Find the Area Challenge

Presh Talwalkar has a neat shortcut in this video.

[tttoastbut]

honestly, the initial problem looks horribly hard. But the solution was actually easy haha. Thanks for the solution :)

[theresamclaird]

I love problems like this because it demonstrates how to take what appears to be a complex/difficult problem, and break it down into simple steps.

[evilmonkey]

what a crazy solution to a problem like this. as someone who doesn't know the formulas to solve these off the top of my head, i see things like this at work all the time (im a contractor and volume and area come up constantly) and i always just end up estimating. but to be able to crank out a real solution would be so satisfying

[jarcuadanantus]

Why am I sick and watching math at 9am? I’m almost 30.

[souravde]

Usually, students panic on seeing these types of figures and give up.
Thanks for simplifying the seemingly complex problem!

[gabithefurry]

Man, your channel shows exactly what all my math teacher once told me: math is not hard, in fact, it is easy, you just need to decompose the complex processes into simpler ones until you solve everything
Really nice content, keep on the great work!

[patrickkeller]

I did it a bit simpler: cut diagonally the shape is made from two white circle segments cut from two red circle segments. The two radius 8 segments are equal, so we can fit them together and the total area is a radius 12 segment, minus a radius 4 segment, or (36pi-72)-(4pi-8)=32pi-64.

[peterpan]

You can also cut the claw along the diagonal of the bottom left square. Then shift the smaller part to the top right corner of the box and turn it by 180 degrees. Then x is just (1-1/3^2)*A, where A is given by the quarter of the area of a disc with radius 12 minus the area of a square with diagonal 24.

[JTCF]

This is truly beautiful. Like these kinds of problems, you mostly just need to break it up into nice pieces and then it all comes together (pun intended) in such a beautiful way. Love your explanation too.

[ZantierTasa]

If you add and subtract the obvious quarter circles, and use a pen to keep track of double counting for each little region, you find that you simply overcount by exactly 4 of the squares :)

Quarter circles: 36pi + 16pi - 16pi - 4pi = 32pi

Subtract the squares: 32pi - 64

[kamionero]

1:33 How do you know the arcs are a quarter circle? Thats an assumption that the drawing doesnt really confirm. It could be a slightly asymptotic line, not a radial.

[thanhhai]

I watched the other video in the description and I have another solution as well: 1/ calculate the A of the right claw by substracting the larger 1/4 circle (r=12) from the smaller 1/4 circle (r=8) and 2 squares on the left. 2/ calculate the A of the left claw by substracting the larger 1/4 circle (r=8) from the smaller 1/4 circle (r=4) and 1 square in the bottom mid. 3/ Add the A of the 2 claw and substract 1 extra left corner square. This is fun! Thanks for the vid!

[IDE_Busmaster]

What a great video! I had no clue how the process for those inner sections would be calculated when I started and was shocked at how intuitive it was that making them quarter circles and removing the triangle came right as you started saying the solution! Very well structured and paced!

[HenkTheUnicorn]

tl;dr: You can shift the white area above into four white squares and a white quarter circle, turning this problem into something elementary.

It can be done even quicker and in a much simpler way (even simpler than the trick in the other video). Since you have two congruent quarter circles, a lot of symmetry can be used. Start with the area of the biggest quarter circle π*6^2, the white area outside it is not necessary.

If you look at the two quarter circles with radius 8 you can actually find two full white squares with them, fit the white area in cell 3 and 6 above the red into the white area above the red in cell 4 and 7. Together with the two cells in the top row you have 4 white squares, so you can subtract 4*4^2 from the total.

Now the very middle cell is left. In that cell, if you see that the red part in the bottom right has the same area as the white in the top left (again due to symmetry of the two congruent circles) then it suffices to just subtract the area of the smallest quarter circle for the solution.
So π*6^2-8^2-π*2^2= 32π - 64.

I think this would be the method with the least amount of calculation.

[DataScienceDIY]

This problem is extremely easy with some basic calculus. Calculating area by integration is one of the most common problems in early calculus courses. For this problem all you need besides integration is the equation for a circle.

[tagnetorare]

If you cut the area along the diagonal from the top right to the bottom left, and rotate that piece on the left around the center of the large square by 180 degree, you can simply your calculation. The area would be a large circular segment minus a small circular segment, which is 36pi-72-(4pi-8)

[balloony]

This is a crazy problem to solve. I thought by clicking on the video I was going to see this being solved with calculus. However, your problem solving solving method stunned me as you were able to make complete sense out such a odd (and complex) question. This is definitely one of the coolest videos I’ve seen lately.

[DuongPham-bdvr]

I used to be afraid of these kinds of problems, until I learned double integrals. Now I can probably solve this in somewhere around 15 mins. Your solution, which only includes basic mathematics, and takes no more than 5 mins, is beautiful

[roybixby]

This assumes all curves are spherical - what if they were aspherical ? ...

[bitandbob]

Thanks so much for this stepped solution. I struggled with algebra and have not done it for nearly 20 years yet watching this really made my memory trigger with how do all that - i understood it and feel like i could apply those principles in other circumstances.