Can you find area of the Yellow Circle? | (Equilateral Triangle) |
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- Опубліковано 30 вер 2024
- Learn how to find the area of the Yellow Circle. Area of the Equilateral Triangle is known. Important Geometry and Algebra skills are also explained: circle theorem; area of the circle formula; area of the triangle formula. Step-by-step tutorial by PreMath.com
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@@PreMath ☺
Great solution. Can you tell me what program you use to record your video lesson?
Nice!
At 4:40, I would have connected triangle AOD, and applied the 30-60-90 factors directly to get the radius.
Instead of 2 triangles AOC & OCD, 1 triangle is faster.
One triangle faster can make all the difference in samosa eating competition 😜
Thanks ❤️
AOD is Rigth triangle (30°; 60°; 90°)
AO=2
So AD=2/2=1 units
AO^2=OD^2+AD^2
2^2=r^2+1^2
So r=√3 unitd.
Yellow circle area=π(√3)^2=3π square units =9.42 square units.❤❤❤thanks sir.
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You are very welcome!
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1/ Let a and r be the side ò the equilateral tríangle and the radius of the circle.
We have: the area of the trisngle = (sqa. sqrt3)/4=sqrt48=sqrt (16x3)--> a=4
2/THe height OC=a.sqrt3/2 = 4sqrt3/2= 2sqrt3
3/ OE= r = 0C/2 ( the triabgle OEC is a special 30-60-90 triangle
r= sqrt3
Area of the circle = 3pi sq units
Excellent!
Thanks for sharing ❤️
The area of an equilateral triangle is a² √3 / 4. So, in our case, a = √(4√48 / √3) = 4.
Then OAD is a 30° - 60° - 90° right triangle whose hypotenuse is 2 so r = √3.
Therefore, the area of the yellow circle is 3π ≈ 9.42 square units.
Thank you PreMath! 🙏
Excellent!
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Very nice writeup.
... I wish the whole instructive " PreMath " channel members a nice and peaceful Easter ... your strategy is always so clear, and personally have nothing significant to add this time around .... thank you sir for all time invested posting your videos .... best regards, Jan-W
So nice of you🌹
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φ = 30°; AO = BO = a; area ∆ ABC = 4√3 = (a^2 )√3 → a = 2 → CO = a√3 → sin(φ) = 1/2 = r/a√3 →
r = √3 → area yellow circle = 3π
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Need do some computation, but not difficult, 1/2 s^2 sqrt(3)/2=sqrt(48), s^2=4sqrt(48/3)=4sqrt(16)=16, s=4, r=s/2 sin 60=2 sqrt(3)/2=sqrt(3), Then the area of the circle is 3 pi.😂😂😂
Excellent!
Thanks ❤️
a solution without trig:
1. side length of triangle ABC = a= 4;
2. AO = a/2 = 2;
3. AC^2=AO^2+OC^2; 4^2 = 2^2 +OC^2; => OC = sqrt (4^2-2^2)= 2*sqrt(3);
4. Area of triangle AOC = AO*OC/2 = AC*OD/2 => 2*2*sqrt(3)/2=4*r/2 => r= sqrt(3);
5. Acirc = pi*r^2 = 3*pi.
Excellent!
Thanks ❤️
Happy easter to the whole PreMath community. And now let's face this challenge:
.
..
...
....
.....
With s being the side length of the equilateral triangle ABC its area turns out to be:
A(ABC) = √3*s²/4
√48 = √3*s²/4
4√3 = √3*s²/4
16 = s²
⇒ s = 4
Since AC and BC are tangents to the circle, we know:
∠ODA = ∠OEB = 90°
∠AOD = 180° − ∠OAD − ∠ODA = 180° − 60° − 90° = 30°
∠BOE = 180° − ∠OBE − ∠OEB = 180° − 60° − 90° = 30°
Therefore the triangles OAD and OBE are 30°-60°-90° triangles and we can conclude:
OD/OA = OE/OB = √3/2
So with r being the radius of the circle we obtain:
r/(s/2) = √3/2
2*r/s = √3/2
⇒ r = (√3/2)*s/2 = (√3/2)*4/2 = √3
Finally the area of the yellow circle turns out to be:
A(circle) = πr² = 3π
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Happy Easter!
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Triangle ∆ABC:
A = √3s²/4
√48 = (√3/4)s²
s² = (4/√3)4√3 = 16
s = √16 = 4
Draw radius OE. As CB is tangent to circle O and OE is a radius, ∠OEB = 90°. As ∠EBO = 60°, as the vertex of an equilateral triangle, ∠BOE must be 30°, and ∆OEB is a 30-60-90 special right triangle. If EB = x, then BO = 2x and OE = √3x.
Triangle ∆OEB:
BO = 2x
2 = 2x
x = 1
OE = √3x = √3 = r
Circle O:
A = πr² = π(√3)² = 3π
Excellent!
Thanks for sharing ❤️
rting from 6:40 min: The area of AOC triangle is 1/2 of a big triangle, OD is a radius, area is a times R/2=sqroot 48/2, aR= sqroot 48. So, R= sqroo 48/a. No need to consider angles.
Thanks for sharing ❤️
Dai dati risulta h(altezza)=2√3 b(base)=4...l=4...R=bh/2l=√3...Ay=π(√3)^2=3π
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I absolutely love it! 😉
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We don't know any of the side lengths of △ABC. But because the triangle is equilateral, the sides are all equal and we can use a different area formula:
A = [(√3)/4]s²
√48 = [(√3)/4]s²
√(2 * 2 * 2 * 2 * 3) = [(√3)/4]s²
4√3 = [(√3)/4]s²
s² = 4√3 * [4/(√3)]
s² = (16√3)/(√3)
s² = 16
s = 4
Draw altitude CO. This forms △AOC & △BOC, two congruent special 30°-60°-90° right triangles. Line CO is the line of symmetry. We'll use △AOC.
Since side AC is the hypotenuse of △AOC, AO = 4/2 = 2, & CO = 2√3.
Draw radius DO. By the Circle Theorem, ∠ADO is a right angle.
Since segment AO is the hypotenuse of the newly formed △ADO, another special 30°-60°-90° right triangle, AD = 2/2 = 1, & DO = √3. But notice we now have the radius of ⊙O.
A = πr²
= π(√3)²
= 3π
So, the area of the yellow circle is 3π square units (exact), or about 9.42 square units (approximation).
Excellent!
Thanks ❤️
I've tried to modify the figure drawing a rhombus, (overturning the equilateral triangle on the low side) getting a socalled tangential quadrilateral. The radius of the inscribed circle is = area/semiperimeter
Once found the side of the triangle with formula Area = s²/4*√ 3 that is s = 4, we can write:
r = 2*√ 48/4*2 = √ 3 (in a rhombus the semiperimeter is twice its side)
then yellow area = 3 pi
💡The formula _r = A / s_ applies to any tangential polygon where _A_ is the area of the polygon and _s_ is the semiperimeter.
Thanks for sharing ❤️
My way of solution ▶
A(ABC)= √48
A(ABC)= √16*3
A(ABC)= 4√3 square units
the area of an equilateral triangle: a²√3/4
⇒
4√3 = a²√3/4
a²= 16
a= 4
if we consider the triangle OBE:
∠ BEO= 90°
∠ EOB= 30°
∠ OBE= 60°
sin(OBE)= OE/OB
OE= r
OB= a/2
OB= 2
⇒
sin(60°)= r/2
√3/2= r/2
r= √3 length units
Aye= πr²
Aye= π*(√3)²
Aye= 3π
Aye ≈ 9,425 square units
b) 2 nd way:
the area of the triangle A(OBC)= 1/2 A(ABC)
A(ABC)= 4√3
A(OBC)= 2√3 square units
A(OBC)= BC*OE/2
BC= a
BC= 4
OE= r
⇒
2√3 = 4*r/2
4√3 =4r
r= √3
Aye= πr²
Aye= π*(√3)²
Aye= 3π
Aye ≈ 9,425 square units !
Excellent!
Thanks for sharing ❤️
(AC)^2*cos(60)sin(60)=sqrt(48)
(AC)^2*sin(120)/2=sqrt(48)=4sqrt(3)
(AC)^2=16
AC=4
AO=2
OD=2sin(60)=sqrt(3)
A=3pi
Thanks for sharing ❤️
As per diagram, with the help of trignometry you can solve this mensuration problem. Means you have to deep knowledge of Trignometry and Mensuration to solve these types of typical questions of mathematics. These types of questions asked in govt. service examination.
Excellent!
Glad to hear that!
Thanks Manoj dear❤️
The area of an equilateral triangle which side length is c is ((c^2).sqrt(3))/4. Here this area is sqrt(48) = 4.sqrt(3), so c^2 = 16 and c = 4.
Let's use now an orthonormal, center O, first axis (OA). Then we have O(0;0) A (2; 0) B(-2; 0) C(0; 2.sqrt(3)) (the hight of the equilateral triangle is
(sqrt(3)/2).c
VectorBC(2; 2.sqrt3)) and the equation of (BC) is: (x +2).(2.sqrt(3) - (y).(2) = 0 or 2.sqrt(3).x -2.y +4.sqrt(3) = 0
The radius of the circle is the distance from O to (BC), it is: abs(2.sqrt(3).0 -2.0 +4.sqrt(3)) / ((2.sqrt(3))^2 + (-2)^2) = 4.sqrt(3)/ 4 = sqrt(3)
And finally the area of the circle is 3.Pi
Le dernier caractère de la première ligne devrait être "4" 😉
@@marcgriselhubert3915 vous pouvez modifier votre commentaire en anglais (et puis nous effacerons notre conversation en français) 😇
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@@marcgriselhubert3915 Lorsque le curseur de la souris se trouve au dessus d'un commentaire, trois points verticaux apparaissent à droite de ce commentaire.
Si vous cliquez sur ces trois points, l'option "signaler" apparaît. Mais si il s'agit de vos propres commentaires, les options "modifier" ou "supprimer" apparaissent.
The height of the triangle is h
The side of the triangle is 2h/✓3
✓48 = h (2h /✓3) ÷ 2
h^2 = ✓144
h = ✓12 = 2✓3
radius = h/2 = ✓3
circle area = (✓3)(✓3)π = 3π
Excellent!
Thanks for sharing ❤️
@@PreMath Thank you for all the fun math puzzles 🙏
Thanks sir i solved it too😊
.......... AB=4
∆ AOD :
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The professor has nicely worked out.
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Glad to hear that!
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Nice! Enjoyed this problem.
Glad you enjoyed it!
Thanks ❤️
1) The Area of Equilateral Triangle [ABC] = sqrt(48) = 4*sqrt(3)
2) Using Herons Formula to find the Side of Equilateral Triangle [ABC]
3) sqrt(3a * a * a * a) / 4 = 4*sqrt(3) ; sqrt(3)*sqrt(a^4) = 16*sqrt(3) ; a^2 = 16 ; a = 4
4) OB = 2
5) OE = R (Radius)
6) EB = unknown
7) Triangle [OBE] is a Rigth Triangle with Angles (30 - 60 - 90); Angle OBE = 60º ; Angle OEB = 90º and Angle BOE = 30º
8) Cos(30º) = R/2 ; sqrt(3)/2 = R/2 ; R = sqrt(3)
9) Yellow Circle Area = (sqrt(3))^2 * Pi = 3Pi
10) Answer: The Yellow Circle Area is Equal to 3Pi Square Units or Approx. Equal to 9,425 Square Units.
Oh My, You could cut this video shortly: 5'.
Equilateral ∆ = (1/2)•a•a•sin60° = (1/4)√3 • a" = √48
So a" = 16 ! (square of a)
Then, ∆ BOC ---> sin60° = h / a
Then, ∆ OCD ---> sin30° = r / h
Multiply it, we get:
(r / h) × (h / a) = sin60° × sin30°
r = (1/4)•a•√3
Circle O = π•r" = π•(1/16)•16•(3) = 3π = 9,4248
Much-much Easier & Shortcut
Thumbs Up for You, anyway^ 👍
29/3/24_6am_North Sumatra
Very nice video ! i did`nt read equilateral triangle only area = square ( 48 ) . My fault . Otherwise I could find a solution for this
problem.But may be is it possible to solve this problem without provided that the triangle is equilateral ?
I have a new method to solve it
Thanks ❤️
First view
Thanks dear❤️
16×(22/7) or 16× 3.14=50.24
3pi
Excellent!
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This one is too easy !
2(30°)=60° 2(30°)=60° 2(30°)=60° {60°+60°+60°}=180° {360°AR/180°}=2 (x+2x-2)
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👍❤✔️❤❤
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16πsq.u