At a quick glance, A is half AB vertical and horizontal from O. The tangents at A and B are 45. The radius is then sqrt((3/4 *30^2 * 2) = 36.74. Angle OAB = 90. The area of the circle is PI * 36.74^2 = 1350 * PI. Angle from A to B is 360 - 270 = 90 and area of circle segment OAB= 1/4 *1350 * PI. Area of triangle AOB = 0.5 *30 *sqrt(3) * 0.5 *36.74 = 477.27. Drawing a horizontal line from O , intersecting the circumference at C . The area of OAC = 0.5 * 36.74 * 0.5 * 30*sqrt(3) =477 .27 . then the difference of the circle segment and triangle segment is segment area of of white shaded area in semicircle = area of green shaded area in semicircle = 1/8 * 1350 * PI - 477.27 = 52.87. Then area of quarter circle segment = 1/4 * 1350 * PI =1060.3 . Then 1060.3 + 52.87 =556 area units.
1/ Notice that the triangle AOC is an equilateral one, so OA.sqrt3/2 =AB/2---> R=30 2/ The area of the dark green = the area of the 120 degree white segment = 1/3 pi sqR - (sqR. sqrt3)/4= 300 pi - 225.sqrt3 sq units
Since it was obtained by folding along the chord AB, the dark green region is congruent with the circular segment AB. Draw the diameter DC bisecting AB at P. Since OD = OC is twice PO = PC then ABD must be an equilateral triangle. So the angle AOB at the center is 2 × 60° = 120° = ⅔ π rad = θ, sin θ = √3 / 2 and the radius of the circumcircle is 30. Then the area of the circular segment is ½ 30² (⅔ π − √3 / 2) which simplifies to 75 (4π − 3√3) ≈ 552.77 square units.
Keep folding the DARK GREEN SHADED REGION until it becomes the deepest shade of Black Green...the ultimate color of India's Pepper Plant. The first time I seen a garden full of this flowering vine, I just can't get enough of it and really could stare at it for hours. 🙂
Reflect the light green region across chord AB. This forms a new circle. They overlap in the middle. Let's name the center of this circle point P. The radius OP is shared by both circles, so they have the same radius. Draw radii AO, AP, BO, & BP. This forms a rhombus AOBP as the radii are congruent. By the Parallelogram & Rhombus Diagonals Theorems, the diagonals of rhombus AOBP bisect each other and are perpendicular. Use the Intersecting Chords Theorem. (3r/2)(r/2) = (15√3)(15√3) 3r²/4 = 675 3r² = 2700 r² = 900 r = 30 Label the intersection of the diagonals of rhombus AOBP as point M. Since one of the diagonals is the radius of both circles, and chord AB bisects it, MP = 15. Since AM = 15√3, △AMP is a special 30°-60°-90° right triangle. Therefore, m∠APM = 60°. By the Rhombus Opposite Angles Theorem, each diagonal bisects a pair of opposite angles. Thus, m∠APB = 120°. Now, notice the dark green shaded region is a segment of circle P. The central angle measure is in degrees. Minor Segment AB Area = Sector PAB Area - △APB Area A₁ = (θ/360°)πr² = (120°/360°) * π * 30² = 1/3 * π * 900 = 300π A₂ = 1/2 * a * b * sinC = 1/2 * 30 * 30 * sin(120°) = 450 * sin(120°) = 450 * (√3)/2 = (450√3)/2 = 225√3 Dark Green Shaded Region = 300π - 225√3 So, the area of the dark green shaded region is 300π - 225√3 square units, a.k.a. 75(4π - 3√3) square units (exact), or about 552.77 square units (approximation).
First we calculate the radius R of the main circle with the Pythagorean theorem in triangle OHA (with H the middle of [A,B] We have OA^2 = OH^2 = HA^2, so R^2 = (R/2)^2 + (15.sqrt(3))^2, giving (3/4).R^2 = 675, then R^2 = 900 and R = 30. The dark green area is the same as the white area inside the main circle. We use an orthonormal center O, first axis (OH); In this orthonormal the equation of the main circle is x^2+ y^2 = 30^2, or y = sqrt(900 - x^2) for the semi circle above the axis (OH) The unknown area is then 2 times the integral from R/2 = 15 to R =30 of sqrt(900 - x^2).dx Let's now use t such as x = 30.sin(t), dt=x = 30.cos(t).dt I do not copy the calculus, it's difficult with the computer and without special characters, but it is simple on a paper with some knowledge on trigonometry. We naturally get the same result than Premaths.
Draw a radius to point C such that OC is perpendicular to AB. As OC is a radius and AB is a chord, OC bisects AB. Let that bisection point be labeled D. Therefore BD = DA = AB/2. As the dark green circular segment is folded over from the circumference, AB also bisects OC, thus CD = OD = OC/2 and OD = r/2. Draw OA and OB. As both are radii, ∆BDO and ∆ODA are congruent triangles. Triangle ∆BDO: OD² + BD² = OB² (r/2)² + (30√3/2)² = r² r²/4 + (15√3)² = r² 3r²/4 = 225(3) = 675 r² = 675(4/3) = 4(225) r = √900 = 30 O/H = sin θ BD/OB = sin θ sin θ = 15√3/30 = √3/2 θ = arcsin(√3/2) = 60° As dark green circular segment AOB is folded over from circular segment BCA, they are congruent. Therefore area of segment AOB is equal to sector encompassed by minor arc BA minus the isosceles triangle ∆AOB. Circular segment AOB: A = (2θ/360)πr² - bh/2 A = (2(60)/360)π(30²) - 30√3(15)/2 A = 900π/3 - 225√3 A = 300π - 225√3
Let's do some math: . .. ... .... ..... Let C be the midpoint of AB. The triangle OAB is a isosceles triangle, therefore we know that the triangles OAC and OBC must be congruent right triangles. Since the dark green area and the white area are also congruent, we can conclude that OC=R/2 with R being the radius of the full circle. We also known that OA=OB=R, so the triangles OAC and OBC are both 30°-60°-90°-triangles and we can conclude that: ∠AOB = ∠AOC + ∠BOC = 60° + 60° = 120° Now we apply the Pythagorean theorem to calculate the radius of the full circle: OA² = AC² + OC² R² = (AB/2)² + (R/2)² R² = AB²/4 + R²/4 3*R²/4 = AB²/4 R² = AB²/3 = (30√3)²/3 = 900 ⇒ R = 30 Finally we are able to calculate the size of the dark green area: A(dark green) = A(white) = A(circle sector OAB) − A(triangle OAB) = π*R²*(∠AOB/360°) − (1/2)*AB*OC = π*R²*(∠AOB/360°) − (1/2)*AB*(R/2) = π*30²*(120°/360°) − (1/2)*(30√3)*(30/2) = 300π − 225√3 ≈ 552.77 Best regards from Germany
It looks like most every commenter also found the result! YAY! Finding (𝒓) radius isn't too bad [1.1] (½𝒓)² + (½ 30√3)² = 𝒓² … expand [1.2] ¼𝒓² + 675 = 𝒓² … move 𝒓 [1.3] 675 = ¾𝒓² … multiply by ⁴⁄₃ both sides [1.4] 900 = 𝒓² … square root to find [1.5] 30 = 𝒓 Wasn't so crazy. The maker of this problem conveniently made the LENS below the 30√(3) axis a folded reflection of the lens above the axis, so the symmetry implies that finding the area of the lens ABOVE the 30√(3) axis is the desired area solution. Area of lenses? [2.1] lens = area-of-arc - area-of-△ inside [3.1] area-of-arc = ½θ𝒓² [3.2] area-of-△ = ½ base⋅height Ah, hyesss… so what might these θ thetas and bases and heights be? [4.1] base = 30√3 … given [4.2] height = ½𝒓 = ½ 30 = 15 [5.1] θ = 2 arcsin( hypotenuse / opposite ) [5.2] opposite = 15√3 [5.3] hypotenuse = 𝒓 = 30 (→ 2 × 15) [5.4] θ = 2 arcsin( (2 × 15) ÷ 15√3 ) … 15 cancels [5.5] θ = 2 arcsin( 2 / √3 ) … ah HA! a 1-2-√3 △ = 30-60-90 [5.6] θ = 2 × 60° [5.7] θ = 120° … and put into radians [5.8] θ = 120° ÷ 360° × 2π ... what fraction of a full circle? [5.9] θ = ⅔π Now to finish up [2.1] (...above...) [2.2] lens = (arc = ½ ⅔π30²) - (△ = ½ (30√(3) 15)) [2.3] lens = 300π - 225√(3) [2.4] lens = 75(4π - 3√(3)) [2.5] lens = 552.77 And that'd be the answer we were looking for. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅
M es punto medio de AB Potencia de M respecto a la circunferencia =(r/2)(r/2 +r)=(30√3 /2)²→r=30→ OM=r/2=15 ; MA=15√3 ; OA=30→ Ángulo AOM=60º→ Área verde oscuro =Área segmento circular ángulo central 120º, cuerda AB y flecha r/2 =(30²π/3)-(15*15√3) =300π -15²√3 =552,7663. Gracias por el vídeo. Un saludo cordial.
Here I am to propose my Resolution Method. FINDING THE RADIUS (R) OF THE GIVEN CIRCLE 1) R^2 - R^2/4 = [15*sqrt(3)] ; R^2 - R^2/4 = 225 * 3 ; R^2 - R^2/4 = 675 ; 4R^2/4 - R^2/4 = 2.700/4 ; 3R^2/4 = 2.700/4 ; 3R^2 = 2.700 ; R^2 = 2.700/3 ; R^2 = 900 ; R = sqrt(900) ; R = 30 2) OA = OB = 30 lin un Let's call the Middle Point between AB, "C" FINDING THE ANGLE AOC (alpha) 1) Angle AOC it's half of angle AOB 2) sin (alpha) = 15*sqrt(3)/30 = sqrt(3)/2 3) arcsin (sqrt(3)/2) = 60º 4) Angle alpha = 60º 5) Angle AOB = 2 * 60º = 120º CALCULATING THE AREA OF CIRCLE SLICE AOB 1) The Area of the Circle is 30^2 * Pi = (900 * Pi) sq un ~ 2.827,433 sq un 2) The Circle Slice is equal to 900Pi/3 = 300Pi ~ 942,478 sq un CALCULATING DARK GREEN FOLDED AREA 1) Area of Triangle AOB = (30*sqrt(3)) * 15 / 2 = 450 * sqrt(3) / 2 = 225 * sqrt(3) ~ 389,711 sq un 2) Subtracting 300Pi - 225*sqrt(3) = 942,478 - 389,711 ~ 552,766 sq un Answer: The Dark Green Folded Area is equal to approx. 552,8 Square Units.
Sir, how you can claim that right side white segment is folded leftwise and both or identical ??? Chord may be in a position displaced from centre....it is just a supposition by you that it is in centre.
👇Respect button for premath
Thanks dear ❤️🌹
At a quick glance, A is half AB vertical and horizontal from O. The tangents at A and B are 45. The radius is then sqrt((3/4 *30^2 * 2) = 36.74. Angle OAB = 90. The area of the circle is PI * 36.74^2 = 1350 * PI. Angle from A to B is 360 - 270 = 90 and area of circle segment OAB= 1/4 *1350 * PI. Area of triangle AOB = 0.5 *30 *sqrt(3) * 0.5 *36.74 = 477.27. Drawing a horizontal line from O , intersecting the circumference at C . The area of OAC = 0.5 * 36.74 * 0.5 * 30*sqrt(3) =477 .27 . then the difference of the circle segment and triangle segment is segment area of of white shaded area in semicircle = area of green shaded area in semicircle = 1/8 * 1350 * PI - 477.27 = 52.87. Then area of quarter circle segment = 1/4 * 1350 * PI =1060.3 . Then 1060.3 + 52.87 =556 area units.
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1/ Notice that the triangle AOC is an equilateral one, so OA.sqrt3/2 =AB/2---> R=30
2/ The area of the dark green = the area of the 120 degree white segment
= 1/3 pi sqR - (sqR. sqrt3)/4= 300 pi - 225.sqrt3 sq units
Excellent!
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Since it was obtained by folding along the chord AB, the dark green region is congruent with the circular segment AB.
Draw the diameter DC bisecting AB at P. Since OD = OC is twice PO = PC then ABD must be an equilateral triangle.
So the angle AOB at the center is 2 × 60° = 120° = ⅔ π rad = θ, sin θ = √3 / 2 and the radius of the circumcircle is 30.
Then the area of the circular segment is ½ 30² (⅔ π − √3 / 2) which simplifies to 75 (4π − 3√3) ≈ 552.77 square units.
Excellent!
Thanks ❤️
Keep folding the DARK GREEN SHADED REGION until it becomes the deepest shade of Black Green...the ultimate color of India's Pepper Plant. The first time I seen a garden full of this flowering vine, I just can't get enough of it and really could stare at it for hours. 🙂
😀
Thanks ❤️
😀
😅
This time, I used trigonometry to find the angle.
Since r= 30 is the hypotenuse and r/2= 15 is the adjacent,
cos x= 15/30
cos x = ½
cos^-1 (x) = 60°
Reflect the light green region across chord AB. This forms a new circle. They overlap in the middle.
Let's name the center of this circle point P.
The radius OP is shared by both circles, so they have the same radius.
Draw radii AO, AP, BO, & BP. This forms a rhombus AOBP as the radii are congruent.
By the Parallelogram & Rhombus Diagonals Theorems, the diagonals of rhombus AOBP bisect each other and are perpendicular. Use the Intersecting Chords Theorem.
(3r/2)(r/2) = (15√3)(15√3)
3r²/4 = 675
3r² = 2700
r² = 900
r = 30
Label the intersection of the diagonals of rhombus AOBP as point M.
Since one of the diagonals is the radius of both circles, and chord AB bisects it, MP = 15. Since AM = 15√3, △AMP is a special 30°-60°-90° right triangle.
Therefore, m∠APM = 60°. By the Rhombus Opposite Angles Theorem, each diagonal bisects a pair of opposite angles.
Thus, m∠APB = 120°.
Now, notice the dark green shaded region is a segment of circle P. The central angle measure is in degrees.
Minor Segment AB Area = Sector PAB Area - △APB Area
A₁ = (θ/360°)πr²
= (120°/360°) * π * 30²
= 1/3 * π * 900
= 300π
A₂ = 1/2 * a * b * sinC
= 1/2 * 30 * 30 * sin(120°)
= 450 * sin(120°)
= 450 * (√3)/2
= (450√3)/2
= 225√3
Dark Green Shaded Region = 300π - 225√3
So, the area of the dark green shaded region is 300π - 225√3 square units, a.k.a. 75(4π - 3√3) square units (exact), or about 552.77 square units (approximation).
Bravo!
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First we calculate the radius R of the main circle with the Pythagorean theorem in triangle OHA (with H the middle of [A,B]
We have OA^2 = OH^2 = HA^2, so R^2 = (R/2)^2 + (15.sqrt(3))^2, giving (3/4).R^2 = 675, then R^2 = 900 and R = 30.
The dark green area is the same as the white area inside the main circle. We use an orthonormal center O, first axis (OH);
In this orthonormal the equation of the main circle is x^2+ y^2 = 30^2, or y = sqrt(900 - x^2) for the semi circle above the axis (OH)
The unknown area is then 2 times the integral from R/2 = 15 to R =30 of sqrt(900 - x^2).dx Let's now use t such as x = 30.sin(t), dt=x = 30.cos(t).dt
I do not copy the calculus, it's difficult with the computer and without special characters, but it is simple on a paper with some knowledge on trigonometry.
We naturally get the same result than Premaths.
Thanks ❤️
Thank you! Required thought on how to define the dark green area. Good challenge!
Glad it was helpful!
You are very welcome!
Thanks ❤️
Area of the dark green shaded region=300π-225√3=552.8 square units.❤❤❤ Thanks sir.
Great job
You are very welcome!
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you also have height and base for triangle AOB (30sqrt(3) x 15)/2 = 225sqrt(3)
Draw a radius to point C such that OC is perpendicular to AB. As OC is a radius and AB is a chord, OC bisects AB. Let that bisection point be labeled D. Therefore BD = DA = AB/2. As the dark green circular segment is folded over from the circumference, AB also bisects OC, thus CD = OD = OC/2 and OD = r/2.
Draw OA and OB. As both are radii, ∆BDO and ∆ODA are congruent triangles.
Triangle ∆BDO:
OD² + BD² = OB²
(r/2)² + (30√3/2)² = r²
r²/4 + (15√3)² = r²
3r²/4 = 225(3) = 675
r² = 675(4/3) = 4(225)
r = √900 = 30
O/H = sin θ
BD/OB = sin θ
sin θ = 15√3/30 = √3/2
θ = arcsin(√3/2) = 60°
As dark green circular segment AOB is folded over from circular segment BCA, they are congruent. Therefore area of segment AOB is equal to sector encompassed by minor arc BA minus the isosceles triangle ∆AOB.
Circular segment AOB:
A = (2θ/360)πr² - bh/2
A = (2(60)/360)π(30²) - 30√3(15)/2
A = 900π/3 - 225√3
A = 300π - 225√3
Thanks ❤️
30√3=270° (180°-270°)=√90° 3^√30 3^√5√,6✓√3^1^3^2 √1^√13^2 (x+2x-3)
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Let's do some math:
.
..
...
....
.....
Let C be the midpoint of AB. The triangle OAB is a isosceles triangle, therefore we know that the triangles OAC and OBC must be congruent right triangles. Since the dark green area and the white area are also congruent, we can conclude that OC=R/2 with R being the radius of the full circle. We also known that OA=OB=R, so the triangles OAC and OBC are both 30°-60°-90°-triangles and we can conclude that:
∠AOB = ∠AOC + ∠BOC = 60° + 60° = 120°
Now we apply the Pythagorean theorem to calculate the radius of the full circle:
OA² = AC² + OC²
R² = (AB/2)² + (R/2)²
R² = AB²/4 + R²/4
3*R²/4 = AB²/4
R² = AB²/3 = (30√3)²/3 = 900
⇒ R = 30
Finally we are able to calculate the size of the dark green area:
A(dark green)
= A(white)
= A(circle sector OAB) − A(triangle OAB)
= π*R²*(∠AOB/360°) − (1/2)*AB*OC
= π*R²*(∠AOB/360°) − (1/2)*AB*(R/2)
= π*30²*(120°/360°) − (1/2)*(30√3)*(30/2)
= 300π − 225√3
≈ 552.77
Best regards from Germany
Excellent!
Thanks ❤️
What about outside of the traingle.
1:30 "And furthermore since these two regions are identical..." But are they? What is the proof that they are?
The white area is folded along AB to give the dark green area. So both areas must be congruent.
@@PreMath This is the supposition, but at school I would have been required to prove it, or to have it stated in the opening conditions.
@@Ensign_CthulhuIf you stick a needle through a folded piece of paper, when unfolded, the holes will always be an equal distance from the crease.
300pi-225sqrt(3)
(4π/3 - √3)(15)^2
Thanks ❤️
It looks like most every commenter also found the result! YAY!
Finding (𝒓) radius isn't too bad
[1.1] (½𝒓)² + (½ 30√3)² = 𝒓² … expand
[1.2] ¼𝒓² + 675 = 𝒓² … move 𝒓
[1.3] 675 = ¾𝒓² … multiply by ⁴⁄₃ both sides
[1.4] 900 = 𝒓² … square root to find
[1.5] 30 = 𝒓
Wasn't so crazy. The maker of this problem conveniently made the LENS below the 30√(3) axis a folded reflection of the lens above the axis, so the symmetry implies that finding the area of the lens ABOVE the 30√(3) axis is the desired area solution.
Area of lenses?
[2.1] lens = area-of-arc - area-of-△ inside
[3.1] area-of-arc = ½θ𝒓²
[3.2] area-of-△ = ½ base⋅height
Ah, hyesss… so what might these θ thetas and bases and heights be?
[4.1] base = 30√3 … given
[4.2] height = ½𝒓 = ½ 30 = 15
[5.1] θ = 2 arcsin( hypotenuse / opposite )
[5.2] opposite = 15√3
[5.3] hypotenuse = 𝒓 = 30 (→ 2 × 15)
[5.4] θ = 2 arcsin( (2 × 15) ÷ 15√3 ) … 15 cancels
[5.5] θ = 2 arcsin( 2 / √3 ) … ah HA! a 1-2-√3 △ = 30-60-90
[5.6] θ = 2 × 60°
[5.7] θ = 120° … and put into radians
[5.8] θ = 120° ÷ 360° × 2π ... what fraction of a full circle?
[5.9] θ = ⅔π
Now to finish up
[2.1] (...above...)
[2.2] lens = (arc = ½ ⅔π30²) - (△ = ½ (30√(3) 15))
[2.3] lens = 300π - 225√(3)
[2.4] lens = 75(4π - 3√(3))
[2.5] lens = 552.77
And that'd be the answer we were looking for.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
Excellent!
Thanks for the nice feedback❤️
Calcolo r,tramite r^2-(r/2)^2=(15√3)^2...r=30..io ho utilizzato gli integrali Adg=r^2(π/3-√3/4)
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M es punto medio de AB
Potencia de M respecto a la circunferencia =(r/2)(r/2 +r)=(30√3 /2)²→r=30→ OM=r/2=15 ; MA=15√3 ; OA=30→ Ángulo AOM=60º→ Área verde oscuro =Área segmento circular ángulo central 120º, cuerda AB y flecha r/2 =(30²π/3)-(15*15√3) =300π -15²√3 =552,7663.
Gracias por el vídeo. Un saludo cordial.
Thanks ❤️
Yes!
Why you say at 1:29 this 2 regions are identicale ? what evidence ?
Becoz he folded the circle
@rasingh-oe9tj ok I hadn't listened the first seconds ( sorry for my english) Thx .
Here I am to propose my Resolution Method.
FINDING THE RADIUS (R) OF THE GIVEN CIRCLE
1) R^2 - R^2/4 = [15*sqrt(3)] ; R^2 - R^2/4 = 225 * 3 ; R^2 - R^2/4 = 675 ; 4R^2/4 - R^2/4 = 2.700/4 ; 3R^2/4 = 2.700/4 ; 3R^2 = 2.700 ; R^2 = 2.700/3 ; R^2 = 900 ; R = sqrt(900) ; R = 30
2) OA = OB = 30 lin un
Let's call the Middle Point between AB, "C"
FINDING THE ANGLE AOC (alpha)
1) Angle AOC it's half of angle AOB
2) sin (alpha) = 15*sqrt(3)/30 = sqrt(3)/2
3) arcsin (sqrt(3)/2) = 60º
4) Angle alpha = 60º
5) Angle AOB = 2 * 60º = 120º
CALCULATING THE AREA OF CIRCLE SLICE AOB
1) The Area of the Circle is 30^2 * Pi = (900 * Pi) sq un ~ 2.827,433 sq un
2) The Circle Slice is equal to 900Pi/3 = 300Pi ~ 942,478 sq un
CALCULATING DARK GREEN FOLDED AREA
1) Area of Triangle AOB = (30*sqrt(3)) * 15 / 2 = 450 * sqrt(3) / 2 = 225 * sqrt(3) ~ 389,711 sq un
2) Subtracting 300Pi - 225*sqrt(3) = 942,478 - 389,711 ~ 552,766 sq un
Answer:
The Dark Green Folded Area is equal to approx. 552,8 Square Units.
Excellent!
Thanks ❤️
How can we say the length OP = PC
The white area is folded along AB to give the dark green area. So both areas must be congruent.
Best regards from Germany
No logic in equating op=pc
FOLDED is the key word when setting up the problem. Otherwise, you are correct
Sir, how you can claim that right side white segment is folded leftwise and both or identical ??? Chord may be in a position displaced from centre....it is just a supposition by you that it is in centre.
Anyone CREATING a problem sets stipulations. This one stipulated that it was a circle with the right side folded back until the edge is at the center