Calculate area of the Green shaded region | Quarter circle radius is 8 | Important skills explained

[@bramont6225]

Buenos días profesor, bonito problema, gracias

[@PreMath]

Good morning! Glad to hear that! Thanks for your feedback! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

[@jaimeyomayuza6140]

Congratulaciones. Desde Bogota D.C.

[@PreMath]

Muy amable de tu parte, Jaime. ¡Gracias por su continuo amor y apoyo! Usted es maravilloso. Sigue sonriendo 👍 ¡Amor y oraciones desde los EE. UU.! 😀

[@aysunmatematika1270]

Very good. Thanks.

[@jamesrogers4761]

That r=R/3 makes me happy. Circles are fab.

[@PreMath]

Excellent! Thanks for your feedback! Cheers! You are awesome, James. Keep smiling👍 Love and prayers from the USA! 😀

[@vara1499]

Nice problem and solution.

[@theoyanto]

Very good, I always seem to have trouble getting started with these, so more like this please, I need the practice 🤓👍🏻

[@kennethstevenson976]

Got the correct answer using the same steps the first try. Good practice.

[@abdessamadsafri8791]

Thank you, Professor 😊❤

[@PreMath]

You are very welcome! Thanks for your continued love and support! You are awesome, Safri. Keep smiling👍 Love and prayers from the USA! 😀

[@bigm383]

Thanks Professor, excellent explanation!🥂👍🍺

[@PreMath]

Glad you think so! Thanks for your continued love and support! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

[@bigm383]

@@PreMath 😀🥂

[@KAvi_YA666]

Thanks for video.Good luck sir!!!!!!!!!!!!

[@PreMath]

You are very welcome! So nice of you. You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

[@murdock5537]

Nice and awesome, many thanks, Sir! ∆ABC → AC = 4 = r → AB = 4 + x → BC = 8 - x → sin⁡(BCA) = 1 → sin⁡(φ) = BC/AB = (8 - x)/(4 + x) → cos⁡(φ) = 4/(4 + x) → sin^2(φ) + cos^2(φ) = 1 → 24x = 64 → x = 8/3 → (π/4)(2r)^2 - (π/2)(x^2 + r^2) = 40π/9

[@PreMath]

You are very welcome! So nice of you. Thanks for sharing! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

[@khalidhashimeh]

Very nice

[@PreMath]

Glad you think so! Thanks for your feedback! Cheers! You are awesome, Khalid. Keep smiling👍 Love and prayers from the USA! 😀

[@mohanramachandran4550]

Marvelous sum A different sum Thanks a lot

[@PreMath]

You are very welcome! Glad you think so! Thanks for your continued love and support! You are awesome, Mohan. Keep smiling👍 Love and prayers from the USA! 😀

[@AmirgabYT2185]

S=40π/9≈13,96

[@NedumEze]

Decent and teaching

[@mraymanhassan]

Great 👍

[@PreMath]

Glad you think so! Thanks for your feedback! Cheers! You are awesome, Hassan. Keep smiling👍 Love and prayers from the USA! 😀

[@mraymanhassan]

@@PreMath thanks a lot

[@birendramaiti5399]

The way of your explanation is like a history teacher teaching maths.

[@raya.pawley3563]

Thank you

[@HappyFamilyOnline]

Very well explained👍 Thanks for sharing😊😊

[@PreMath]

Thanks for your feedback! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

[@mahdiali4218]

Thanks premath

[@PreMath]

You are very welcome! Thanks for your continued love and support! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

[@bentels5340]

Hah! Solved it on my own! I usually don't get geometry problems because I had little geometry in high school, but this one I solved (in the same way you did)! 😁

[@PreMath]

No worries. We are all lifelong learners. That's what makes our life exciting and meaningful! Thanks for your feedback! Cheers! You are awesome, Ben. Keep it up 👍 Love and prayers from the USA! 😀

[@richardblackmore527]

A great problem.

[@Copernicusfreud]

Yay! I solved the problem.

[@PreMath]

Bravo! Thanks for sharing! Cheers! You are awesome, Mark. Keep smiling👍 Love and prayers from the USA! 😀

[@MrPaulc222]

Quadrant area = 16pi Lower semicircle is 8pi. Green area is 8pi - upper semicircle. Find its radius, BE or equivalent. AB is 4 + r. CB is 8 - r. CA = 4. 4^2 + (8 - r)^2 = (4 + r)^2 16 + 64 - 16r + r^2 = 16 + 8r + r^2 Simplify to 64 - 16r = 8r 64 = 24r r = 64/24 = 8/3 (8/3)^2 * pi = 64pi/9. Halve it for (32/9)pi. Green area = 8pi - (32/9)pi. (72/9)pi - (32/9)pi = (40/9)pi (40/9)(22/7) = 880/63 un^2 = 13.97 un^2 (rounded). A slight variation with rounding due to using 22/7.

[@pralhadraochavan5179]

Good morning sir

[@PreMath]

Hello dear You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

[@pranavamali05]

Thnku

[@PreMath]

You are very welcome! So nice of you, Pranav Thank you! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

[@shadmanhasan4205]

Very tricky -> CD(A) = 8 units, while side B = ~ 5.3 units. We get 2 semicircles (blue) covering some of the main quarter-circle (green). We can approximate it as: Green - Blue1 - Blue2 (Pi*(8)^2)/4 - (Pi*(4)^2)/2 - (Pi*(~2.6)^2)/2 = ~14.5 units square.

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@gorgioprio2629]

you must prove that A B and the intersection between the two semi cercles is aligned

[@maxxie8058]

Yes ! It may seem trivial to some, but it is the crux of the problem, so it was definitely a mistake to skip this step.

[@johnbutler4631]

True. This can be done by the fact that the circles share a point of tangency, and that the common tangent is at a right angle to both radii. Therefore, their radii must be collinear. It could probably be formally proven by first establishing the right angle between the tangent and radius of one of the semicircles, then continuing that radius into the other semicircle, then establishing that the continued line must go through the center of the other semicircle.

[@anuragk6544]

How did you assume that AB is a straight line. ?

[@bienvenidos9360]

He says the 2 semicircles are tangent to each other at one point, so we take that as a given. When 2 circles are tangent to each other, the point of tangency is at a 90° angle from the radius to the center of the circle. So the line of tangency is perpendicular to the 2 radii to the point of tangency and the line to the 2 centers is a straight line that passes through the point of tangency.

[@anuragk6544]

@@bienvenidos9360 ok thankyou

[@eckhardfriauf]

Looks complex, yet is not. I solved it the same way as you did.

[@PreMath]

Excellent! Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@leslievincent20]

That was a difficult one to calculate without watching the video first. Damm!

[@soli9mana-soli4953]

But how intrusive is this Pythagorean Theorem!!! It's there all the time :))))

[@PreMath]

It's the main building block in Mathematics! Thanks for your feedback! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

[@himo3485]

larger semicircle : 8/2 = 4 smaller semicircle : r 4^2+(8-r)^2=(4+r)^2 16+64-16r+r^2=16+8r+r^2 24r=64 r=8/3 8＊8＊π＊1/4 - 4＊4＊π＊1/2 - 8/3＊8/3＊π＊1/2 = 16π - 8π - 32π/9 = 72π/9 - 32π/9 = 40π/9 Green shaded region : 40π/9

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@marcello3621]

ABC is rectangle, R is the radius of the smaller semicircle. AB = 4 + R, AC = 4, let's calculate BC in terms of R, with pythagorean theorem: (4 + R)² = 4² + BC² 16 + 8R + R² = 16 + BC² 8R + R² = BC² *BC = √8R + R²* But BC + BE = 8, so: √(R² + 8R) + R = 8 √(R² + 8R) = 8 - R R² + 8R = 64 - 16R + R² 24R = 64 *R = 8/3* Now, the areas: Greater semicircle = 4²pi/2 = 8pi Smaller semicircle = (8/3)²pi/2 = 32/9pi Quarter circle = 8²pi/4 = 16pi Now, the green area: 16pi - (8 + 32/9)pi 16pi - 104/9pi *40/9 pi*

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@giuseppemalaguti435]

Detto r il Raggio del semicerchio in alto a sinistra risulta (r+4)^2=4^2+(8-r)^2...r=8/3...Agreen=40/9pi

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@鈞齊]

Formula for the difference of square: a²-b²=(a+b)(a-b) ------------------------- R: the biggest radius, and R=8 r: the smallest radius Area of green shaded region: πR²/4-[π(R/2)²/2+πr²/2] = π/2[(R/2)²-r²] = (4+r)(4-r)π/2 (R/2)²+(R-r)²=(R/2+r)² => 4²+(8-r)²=(4+r)² => (4+r)²-(8-r)²=4² => 12(2r-4)=4² => r=8/3 = (20/3)(4/3)π/2 = 40π/9

[@quigonkenny]

Let r be the radius of the smaller semicircle at B. We are given the radius of the encompassing quarter circle as 8, and by observation, the radius of the larger semicircle at A is 8/2 = 4. As EB = r and EC = 8, BC = 8-r. Let F be the point on AB where the two semicircles are tangent. By observation AB = 4+r. Triangle ∆BCA: a² + b² = c² 4² + (8-r)² = (4+r)² 16 + 64 - 16r + r² = 16 + 8r + r² 80 - 16r = 16 + 8r 24r = 64 r = 64/24 = 8/3 Green area: A = π(8²)/4 - π(4²)/2 - π(8/3)²/2 A = 16π - 8π - 32π/9 = 72π/9 - 32π/9 A = 40π/9 ≈ 13.96

[@s.j.r7656]

~ 4.45 π

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@mohamadtaufik5770]

40pi/9=13,962

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@Lemda_gtr]

Why assumed A as the middle, like this always confused me in ur vids