Buenos días profesor, bonito problema, gracias

@PreMath

Жыл бұрын

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

Congratulaciones. Desde Bogota D.C.

@PreMath

Жыл бұрын

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.! 😀

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

@PreMath

Жыл бұрын

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

Very good. Thanks.

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

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

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

Жыл бұрын

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

Nice problem and solution.

Thank you, Professor 😊❤

@PreMath

Жыл бұрын

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

Thanks Professor, excellent explanation!🥂👍🍺

@PreMath

Жыл бұрын

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

@bigm383

Жыл бұрын

@@PreMath 😀🥂

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

Thank you

Very well explained👍 Thanks for sharing😊😊

@PreMath

Жыл бұрын

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

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

@PreMath

Жыл бұрын

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

Decent and teaching

A great problem.

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

Жыл бұрын

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! 😀

Marvelous sum A different sum Thanks a lot

@PreMath

Жыл бұрын

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! 😀

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

Жыл бұрын

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

Very nice

@PreMath

Жыл бұрын

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

Thanks premath

@PreMath

Жыл бұрын

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

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

@maxxie8058

Жыл бұрын

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

Жыл бұрын

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.

S=40π/9≈13,96

The radius of smallest circle can be found by considering the right angled triangle on the left hand side, 4^2+(8-r)^2=(r+4)^2, so 16+64-16r+r^2=r^2+8r+16, so r=64/24=8/3. Thus the area is pi(64/4-16/2-(8^2/3^2)/2)=40/9 pi=13.96 approximately, done.🙂

@PreMath

Жыл бұрын

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

Great 👍

@PreMath

Жыл бұрын

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

@mraymanhassan

Жыл бұрын

@@PreMath thanks a lot

Yay! I solved the problem.

@PreMath

Жыл бұрын

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

Thnku

@PreMath

Жыл бұрын

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

Good morning sir

@PreMath

Жыл бұрын

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

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

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

Жыл бұрын

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

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

@PreMath

Жыл бұрын

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

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

Жыл бұрын

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

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

Жыл бұрын

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

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

@PreMath

Жыл бұрын

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

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

~ 4.45 π

@PreMath

Жыл бұрын

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

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

@bienvenidos9360

Жыл бұрын

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

Жыл бұрын

@@bienvenidos9360 ok thankyou

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

40pi/9=13,962

@PreMath

Жыл бұрын

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