Calculate area of the Yellow shaded circle | Semicircle | (Important Geometry skills explained)

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Learn how to find the area of the Yellow shaded circle. A blue semicircle has an area 2pi . Important Geometry and Algebra skills are also explained: Pythagorean Theorem; Area of a circle formula. Step-by-step tutorial by PreMath.com
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Calculate area of the Yellow shaded circle | Semicircle | (Important Geometry skills explained)
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Пікірлер: 60

@math-physics3329 11 ай бұрын

first view and like

@PreMath 11 ай бұрын

Thank you! Cheers! 😀

@thewolfdoctor761 11 ай бұрын

I created a triangle OPF, where F is on the line OQ, and PF is perpendicular to OQ. r is radius of yellow circle. So (2+r)^2 = (2-r)^2 + (2-r)^2. Solve for r , etc.

@PreMath 11 ай бұрын

Thank you! Cheers! 😀

@marioalb9726 11 ай бұрын

Blue Area= ½ π R² = 2π cm² R² = 4 R = 2 cm Equalling over diagonal OA: R + r + r/cos45° = R / cos45° r (1 + 1/cos45°) = R (1/cos45° - 1) r = R (√2-1) / (1+√2) r = 0,343 cm Area = π r² Area = 0,37 cm² ( Solved √ )

@redfinance3403 11 ай бұрын

Thanks for posting! Solved this one 😊 using the relationship : (2+r)^2 = 2(2-r)^2

@PreMath 11 ай бұрын

Thank you! Cheers! 😀

@wackojacko3962 11 ай бұрын

This problem is best one yet for making observations and following through with pts. of tangency...circle theorem...Pythagorean theorem...area of circle formula too find radius...right triangle properties...comparing equations...rationalizing fractions ...and calculating area of circle. Awesome way to start a day! 🙂

@PreMath 11 ай бұрын

Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

@Abby-hi4sf 11 ай бұрын

As usual, it is pleasant to see how you are teaching different methods!

@PreMath 11 ай бұрын

Thanks for your feedback! Cheers! 😀

@HappyFamilyOnline 11 ай бұрын

Amazing👍 Thanks for sharing 🌺

@PreMath 11 ай бұрын

Thanks for visiting

@prabhatchandramondal6774 10 ай бұрын

Nice way of explanation

@raya.pawley3563 6 ай бұрын

Thank you

@murdock5537 11 ай бұрын

Nice! (π/2)(OD)^2 = 2π → OD = 2 → OA = 2√2 → 2√2 - 2 = 2(√2 - 1) = r(√2 + 1) → r = 2(√2 - 1)/(√2 + 1) = 2(√2 - 1)^2 → πr^2 = 4π(17 - 12√2) ≈ 0,3699

@PreMath 11 ай бұрын

Thank you! Cheers! 😀

@santiagoarosam430 11 ай бұрын

Área azul =2Pi》Radio azul =2 》ABCD rectángulo de 4×2》Potencia de A respecto a la circunferencia azul =2×2 =r(1+sqrt2)(2+2sqrt2)》r=6-4sqrt2 》Área amarilla =68-48sqrt2 =0.117749 Gracias y saludos.

@MrPaulc222 11 ай бұрын

Great explanation. I need to go through this again more slowly. Without the video I got as far as (2+r+r*root2)^2=8, but went off in a wrong direction after that.

@KAvi_YA666 11 ай бұрын

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

@PreMath 11 ай бұрын

Thank you too

@pranavamali05 11 ай бұрын

👌👌

@PreMath 11 ай бұрын

Thank you! Cheers! 😀 You are awesome. Keep it up 👍

@amitsinghbhadoriya6318 11 ай бұрын

Thanks 👍

@PreMath 11 ай бұрын

Thank you too

@musicsubicandcebu1774 11 ай бұрын

Draw tangent thru T then calculate radius of in-circle of resulting right triangle using (AT+AT - hypotenuse)/2

@vidyadharjoshi5714 11 ай бұрын

DO = 2. Let PT = r. AP = r*sqrt2. AT = r + r*sqrt2 = 2*sqrt2 - 2. r = 2/(3+2*sqrt2) = 0.343. Yellow Area = 0.37

@ybodoN 11 ай бұрын

Generalized: Y = 2B (17 − 12√2) where Y is the area of the yellow circle and B is the area of the blue semicircle.

@PreMath 11 ай бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@kxllxr 11 ай бұрын

I solved it in a different way. OT is a radius: pi*R^2/2=2 cm. I connected points O and Q. Then I calculated a diagonal of the square ODAQ using this formula: d=sqrt2*a=sqrt2*OQ. OQ is a radius as well as OT and equaled to 2 cm. So, OA is equalled to 2*sqrt2 cm. Then TA is OA-OT=2sqrt2-2 which is approximately 0.8284 cm. EPFA is a square, point P is a center of the yellow circle (2 tangent theorem, so EA=EF and EP=PF as radius). So, AP is a diagonal of EPFA and it equals to sqrt2*r. TP is a radius of the yellow square. AT=AP+TP= sqrt2*r+r=r(sqrt2+1)=0.8284 cm. r=0.8284/(sqrt2+1)=0.3431 cm. And the final step: A(yellow circle)=pi*r^2=3.14*0.1177=0.37 square cm.

@williamwingo4740 11 ай бұрын

Nothing new to add this time: did it pretty much the same way, right down to rationalizing the denominator. Maybe we're beginning to think alike.... 🤠

@Patrik6920 11 ай бұрын

Nice to see an update to the earlier one... keep going.... maby u should take that one down or write an disclamer (its the one with incorrect tangent assumptions)

@PreMath 11 ай бұрын

Thank you! Cheers! 😀

@unknownidentity2846 11 ай бұрын

The line through O and Q and the line through E and P intersect at point R. Then the triangle OPR is a right triangle and we can use the Pythagorean Theorem in the following way: (2−r)² + (2−r)² = (2+r)² 2(2−r)² = (2+r)² √2(2 − r) = 2 + r # r < 2 ⇒ 2 − r > 0 and 2 + r > 0 2√2 − r√2 = 2 + r 2√2 − 2 = r + r√2 2(√2 − 1) = r(√2 + 1) r = 2(√2 − 1)/(√2 + 1) r = 2(√2 − 1)² r = 6 − 4√2

@PreMath 11 ай бұрын

Thank you! Cheers! 😀

@soli9mana-soli4953 11 ай бұрын

I try to solve considering the diagonal of AQOD square that is 2√ 2) and that one of AFPE square that is r√ 2, then r√ 2 = 2√ 2 - 2 - r r = (2√ 2 - 2)/(√ 2 + 1)

@marioalb9726 11 ай бұрын

Blue Area= ½ π R² = 2π cm² R² = 4 R = 2 cm Equalling over diagonal AT : r + r/cos45° = R/cos45° - R r + √2 r = √2 R - R r ( 1+√2) = R ( √2 - 1) r = R (√2-1) / (1+√2) r = 0,343 cm Area = π r² Area = 0,37 cm² ( Solved √ )

@PreMath 11 ай бұрын

Thank you! Cheers! 😀

@marioalb9726 11 ай бұрын

Blue Area= ½ π R² = 2π cm² R² = 4 R = 2 cm Equalling over line OP: R + r = (R -r) / cos45° R + r = √2 (R-r) R + r = √2 . R - √2 . r r + √2 r = √2 R - R r ( 1+√2) = R ( √2 - 1) r = R (√2-1) / (1+√2) r = 0,343 cm Area = π r² Area = 0,37 cm² ( Solved √ )

@redeyexxx1841 11 ай бұрын

Is there any way to prove that the linr joining the 2 centres passes through point A??

@devondevon4366 11 ай бұрын

0.376 cm^2 answer The circle's radius is 2 since its area is 4 pi (twice the semi-circle.) Let's label the yellow circle's center r. From the center of this circle to A is the hypotenuse, h. h^2 = r^^2 + r^2 (Pythagorean) Hence h^2 = 2 h= sqrt 2r^2 or 1.41 r The distance from thethe center of the yellow circle to where its touches the blue semi-circle is also r. Hence the distance from A to where it touches the blue is 2.41 (1.41 r + r). Hence the distance from that to the center of the blue cirlce is 2.41 r + 2 . Construct a triangle using 2.41r + 2 as the hypotenuse; hence the other two sides are 2 and 2 ( the cirlce's radius) Using Pythagorean (2 + 2.41 r )^2 = 2^2 + 2^2 5.8081 r^2 + 9.64 r +4 =8 5.8081 r^2 + 9.64 r -4 = 0 r = .34376 Circle area hence is 0.376 cm^2

@prollysine 9 ай бұрын

Hi, large circle radius R=2 cm, let r=small circle radius, large circle center point to square corner distance with the circular R, r,: R+r+sqrt(2)*r = sqrt(2)*R --> 2+r+sqrt(2)*r = sqrt(2)*2 , r*(1+sqrt(2))=sqrt(2)*2-2 , r=(sqrt(2)*2-2)/(1+sqrt(2) =~ 0,3431 cm , small circle area = r^2*pi , T = (0,3431^2)*pi = 0,3699 cm^2 , ok ...

@quigonkenny 5 ай бұрын

Let R be the radius of the large blue semicircle and r be the radius of the small yellow circle. As T is tangent both to Semicircle O and Circle P, points O, T, P, and A are colinear. Therefore ∆AFP and ∆AQO are similar. Triangle ∆AQO: a² + b² = c² R² + R² = OA² OA² = 2R² OA = √2R² = R√2 OA = OT + TA R√2 = R + TA TA = R√2 - R = R(√2 - 1) ---- (1) Triangle ∆AFP: PA/AF = OA/AQ PA/r = R√2/R = √2 PA = r√2 TA = TP + PA TA= r + r√2 = r(1+√2) ---- (2) r(1+√2) = R(√2 -1) r = R(√2 -1)/(1+√2) ---- (3) Semicircle O: A = πR²/2 2π = (π/2)R² R² = 4 R = 2 Circle P: r = R(√2 -1)/(1+√2) r = 2(√2 -1)/(1+√2) r = 2(√2 -1)(1-√2)/(1+√2)(1-√2) r = 2(√2 -1)(1-√2)/(1-2) r = 2(1-√2)(1-√2) r = 2(1 - 2√2 + 2) r = 2(3-2√2) A = πr² = π(2(3-2√2))² A = 4π(9-12√2+8) A = 4π(17-12√2) ≈ 0.37 cm²

@Algebronic_Animations24 11 ай бұрын

7:26 why OA is not equal to only 2+ r√2

@mohanramachandran4550 11 ай бұрын

πR² ÷ 2. = 2π² R. = 2 cm Diameter of the Rectangle = 2 * √2 Yellow circle area = ( 2 √2 -- 2 ) Radius of yellow circle = r r = (2√2-2) ÷ (1+√2) = 0.34314575 Area of the Yellow circle = πr² Area = Π × ( 0.34314575)² Area = 0.36991941250

@PreMath 11 ай бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@devondevon4366 11 ай бұрын

0.376 cm^2

@vierinkivi 11 ай бұрын

Pisteestä A sinisen täydennetyn ympyrän ulkolaitaan 2√2+2,keltaisen ulkolaitaan 2√2-2. Keltaisen ala ((2√2-2)/(2√2+2))^2*4π

@thomakondaciu6417 11 ай бұрын

A=0.36×3.14

@gelbkehlchen 11 ай бұрын

Solution: R = radius of the blue circle, r = radius of the yellow circle. It shall be: π*R²/2 = 2π |*2/π ⟹ R² = 4 |√() ⟹ R = 2 ⟹ Pythagoras: (R-r)²+(R-r)² = (R+r)² ⟹ (2-r)²+(2-r)² = (2+r)² ⟹ 4-4r+r²+4-4r+r² = 4+4r+r² |-4-4r-r² ⟹ r²-12r+4 = 0 |p-q-formula ⟹ r1/2 = 6±√(36-4) = 6±√32 = 6±4*√2 ⟹ r1 = 6+4*√2 > R = 2 [that cannot be] and r2 = 6-4*√2 ≈ 0,3431 < R = 2 [that is o.k.] ⟹ area of the yellow circle = π*r2² = π*(6-4*√2)² = π*(36-48*√2+32) = π*(68-48*√2) ≈ 0,3699[cm²]

@PreMath 11 ай бұрын

Thank you! Cheers! 😀

@amitavadasgupta6985 11 ай бұрын

Ar small circle=pi/2when pi =22/7

@mohamedgamal-ze1gb 11 ай бұрын

سؤال إذا كانت مساحة الدائرة تساوى ٢باى فنصف القطر يساوى جذر ٢ باى وليس ٢ ارجو التوضيح

@PreMath 11 ай бұрын

معطى: مساحة نصف الدائرة 2pi. لذا ، مساحة الدائرة الكاملة ستكون 4 نقطة في البوصة. هتافات

@mohamedgamal-ze1gb 11 ай бұрын

@@PreMath اسف عرفتها بعد كتابة التعليق وشكرا على التوضيح

@misterenter-iz7rz 11 ай бұрын

First note that 2 is the radius of the large circle, let r be the radius of the small circle, then 2root2=2+2r, r=root 2-1, the,answer is (root 2-1)^2pi=(3-2root2)pi=0.539 approximately. 😊 Oh, I make a mistake, 2root2=2+r+(root 2)r, so r=(2root 2-2)/(1+root 2)=(2root2-2)(root2-1)=2(root2-1)^2=2(3-2root2), the answer should be 4(3-2root 2)^2pi=4(17-12root2)pi=0.37 approximately. 😅

@krislegends 11 ай бұрын

You can't do 2r, because the radius doesn't extend all the way to

@PreMath 11 ай бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@marioalb9726 11 ай бұрын

Blue Area= ½ π R² = 2π cm² R² = 4 R = 2 cm Pitagorean theorem: (R+r)² = 2.(R-r)² R²+2Rr+r² = 2 . (R²-2Rr+r²) R²+2Rr+r² = 2R²- 4Rr + 2r² r²- 6Rr +R² = 0 r²- 12r + 4 = 0 r = 0,343 cm Area = π r² Area = 0,37 cm² ( Solved √ )