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.
To calculate r, you don't need to resort to Pythagoras. The perpendicular to BC through A is a median, O is the intersection of the medians and divides the perpendicular in the ratio 2:1...
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 √ )
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)
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.
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
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
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.
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=OA^2+OB^2-2(OA)(OB)cos(120°) AB^2=2^2+2^2-2(2)(2)cos(120°) So AB=2√3 Area of triangle=1/2(2√3)^2sin(60)=3√3 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.❤❤❤❤
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
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.
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😅
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
STEP-BY-STEP RESOLUTION PROPOSAL : 01) Pi * R^2 = 4 * Pi ; R^2 = 4 ; R = +/- sqrt(4) ; R = +/- 2 ; R = 2 02) OA = OB = OC = R = 2 lin un 03) Inscribed Angles : Angle ABC = Angle BCA = Angle CAB = 120º 04) Inner Angles : Angle AOB = Angle BOC = Angle COA = 60º 05) Tangency Points between Equilateral Triangle and Small Circle : D ; E ; F 06) OD = OE = OF = r lin un 07) Now we have 3 Isosceles Triangles : [AOB] ; [BOC] ; [COA] subdivided in 6 Rigth Triangles (30º - 60º - 90º) : [AOD] ; [BOD] ; [BOE] ; [COE] ; [COF] ; [AOF] 08) Let's call the Big Cathetus X and Small Cathetus Y. 09) cos (30º) = X/2 ; X = 2 * cos (30º) ; X = 2 * (sqrt(3)/2) ; X = sqrt(3) lin un 10) sin (30º) = Y/2 ; Y = 2 * sin (30º) ; Y = 2 * 1/2 ; Y = 1 lin un 11)OD = OE = OF = r = 1 lin un 12) Small Circle Area (A) = Pi sq un 13) Blue Triangle Area with Sides equal to 2*sqrt(3) = 3*sqrt(3) ; BTA ~ 5,19615 sq un (Using Heron's Formula). 14) Yellow Area = (BTA - SCA) sq un ; YA = [3*sqrt(3) - Pi] sq un ; Yellow Area ~ 2,055 sq un Thus, OUR BEST ANSWER IS : The Yellow Area is equal to approx. 2,055 Square Units. Exact Form : [(3*sqrt(3)) - Pi] Square Units. Best Wishes from Islam (Peace on Earth) - Al Andalus.
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
Solution: Yellow Area = Equilateral Triangle Area - Small Circle Area ... ¹ Area of Large Circle = 4π A = π R² 4π = π R² (÷π) R² = 4 R = 2 Therefore, when connecting O to B, we have R = 2 and we form a 30°- 60° - 90° Special Triangle and, thus, r = 1 and AB = BC = AC = 2√3 Area of Small Circle = π r² Area of Small Circle = π (1)² Area of Small Circle = π Area of Equilateral Triangle = ½ AB . r + ½ BC . r + ½ AC . r Area of Equilateral Triangle = ½ 2√3 . 1 + ½ 2√3 . 1 + ½ 2√3 . 1 Area of an Equilateral Triangle = √3 + √3 + √3 Area of an Equilateral Triangle = 3√3 Substituting in ¹ Yellow Area = 3√3 - π Area Units ✅ Yellow Area ≈ 2.0545 Area Units ✅
Stretching out a basic idea is silly. The maths should be outlined in your plan . People doing maths do not need this laborious step by step as they need guidance not be treated as totally incapable of following your method. 10:53
Tips and Tricks...clickbait for those needing too acquire Algebra and Geometry skills i.e area of circle formula, Perpendicular Bisector Theorem, Area of the Triangle Formula, Circle Theorem, Equalateral Triangle...I know a lot of Tricks cuz I am a magician and can pull a rabbit out of a hat and make things disappear. 🙂
S=3√3-π≈2,06. First, pin me please
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Nice sharing sir❤🎉❤
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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.
This way no need tu use trigonometry…
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To calculate r, you don't need to resort to Pythagoras. The perpendicular to BC through A is a median, O is the intersection of the medians and divides the perpendicular in the ratio 2:1...
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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 √ )
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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
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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)
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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)
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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.
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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
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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
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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.
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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=OA^2+OB^2-2(OA)(OB)cos(120°)
AB^2=2^2+2^2-2(2)(2)cos(120°)
So AB=2√3
Area of triangle=1/2(2√3)^2sin(60)=3√3
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.❤❤❤❤
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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
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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.
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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😅
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Non-geometry problems, PLZZZZZZZZ!
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
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
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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)
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R=2...l=2√3...r=1.. Ayellow=(1/2)(2√3)^2sin60-π(1)^2=3√3-π.
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STEP-BY-STEP RESOLUTION PROPOSAL :
01) Pi * R^2 = 4 * Pi ; R^2 = 4 ; R = +/- sqrt(4) ; R = +/- 2 ; R = 2
02) OA = OB = OC = R = 2 lin un
03) Inscribed Angles : Angle ABC = Angle BCA = Angle CAB = 120º
04) Inner Angles : Angle AOB = Angle BOC = Angle COA = 60º
05) Tangency Points between Equilateral Triangle and Small Circle : D ; E ; F
06) OD = OE = OF = r lin un
07) Now we have 3 Isosceles Triangles : [AOB] ; [BOC] ; [COA] subdivided in 6 Rigth Triangles (30º - 60º - 90º) : [AOD] ; [BOD] ; [BOE] ; [COE] ; [COF] ; [AOF]
08) Let's call the Big Cathetus X and Small Cathetus Y.
09) cos (30º) = X/2 ; X = 2 * cos (30º) ; X = 2 * (sqrt(3)/2) ; X = sqrt(3) lin un
10) sin (30º) = Y/2 ; Y = 2 * sin (30º) ; Y = 2 * 1/2 ; Y = 1 lin un
11)OD = OE = OF = r = 1 lin un
12) Small Circle Area (A) = Pi sq un
13) Blue Triangle Area with Sides equal to 2*sqrt(3) = 3*sqrt(3) ; BTA ~ 5,19615 sq un (Using Heron's Formula).
14) Yellow Area = (BTA - SCA) sq un ; YA = [3*sqrt(3) - Pi] sq un ; Yellow Area ~ 2,055 sq un
Thus,
OUR BEST ANSWER IS :
The Yellow Area is equal to approx. 2,055 Square Units. Exact Form : [(3*sqrt(3)) - Pi] Square Units.
Best Wishes from Islam (Peace on Earth) - Al Andalus.
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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
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I solved it using my pink color muscles 💪 😏
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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
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2.056
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bill cipher
Solution:
Yellow Area = Equilateral Triangle Area - Small Circle Area ... ¹
Area of Large Circle = 4π
A = π R²
4π = π R² (÷π)
R² = 4
R = 2
Therefore, when connecting O to B, we have R = 2 and we form a 30°- 60° - 90° Special Triangle and, thus, r = 1 and AB = BC = AC = 2√3
Area of Small Circle = π r²
Area of Small Circle = π (1)²
Area of Small Circle = π
Area of Equilateral Triangle = ½ AB . r + ½ BC . r + ½ AC . r
Area of Equilateral Triangle = ½ 2√3 . 1 + ½ 2√3 . 1 + ½ 2√3 . 1
Area of an Equilateral Triangle = √3 + √3 + √3
Area of an Equilateral Triangle = 3√3
Substituting in ¹
Yellow Area = 3√3 - π Area Units ✅
Yellow Area ≈ 2.0545 Area Units ✅
R=2, R=a/√3, r=a/(2√3)= R√3/(2√3)=1.
As simple as that. The rest is pure Math Routine.
@@LuisdeBritoCamacho Different countries have different math programs. We study these formulas in the geometry course at school.
@@ОльгаСоломашенко-ь6ы , I believe you are from Russia. I am not aware of Math Programs there. I am from Portugal - Lisbon. Very far away from Moscow.
@@LuisdeBritoCamacho No. I am from Belarus.
@@LuisdeBritoCamacho 3160 km from where l live.
R=2=c ; a=1=r ; b=/3 ; base=/3+/3=2/3 ; high=R+r=2+1=3
You have certainly stirred the mathematical pot so even if your podcasts do nothing else they cause an awful lot of Mathematics to be done.
Stretching out a basic idea is silly. The maths should be outlined in your plan . People doing maths do not need this laborious step by step as they need guidance not be treated as totally incapable of following your method. 10:53
Tips and Tricks...clickbait for those needing too acquire Algebra and Geometry skills i.e area of circle formula, Perpendicular Bisector Theorem, Area of the Triangle Formula, Circle Theorem, Equalateral Triangle...I know a lot of Tricks cuz I am a magician and can pull a rabbit out of a hat and make things disappear. 🙂
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