Find Area Of Circle Inscribed in An Equilateral Triangle with side length 10 | 2 methods

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[murdock]

Nice! ∆ QBO → φ = BOQ → sin⁡(φ) = 1/2 =(5√3/3)(10√3/3) → r = √(25/3) →
area circle = 25π/3

[bikashmohanty]

Good morning sir.your all videos are very interesting and helpful to enhance our knowledge. Please put more videos like this. Warm regards 🙏🙏.

[sambhavbhalla]

ayee nice method
i did this personally by using solutions of triangle
radius of incentre = area of big triangle/semiperimeter

[marioalb]

Height of triangle:
H = 10. cos 30° = 8, 66 cm

Radious of circle (Inside equilateral triangle):
R = H/3 = 2, 8867 cm

Area = π R²
Area = 26, 18 cm² ( Solved √ )

[alricmcclarty]

I used trigonometry. Seeing that it is an equilateral triangle, all lines drawn from the vertex of the triangle to centre O bisects all 3 angles of the triangles making the angles 30 degrees since it's an equilateral triangle with all 3 angles being 60. I then ended up with a right angle triangle QOB with length QB= 5 and angle OQB= 30 degree. OB is the radius which would be opposite to the 30 degree angle and QB is adjacent to the angle so that would be the tan ratio because we have the value of the adjacent side and want to find the length of the opposite SOH CAH TOA.
tan30= OB/5
5tan30= OB
Therefore area of circle = π(5tan30)^2= 26.18

[solimana-soli]

in an equilateral triangle it can easily be shown that its height is three times the radius of the inscribed circle. And the height is half the side moltiplied the square root of three. With these data it is easy to arrive at the solution

[adgfx]

25(3^1/2)=15R=>R=5/3^0.5.Area of inscribed circle=25.pai/3 .where pai =22/7

[adgfx]

Area of inscribed circle=25.pi/3 unit when pi=22/7.(Radius =5/3^0.5)okay?

[KamleshKumar-yxog]

Equilateral triangle is given
So, tan 30=r/5

[marioalb]

tan α = R / (½S)
R = ½S tan α
R = 5 tan 30°
R = 2, 8867 cm

Area = π R²
Area = 26, 18 cm² ( Solved √ )