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

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

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[jimlocke]

Other ways to compute the area of ΔABC: At about 6:40, we have the side lengths of ΔOAD. Using OD as the base and AD as the height, the area is (1/2)bh = (1/2)(1)(√3) = (√3)/2. ΔABC is composed of ΔOAD and 5 more triangles congruent to it, so has 6 times the area of ΔOAD, or 6(√3)/2 = 3√3.

Alternatively, once the side of the equilateral triangle is found to be 2√3, one can apply the formula A = (√3)a²/4, where a is the equilateral triangle's side length. Here, a = 2√3, therefore a² = (2√3)² = 12 and A = (√3)(12)/4 = 3√3.

[himo]

R*R*π=4π R=2
radius of smaller circle : r r=R/2=2/2=1 smaller circle area = 1*1*π=π
√]2²-1²]=√3 △ABC=2√3*3*1/2=3√3

Yellow area = 3√3 - π

[prossvay]

Connect O to A ; B and C
So AOB=AOC=BOC=120°
Big circle area=4π
So R=2
In triangle AOB
Law of cosines

AB^2=2^2+2^2-2(2)(2)cos(120°)
So AB=2√3
Area of
3√3=3(1/2(2√3)(r)
So r=1
Yellow area=3√3-π(1^2)=3√3-π square units=2.05 square units.❤❤❤❤

[phungpham]

Alternative solution:
1/ R=2
2/ The three arcsAB= AC=BC so the angle at the center of the 3 sectors = 120 degrees
So by using the cosine law
Side of the equilateral = sqR+sqR -2 RxR cos (120 degrees) = 4+4 -2 x2x2(-1/2)= 12
side of the equilateral = 2 sqrt3.
2/ The distance from O to one side= r= 1/2R=1
3/ Area of the yelllow region = sq(2sqrt3).sqrt3/4 - pi=3sqrt3- pi=2.05 sq units

[devondevon]

2.05615
Draw a line inside the equilateral from the circle's center to B and another from its center to A, forming a 120-degree triangle
Draw a line from the circle's center to B and one from the center to the point of tangency on line AB to form a 30-60-90
right triangle, BOP.
Since the radius of the large circle = 2 (given its area of 4 pi), then the radius of the small circle is 1 (since in 30-60-90
right triangle the ratio is a, 2a, and a sqrt 3 ), and BP = sqrt 3 (since a =radius =1)
Hence, length AB = 2 sqrt 3 since it is twice BP. Hence, the length of the equilateral triangle = 2 sqrt 3
Hence, the area of the equilateral triangle =
sqrt 3/4 * 2 sqrt 3 * 2 sqrt 3 = 5.19615
The radius of the small circle = pi or 3.14 (since its radius =1, see above)

Hence, the area of the yellow region is 5.19615 - 3.14 = 2.05615

[marcgriselhubert]

Be c the side length of the triangle ABC, R = 2 the radius of the big circle and H the orthogonal projection of a on (BC).
We have 2 = R = (2/3).OH, so OH = 3
3 = OH = (sqrt(3)/2).c, so c = 6/sqrt(3) = 2.sqrt(3)
The area of ABC is (c^2).(sqrt(3)/4) = 3.sqrt(3)
the radius of the little circle is OH =(1/2).OA = R/2 = 1
The area inside the little circle is then Pi and finally the yellow area is 3.sqrt(3) - Pi. (No difficulty)

[phungpham]

1/ Let R, r and a be the radius of the big, small and side of the equilateral triangle.
Extend AO intersecting AC at point H and the big circle at A’.
We have AA’ is the perpendicular bisector of BC and the triangle BHO is a 30-90-6O special one.
R= 2 —> OH=1 and BH=sqrt3
So, BC= 2BH = a =2 sqrt3
and r= OH= 1
Area of the yellow region = Area of the equilateral - Area of the small circle
= sq (2sqrt3).sqrt3/4 - pi
= 3sqrt3- pi= 2.05 sq units😅

[georgebliss]

After the preliminaries.
Triangle AOD.
Angle AOD = 60 degrees.
So sector area of small circle = Pi x 1^2 / 6 = Pi / 6.
Area triangle OAD = 1/2 x 1 x root 3 = 1/2 x root 3.
Small yellow area subtended by angle OAD = area of triangle minus area of sector.
Small yellow area = 1/2 x root 3 - Pi / 6.
There are six of these small congruent yellow areas.
Total yellow area = 3 root 3 - Pi.

[MrPaulc]

R=2
Triangle comprises 6 30, 60, 90 triangles that have hypotenuses of 2.
They each have short sides of 1 and sqrt(3).
This gives the area of the equilateral as 3*sqrt(3) un^2.
r is the shortest leg of one of those triangle, so r=1.
Small circle area is pi un^2.
Yellow area is 3*sqrt(3) - pi
3*1.732 - 3.142 = 5.196 - 3.142
2.054 un^2 (rounded).
Now looked: okay our answers were 0.001 apart, which is due to rounding, so no problem.

[nandisaand]

I didnt bother with trig, I just saw
Height=r+R
=(1)+(2)=3
Therefore Area of Triangle:
=½BH
=½(2SQRT(3))•3
=3SQRT(3)

[juanalfaro]

4 π = π*R^2 -> R=2. Now COB angles are 30, 30, 120 and BC/sin 120 = 2/sin (30) --> BC=AB=AC=2*sqrt (3). Now [ABC} = [2*sqrt (3)] ^2 *sqrt (3)/4 = 12*sqrt (3)/4 = 3*sqrt (3)
Semi-Perimeter = 3*2sqrt (3)/2 = 3*sqrt (3) = S. Now [ABC] = 3*sqrt (3) = S*r = 3*sqrt (3) * r --> r=1 unit (Thus A (small circle = π*1^2 = π)
Yellow Area = [ABC] - π*r^2 = 3*sqrt (3) - π sq. units

[santiagoarosam]

Radio del círculo exterior =√4=2→ El centro de ambos círculos coincide con el baricentro del triángulo equilátero de lado b y altura h→ h=2+1=3→ b/2=√3→ Radio del círculo interior =r =h/3=1→ Área amarilla =(h*b/2)-πr² =3√3 -1*π → 3√3 -π.
Gracias y saludos.

[MaximeDube-whzr]

3r=h
h=(√3/2)x
r=(√3/6)x
(x/2)^2 +((√3/6)x)^2 =4.... x=2√3
.
..
...
h=3
r=1
A=3√3-π

[marioalb]

Area and radius of big circle:
A = πR² = 4π cm²
R = √4 = 2 cm
Chord / Side of equilateral triangle:
c = 2R.cos30° = 2 . 2 . √3/2
c = 2√3 cm

Circular ring area :
A = ¼.π.c² = ¼.π.(2√3)²
A = 3π cm²

Círcular segment area x 3:
A = 3 . ½R²(α-sinα)
A = 3 . ½. 2². (120° - sin 120°)
A = 3 . 2, 4567 cm² = 7, 37 cm²

Yellow area:
A = A₁ - A₂ = 3π - 7, 37
A = 2, 055 cm² ( Solved √ )

[misterenter-izrz]

Simply 1/2(2sqrt(3))^2 sin 60-1^2pi=3sqrt(3)-pi.😁

[toninhorosa]

I solve a little different.
Big circle area = 4π
πR^2 = 4π
R^2 = 4π/π
R^2 = 4
R = 2
Equilateral blue triangle ABC have 3 equal sides and 3 equal angles = 60° each one.
From point "O" we draw a perpendicular until we reach straight AB at point D. Forming a right angle at this point D. And we join point A until we reach straight BC where we mark point E. This straight AE passes through point "O " and will form ∆AOD which has a right angle at D. And angle A is 30° (1/2 of 60°).
∆AOD is right triangle
Angle A = 30°
Angle D = 90°
Angle O = 30°
AO = R = 2
OD = (1/2)AO = (1/2)*2 = 1
AD = √3
BD = √3
AB= AC = BC = 2√3
Equilateral ∆ Area = 1/2*base*height
Height = AE = AO+OE=R+r
Height = 2+1 = 3
∆ABC area = (1/2)*2√3*3
∆ABC area = 3√3
Small cicle area = πr^2 = 1π
Yellow area = 3√3 - π
= 2, 054559769 unit.^2

[Nothingx]

I solved it using my pink color muscles 💪 😏

[thewolfdoctor]

I did it the same way except I calculated the area of triangle BAC to be 6 * area of small triangle AOD = 6* (1/2)*(SQRT(3)*1) = 3*SQRT(3)

[quigonkenny]

Outer circle O:
Aᴏ = πR²
4π = πR²
R² = 4
R = √4 = 2

Draw OA, OB, and OC. As ∆ABC is equilateral, ∠AOC = ∠COB = ∠BOA = 360°/3 = 120°.

Let M, N, and P be the points of tangency between inner circle O and AB, BC, and CA respectively. Draw OM, ON, and OP. As AB, BC, and CA are tangent to inner circle O, ∠AMO = ∠BNO = ∠CPO = 90°, so OM bisects AB, ON bisects BC, and OP bisects CA. As OM, ON, and OP are all radii of inner circle O and length r, and as ∆ABC is equilateral and thus AM = MB = BN = NC = CP = PA, and as OA, OB, and OC are all radii of outer circle O and length 2, then ∆AMO, ∆OMB, ∆BNO, ∆ONC, ∆CPO, and ∆OPA are all congruent triangles.

As ∆ABC is equilateral and all 3 internal angles equal 60°, and as OA bisects ∠CAB by symmetry, then ∠PAO = ∠OAM = 30° and ∆AMO (and all congruent triangles) is a 30-60-90 special right triangle, OM (= ON = OP) = OAsin(30°) = 2(1/2) = r = 1, and AM (= MB = BN = NC = CP = PA) = OAcos(30°) = 2(√3/2) = √3.

The yellow area is equal to the area of triangle ∆ABC minus the area of inner circle O.

Yellow area:
Aʏ = bh/2 - πr² = (BN+NC)(OA+ON)/2 - π(OM)²
Aʏ = (√3+√3)(2+1)/2 - π(1)²
Aʏ = 2√3(3)/2 - π = 3√3 - π ≈ 2.05 sq units

[unknownidentity]

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


First of all we calculate the radius R of the circumscribed circle:

A = πR²
4π = πR²
4 = R²
⇒ R = 2

May s be the side length of the equilateral triangle. Then we can apply the relation between the area A of the triangle, its side length and the radius of the circumscribed circle:

R = AB*AC*BC/(4*A(ABC)) = s*s*s/(4*(√3/4)*s²) = s/√3 ⇒ s = √3*R

The radius r of the inscribed circle can be calculated from the triangles area A and perimeter P:

r = 2*A(ABC)/P(ABC) = 2*(√3/4)*s²/(3*s) = s/(2√3)

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

A(yellow)
= A(triangle ABC) − A(inscribed circle)
= (√3/4)*s² − πr²
= (√3/4)*s² − π*(s/(2√3))²
= (√3/4)*s² − π*s²/12
= (√3/4 − π/12)*s²
= (√3/4 − π/12)*3*R²
= (√3/4 − π/12)*3*2²
= 3√3 − π
≈ 2.055

Best regards from Germany