Find the Red Area Area

[AndersRisager]

Dude, I wanted to kick my self right after you drew the square... Fun assignment, my students would enjoy this one 🙏

[Grim]

Interesting method. I drew two more semi-circles, and that gave me 4 such regions of overlap. I then simply subtracted 100 from the area of the 4 semi-circles combined and divided it by 4 to get the are of one overlapping.

[connorgabriel]

i got the same answer differently!

first i found the area of the semi circle, and times it by 4. if you put all 4 semicircles on each side you can see there are 4 equal intersections. so if you take the area of all 4 semicircles and minus it by the area of the square, you get the area of 4 intersections. divide that by 4 and you get the area of 1 intersection. then all you have to do is find the area of the original 2 semi circles, minus it by the intersection, and then minus that total from the area of the square which is 100.

[IAmOOFy]

Wow, what a simple method. What I did was took the area of the square, then subtracted the area of the two semicircles, which make up one full circle. Then I added the geometric lense’s area formed by the overlap of the semicircles.

[richardngo]

I found the area of a circle with a radius of 5 to be 78.54, half of it is 39.27. (White area)
Though 39.27 is enough to continue, I double checked by dividing it in half to get the area of the circle in those corners.
Box is 10x10, four corners of 5x5, top left is going to be 25 units (White area)
Add the top left corner (25), top right corner, and bottom left corner (2x19.635), gives 64.27 (Total white area)
Now, that last step could have been skipped by using the 39.27 found in line 2, by adding with 25 to get the same result.
Subtract total white area from the whole box to get 35.73 (Red area)

[OmarCowley]

You’re my favourite channel to watch right now. I’m really bad at math but by watching your videos, its really helped me approach and solve a lot of problems at work. Thank you Andy Math

[Alpha_]

Your method is simple
I did it like : creat 2 more semicircles from the sides of squares which doesn't have one, which will create 4 lens like structures
Now
area of 4 semi circles -4(area of lens)= area of the square
Find the area of lens and then use to to calculate the required area

[rajachan]

Elegant. Love your work. And your delivery style. And the “How exciting” at the end!

[detroitstudios]

I have been watching this channel for a month now, this problem is the first one I could solve, I did the method as you, but my answer came out as 35.71. If you arrange it like this:
area of square - (area of a semicircle as the two quarters make a semicircle + area of small square)
so,
100 - (25*pi/2 + 25)
1400-900/14
500/14
35.714

[akra]

i have been watching you for the past two months or so, and this is the first challenge i could actually solve :D. I defined the shape the two semicircles formed as s, and the "triangle" as t. A circle then could be defined as 4s + 2t, so i considered pi as 3 and went πr² = 4s + 2t.

3.5² = 4s + 2t

75 = 4s + 2t

By looking at the square i could it break down to 4s + 4t, and since the side is 10 we get 100 = 4s + 4t.

100 = 4s + 4t
75 = 4s + 2t

25 = 2t
12, 5 = t

75 = 4s + 25
50 = 4s
12, 5 = s

The area of X can be broke down to s + 2t, so

x = 12, 5 + 25 = 37, 5


if we go back to your video and consider pi as 3 we can get by your result that

75 - 75/2 = x

75 - 37, 5 = 37, 5

So, basically solved ^^

[RunstarHomer]

Way more elegant than my solution. I calculated the overlap of the two semicircles, then did A U B = A + B - (A n B) to find the total white area. Then subtracted it from the whole

[gerardacronin]

I drew diagonal lines across the square. It was immediately apparent that the “lenses” closest to the red area were identical to the overlap between the semicircles. The area of the lenses can be determined by calculating the sides of the half diagonals and subtracting the area of those triangles from the area of the semicircles. I didn’t do the calculations but the method should work.

[ehudkotegaro]

Mark the Points of the square as ABCD, their Middles as E, F, G, H, and O as the Intersection of the two circles. X is the area of OFCG (5*5=25), plus the area of the shape in HOGD (which is HOGD - the quarter circle 5^2*pi/4), plus the area of the shape in EBFO (the same. All in the area should be 75-12.5 pi.
Is that right?

[Clappybara.]

Hey, I'm in grade 9 and this is my favorite math problem by far. I found the answer immediately and it was so fun!

[Phoenix-nhkt]

Use definite integration and take two integrals with limits 0 to x coordinate intersection and x coordinate intersection to 10. Easy

[PulseTriangle]

Oddly proud to have solved it the exact same way and relatively quickly despite not doing any math for like 10 years. I liked math, and wish there was an application of it in my day to day, but sadly it's just not something that comes up a lot!

[brrrrr]

Neat, ended up getting this right with heuristics. In essence: the area of half of the parent square (100/2=50) plus (the area of a circle with diameter 10 (25pi) minus the area of a square with side 5 (25) to derive the area of the circle with a square-sized portion removed. Divide that by 2 to get the area of just two of those slivers). You end up with the area of half the square plus the area of two circle slivers, or 35.73 approximately.

I'm sure the method in the video is more precise, but not having that formal methodology in my head, this is what I had to bank on.

[AyushTH]

Since the two circles intersect at the middle of the box, we can partition a segment of x that is 1/4th of the whole square. We can then use the other 1/4th segments adjacent to the one that is a part of x and subtract it by exactly one half of the semicircle's area since that is the extra area that is not a part of x. So the answer would be 1/4th of the square (25), plus 2/4ths of the square minus the area of the semi circle (50 - 5^2pi/2). Making the full solution (50 - 5^2(pi)/2) + 25. I rate this problem a 7/10.

[abhirampaku]

I think of a different approach using symmetry fold this paper with symmetry fold line across point between tens and the diagonal point
Area of red region is half of area of current region
A triangle is formed divie the semi circle into two parts on quater circle and triangle find area minus it from total area and double it you'll get your answer

[massivecowbreakout]

Struggled at first, instantly got it after you drew a square