I must admit that I usually find videos that have a catch phrase at the end super annoying. That said, whenever something cool happens in my life I now find myself saying “how exciting!” So well played sir! Keep up the great work. ❤
Nope. You're the one making yourself smart by trying to solve the problem. He's just there to help when you give up, succeed or fail. Respect to the guy obviously, his presentation is great.
Was not expecting to hop onto youtube today and learn more math from a stereotypical looking college frat bro in under 4 minutes than I did all semester in my calculus class, but very informative!
Look, I like to dunk on how unhelpful maths classes sometimes are, but this is literally just familiar triangles, pythagorean theorem and simple arythmatic. Any 15 year old with a B in maths should solve this in 5 minutes tops.
I'm not sure if this is just the perspective of Andy's camera or the room or if he is tall, but it looks like he is about to hit the ceiling, no? If this is some secret joke to make it seem he is 9ft tall that'd be amazing. That was another fun problem, I adore your enthusiasm, and humour, its good fun! So good on ya my man! Have a nice day! :)
Idk anything about math, in fact I've stop doing math at 15, but I really enjoyed learning this topic back in school, now I'm completly washed up ofc. But god, I love this channel !
@@SiddharthHarsh yar Mera abhi sirf civics hua hai, economics aadha ho gaya aaj kar lunga. Kal history pura + geography thoda aur parso geography khatam kar dunga. Fir 1 sample paper dunga Digraj sir ka
<a href="#" class="seekto" data-time="110">1:50</a> The way I would've done it from here is this: Make b = 4a -13 Plug in: a + 4(4a - 13) = 4a 16a + a - 52 = 4a 17a - 4a = 52 13a = 52 a = 4 Plug in a = 4 for the equation for b 4(4) - 13 = b 16 - 13 = b 3 = b Solve for the smallest triangles hypotenuse (3^2 + 4^2 = c^2) The smallest triangle is a 3 4 5 triangle (special right triangle), so c = 5 Solve for the squares 5 * 5 = 25 Since there's 4 squares: 25 * 4 = 100 That's our answer, I'm not sure if my way takes any longer, I just don't like dealing with fractions at all
Nice. My solve was similar, but I labeled the shorter leg a and longer leg b, and I multiplied by 4 rather than dividing by 4 and avoided having to use any fractions.
awesome question, and well-solved and explained. i thought they asked for the shaded area that is within the square spoilers: . . . . . . . . . . . . . . . . . it's 725/8, if anyone cares
Solving this before watching the video. So, the square on the bottom combined with the angles on either side of it on the bottom line is 180 degrees. The square has a 90-degree angle which is translated to the two 90-degree angles on the bottom edges of the large square. The angles on the sides of the bottom square also translate to the opposite angles of the two triangles on either side of the shaded area, showing that the triangles are the same, just differently scaled. The hypotenuse of both triangles also matches up with the shaded area, with the smaller triangle being the same as the side of one square, while the larger triangle matches with the side of all four squares. Labeling the sides of the triangle similar to the Pythagorean theorem (a^2+b^2=c^2), the side of the square is c, while the rectangle made up of four squares is 4c. This also shows that the larger triangle is 4x as big as the smaller one, so we'll also label the longer ends a and 4a, and the shorter ends b and 4b. This then shows that the sides of the large square are equal to 4a, 4b+a, and 13+b. So, if we start with 4b+a=13+b and subtract 4b from both sides, we get that a=13-3b. Multiply both sides by 4 and you get 4a=52-12b. Now, since 4a also = 13+b, we can say 13+b=52-12b. subtract 13 from both sides and add 12b to both sides and you wind up with 13b=39. Then you divide both sides by 13 to get b=3. Plug that back into 4a=13+b to get 4a=13+3 or 4a=16. Divide both sides by 4 to get a=4. Now, by using the Pythagorean theorem, we can determine that these are 3, 4, 5 right triangles, meaning that c=5 (a^2+b^2=c^2, 3^2+4^2=c^2, 9+16=c^2, 25=c^2, c=5) Now, the area of the small square is the square of the side, being 25. Multiply that by four squares, and you get the total area of the shaded rectangle to be 100. (conversely, you could multiply the side by 4 to get 20 for the long leg of the rectangle, and then multiply that by the shorter side of 5, to get the area of 100 as well)
@@kaitek666imgur.com/a/AxMbvaH Here's the image and how I labelled it. So I started by proving congruent triangles the same way. △ABC ᔕ △A'BC', therefore, AB/A'B = BC/BC' (x-13)/sqrt(16y^2-x^2) = y/4y = 1/4 4(x-13) = sqrt(16y^2 - x^2) (*) We have: x = AC + CA' = sqrt(y^2 - (x-13)^2) + sqrt(16y^2 - x^2) Substitution with (*) we get: x = sqrt(y^2 - (16y^2 - x^2)/16) + 4(x - 13) x = sqrt(x^2/16) +4x -52 3x + x/4 = 52 -> x = 16 -> y = 5 -> Area of shaded area = 4y^2 = 100 (sq units) Hope this helps!
<a href="#" class="seekto" data-time="57">0:57</a> you missed one piece of information (but obviously it didn’t make much of a difference), the side length of blue-green on the larger triangle has a length of 13 + b of the smaller triangle.
Yeah so I had a way worse way of solving it and I still can't confirm if it works because I tried it on the calculator twice and it failed :) Anyway, I used the two similar triangles that you used but then decided that I would call the side of the big square x and write the side of a red square in terms of x by using the small triangle in the bottom left corner. I then created a new triangle by connecting the point the shaded area intersects on the left side and the top right corner. That triangle would have the sides x, 13, and sqrt((85x^2)/16-130x+845) If it's already wrong at this point or if I messed something up beyond that I do not know but tl;dr my method could probably work but it's too messy and I'm too lazy to figure out what the problem is
I neglected to use the ratio <a href="#" class="seekto" data-time="100">1:40</a> of the sides, so I went straight into trigonometric ratios... and solved it, but in a slightly harder way, but without Pythagoras.
There are math channels on RU-vid, which make from a simple x-equation a 5 minute video clip. Bummer. And your stuff? I'm not the dumbest in using math, but when i see one of your problems, i usually go: huh? what? HOW? And then you solve it in 3-4 minutes, using just simple algebra and geometry. Awesome!
Yeah. The few times I've been punished for assuming a figure is to scale unless stated otherwise just won't let any geometrical intuition rest well within me 😂
The shaded part that sticks out at the top is also one of those similar triangles, right angle pointing up this time. You know that one side of that triangle is 5 and you also know that another side is 3/4 times 5. Both these sides are next to the right angle so you just multiply them with each other and divide by 2 to get the area of that triangle. 5*(3/4)*5 =5*(15/4) =75/4 (75/4)/2=75/8 Then subtract that from 100 (the whole shaded area). 100-(75/8) =(800/8)-(75/8) =725/8
when i see a colorful handmade drawing, i can instantly see that is a Catriona Agg question.. bc nobody redraw the question digitally .. dont know why tho :))))
In middle Egypt they knew "the area of a square of 100 is equal to that of two smaller squares. The side of one is ½ + ¼ the side of the other." I thought it interesting the shaded areas is 100 and you got b=3/4a on your working
So, knowing this here's another way of solving once you know the scale factor is 4 and have worked out the bottom left triangle: You know the length given is 13, you also know the small triangle at the bottom left hand corner is 3,4,5. So the length of the side of the black square is 16. So because we now know the length is 16, that means the side opposite it is also 16. So 3/4 of 16 is 12 which means the other side of the triangle is length 12. 12^2 plus 16^2 is 400. Divide 400 by 4(the scale factor) and you get 100. because 3,4,5 is also equivalent to 6,8,10 and therefore 12,16,20 then you now know the hypotenuse of the triangle on the right is 20 and the hypotenuse can be divide into 4 lengths, 20/4 is 5 which gives you the hypotenuse of the bottom left triangle and then you get the areas of 25 for one square. Then you go 25*4 which is 100. which proves the Egyptians were right about a side being 3/4 the length of another
Would it be 96.625. By using the same method the top left triangle is also similar and we can do the area of the square minus the area of all of the triangles.
Great problem! Thanks for showing so many people the fun in math! To see a similar problem solved using a free, browser-based modeling tool see ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-haS2FMwA1ZM.html or ru-vid.comq8AL39jCUGI?feature=share.
