At last week's math department meeting (I teach Spec Ed at a ~600 student high school) I mentioned this fun new channel I discovered called Andy Math. This video is going in an email to the department. Thanks, these are a lot a fun.
I JUST REALIZED, I GOT THIS EXACT PROBLEM ON MY MATH TEST! I’m so gonna watch all your videos looking for more problems they put on my tests, sneaky teachers…
Pretty sure I’ve looked at Brilliant. I will definitely look again. Anything that demonstrates the deep beauty of mathematics is truly exciting to me. Thanks.
Fun, I took a different approach. Since all of the 3 small RTs are similar, they're all 30-60-90. So I made my own not-to-scale version, and I strategically but arbitrarily assigned my square's side a length of sqrt(3). Then I found the left portion of my hypotenuse to be 1 and the right one to be 3. So I know my hypotenuse measures 4 + sqrt(3), and my side divided by my hypotenuse works out to be (4sqrt(3) - 3)/13. Since I know the given hypotenuse is 13, the given square side must be 4sqrt(3) - 3. Then I squared it.
I am a designer so sometimes I use 3d model to solve these questions. But I also do this with my own. And usually the number is without any decimals so I solve it 2 times reaching same answer 15.430 and then give up to look for your explanation and get the same answer (never doubting me again)
I did it using trigonometry. Let: x be the length of the lower side of the right. y be the length of the side of the square. z be de length of the lower side of the left. Using tan(), we can get this: x = y÷tan(60°) z = y÷tan(30°) Then, we can solve for y: x + y + z = 13 cm (y÷tan(60°)) + (y÷tan(30°)) + y = 13 cm y(1÷tan(60) + 1÷tan(30°) + 1) = 13 cm y = 13 cm÷(1÷tan(60°) + 1÷tan(30°) + 1) I plugged it into my calculator and gave me y = 4√3 - 3. y² = 57 - 24√3.
I have this math challenge I’ve tried for a while but still can’t get it. Is there a way you can try it? Also how would I be able to give the challenge? Edit: I don’t know if it’s possible.
I had an extremely valid solution method going on, but, right at the opening, I made a rookie mistake. I saw 30° 60° 90°, and so I said... _"A-HA! a standard Right Triangle!"_ I then saw *13 cm* , and so I said... _"A-HA! it must be the ol' standard_ *5-12-13* _Right Triangle!"_ Per the valid solution method that I came up with, I used a length of 5 cm (per the idea of a 5-12-13 right triangle) WHEN WHAT I REALLY NEEDED was *6.5 cm* (per a 1-2-√3 right triangle!) I feel certain that the people who stay awake, late at night, designing these problems, are well aware of the standard STUPID ASSUMPTIONS that careless people usually make (such as automatically assuming that, when you see a *13 hypotenuse* on a right triangle, the triangle must automatically be a *5-12-13* right triangle!) Now I know how Sheldon felt when he got caught making an arithmetic error in front of Stephen Hawking. Anyways, my solution method is as follows... [ *note:* The following solution approach inscribes all possible rectangles inside of the triangle, and finds the independent variable value, i.e. 𝑤 , where all the sides of the inscribed rectangle have equal length, thus producing an inscribed square.] ( 𝑤 / cos 60° ) = sin 60° ( 6.5 - 𝑤 ) ( 𝑤 / 0.5 ) = 0.866025 ( 6.5 - 𝑤 ) 2𝑤 = 0.866025 ( 6.5 -𝑤 ) 2𝑤 = 5.6291625 - 0.866025𝑤 2𝑤 + 0.866025𝑤 = 5.6291625 2.866025𝑤 = 5.6291625 𝑤 = 5.6291625 / 2.866025 𝑤 = 1.964101 Substituting the value obtained for 𝑤 into... sin 60° ( 6.5 cm - 𝑤 ) ...yields... *𝑥 = 3.928202 cm* [ *note:* 𝑥 is the length of a side of the square, as shown in Andy's drawing] Squaring 𝑥 yields the area of the square, which is... *𝑥² = 15.430771 cm²*
Because that is just a decimal approximation of the area. You should always try to arrive at an exact form of the solution first, rather than approximating as you go. 3.93² actually gives you 15.44 (rounded down), not the correct rounded down approximation of 15.43