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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.
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)! 😁
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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.🙂
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
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.
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
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.
Buenos días profesor, bonito problema, gracias
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Congratulaciones. Desde Bogota D.C.
Muy amable de tu parte, Jaime.
¡Gracias por su continuo amor y apoyo!
Usted es maravilloso. Sigue sonriendo 👍
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Got the correct answer using the same steps the first try. Good practice.
That r=R/3 makes me happy. Circles are fab.
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Thank you, Professor 😊❤
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Nice problem and solution.
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
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Very good. Thanks.
Very good, I always seem to have trouble getting started with these, so more like this please, I need the practice
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Thanks Professor, excellent explanation!🥂👍🍺
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@@PreMath 😀🥂
Decent and teaching
Thanks for video.Good luck sir!!!!!!!!!!!!
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Marvelous sum
A different sum
Thanks a lot
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A great problem.
The way of your explanation is like a history teacher teaching maths.
Very well explained👍
Thanks for sharing😊😊
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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.
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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)! 😁
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Thank you
Very nice
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Thanks premath
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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.🙂
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Yay! I solved the problem.
Bravo!
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Great 👍
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@@PreMath thanks a lot
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
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Good morning sir
Hello dear
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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*
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Thnku
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That was a difficult one to calculate without watching the video first. Damm!
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
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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
you must prove that A B and the intersection between the two semi cercles is aligned
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.
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.
Looks complex, yet is not. I solved it the same way as you did.
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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
How did you assume that AB is a straight line. ?
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.
@@bienvenidos9360 ok thankyou
But how intrusive is this Pythagorean Theorem!!! It's there all the time :))))
It's the main building block in Mathematics!
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~ 4.45 π
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40pi/9=13,962
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Why assumed A as the middle, like this always confused me in ur vids