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 √ )
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.
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 √ )
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 √ )
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
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! 🙂
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.
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
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 ...
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
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.
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)
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.
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. 😅
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
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)
first view and like
Thank you! Cheers! 😀
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 √ )
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.
Thank you! Cheers! 😀
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 √ )
Thank you! Cheers! 😀
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 √ )
Thanks for posting! Solved this one 😊 using the relationship : (2+r)^2 = 2(2-r)^2
Thank you! Cheers! 😀
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
Thank you! Cheers! 😀
Generalized: Y = 2B (17 − 12√2) where Y is the area of the yellow circle and B is the area of the blue semicircle.
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
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
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
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]
Thank you! Cheers! 😀
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! 🙂
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
Amazing👍
Thanks for sharing 🌺
Thanks for visiting
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 √ )
As usual, it is pleasant to see how you are teaching different methods!
Thanks for your feedback! Cheers! 😀
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.
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
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 ...
Thanks 👍
Thank you too
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
Thanks for video.Good luck sir!!!!!!!!!!!!!
Thank you too
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.
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)
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.
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.... 🤠
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
Draw tangent thru T then calculate radius of in-circle of resulting right triangle using (AT+AT - hypotenuse)/2
Pisteestä A sinisen täydennetyn ympyrän ulkolaitaan 2√2+2,keltaisen ulkolaitaan 2√2-2.
Keltaisen ala ((2√2-2)/(2√2+2))^2*4π
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. 😅
You can't do 2r, because the radius doesn't extend all the way to
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Is there any way to prove that the linr joining the 2 centres passes through point A??
Ar small circle=pi/2when pi =22/7
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
Thank you! Cheers! 😀
7:26 why OA is not equal to only 2+ r√2
Nice way of explanation
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)
Thank you! Cheers! 😀
👌👌
Thank you! Cheers! 😀
You are awesome. Keep it up 👍
Thank you
0.376 cm^2
A=0.36×3.14
سؤال إذا كانت مساحة الدائرة تساوى ٢باى فنصف القطر يساوى جذر ٢ باى وليس ٢ ارجو التوضيح
معطى: مساحة نصف الدائرة 2pi. لذا ، مساحة الدائرة الكاملة ستكون 4 نقطة في البوصة.
هتافات
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