Can you find area of the Brown circle? | (Rectangle) | #math #maths | #geometry

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Can you find area of the Brown circle? | (Rectangle) | #math #maths | #geometry

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[yalchingedikgedik]

A wonderful method of solve
Thanks Sir
Very useful
Good luck with glades
❤❤❤❤

[yesterwilliamquebradaholgu]

(R+r)²=(18-R-r)² + (25-R-r)², we can to know R with πR²=64π; R²=64 and R=8;

Then:

(8+r)²=(10-r)² +(17-r)²; solving this equation we obtained r.

With r we calculate area's circle.

[MDH.SHameem]

I solved it in another way.
1) First of all from the square KOFB from your figure (3:01 video time) I find BO = 8√2
2) Taking radius of brown circle= r. From square JPHD from your same figure DP = r√2
3) Then the length of full diagonal will be r√2 + r + 8 +
4) from the length and width of rectangle diagonal =
5) From (1) and (2) i solved for r and got r = 4.76 but not 5 why is that?

[marcgriselhubert]

We use an orthonormal center B and first axis (BA)
We have O(8; 8) and P(18 -R); 25 -R) with R the radius of the brown circle. (8 beeing the radius of the blue one)
Then VectorOP(10 - R; 17 - R)
OP^2 = (10 - R)^2 + (17 -R)^2 = 2.(R^2) - 54.R +389
In an other side OP = 8 + R (sum of the radius),
so OP^2 = (8 +R)^2 = R^2 + 16.R +64
Then we have: 2.(R^2) - 54.R + 389 = R^2 + 16.R + 64
or R^2 -70.R + 325 = 0
Deltaprime = 35^2 - 325 = 900 = 30^2
So R = 35 + 30 = 65 (rejected) or R = 35 - 30 = 5
Finally the area of the brown circle is 25.Pi

[murdock]

Nice! ∆ PQR → PQ = 10 -r; QO = 8 + (9 - r) = 17 - r; PO = r + 8; sin⁡(PQO) = 1 →
(8 + r)^2 = (17 - r)^2 + (10 - r)^2 → r1 = 5; r2 = 65 ≠ solution, x < 8

[devondevon]

25 pi cm^2
The radius of the blue =8
Let the radius of the yellow = r

Draw a line from the center of one circle to the other. This length is 8+r.
Draw a horizontal and vertical line to form a right triangle
The horizontal line = 18 minus the radius of the blue circle + the radius of the yellow circle
=10 - r

The vertical line = 25 minus the radius of the blue circle + the radius of the yellow circle

= 17 -r

Hence, the sides of the triangle are 8+ r (hypotenuse), 10 -r, and 17 -r

Hence, (8 + r)^2 = (17-r)^2 + (10 -r )^2

r^2 + 16 r - 64= r^2 - 34 r + 289 + r^2 - 20r + 100

0 = r^2 - 70 r + 325
0 = (r- 5)(r-65)
r=5 reject 65 since the radius cannot be larger than the dimension of the 18 * 25 rectangle

Area = 25 pi

[misterenter-izrz]

(8+r)^2=(10-r)^2+(17-r)^2, solving r=5, therefore the area is 25pi.😊

[petersaunders]

I seem to be missing something, if O to F = 8, then O to T must be > 8 yet we have 10-r 5 :-(

[santiagoarosam]

Área azul =64π---> Radio azul =√64=8.
En el triángulo rectángulo cuya hipotenusa es el segmento que une los centros de los círculos y sus catetos las proyecciones horizontal y vertical de ese segmlento: (r+8)²=(10-r)²+(17-r)² ---> r=5 ---> Área del círculo marrón =π5² =25π ud².
Buen problema. Gracias y saludos.

[jacquespictet]

You may create an interesting exercice by adding a circle radius 4 on the upper left corner, using two similar triangles.

[sergioaiex]

Solution:

A = π R²
64π = π R²
R² = 64
R = 8

Joining the centers of the blue and brown circles, we have, as hypotenuse, 8 + r, and, as cathetes, we have 10 - r and and 17 - r, so, we are going to use The Pythagorean Theorem:

(10 - r)² + (17 - r)² = (8 + r)²
100 - 20r + r² + 289 - 34r + r² = 64 + 16r + r²
389 - 54r + 2r² = 16r + 64
r² - 70r + 325 = 0

r = 70 ± √4900 - 1300/2
r = 70± 60/2

r' = 65 Rejected
r" = 5 Accepted

A = π r²
A = π 5²
A = 25π Square Units

A ~= 78, 539 Square Units

[quigonkenny]

Let R be the radius of blue circle O and r be the radius of brown circle P.

Circle O:
Aᴏ = πR²
64π = πR²
R² = 64
R = 8

Let M and N be the points of tangency for circle O on AB and BC respectively, and J and K be the points of tangency for circle P on CD and DA respectively.

As OM and ON are radii of circle O, ∠BMO = ∠ONB = 90°. As ∠NBM = 90° as well and adjacent sides OM and ON equal 8, ∠MON = 90° as well and ONBM is a square.

As PJ and PK are radii of circle P, ∠PJD = ∠DKP = 90°. As ∠JDK = 90° as well and adjacent sides PJ and PK equal r, ∠KPJ = 90° as well and PJDK is a square.

Draw OE, where E is the point on CD where OE is parallel to BC, and draw PF, where F is the point on BC where PF is parallel to CD. As ABCD is a rectangle and all internal angles are 90°, PF is perpendicular to BC and OE is perpendicular to CD and thus OE and PF are perpendicular to each other. Let T be the point of intersection of the two line segments. Draw OP.

As OM = R = 8, TE = PJ = r, and ME = AD = 25, OT = ME-OM-TE = 25-8-r = 17-r. As PK = r, TF = ON = 8, and KF = AB = 18, PT= KF-PK-TE = 18-r-8 = 10-r. As the point of tangency between two circles is collinear with the two centers of the circles, OP = R+r = r+8.

Triangle ∆PTO:
PT² + OT² = OP²
(10-r)² + (17-r)² = (r+8)²
100 - 20r + r² + 289 - 34r + r² = r² + 16r + 64
r² - 54r + 389 = 16r + 64
r² - 70r + 325 = 0
(r-65)(r-5) = 0
r = 65 ❌ | r = 5 --- r < 9

Circle P:
Aᴘ = πr² = π5² = 25π cm² ≈ 78.54 cm²

[himo]

64=8*8 (8+r)²=(18-8-r)²+(25-8-r)²
r²-70r+325=0 (r-5)(r-65)=0 r=65 is rejected, thus r=5
Brown Circle area : 5*5*π = 25πcm²

[Birol]

My way of solution ▶
Ablue= 64 π cm²
πr²= 64 π
r²= 64
r= 8 cm

if we draw a line and connect the radius of the both circles [OP] this would the the "Hypotenuse" of the right triangle ΔPKO
K ∈ [OP]
let's find [PK] and [KO]
[PK]= 18-r-8
[PK]= 10-r

[KO]= 25 - 8 - r
[KO]= 17-r

let's apply the Pythagorean theorem for the ΔPKO :
[OP]²= [PK]²+[KO]²
[OP]= 8+r
⇒
(8+r)²= (10-r)² + (17-r)²
64+16t+r²= 100-20r+r²+289-34r+r²
⇒
r² - 70r + 325=0
Δ= 70²-4*1*325
Δ= 3600
√Δ= 60

r₁= (70-60)/2
r₁= 5 cm

r₂= (70+60)/2
r₂= 65 cm

if r₂= 65 cm [KO] and [PK] would be negative ❌
⇒
r= 5 cm

Area of the brown circle, Acircle
Acircle= π*r²
Acircle= π*5²
Acircle= 25π
Acircle= 78, 54 cm²

[prossvay]

Big circle
πR^2=64π
R^2=64
So R=8cm
Connect P to O ; PQ and OQ
In ∆ OPQ
So (17-r)^2+(10-r)^2=(8+r)^2
So r=5cm
So area of the brown circle=π(5^2)=25π cm^2=78.54cm^2.❤❤❤

[giuseppemalaguti]

Rblue=8..R+8+x=18, R+8+y=25..R+x=10, R+y=17, con (R+8)^2=x^2+y^2....sommo e sottraggo le due equazioni, risulta 2R+(x+y)=27..y-x=7...elevo al quadrato le due equazioni e quindi una equazione in

[wackojacko]

I'm not unwilling too express my opinion and quite confident about doing so. ...and, I don't isolate myself in fear of what any group's perception is of my own judgment. Beginning @ 7:31 is the only Cognitive Cancel Culture I belong to. Actually it's Analytic Philosophy, the Metaphysics of our times. 🙂

[MrPaulc]

Mine was much messier than yours:
I ended up with 10^2 + 17^2 = (8 + r + r*sqrt(2))^2.
If only I had picked out that I could have shortened that triangle :)
389 = (8 + r + r*sqrt(2)) (8 + r + r*sqrt(2))
Ouch! :)