A very Nice Geometry Problem | You should be able to solve this! | 3 Methods

[WahranRai]

0:31 What you call lowercase r is called GAMMA in Europe, lowercase r is r !

[TheAlavini]

Congrats. I liked so much the second metrod. Its more elegant. Thank you.

[michaeldoerr]

I almost forgot mention that I think that I have understood the third method better than the second. It has been a while since I have learned of a three method geometry problem!!!

[KipIngram]

The radius of the small circle is not specified precisely, so we can choose it within reason. Let's make it zero. Then 8 is the radius of the circle and the sought after area is the area of the entire quarter circle. So, pi*R^2/4 = 16*pi.

[michaeldoerr]

The answr is 16pi. Both methods are barely any different. The first method is parcticed within the quarter circle and the second method makes use of the hypothetical circle that contains the quarter circle. I like the first method. I hope that this means that I got this!!!

[himo]

radius of quarter circle : R
radius of smaller circle : r


Shaded area = R^2π/4 - r^2π = 16π

[olesgaiduk]

Mirror line ACO add a quarter circle ACOD’B’/ We construct an isosceles triangle, where DD’ = 2CD = 8*2 =16; ; AC = 6; AD =10 . The area of ​​this triangle S_ АDD’ = 0.5*AC * DD’ = 0.5*6* 16 = 48. The radius of the great circle R = (AD’ * AD *D’D) /4S_ ADD’ = 10*10*16 / 4*48 = 25/3 ; The area of ​​the quarter of the great circle Sb = 0.25 π R^2 =0.25 π (25/3)^2 = 625 π /36; The radius of the small circle r = 0.5(R – AC) = 0.5 [(25/3) – 6) = 7/6; The area of ​​the small circle Sr = π r^2 = π (7/6)^2 = 49 π/36. The area of ​​the pink circle Sp = Sb – Sr = (625 π /36 ) – (49 π/36) = 16 π.

[santiagoarosam]

El área sombreada es igual a la cuarta parte de la superficie de la corona circular delimitada por la circunferencia exterior y otra concéntrica de radio el diámetro del círculo interior---> Área sombreada =8²π/4=16π.
Gracias y saludos.

[devondevon]

16 pi
Let the radius of the quarter circle = T, then
its area = pi T^2/4 (since the area of a quarter circle is 1/4 the circle)
Let the radius of the small circle = m, then
its area = pi m^2

Hence, the area of the shaded region is the difference between the quarter circle and the smaller circle.
Hence, the area of the shaded region is ( pi T^2)/4 - pi m^2
Let's give both a common denominator. Hence, (pi T^2/)4 - (4 pi m^2)/4'

Hence, pi/4 ( T^2 - 4 m^2) factors out pi/4 (This is the area of the shaded region in terms of T and m). Figure A

Draw a line from D to O to form the right triangle, CDO.
DO is the radius of the circle and also the hypotenuse.
Hence, DO= T
CD = 8 given
CO = 2m (Since m is the radius of the circle and C diameter and the diameter is twice the radius)
Let's use Pythagorean
T^2 = 8^2 + ( 2m)^2
T^2 =64 + 4m^2
T2 - 4m^2 = 64 Equation 1

pi/4 (64) substitute equation 1 into Figure A (pi/4 ( T^2 - 4m^2)
pi (64/4) = 16 pi Answer

[solimana-soli]

I think that the three methods proposed in the video are essentially equivalent, they all lead to the same identity R²-4r=64. The solution using Euclid's theorem could also have been added with the same result.
I solved in a different way with trigonometry:
8 = R cos alpha => R = 8/cos alpha
2r = R sen alpha => r = R sen alpha/2
Shaded Area = (R²/4 - r²)*pi
shaded area = ( (8/cos alpha)²*1/4 - (R sin alpha/2)²)*pi
shaded area = (64/cos²alpha - R²sin²alpha)*1/4*pi
shaded area = (64/cos²alpha - 64 sin²alpha/cos²alpha)*1/4*pi
shaded area = (64*(1 -sin²alpha)/cos²alpha)*1/4*pi
shaded area =
shaded area = 64*1/4*pi = 16pi

Moreover there is a logical approach. If you think to reduce the smaller circle untill its radius = 0 then R = 8 and the quarter is 64/4*pi = 16 pi

[hongningsuen]

Method using properties of circle and Pythagoras theorem:
1. Let radius of big circle be R and radius of small circle be r.
2. Area of shaded region = (pi/4)R^2 - (pi)r^2 = (pi/4)(R^2 - 4r^2)
3. In triangle CDO, by Pythagoras theorem CO = sqrt(OD^2 - CD^2) = sqrt(R^2 - 8^2)
From construction, CO = diameter of small circle = 2r
Hence 2r = sqrt(R^2 - 8^2)
4r^2 = R^2 - 64
R^2 - 4r^2 = 64
4. Hence area of shaded region = (pi/4)64 = 16(pi)

[kateknowles]

Let r = radius of black circle and let R = OA=OD=OC.
Using Pythagoras' theorem in triangle ODC : OC.OC= OD.OD - CD.CD = R.R - 8.8 and R.R - OC.OC=64

( Also extending DC to the left across the next quadrant: CD.CD= 64 = OA(2R-OA) (intersecting chords into four parts into two equal products)
seems superfluous now ) ( This would not have been so nice with CD given equal to any other value)

Required shaded area = pi/4 R.R - pi r.r. = pi/4(R.R - 4 r.r) 2r is also OC so R.R - 4r.r = 64
So required area = pi/4(64)
= 16 pi square units

[RealQinnMalloryu]

(8)^2 =64 90°/64=1.26 1^1.2^13 2^1^1 2^1 (x ➖ 2x+1). ❤ it looks properly