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

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

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[adept]

Intersecting chords property:
AT•TB=CT•TD.
2R•2r=50•50.
R•r=625.

[neilcourse]

There's an easier way. Since the answer doesn't depend on TP, you might as well assume that TP=0. Then the big circle has a radius of 50 and the two white circles each have a radius of 25.

[marcelowanderleycorreia]

Very smart question!! Congrats teacher!! 👏👏👍👍

[marioalb]

Shaded area value doesn't depend on circle radius
We can match both inner circles, and all original conditions are still being fulfilled.
Therefore, R=2r, R=50cm and r=25cm

A = πR² - 2πr² = π(R² - 2r²)
A = π (50² - 2*25²)
A = 1250π cm² ( Solved √ )

[alanthayer]

Cool break down ! I like these type of Problems which is finding Area of spherical curvatures! Thanks as usual sir !

[GilmerJohn]

Since the only information given was the 100 units it's likely that so long as every thing fits, it's possible that the two inner circles are the same size. If that's the case, each inner is 1/4th of the big circle. The green area is 1/2 (50^2)xpi.

[jackwhite]

Very nice, but as a conclusion it must be emphasized that the result doesn't depend on the position of the T point (between C and D as limits ) . An animated displacement would be very interesting. The easiest calculation is for the position in P.

[marioalb]

A= ½(¼πc²) = ⅛π100²
A = 1250π cm² ( Solved √ )


Shaded area is equal to half of the circular ring area, respect to the chord tangent to the equivalent inner circle.
This chord is same chord that separates both given internal circles. ( given data c= 100 cm)

[himo]

CO=OT=x TQ=QD=y CP=PD=z AT=TB=50
50*50=2x*2y 4xy=2500 xy=625
(2x+2y)/2=z 2(x+y)/2=z x+y=z (x+y)²=x²+2xy+y²=z²

Green shaded area = z²π - y²π - x²π = x²π + 2xyπ + y²π - y²π -x²π = 1250π

[quigonkenny]

Let a be the radius of circle O, b the radius of circle Q, and r the radius of circle P. The shaded area will be equal to the area of circle P minus the areas of circles O and Q.

A = πr² - (πa²+πb²)
A = π(r²-(a²+b²)) --- [1]

As circles O and Q are tangent to each other and to circle P, then their center points and points of tangency are collinear, so O, T, P, and Q are on CD.

As AB is tangent to circles O and Q at T, then AB is perpendicular to CD, and as CD is a diameter of circle P, then T is the midpoint of AB. AT = TB = 100/2 = 50.

By the intersecting chords theorem, AT(TB) = CT(TD).

AT(TB) = CT(TD)
50(50) = (2a)(2b)
2500 = 4ab
ab = 2500/4 = 625 --- [2]

CD = CT + TD
2r = 2a + 2b
r = a + b
r² = (a+b)²
r² = a² + b² + 2ab
r² = a² + b² + 2(625)
r² = a² + b² + 1250
a² + b² = r² - 1250 --- [3]

A = π(r²-(a²+b²))
A = π(r³-(r²-1250))
A = π(r²-r²+1250)
A = 1250π ≈ 3926.99 sq units

[andrepiotrowski]

Pythagoras ad nauseam. There are also other theorems, such as Thales' theorem and the altitude theorem (or intersecting chords theorem, even simpler). It follows immediately: 2r * 2R = 50^2

[shrawankumarmishra]

Nice sir please provide some other topic video

[unknownidentity]

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


All three circles have exactly one intersection point in pairs. Therefore we know that the centers of all these circles are located on the same line (CD). We also know that AB is a tangent to the smaller and the bigger white circle. From this we can conclude that AB is perpendicular to CD. This also means that the common intersection point (T) is the midpoint of AB. Now we can apply the intersecting secants theorem. With R₁>R₂>R₃ being the radii of the circles we obtain:

AT*BT = CT*DT
(AB/2)*(AB/2) = (2R₂)*(2R₃)
AB²/4 = 4R₂R₃
⇒ 2R₂R₃ = AB²/8 = 100²/8 = 10000/8 = 1250

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

A(green)
= π(R₁)² − π(R₂)² − π(R₃)²
= π[(R₁)² − (R₂)² − (R₃)²]
= π[(R₂ + R₃)² − (R₂)² − (R₃)²]
= π[(R₂)² + 2R₂R₃ + (R₃)² − (R₂)² − (R₃)²]
= π*2R₂R₃
= 1250π

Best regards from Germany

[yakovspivak]

R = r1 + r2 is an outer circle radius.
50^2 = 4(r1 × r2) - geometric mean
50^2 = (r1 + r2)^2 - (r1 - r2)^2
Or:
r1+ r2 = 50 and r1 - r2 = 0
r1 = 25, r2 = 25
S = 1250 × pi

[iosifbitenskiy]

2nd way. Let consider a triangle CBD inscribed into a big circle. One side of the triangle is a diameter CD, therefore the angle CBD=90°. Let diameters of the circles be a = 2r, b = 2R, CD = d.
In the right angle triangle CBD
CD² = CB² + BD² (1)
From the right angle triangles CBT and TBD: CB² = CT² + TB² or CB² = a² + 50², and
BD² = TD² + TB² or BD² = b² + 50².
Inserting into (1)
d² = a² + 50² + b² + 50² or d² - a² - b² = 5000.
Multiplying both sides by π/4 one gets the shaded area on the left equal 1250π on the right.

[cyruschang]

r is the small white circle radius, R is the large green circle radius
R - r = the radius of the larger white circle
50^2 + (R - 2r)^2 = R^2
50^2 + 4r^2 - 4rR = 0
r(R - r) = 25^2
Green area = π[R^2 - r^2 - (R - r)^2] = π[2rR - 2r^2] = π(2)(25)(25) = (50)(25) = 1250π

[maroonshaded]

So if b = length of AB, a general expression for the shaded area could be πb^2/8.

[wackojacko]

Another reason why the 'red and green should never be seen' rule should be left behind. Why are barns painted red? ... green pastures! Complementary colors! In this case the red line and green area gives the problem a vibrant look! Happy Holidays 😊

[imetroangola]

*Solução:*

Seja k o raio da circunferência maior.

Quando as cordas são perpendiculares, vale a relação:

AT² + TB² + TD² + TC² = (2k)²

Como T é ponto médio do segmento AB, logo TB=AT= 50.

Daí,

50² + 50² + (2r)² + (2R)² = (2k)²

5000 + 4r² + 4R² = 4k² × (π/4)

1250π + πr² + πR² = πk²

πk² - (πr² + πR²) = 1250π

Portanto, a área procurada é:

*1250π unidades quadradas.*

[donfzic]

Thank you.
Is it possible to specify also the values of r and R ?