Find area of Green shaded circle inscribed in an equilateral triangle with side length 8 | 2 methods

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Find area of Green shaded circle inscribed in an equilateral triangle with side length 8 | 2 methods

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[pwmiles]

You can also use the fact that the lines between the vertices and the mid-points of the opposite sides divide each other in the ratio 2:1. In this case the line lengths are 4 sqrt(3) so r = 4/sqrt(3)

[ConorChewy]

I approached this slightly differently.
Where you made the triangle AOD, I observed that the total area of the whole triangle is made up of six of these identical triangles; AOD, AFO, BOD, BOE, COF, COE.
Each one has an area equal to 0.5×4×r = 2r, and thus the total area of ABC is 12r.
So then you can easily find the area of ABC via calculating the height CD, set the two exoressions for the area of ABC as equal to each other, rearrange for r, and then calculate the circle area.

[ybodoN]

The general formula to find the inradius of the incircle in any triangle also works.
In this case, it simplifies to _a √3 / 6_ where _a_ is the side of the equilateral triangle.
Thus, the formula for the area of the incircle in an equilateral triangle is _π a² / 12._

[ОльгаСоломашенко-ьы]

Можно воспользоваться теоремой о биссектрисах треугольника. Делится точкой пересечения в отношении 2:1, считая от вершины. Поделить высоту, которая является биссектрисой на 3. Это и будет радиус вписанной окружности.

[notovny]

Clone triangle ABC, rotate it 180° around O. Now you've got an regular hexagram, with six equilateral triangles surrounding a regular hexagon, all with side length 8/3, and the hexagon has the green circle inscribed. Radius of the circle is equal to the minimal radius of the hexagon, equals 4/sqrt(3) in this case. Formula for area of a circle resulted in 16 * pi / 3

[phungpham]

In an equilateral triangle, the height is also the median, and the bisector of the the 3 corner angles. So r=OB=CD /3=1/3 x 8sqrt3/2=4sqrt3/3
Area of the inscribed circle= pi .( 16x3)/9= 16pi /3 sq units

[dushyantprajapati]

we just can calculate the area of triangle first and then divide this triangle in three part from center and height of each part will be radius and side will base and just using formula area(it will be 1/3 of that) 1/2 * base * height we can get r easily
(please reply sir)

[sandanadurair]

h =4sqrt(3) using pythagoras.

h.(h-2r) = 4.4
As per Tangent secant rule
h^2-2hr = 16
48-8sqrt(4)r = 16
r = 4/sqrt(3)
A = 16pi/3

[vidyadharjoshi]

DE = EF = FD = 4. Extend DO till edge of circle at H via G. GH*HD = 2*2 = 4. GH = 4/sqrt12. So Area = 16.755

[bigm]

Very elegant solutions. Thanks Professor!❤🥂

[santiagoarosam]

El centro del círculo coincide con el baricentro del triángulo equilátero y dista de la base un tercio de la altura “h” → h=(8√3)/2 =4√3 → Radio r=h/3=(4√3)/3 → Área verde =πr² =π16x3/9 =π16/3 =16.7552
Gracias y un saludo cordial.

[saltydog]

As the triangle is equilateral, the triangle FOA will have the angles 30:60:90. The hypotenuse of such a triangle is 2 times the shortest side. The shortest side is the radius of the circle, so the length OA is 2r. This makes the whole length of the line AOE 3r which is the height of the triangle. Using Pythagoras, the height of the triangle is also the sq root of (8^2 - 4^2) = sq root of (64-16) =the sq root of 48. So the radius is 1 third of the square root of 48. The area is pi * r^2 so the area of the circle is 48/9*pi.

[prabhagupta]

Find area through heron's formula of ABC --(1)
Also, area of ABC=r×(8+8+8)/2--(2)
Equating (1) &(2)
you will get r

[solimana-soli]

The height of the equilateral triangle is three times the radius. Considering the height AE and the triangle AOD which has angles of 30°, 60° and 90°, therefore the hypotenuse is twice the minor cathetus. But the minor cathetus OD is the radius, so along the height AE we have AO which is twice the radius and OE which is the radius

[mahatmapodge]

The solution can be intuited from similar triangles

ACD and AOD are similar

therefore

Tan 30 == AD/CD == OD/AD

r = OD = 4/4 * (4sqrt(3))

[MarieAnne.]

Equilateral triangle ABC has area = 1/2 × 8 × 8 × cos(60°) = 8√3
Each of triangles AOB, BOC, and AOC have base = 8, height = r, and area = 1/2 × 8 × r = 4r.
So area of triangle ABC = 3 × 4r = 12r = 8√3 → r = 4/√3
Area of circle = πr² = 16π/3

[hkulachandrasingha]

The centre O of the inscribed circle is, also the centroid of the equilateral triangle ABC which divides CB in 2:1. So, OD=1/3 of CB=1/3 of 4√3.

[Ravis_funn_study]

Angle(EBD) =60°

angle(EOD) =120°

Angle(EBO) =30°

EB=4cm.

tan30°=1/√3 =r/4

r=4/√3

Green 🟢=π(4/√3)^2
=16π/3 unit

[rycvksh]

You can make a 120 degree sector by drawing lines through the centre to 2 of the midpoints and use cosine rule to find the radius (knowing the line connecting the midpoints has to be 4 because it creates a mini equilateral triangle)

[HappyFamilyOnline]

Very well explained👍
Thanks for sharing😊😊