1/ Let R and r be the radius of the blue and pink circle respectively. Just name the 2 centers of the blue as O and O’. 2/ Notice that OO’ is perpendicular to the width of the rectangle. And, PQ is the perpendicular bisector of OO’ and AD and BC. Similarly, OO’ is the perpendicular bisector of PQ ,AC and BD. So, OO’ , PQ and the diagonal intersect at point H which is the midpoint of the three line segments. 2/ R= 2 -> AD= 8-> AC=6 ( the triangle ABD is a 3-4-5 triple) OO’ and PQ intersect at point H Consider the triangle PHO sq PH+sqHO=sqPO sq(3-r)+sqR=sq(R+r) sq(3-r)+4=sq(2+r)-->r=9/10 Area = 2 x pi x81/100=pi x81/50 sq units
R=√4=2---> Base del rectángulo=4R=8 ---> Altura del rectángulo=2*3=6 ---> En el triángulo de vértices, centro azul, punto de tangencia y centro rosa, podemos escribir 2²+[{6/2)-r]²=(2+r)² ---> r=9/10 ---> Área rosa =2*81*π/100 =81π/50. Gracias y saludos
If R is the radius of the blue circles and r is the radius of the pink circles, then the width of the rectangle is 4R, as the centers and point of tangency of two tangential circles are collinear, and the blue circles are centered vertically along the center line of the rectangle. Blue circle: A = πR² 4π = πR² R² = 4 R = 2 Therefore AD = BC = 4(2) = 8. As AB = 10, ∆ADB and ∆BCA are congruent 2:1 ratio 3-4-5 Pythagorean triple right triangles and DB = CA = 6. Let O be the center of the right blue circle and P be the center of the top pink circle. Let T be the point of tangency between the two blue circles. Draw PT, which can be seen to be tangent to circle O, and thus ∠PTO = 90°. Draw OT and OP. OT = R = 2 and as the centers and point of tangency of two tangential circles are collinear, OP = R+r = 2+r. As the circles are inscribed symmetrically vertically and horizontally, PT is parallel to CA and DB, so PT = (6/2)-r = 3-r. Triangle ∆PTO: PT² + OT² = OP² (3-r)² + 2² = (r+2)² 9 - 6r + r² + 4 = r² + 4r + 4 10r = 9 r = 9/10 Pink circles: A = 2(πr²) = 2π(9/10)² = 81π/50 cm²
Once I calculated the rectangle as 8 by 6, I made a right triangle of 2^2 + (3-r)^2 = (r+2)^2 My final answer was slightly different though due to using 22/7 as pi.
Let's find the area: . .. ... .... ..... First of all we calculate the radius R of the blue circles: A = πR² 4πcm² = πR² 4cm² = R² ⇒ R = 2cm Therefore the side length AD=BC turns out to be: AD = BC = 4*R = 8cm Since ABC is a right triangle, we can apply the Pythagorean theorem: AC² + BC² = AB² AC² + (8cm)² = (10cm)² AC² + 64cm² = 100cm² AC² = 36cm² ⇒ AC = 6cm Let E be the center of one of the blue circles, let F be the center of one of the pink circles and let G be the point where the two blue circles and AB intersect. The triangle EFG is also a right triangle and we can apply the Pythagorean theorem again. Since each pink circle has exactly one point of intersection with each blue circle, the distance EF of their centers is equal to the sum of their radii. So with r being the radius of the pink circles we obtain: EG² + FG² = EF² R² + [(AC − 2*r)/2]² = (R + r)² (2cm)² + [(6cm − 2*r)/2]² = (2cm + r)² (2cm)² + (3cm − r)² = (2cm + r)² 4cm² + 9cm² − (6cm)*r + r² = 4cm² + (4cm)*r + r² 9cm² = (10cm)*r ⇒ r = 0.9cm Now we are able to calculate the total area of the pink circles: A(pink) = 2*πr² = 2*π*(0.9cm)² = (1.62*π)cm² ≈ 5.089cm² Best regards from Germany
STEP-BY-STEP RESOLUTION PROPOSAL : 01) AB = 10 02) Blue Circle Radius = R = 2 03) AD = 8 04) 100 - 64 = 36 05) AC = 6 06) Let's call X the Radius of Small Circles. 07) Draw a Vertical Line passing through the two centers of the Small Circles. 08) The Distance between the two Small Circles is equal to : (6 - 4X). Half of this is equal to : (3 - 2X) 09) Now look at the Red Line. It intercepts the Point of Tangency of the two Blue Circles. 10) From that Point to the Center of the Small Circle is : (3 - 2X) + X = (3 - X) 11) Let's give a warm welcome to Mr. Pythagoras, now! 12) (3 - X)^2 + 2^2 = (2 + X)^2 13) 9 - 6X + X^2 + 4 = 4 + 4X + X^2 14) 9 - 10X = 0 15) 10X = 9 16) X = 9 / 10 17) So, the Radius of Small Circle is equal to 9/10!! 18) Total Area of Pink Circles : 19) A(2 Circles) = 2 * (Pi * X^2) 20) A(2 Circles) = 2 * (Pi * (9/10)^2) 20) A(2 Circles) = 2 * (Pi * 81/100) 21) A(2 Circles) = (81 * Pi) / 50 22) A(2 Circles) ~ 16,62 * Pi 23) A(2 Circles) ~ 5,0894 ANSWER : The Area of the Two Pink Circles is approx. equal to 5,0894 Square Units Best Regards from Cordoba Caliphate - International Center for Advanced Studies in Ancient Thinking, Knowledge and Wisdom. Department of Mathematics and Geometry.
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merci sire , excellente vidéo , je vous souhaite de la bonne continuité
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1/ Let R and r be the radius of the blue and pink circle respectively. Just name the 2 centers of the blue as O and O’.
2/ Notice that OO’ is perpendicular to the width of the rectangle.
And, PQ is the perpendicular bisector of OO’ and AD and BC.
Similarly, OO’ is the perpendicular bisector of PQ ,AC and BD. So, OO’ , PQ and the diagonal intersect at point H which is the midpoint of the three line segments.
2/ R= 2 -> AD= 8-> AC=6 ( the triangle ABD is a 3-4-5 triple)
OO’ and PQ intersect at point H
Consider the triangle PHO
sq PH+sqHO=sqPO
sq(3-r)+sqR=sq(R+r)
sq(3-r)+4=sq(2+r)-->r=9/10
Area = 2 x pi x81/100=pi x81/50 sq units
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In my post, P and Q are the centers of the red circles.
Very nice, many thanks, Sir!
R=√4=2---> Base del rectángulo=4R=8 ---> Altura del rectángulo=2*3=6 ---> En el triángulo de vértices, centro azul, punto de tangencia y centro rosa, podemos escribir 2²+[{6/2)-r]²=(2+r)² ---> r=9/10 ---> Área rosa =2*81*π/100 =81π/50.
Gracias y saludos
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Let r is Radius of small circle
(3-r)^2+2^2=(2+r)^2
So r=0.9cm
So Total area of the pink circle =2(π)(0.9)^2=81π/50=5.089 cm^2.❤❤❤
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Very nice problem. Thank you.
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If R is the radius of the blue circles and r is the radius of the pink circles, then the width of the rectangle is 4R, as the centers and point of tangency of two tangential circles are collinear, and the blue circles are centered vertically along the center line of the rectangle.
Blue circle:
A = πR²
4π = πR²
R² = 4
R = 2
Therefore AD = BC = 4(2) = 8. As AB = 10, ∆ADB and ∆BCA are congruent 2:1 ratio 3-4-5 Pythagorean triple right triangles and DB = CA = 6.
Let O be the center of the right blue circle and P be the center of the top pink circle. Let T be the point of tangency between the two blue circles. Draw PT, which can be seen to be tangent to circle O, and thus ∠PTO = 90°. Draw OT and OP. OT = R = 2 and as the centers and point of tangency of two tangential circles are collinear, OP = R+r = 2+r. As the circles are inscribed symmetrically vertically and horizontally, PT is parallel to CA and DB, so PT = (6/2)-r = 3-r.
Triangle ∆PTO:
PT² + OT² = OP²
(3-r)² + 2² = (r+2)²
9 - 6r + r² + 4 = r² + 4r + 4
10r = 9
r = 9/10
Pink circles:
A = 2(πr²) = 2π(9/10)² = 81π/50 cm²
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I used same approach and got the same result.
I am getting better day by day
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Thank you for the interesting problem.
Can you please suggest a book for learning calculation of such things? thank you in advance.
(sqrt((2+r)^2-4)+r)^2+16=25, (sqrt((2+r)^2-4)+r)^2=9, sqrt((2+r)^2-4)+r=3, (2+r)^2-4=(3-r)^2, 4+4r+r^2-4=9-6r+r^2, 10r=9, =9/10, therefore the answer is 2*(9/10)^2 pi=81/50pi.😅
(2+r)²=2²+(3-r)²
A=2*π*r²=1.62π
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Once I calculated the rectangle as 8 by 6, I made a right triangle of 2^2 + (3-r)^2 = (r+2)^2
My final answer was slightly different though due to using 22/7 as pi.
Let's find the area:
.
..
...
....
.....
First of all we calculate the radius R of the blue circles:
A = πR²
4πcm² = πR²
4cm² = R²
⇒ R = 2cm
Therefore the side length AD=BC turns out to be:
AD = BC = 4*R = 8cm
Since ABC is a right triangle, we can apply the Pythagorean theorem:
AC² + BC² = AB²
AC² + (8cm)² = (10cm)²
AC² + 64cm² = 100cm²
AC² = 36cm²
⇒ AC = 6cm
Let E be the center of one of the blue circles, let F be the center of one of the pink circles and let G be the point where the two blue circles and AB intersect. The triangle EFG is also a right triangle and we can apply the Pythagorean theorem again. Since each pink circle has exactly one point of intersection with each blue circle, the distance EF of their centers is equal to the sum of their radii. So with r being the radius of the pink circles we obtain:
EG² + FG² = EF²
R² + [(AC − 2*r)/2]² = (R + r)²
(2cm)² + [(6cm − 2*r)/2]² = (2cm + r)²
(2cm)² + (3cm − r)² = (2cm + r)²
4cm² + 9cm² − (6cm)*r + r² = 4cm² + (4cm)*r + r²
9cm² = (10cm)*r
⇒ r = 0.9cm
Now we are able to calculate the total area of the pink circles:
A(pink) = 2*πr² = 2*π*(0.9cm)² = (1.62*π)cm² ≈ 5.089cm²
Best regards from Germany
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S=81π/50=1,62π≈5,09
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STEP-BY-STEP RESOLUTION PROPOSAL :
01) AB = 10
02) Blue Circle Radius = R = 2
03) AD = 8
04) 100 - 64 = 36
05) AC = 6
06) Let's call X the Radius of Small Circles.
07) Draw a Vertical Line passing through the two centers of the Small Circles.
08) The Distance between the two Small Circles is equal to : (6 - 4X). Half of this is equal to : (3 - 2X)
09) Now look at the Red Line. It intercepts the Point of Tangency of the two Blue Circles.
10) From that Point to the Center of the Small Circle is : (3 - 2X) + X = (3 - X)
11) Let's give a warm welcome to Mr. Pythagoras, now!
12) (3 - X)^2 + 2^2 = (2 + X)^2
13) 9 - 6X + X^2 + 4 = 4 + 4X + X^2
14) 9 - 10X = 0
15) 10X = 9
16) X = 9 / 10
17) So, the Radius of Small Circle is equal to 9/10!!
18) Total Area of Pink Circles :
19) A(2 Circles) = 2 * (Pi * X^2)
20) A(2 Circles) = 2 * (Pi * (9/10)^2)
20) A(2 Circles) = 2 * (Pi * 81/100)
21) A(2 Circles) = (81 * Pi) / 50
22) A(2 Circles) ~ 16,62 * Pi
23) A(2 Circles) ~ 5,0894
ANSWER : The Area of the Two Pink Circles is approx. equal to 5,0894 Square Units
Best Regards from Cordoba Caliphate - International Center for Advanced Studies in Ancient Thinking, Knowledge and Wisdom. Department of Mathematics and Geometry.
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Cordoba Caliphate - International Center for Advanced Studies in Ancient Thinking, Knowledge and Wisdom. Department of Mathematics and Geometry❤️🌹
@@PreMath, thank so much for your kindness. You are a fabulous promoter of Mathematical Thinking through these Challenges.
Fine, i have not found a very different solution.
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Whatever the the value of the result from now on, I want $100 dollars times that amount. 😎
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First comment and first like
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