Find the Area of the Green Region in a Quarter Circle | Learn 2 Easy Methods

Can you calculate the area of the green shaded region in a 1/4 circle when MN=2? Learn how to use the area of a circle formula, tangent to a circle theorem, the pythagorean theorem, and the intersecting chords theorem. Step-by-step explanation by PreMath.com

[@aravindkramesh7]

Great work, Sir ❤

[@PreMath]

Thanks Aravind dear for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃

[@rogerclarke7407]

I figured this out by imagining the constraints on circle O. Finding there was nothing keeping the diameter fixed, I made the diameter 0, which gave the Quarter circle a radius of 2 and figure out the area from there.

[@PreMath]

Great Thanks Roger for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard

[@stevenbastien9028]

Yes, I solved it the same way. The glaring thing when you first look at the problem is that only one dimension is shown with value 2. I kept thinking there must be missing information, but if more information is not needed, then various choices for the two circle radii are possible. So, why not make the small circle radius zero and then the diameter of the big circle is 2. Trivial at that point.

[@joeleungcf]

but logically your approach is only solved the particular situation is when diameter of inner circle is 0, it cannot project to all cases in general

[@soli9mana-soli4953]

Great solution!!!

[@charlesbromberick4247]

Sometimes your solutions are downright elegant - I really enjoyed this video - thanks, profe

[@batchrocketproject4720]

Wow. I could not see where this was going until it all fell into place at 8:00. It shows how even when you don't recognise a route the solution, doing calculations with the known data can reveal the secret. Likewise with the intersecting chord approach. Both solution relied on writing down the somewhat trivial expression for "quarter big circle minus small circle".

[@Abby-hi4sf]

Love your outside the box methods! beautiful

[@PreMath]

Thank you! Cheers!

[@tomcruise6738]

One of the finest questions of Geometry.😎 And the solution is outstanding, as usual. ❤️ Do always bring these types of questions, not the simple questions, as you were uploading in the last few days. Greetings from India! 🇮🇳❤️

[@Awesome-ct7vr]

This is so satisfying when you use an equation to find what the whole equation equals to. I could easily lose track by trying to isolate R or r. Brilliant

[@ramanivenkata3161]

Very impressed. The basic thought - that is how to start is very important in Mathematics. This is where the Maths Master is Great.👍

[@PreMath]

Thanks Ramani for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard

[@easy_s3351]

Or with the 1st method at step 4 you could stop at R²=4+4r²=4(1+r²). The area of the quarter circle is 1/4*π*R²=1/4*π*4(1+r²)=π(1+r²). The area of the small circle is πr² and so the green shaded area is π(1+r²)-πr²=π+πr²-πr²=π.

[@ВладимирСтрельников-ф8м]

Всё понятно и очень подробно! Respect!👍🙏‼

[@timeonly1401]

Nifty how R & r both went away together! Thx.

[@jarikosonen4079]

I can not see if R & r are solvable also. How to prove in this solution that the smaller circle is completely inside the larger circle quarter? Ratio of the R/r=1+sqrt(2) might help get it checked. It looks like working well in range 0

[@JLvatron]

Very nice! I solved with the 1st method, but liked the 2nd method you showed. Thank you!

[@swinkscalibur8506]

The lower bound of r (the diameter of the small circle) is zero. If we calculate the value of the Area when r=0 we get R=2 and therefore A= pi. The upper bound for r=1 as that will cause the small circle to be tangent to the right wall of the quarter circle. If we set r=1 then R=2 root 2. Again solving for A gives A= pi. The question implies a single answer, not a function of r, therefore the answers for the lower and upper bound should be the answer in general. :)

[@rajendramisir3530]

Amazing result. I like the Outside the Box method that uses the Chords Theorem.

[@Osama-js2gw]

You always impress me! May Allah bless you!

[@PreMath]

Wow! You are so generous my friend. Thanks for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard

[@242math]

understand both solutions, great job bro

[@AnonimityAssured]

This is what I call a minimal-information problem. The paucity of information allows us to solve it in seconds without introducing any variables. Spoiler alert. Without loss of generality, one can imagine the circle shrinking to a point, leaving 2 as the radius of the quarter circle. That makes the calculation very simple. Area of the green shaded region: π (2²) / 4 = π.

[@alexniklas8777]

The funny thing is that if we drop the perpendicular (MF) to AB from the point M, we get MF=MN= 2. Radius sector R=2√2, circle radius r= 1, required area A= 2π-π= π😊

[@001Mahler]

So the value of r and R do not impact the answer. Hence the shaded area = pi as long as MN= 2.

[@sheungmingchoi6804]

This one brings deep insight: seems that R and r could be any +ve value pairs satisfying R^2 - 4 r^2 = 4.

[@mukeshsai2710]

I love your usage of pythagorean theorem

[@alanbayliss7239]

There is something wrong with method 1, deriving equation 2. 4/4 + 4r^2/4 =R^2/4 does not simplyfy to 1 + r^2 =R^2/4. I have also drawn this up on Fusion and R is not fixed. As R approaches 2 from a larger radius, so the area of the small circle approaches zero.

[@williamwingo4740]

No peeking: Let BN = x; then radius of white circle = OD = OE = x/2. Area of white circle = π (x/2)^2 = π (x^2)/4 = (π/4) (x^2); MB = radius of big quadrant. Then by Pythagoras, MB^2 = 4 + x^2; and area of big quadrant = (π/4) (4 + x^2); So the desired green area is (π/4) (4 + x^2) - (π/4) (x^2) = (π/4) (4 + x^2 - x^2); = (π/4) (4) = π. The diameter of the white circle, x, is irrelevant. It drops out and the problem becomes very simple.

[@misterkite7712]

This problem is really amazing! First, I thought that there was no solution with this amount of data. Secondly I saw the answer: pi and I was popeyed

[@danilok.m.2092]

Easyest way: MB^2=4+NB^2 MB=R NB=2r then R^2=4+4r^2 (R^2)/4=1+r^2 Ga=pi((R^2/4-r^2)) Ga=pi(1+r^2-r^2)

[@acejet6797]

if r = (sqrt(sq(R)-1)/2 then the ratio of R:r is nearly 2:1. The radius t of a circle inscribed in a quarter circle of radius R is t = R/(1+sqrt(2)). The ratio of R:t would be approx. 2.41:1. So, circle radius r would be significantly larger than inscribed circle radius t. If circle radius r does not fit inside quarter circle radius R, then the green shaded area would not be the area of the quarter circle minus the smaller circle. Please explain how a circle approximately half the radius of the larger circle can fit inside the quarter circle.

[@davidseed2939]

Before looking'at anything.. mt thought is... if the circle dimension is not given then it must be irrelevant. So could be zero. Green area = π (2^2)4

[@AnonimityAssured]

That should be π (2^2) / 4, but the basic method is exactly the one I used. PreMath quite often presents these minimal-information problems, and they are always my favourites.

[@DevKumar-xj4ys]

Superb !

[@musharrafali455]

Wonderful explanation.

[@rishiksarkar9293]

Excellent explanation!!🔥🔥

[@PreMath]

Glad you liked it! Thanks Rishik for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards Greetings from the USA!

[@India-jq7pi]

Thank you sir for your valuable videos.

[@PreMath]

So nice of you dear! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃

[@misterenter-iz7rz]

Clearly insufficient conditions to determine r and R radii of circles, but the area of region.144=R^2-4r^2, therefore the answer is R^2pi/4-r^2pi=(1/4)(R^2-4r^2)pi=36pi 😊

[@mohanramachandran4550]

So much interesting Thanks

[@GetMeThere1]

One of my favorites!

[@Mete_Han1856]

Super question super solutions.

[@claudeabraham2347]

Very good!

[@Benzine.Ae1]

👍👍 bring up like these questions

[@hjorleifuringason2778]

This can have several solutions. You can move 'MN' up or down and still meet the criteria of the small circle tangency

[@tombell2904]

Does this solution the require R = 2*sqrt2, r = 1?

[@mohammadw.alomari1322]

yes, very good

[@ramanivenkata3245]

Wonderful explanation 👍

[@grahamdolby5631]

Collapse the circle to a redundant point, by which the radius of the quarter circle is now 2, so the area is π.

[@satyapalsingh4429]

I enjoyed your video . Keep it up !

[@ΚΑΝΑΒΟΣΜΑΝΩΛΗΣ]

Dear sir, as I can imagine, the trick is number 2. So it can be deleted by 2r/2=r. I didn't try to figure a further relation r and the measure of tangent line. Any ideas? Thanks for giving us mind feeding... Emmanuil Kanavos

[@mukeshsai2710]

Thanks a lot sir day gonna end with great video

[@deepasingh5457]

Thank you🙏🙏

[@marioalb9726]

Being C the chord, C = 4 Area = ½ ⅛ π C² Area = ½ ⅛ π 4² Area = π cm² ( Solved √ )

[@joelleking7194]

Fantastic

[@johnbrennan3372]

Great method

[@PreMath]

Thanks John for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃

[@hexacene]

This is an extremely simple problem, looking at triangle BNM one can directly write R^2 - 4*r^2 = 4. No further magic needed.

[@doroundrainer]

that´s the third method and definately the best!!!

[@tahasami597]

Thank for premath

[@2csmoke]

wow ... logical

[@ashoksatija678]

Marvellous

[@EPaozi]

∀ r aire est constante si r=0 ===>R=2 ; l'aire est A=pi(2^2/4)=pi

[@azjumbo]

Angle BMN is 45 degrees, quickest solution with this info

[@سيدعدنان-ص7ه]

من يقول انBmيمر بمركز الدائره الصغيره

[@del66404]

👏👏👏👏👏

[@jaaaayt.20]

i like the video

[@Paul-tb3sk]

Woh !

[@aram5642]

Wait wait. While building the right triangle BNM and using Pythagorean theorem around 7:00 - why do you assume that the hypotenuse is equal the white circle diameter? It doesn't have to be.

[@gregorvolk5898]

As always far too complicated. Let r-》0, then R=2, A=2×2×pi/4=pi.

[@parthosaha4170]

Wow😮😮😮

[@saritdas3833]

π is the area of the green region.

[@dr.akramshalabi2855]

I hate repeatition of the theorems.

[@Robert-uw5kq]

That’s no solution. It’s a space station…

[@اممدنحمظ]

drive.google.com/file/d/1CPY5EPLZ_ppNfi1jkecM5Dp9eervNJjT/view?usp=drive_webتمرين جيد جميل . شرح واضح مرتب. شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم. تحياتنا لكم من غزة فلسطين .