Can you find area of the Green shaded region? | (Annulus) | #math #maths | #geometry

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Can you find area of the Green shaded region? | (Annulus) | #math #maths | #geometry

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[ashutoshkumardalei]

Sir your videos are really interesting 👌🏻👌🏻👌🏻🙏🙏🙏

[DorothyMantoothIsASaint]

Intersecting chords:
draw diameter D through pt C.
Long segment is R + r
Short segment is R - r
(R + r)(R -r) = 4 * 12
R^2 - r^2 = 48
Boom.

[yatesfletcher]

If R is the radius of the larger circle, r is the radius of the smaller, and x is the perpendicular distance from O to the line, then by the Pythagorean Theorem
x^2 + 4^2 = r^2 and
x^2 + 8^2 = R^2

=> pi(R^2) - pi(r^2) = 48pi

[tanyaerskine]

I really enjoyed watching you solve this. Thank you!

[jimlocke]

At first, it seems that the radii of the inner and outer circles, let's designate them as r and R, must be found, from which the inner's circle's area can be deducted from the outer circle's, to determine the green area. However, further examination reveals that there is a range of values for r and R which will satisfy the problem's givens. One of them is the special case of O being on BC. In that case, r = 4 and R = 8, the green area is πR² - πr² = π(8)² - π(4)² = (64)π - (16)π = 48π. If this were a multiple choice test, you could make the assumption the the problem statement implies that the solution for the special case applies to all cases and select 48π as the correct answer. As for proving that 48π is the correct answer for all valid cases, PreMath does excellent work in the video, where he shows that (R² - r²) is constant over the valid range of values for r paired with a value of R, with (R² - r²) having a value of 48. (Actually, for every value of r>=4, there is a corresponding value of R.)

As an aside, the spelling of donut was discussed on the TV program "Wheel of Fortune". They spelled it "doughnut" in a puzzle, presumably to make the puzzle more challenging to solve. The original spelling was doughnut, which is still considered a correct and preferred spelling in many dictionaries. It got shortened to donut, which is the spelling I've seen almost universally in the USA.

[prossvay]

Let R is Radius big circle
and r is Radius small circle
Green area=πR^2-πr^2=π(R^2-r^2)
in small circle:
Let OE=x
(r+x)(r-x)=16
r^2-x^2=16 (1)
in big circle
(R+x)(R-x)=64
R^2-x^2=64 (2)
(2) - (1)
R^2-x^2-r^2+x^2=64-16=48
So R^2-r^2=48
So Green area=48π square units=150.80 square units.❤❤❤ Thanks Sir

[allanflippin]

Starting at 4:30 in the video, the intersecting chords theorem can also be used to find the same results. Extend DE both ways to intersect the edges of the larger circle. Then (R+h) * (R-h) = 8*8. or R^2 - h^2 = 64. Following the pythagorean formula on OEC, h^2 + 4^2 = r^2, or r^2 - h^2 = 16. Subtracting the two equations produces R^2 - r^2 = 48.

[santiagoarosam]

Con centro en "O" trazamos un círculo interior tangente a la cuerda BC→ Área de la corona verde =(Área de la corona delimitada por el círculo exterior y el nuevo círculo) - (Área de la corona delimitada por el círculo interior de la figura propuesta y el nuevo círculo) =π(AD/2)²-π(BC/2)² =π(8²-4²) =π(64-16)=48π ud².
Interesante acertijo. Gracias y un saludo.

[yalchingedikgedik]

That’s very good thanks for you Sir
Very good method for solve ❤❤❤❤

[phungpham]

1/ Area of the green region= pi(sqrR-sqr)
2/ Build the diameter AOA’ intersecting the small circle at points E and F.
We have: AExAF=ABxAC (secant theorem)
or (R-r)x(R+r)=4x12=48
—> sqR-sqr=48
—> Area of the green region=48 pi sq units😅

[alscents]

Since there are no variables provided in the problem that the answer should be expressed in terms of, it means that if the problem is correct, then the area of the green circle is a constant. You can then just choose the special case where AD is a diameter of outer circle.
Meaning that the green area should be pi * 8^2 - pi * 4^2 = pi * (64 - 16) = 48 pi

[juanalfaro]

I solved the problem by the same method you used. Area = Pi*(R^2 - r^2) = 48*Pi

[sergioaiex]

Green Shaded Region = π.R² - π.r² = π (R² - r²) ...¹

By The Chords Theorem

(R + r) (R - r) = 12 . 4
R² - r² = 48

Green Shaded Region = π (R² - r²) = 48π Square Units

[robertlynch]

Isn't it really easier to state:

R² - 8² = k² = S² - 4² . ... where
R is radius of large circle
S is radius of small circle
K is distance from center to middle of line AD

Then, having that, just rearrange things...

(R² - S²) = (8² - 4²) ... = 64 - 16 = 48

Having that ... then turn it back into circles ... Area = pi * radius^2 and the difference is as the video shows

RING = πR² - πS² ... = π(R² - S²) ... = π(48) = 48π which is approximately 150.80

And we're done.

[marioalb]

A = ¼πC² - ¼πc² = ¼π (C² - c²)
A = ¼π (16² - 8²)
A = 48π cm² ( Solved √ )

[jarikosonen]

R=sqrt(73), r=5, h=3
If it can be set h=0, it could be easier to solve with R=8, r=4.

[Ddntitmattrwhtuthnk]

Just a suggestion... What about expanding this equation to show how to find the other unknowns after the area was found? I am always trying to find all unknowns, if possible, because I am very rusty, for professional engineering/programming uses and thought it could be a bonus? I really enjoy the equations.

[unknownidentity]

Let's find the area:
.
..
...
....


Let M be the midpoint of AD. In this case OM is perpendicular to AD and BC as well. Therefore M is also the midpoint of BC and the triangles AMO/DMO and BMO/CMO are right triangles. With R and r being the radii of the outer and inner circle, respectively, we obtain by applying the Pythagorean theorem:

AD = AB + BC + CD = 4 + 8 + 4 = 16
AM = AD/2 = 16/2 = 8
BM = BC/2 = 8/2 = 4

R² = AO² = AM² + MO²
∧ r² = BO² = BM² + MO²
⇒ R² − r² = AM² − BM² = 8² − 4² = 64 − 16 = 48

Now we are able to calculate the area of the green region:

A(green) = A(outer circle) − A(inner circle) = πR² − πr² = π(R² − r²) = 48π

Best regards from Germany

[sergioaiex]

Pythagoras was the greatest mathematical genius

[pedllz]

...if those are the only given values and we know there is a unique answer, the distance from the center is irrelevant, so you can just do (8sq - 4sq) Pi