Math Olympiad | Find area of the Yellow shaded region? | (Quarter circles and square) |

[@nunoalexandre6408]

Love it!!!!!!!!!!!!!!!

[@PreMath]

Thanks a lot ❤️

[@murdock5537]

This is great, many thanks, Sir!

[@lwels49]

Well done

[@marcelowanderleycorreia8876]

Beautiful question. Congrats professor!

[@maxforsberg8852]

I had a totally different way of solving this. First calculate the area of the quarter circle BAD. Then subtract the identical quarter circle complementary areas ADC and ABC. However now we have subtracted the identical “tent” shapes DEC and BFC which I had not included in the first step so these must be added back in. In other words BAD - 2*ABC + 2*DEC.

[@Allyballybean]

How do you get DEC?

[@DB-lg5sq]

شكرا لكم نعتبرa مساحة ربع دائرة ذات الشعاع r وb مساحة القطاع الزاوي الذي زاويته 60درجة والشعاع r و c مساحة المثلث المتساوي الاضلاع ضلعه r المساحة المطلوبة هي X=2b-2c-a =r^2(5pi/12 - (نصف جذر 3))

[@flash24g]

My way was essentially the same, though I had a headstart in knowing the segment area formula: (θ - sin θ) r² / 2, with θ given in radians. It appears that you are squaring 2√3 afresh each time you use it. Why not just remember that r² = 12 from when you originally derived it?

[@aljawad]

Excellent! I used integral calculus after obtaining the coordinates of points E & F to find the area, and reach the same result. BTW, I enjoy watching your videos, which I usually solve as I have coffee in the morning by scribbling on the margins of the daily paper. I consider them part of my mental daily exercise ritual, so keep them coming please!

[@AmirgabYT2185]

S=5π-6√3≈5,32 cm²

[@MrPaulc222]

I had a good go at this one but didn't manage to solve it as I encountered several dead ends.

[@Ibrahimfamilyvlog2097l]

very nice video sar❤❤❤❤❤❤❤

[@PreMath]

Thanks and welcome ❤️

[@santiagoarosam430]

AB=sqrt12=2(sqrt3)=Radio r Amarillo =(Sector circular 30°)+2[(Sector 60°)-(Triángulo equilatero de lado r)] =a+2(2a-q) =a+4a-2q =5a-2q =5Pi -6sqrt3 Gracias y saludos.

[@ganymed1236]

So easy, but sadly I didn't see it. I did hard tour with functions of the 3 graphs and 2 intersections by root 3 and 3 (Points E and F). Then building integrales with help by WolframAlpha. It worked, but needless complicated. On the otherside, it was a good exercise.😁

[@User-jr7vf]

Oh, it must have been a pain. In the past I have also gone through similar lengths trying to solve a problem, just to find out in the end that it was much simpler.

[@JSSTyger]

I tried but came up short. I was able to find the intersection points (x,y) but did not get much further. I believe the intersection points (x,y) are (sqrt(3),1) at point E and (3,sqrt(3)) at point F, all presuming that the origin (0,0) is at point D. I ended up guessing a value of about 5.4.

[@wackojacko3962]

If anything learned from this channel , I label everything when solving problems.. for some reason radians have left the building and seemingly ignored and I am not quite satisfied. But that's okay! I absolutely love this problem and not even mad. 🙂

[@PreMath]

Thanks dear ❤️🌹

[@flash24g]

I agree, especially given that using radians makes sector and segment area more straightforward. See my comment.

[@flash24g]

​@@GordonSimpson-hr4yf​I think it's down to academic level. Probably at A level and above, one pretty much always uses radians, to the point that the argument to trig functions is treated as dimensionless. This is particularly of value when doing calculus or other things with trig functions besides calculating angles in triangles. I'm not sure what academic level these videos are aimed at. Maybe the formulae used are using degrees to be easier for people below this level to understand. The drawback is that those at this level or above are liable to be shouting at the screen, "Why all this palaver with degrees? Just use radians!"

[@nsgaraslan334]

Iove math

[@PreMath]

Thanks dear ❤️

[@giuseppemalaguti435]

Ho usato le equazioni delle circonferenze,e calcolato i due punti intersezione l'area risulta Ay=3(π-√3)...spero di non aver fatto errori..merry Christmas

[@misterenter-iz7rz]

Rather clumsy, 3 parts one sector two identical segments, r^2=12, r radius, 2r-r pi/2=r(2 pi/3-pi/2)=pi r(2/3-1/2)=1/6 pi r arc length, area =1/2 1/6 pi r^2=1/12 pi r^2=pi segment area is 1/6 r^2 pi-1/2 r^2 sqrt(3)/2=r^2(pi/6-sqrt(3)/4)=2pi-3sqrt(3), so area is pi+4pi-6sqrt(3)=5pi-6sqrt(3)=5.315658 approximately. 😊

[@PlumbuM871]

Cool 👍

[@PreMath]

Thanks ❤️

[@shaozheang5528]

I’m finding a method without trigonometry

[@robertlynch7520]

… talk about 'doing it the harder way' … sigh. I seem to fall for the trap every time. Anyway, First, I normalized this to the 1×1 unit square. Why? Because it seems easier. (𝒓 = 1); Then, using formula for circle (𝒚 = √(𝒓² - 𝒙²)), it is easy to see that 2 arcs cross at (𝒙 = 0.5). 𝒚 = √(1² - (½)²) = √(0.75) = 0.866025 Well, that'd be for arcs going upwards from origin. These are the inverses, so [1 - 0.866025 = 0.133975]. Then noting again (symmetry) that the right hand crossing is at (𝒙, 𝒚) = (0.866025, 0.5), we now have an inner triangle, inside the yellow region. Left out are the 3 lenses, 2 of which are symmetric in area. Congruent. There is a 60°-60°-60° big triangle between lower-left origin, right hand midpoint and upper left corner of the unit square. Cool. The area of a larger lens is 𝒛 = (area of pie - area of triangle) = (π/3 × ½ × 1²) - (√3/4 × 1² ); 𝒛 = 0.090586 u² So now we have a pair of lenses and the inner pie-arc of 30° Area = ½ π/6 × 1² ⊕ 2𝒛 Area = 0.44297. The SCALED area is just 'multiply by 12'. Scaled = 5.3157 And there's the solution. The hard way, but a way that's pretty logical if but laborious. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅