Even Adults Cannot Solve This!

Fun Math Problems. Find the Area of Blue Shaded Region.

If you're reading this, drop a comment using the word "blue". Have an amazing day, you all are awesome!

[WhitentonMike]

This can be simplified by drawing a line between the 2 endpoints of the quarter circle. The nonblue area above the hypotenuse happens to be same area as x. Now you only need to find the area of the quarter circle minus the area of the nonblue triangle. Once again we end up with the blue area = 4π-8

[benlap]

You'd be surprised how much kids can learn when you're not passing into them your own fears.

[tom]

This is one of those situations where you can "train" primary school students into solving seemingly high-level problems. The issue is that they don't actually understand the concepts and can only solve specific subsets of problems in the specifically trained way. This allows the schools, principals, and teachers to look good because "look how advanced our students and teaching methods are!" but in reality, the students didn't actually learn anything.

[CyberSlugGump]

Due to symmetry, you can split the lens-shaped part in half and rotate it to fill in the gaps in the circular edge. That just leaves you with finding the area of a circular segment: Area of quarter circle minus area of right triangle = 0.25*pi*4^2 - 0.5*4*4 = 4*pi - 8

[StereoSpace]

They do these in Chinese primary school? That's nothing. In Peru, they do these problems in kindergarten.

[rahulbanerjee]

Area of quater circle- (π4²/4)
Area of trangle - (½ * 4*4)

Now substract the trangle area from the quater circle area.

(π4²/4)-(½ * 4*4)= 4(π-2)

Don't make it complicated..

[engineer]

I solved it the following way:
Imagine square of length 2 at corner
Calculate half of a semicircle
Subtract it from square
Subtract difference from square twice
Thats the first blue area

Calculate total area
Subtract both semicircles
Add first blue area
Thats your second blue area

Add them. Done.

[rudolfgernd]

There it is. I point at the blue area. Found it!

[varoonnone]

I would have failed Chinese primary school because I don't have any logic

[jojojo]

How I did : Took a square of 4x4 in my head. If I draw semi circle of radius 2 on every side of the square I can write: Surface of square - semi-circle - semi-circle + area I look for ( 4 times ) =0. I got the rugby ball = 2PI-4. I took a paper and a pen at that moment....

Then, I took the quarter circle of radius 4 which surface is 4PI, remove semi-circle on a side ( minus 2PI ), remove semi-circle on the other side ( minus 2PI ) and put back the rugby ball that I remove twice ( + 2(2PI-4 ) ) : gives : 4PI-2PI-2PI + 4PI - 8 = 4PI-8

[towguy]

So much simpler:
Draw line between endpoints of quadrant, making a prime right triangle with measurements 4, 4, 4sqrt 2. Bisect quadrant, creating 2 similar triangles and unshaded lens halves which are congruent to the shaded lens halves labeled x. Transposing those lens halves results in the shaded area being the area of the quadrant outside the prime triangle.
Quadrant measure 4pi and prime triangle=8. 4pi-8.

[mweskamppp]

How to take down a complicated looking problem into a couple of easy problems. a good training and its confidence building too.

[cmilkau]

0:35 at this point you can rotate the x surfaces so they fit the z surfaces, and then the y surfaces become just triangles. Then you see the blue area is just a quarter circle minus a quarter square

[A]

This is a primary school question in Singapore, actually not very tough. This can be solved easily by moving tje smaller blue area adjucent to bigger blue area. Afyer moving u have 1/4 th of the circleinus aclean white traingle... i. E, 4pi - 1/2*4*4 which is 4pi- 8 = 4*(pi-2) = 4*1.14 = 4.56

[APprojection]

1/4 of white circle is Pi, we have 2x2 square, white pieace of 2x2 square is 4-Pi, so total white area is 8. Take away 8 from total area of circle which is 4Pi

[anubhavos]

As an indian I may assure I literally solved this problem in my mind withing seconds of looking at it easy like I got a similar question in class 6 ICSE school question

[gelbkehlchen]

Solution:
You can divide the blue "rugby ball" at the bottom left into 2 circular segments of a quarter circle, then place them at the top and bottom in the white bulge of the respective semicircles, then you get the blue area of ​​a circular segment of a quarter circle with a radius of 4.
Then the blue area is
= the area of ​​a quarter circle - the area of ​​a right-angled, isosceles triangle with the two legs of length 4
= π*4²/4-4²/2 = 4π-8 ≈ 4.5664

[gabrieljuliano]

"Teacher, my answer is X, you can find what the true value is by calculating your own exercise"

[BondiAV]

It found it pretty easy to figure that areas of x and z are equal. So I calculated the area of x and multiplied it by 4.
The other method I found below (which involves flipping the x portion until its convex bowed side mates with the concave side of z, making it obvious that the answer is a quarter of the circle of radius 4 minus a right angle triangle that has is legs' lengths equal to 4) is even nicer.

[josephrubalema]

Mathematics creates moulds in our brains into which to fill other intellectual matter like arts, quantum theory etc