^2=A^2+B^2, 45=x^2+4x^2 => 45=5x^2 | :5, => 9= x^2, x=3 (x is a small catheter, a larger catheter, which is also the side of the desired square 2 times larger, that is, the side is 6, the area is 36 sq units). The whole solution is based only on knowledge of the Pythagorean theorem.
Nice video. I solved the problem in a similar manner: Note that the side lengths of the three black squares are sqrt(5) times 1, 2 and 3. Therefore the short catheters of all the similar triangles are half the size of the corresponding long catheters. And the hypotenuses are sqrt(5) times the short catheters. Then the short catheter of the pink triangle must be 3, and thus the long catheter is 6, and the pink square has area 36.
The first blue triangle has ratio 1:2:sqrt(5) and is congruent with the first pink triangle, of which the hypotenusa is 3sqrt(5). Hence the pink square has side 6, and thus an area of 36.
I solved it by using alpha = arctan(/sqrt(5) / /sqrt(20)), then s = cos(alpha) * (/sqrt(5)+/sqrt(20)) and then area = s^2 but your method was also cool
I solved it by setting a system of equations with what he had as the pink triangle netting the first equation: a^2+(b+5)^2=3sqrt(5) and a second triangle in the top left of the big box that determined the diagonal distance must be sqrt(10), so the second equation was: a^2+b^2=10 not necessarily a better way but brings more information about the dimensions of the pink square once you solve the system of equations (either a=3 or -3 overlapping satisfies) :)
Pretty much took those steps. One thing I noticed: In one of the last steps you cross-multiplied both sides, only to divide by one of the terms immediately after. Given that you only needed to multiply by one divisor to get the variable alone that "multiply-divide" step is unnecessary.
This one feels conceptually pretty straightforward if you imagine a coordinate plane over the shapes. We know the slope of both bottom line segments on the square, and could calculate the intersection of those lines. But it does seem like some hairy algebra.
Sidelength of the given squares are √5 and √20 = 2√5. I called s = sidelength of the pink square and e = hypotenuse of right triangle on top of the 20-square. Then e = √[(2√5)² + (√5)²] = √[20 + 5] = 5. Due to similarity of triangles we have (2√5)/e = s/(3√5), hence s = (2√5)·(3√5)/5 = 6·5/5 = 6, therefore area = 6² = 36 square units.
I was thinking of using trigonometry to find the angles in the blue triangle and back again to get the relevant side of it's red counterpart... Much more elegant and efficient your way ^^ However for the length of the bigger square's size, I found calculating the total area and square rooting it to be faster more satisfying than the way you did it, even if it's not faster ^^ You easily find that the area is 45, meaning sides are sqrt(9*5), which is 3sqrt(5) ^^
In India, we learn algebra like this- when any operator on one side goes to the other side of the equation, it changes to the opposite operator. Like + to -, * to / and vice versa. I think it’s a bit faster that doing it on both sides.
Him, a mathematician: *does complex algebra* Me, a rainman level autistic person: just rotate the pink square and see if it matches the white square if it does then pink = white
You could just calculate the sides of the 5 and 20 square then the side of the white square and then you reverse Pythagoras to find out the right triangle sides which have also pink square side
