At a quick glance, draw a line from, m, midpoint of AB to center, o,.calculate the radius OA =0.5 AB / cos(30) = 30, and OM = 0.5 AB * Tan(30)=15. Now calculate angle MOB = 180-90-30 =60 degrees.so angle BOC = 60 degrees.So the area of part of the green segment BOC = pi *radius ^2 * 60/360 =150 * Pi. The triangular part of the green area OAB =half base * height * 2= 15*sqrt(3) *OM =225*sqrt(3) . The green area segment = Area OAB + Area BOC =860.95.
If the inscribed angle BAC = 30° then the central angle BOC = 60° (inscribed angle theorem) Let's draw BC to complete the 30° - 60° - 90° right triangle ABC (Thales's theorem) So BC = AB / √3 = OC = OA = OB = r (the radius of the circle). The green shaded area is r² (π / 6 + √3 / 4).
2R cos 30° = 30√3 R = 30√3 / 2.(√3/2) R = 30 cm Chord BC = h = 30 cm (Height of right triangle) Green area = Right triangle area + circular segment area A = A₁ + A₂ A = ½b.h + ½R²(α- sinα) A = ½.30√3.30 + ½.30².(60°- sin 60°) A = 779,42 + 81,53 A = 860.95 cm² ( Solved √ )
2R cos 30° = 30√3 R = 30√3 / 2.(√3/2) R = 30 cm Green area = Semicirle área - circular segment area A = A₂ - A₁ A = ½πR² - ½R²(α- sinα) A = ½π30² + ½.30².(120°- sin 120°) A = 1413,72 - 552,77 A = 860.95 cm² ( Solved √ )
second approach join BO hence bo=ao=oc=R hence angle BAO= angle OBA =30 and angle AOB=120 and BOC=60 hence triangle OBC is equilateral apply sine rule in triangle AOB and obtain radius =30 hence area of triangle ABO =1/2 R^2 *SIN120 = 223 root3 and area of OBC PART = angle BOC/360*PI R^2 =150 PI hence total area is 150pi + 225 ROOT 3
Clearly we have a right angled triangle with sides 30 60 30sqrt(3), then the radius is 30, OBC is an equilateral triangle, therefore the area is 900 pi/6+(1/2)900×sqrt(3)/2=150pi+225sqrt(3)=861 approximately. 😊
Join BC Will got 30-60-90 triangle Then from this we may get diameter Then area of circle BOC sector =1/6 of the circle area of triangle AOC by sine rule (1/2absunC)
Again a very lovely challenge. Here we go: . .. ... .... ..... Since AC is the diameter, according to Thales theorem the triangle ABC is a right triangle with the 90°-angle at B. So the radius R of the circle is not hard to find: AB / AC = cos(∠BAC) 30√3 / (2R) = cos(30°) 15√3 / R = √3/2 ⇒ R = (15√3)(2/√3) = 30 The green area can be divided into two parts: the triangle OAB and the circular sector OBC. The triangle OAB is an isosceles triangle (OA=OB=R), so we have: ∠BAO = ∠ABO = 30° ⇒ ∠AOB = 180° − ∠BAO − ∠ABO = 180° − 30° − 30° = 120° The area of this triangle can be calculated as follows: A(OAB) = (1/2)*OA*OB*sin(∠AOB) = (1/2)*R*R*sin(120°) = (1/2)*R²*(√3/2) = (√3/4)*R² Now we can determine the area of the circular sector OBC: ∠AOB = 120° ⇒ ∠BOC = 180° − ∠AOB = 180° − 120° = 60° ⇒ A(OBC) = πR²*(∠BOC/360°) = πR²*(60°/360°) = πR²/6 Thus the size of the green area is: A(green area) = A(OAB) + A(OBC) = (√3/4)*R² + πR²/6 = (3√3/12 + 2π/12)*R² = (3√3 + 2π)*R²/12 = (3√3 + 2π)*30²/12 = 75(3√3 + 2π) ≈ 860.95 Best regards from Germany
Mein Solution is: the angle a(ABC)= 180/2= 90° so, ABC is a triangle with a right angle. cos(30)= √3/2= 30√3/2r r= 30 unit length OB= r for the ABO triangle: A₁= (1/2) sin(30)*30√3*30 A₁= 0,5*0,5*900√3 A₁= 225√3 for the OBC piece, a(BOC)= 2*30° a(BOC)= 60° A₂= πr²*(60°/360°) A₂= π*30²*(1/6) A₂= 150 π Agreen= A₁ + A₂ Agreen= 225√3 + 150 π square units 🙂
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Excellent math problem for beefing up reasoning power.
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Got it! A good exercise.
Yay! I solved the problem.
At a quick glance, draw a line from, m, midpoint of AB to center, o,.calculate the radius OA =0.5 AB / cos(30) = 30, and OM = 0.5 AB * Tan(30)=15. Now calculate angle MOB = 180-90-30 =60 degrees.so angle BOC = 60 degrees.So the area of part of the green segment BOC = pi *radius ^2 * 60/360 =150 * Pi. The triangular part of the green area OAB =half base * height * 2= 15*sqrt(3) *OM =225*sqrt(3) . The green area segment = Area OAB + Area BOC =860.95.
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If the inscribed angle BAC = 30° then the central angle BOC = 60° (inscribed angle theorem)
Let's draw BC to complete the 30° - 60° - 90° right triangle ABC (Thales's theorem)
So BC = AB / √3 = OC = OA = OB = r (the radius of the circle).
The green shaded area is r² (π / 6 + √3 / 4).
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2R cos 30° = 30√3
R = 30√3 / 2.(√3/2)
R = 30 cm
Chord BC = h = 30 cm (Height of right triangle)
Green area = Right triangle area + circular segment area
A = A₁ + A₂
A = ½b.h + ½R²(α- sinα)
A = ½.30√3.30 + ½.30².(60°- sin 60°)
A = 779,42 + 81,53
A = 860.95 cm² ( Solved √ )
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2R cos 30° = 30√3
R = 30√3 / 2.(√3/2)
R = 30 cm
Green area = Semicirle área - circular segment area
A = A₂ - A₁
A = ½πR² - ½R²(α- sinα)
A = ½π30² + ½.30².(120°- sin 120°)
A = 1413,72 - 552,77
A = 860.95 cm² ( Solved √ )
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Sir I love your videos. I am preparing my Olympiad from this the problems and more.sir next time can you please make a linear equation question
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BAC = 30° = φ; sin(CBA) = 1 → AB = 30√3 → BC = 30 → AC = 60 = AO + CO = 30 + 30;
BD = h = (AB)(BC)/AC = 15√3 → area ∆ AOB = (30/2)h 225√3
OCB = πr^2/6 = 150 π
r = 30 → green shaded region: 75(2π + 3√3)
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second approach join BO hence bo=ao=oc=R hence angle BAO= angle OBA =30 and angle AOB=120 and BOC=60 hence triangle OBC is equilateral apply sine rule in triangle AOB and obtain radius =30 hence area of triangle ABO =1/2 R^2 *SIN120 = 223 root3 and area of OBC PART = angle BOC/360*PI R^2 =150 PI hence total area is 150pi + 225 ROOT 3
Clearly we have a right angled triangle with sides 30 60 30sqrt(3), then the radius is 30, OBC is an equilateral triangle, therefore the area is 900 pi/6+(1/2)900×sqrt(3)/2=150pi+225sqrt(3)=861 approximately. 😊
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@ 7:14 The pending satisfaction of completion is realized. 🙂
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AC = 30√ 3/√ 3/2 then
AC = 60
AO = 60/2 = 30
Area ABO = 30√ 3*30*1/2*sin 30° = 225√ 3
Green area = 30² π*(60/360) + 225√ 3 = 150 π + 225√ 3
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In polar coordinates with A as the pole r=60cosθ θ=[0, π/6] S=(1/2)∫r^2dθ=(1/2)∫(60cosθ)^2dθ=1800[(sin2θ)/4+θ/2]=1800(√3/8+π/12)=225√3+150π
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S=(25(2π+9√3)/3≈182,08
Join BC
Will got 30-60-90 triangle
Then from this we may get diameter
Then area of circle
BOC sector =1/6 of the circle
area of triangle AOC by sine rule (1/2absunC)
ABC es la mitad de un triangulo equilatero 》Radio r=30》 Área verde = 30×15sqrt3/2 +30^2×Pi/6 =225sqrt3 +150 Pi
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I got this answer. I did it the same way you did.
Done it, but without trigonometry (i used Euclid's theorem and Pythagora of course)... : )
Again a very lovely challenge. Here we go:
.
..
...
....
.....
Since AC is the diameter, according to Thales theorem the triangle ABC is a right triangle with the 90°-angle at B. So the radius R of the circle is not hard to find:
AB / AC = cos(∠BAC)
30√3 / (2R) = cos(30°)
15√3 / R = √3/2
⇒ R = (15√3)(2/√3) = 30
The green area can be divided into two parts: the triangle OAB and the circular sector OBC. The triangle OAB is an isosceles triangle (OA=OB=R), so we have:
∠BAO = ∠ABO = 30°
⇒ ∠AOB = 180° − ∠BAO − ∠ABO = 180° − 30° − 30° = 120°
The area of this triangle can be calculated as follows:
A(OAB)
= (1/2)*OA*OB*sin(∠AOB)
= (1/2)*R*R*sin(120°)
= (1/2)*R²*(√3/2)
= (√3/4)*R²
Now we can determine the area of the circular sector OBC:
∠AOB = 120°
⇒ ∠BOC = 180° − ∠AOB = 180° − 120° = 60°
⇒ A(OBC) = πR²*(∠BOC/360°) = πR²*(60°/360°) = πR²/6
Thus the size of the green area is:
A(green area)
= A(OAB) + A(OBC)
= (√3/4)*R² + πR²/6
= (3√3/12 + 2π/12)*R²
= (3√3 + 2π)*R²/12
= (3√3 + 2π)*30²/12
= 75(3√3 + 2π)
≈ 860.95
Best regards from Germany
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Join BC angle ABC will be 90 use costheta Ac will be found and we can do it
Similar here. I calculated the triangular part as 15root3 * 15 (the height) so 225root3. 225root3 + 150 pi.
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Mein Solution is:
the angle a(ABC)= 180/2= 90°
so, ABC is a triangle with a right angle.
cos(30)= √3/2= 30√3/2r
r= 30 unit length
OB= r
for the ABO triangle:
A₁= (1/2) sin(30)*30√3*30
A₁= 0,5*0,5*900√3
A₁= 225√3
for the OBC piece,
a(BOC)= 2*30°
a(BOC)= 60°
A₂= πr²*(60°/360°)
A₂= π*30²*(1/6)
A₂= 150 π
Agreen= A₁ + A₂
Agreen= 225√3 + 150 π square units 🙂
225√3+150π
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Mmmmmm😊
225sqrt(3)/2+150pi
If I'm wrong its because I didnt write anything down.
And im wrong.
It's supposed to be: 225sqrt(3)+150pi
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