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!
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
Ask 10,000 US high school graduates and they couldn’t figure this out. Great school system, time to year it down & restart…starting with MN Gov Walz…. He pulls his ideas from the same location he inserts his tampons…
Using integration for this problem would only make it at least ten times harder 😂 I wouldn't use integration in an integration exam either and just tell the instructor to deal with it.
if i consider the symmetry i will have to solve 2 equations with 2 unknown numbers, and if i don't, i will have to swap x and y and numbers, see line 100 and line 150: 10 vdu 5:gcol8:print "brainstation-find blue area" 20 for a=0 to 15:gcola:print a;:next a:vdu0:gcol 8:r1=4:r2=r1/2:xml=0:yml=r2 30 xmr=r2:ymr=0:zoom%=1.4*zoom%:nu1=155:nu2=nu1:masy=1200/r1:masx=850/r1 40 if masx run in bbc basic sdl and hit ctrl tab to copy from the results window
You find the equations of both the semicircles and the big quarter circle (and express them in terms of y=f(x) and also x=g(y)). Say the function (in terms of y) for lower semicircle (center in x-axis) is f1, for above one is f2 (center in y-axis) and big quarter circle is f3. Say the function (in terms of x) for lower semicircle (center in x-axis) is g1, for above one is g2 (center in y-axis) and big quarter circle is g3. then for the lower blue region (intersection of both the semicircles), you integrate (f1 - f2) going from x=0 to x=2 (point of intersection) and then for the upper blue region you integrate (g3 - g2) going from y=4 to y=2 and then integrate (g3 - g1) going from y=2 to y=0 Add all the answers and you get the total blue area
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
How to make this animation,
Please help
I already made it on geogebra. PFA link here
www.geogebra.org/calculator/zqnhj42g
Hope it helps :)
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
Nice approach
They do these in Chinese primary school? That's nothing. In Peru, they do these problems in kindergarten.
fr?
thats nothing, in north korea, they do these problems in the womb.
Hah we do these in ball
As someone who lives in Perú I agree
that's nothing, in india we do these before reincarnating
Ask 10,000 US high school graduates and they couldn’t figure this out. Great school system, time to year it down & restart…starting with MN Gov Walz…. He pulls his ideas from the same location he inserts his tampons…
I solved it both ways. I think it is actually easier without using calculus. The take away is that the two blue areas are equal.
Blue axe shape = A
Blue leave shape = L
1/4 Big circle = Q
1/2 Small circle = H
Total blue area = A + L = Q - H + L - H + L = Q - 2H + 2L
x+y+z=2x+y=2pi
x+z=2x, and…2x+2z=4x
So the question becomes simple, we need to solve the x, and then we’ll get the answer
Using integration for this problem would only make it at least ten times harder 😂 I wouldn't use integration in an integration exam either and just tell the instructor to deal with it.
😂
When I saw this figure, I immediately thought of integration. But you know your viewers quite well!
In America we are still confused about what gender we are
😶
This is why I love Algebra
if i consider the symmetry i will have to solve 2 equations with 2 unknown numbers, and if i don't, i will have to swap x and y and numbers, see line 100 and line 150:
10 vdu 5:gcol8:print "brainstation-find blue area"
20 for a=0 to 15:gcola:print a;:next a:vdu0:gcol 8:r1=4:r2=r1/2:xml=0:yml=r2
30 xmr=r2:ymr=0:zoom%=1.4*zoom%:nu1=155:nu2=nu1:masy=1200/r1:masx=850/r1
40 if masx
run in bbc basic sdl and hit ctrl tab to copy from the results window
U hv written a code for it! 🔥👏
Very clever.
Don’t worry Singapore also can do it
we deal with heuristics since 3rd grade
how do you do this with integration?
You find the equations of both the semicircles and the big quarter circle (and express them in terms of y=f(x) and also x=g(y)).
Say the function (in terms of y) for lower semicircle (center in x-axis) is f1, for above one is f2 (center in y-axis) and big quarter circle is f3.
Say the function (in terms of x) for lower semicircle (center in x-axis) is g1, for above one is g2 (center in y-axis) and big quarter circle is g3.
then for the lower blue region (intersection of both the semicircles), you integrate (f1 - f2) going from x=0 to x=2 (point of intersection)
and then for the upper blue region you integrate (g3 - g2) going from y=4 to y=2 and then integrate (g3 - g1) going from y=2 to y=0
Add all the answers and you get the total blue area
pawel zwolnij ale ten taki mol kierownik lol zwolnic tadany robia szybko zakaczyli szkolenie w szoku no ja tez
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