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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…
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.
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.
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 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 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.
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