Height in cm=

$ Area of equilateral triangle=\frac{\sqrt{3}}{4}a^{2} $

Construction \:of \:Equilateral\: triagle\: \:if\: height\: is \:given

You can also  input value of height in input box below in the range 1cm to 10cm

Best viewed in in wide screen monitor.

Author- Tharanath
License- CC BY SA 4.0

a = 7.8 (slider from 1 to 10, step 0.1)
A = (7.84, -2.14)
B = Point(Circle[A, a])
Height = Segment(A, B)
c = OrthogonalLine(A, Height)
A' = Rotate(A, 30°, B)
α = Angle(A, B, A')
d = Ray(B, A')
C = Intersect(c, d)
e = Segment(B, C)
A'_1 = Rotate(A, 330°, B)
β = Angle(A'_1, B, A)
f = Ray(B, A'_1)
D = Intersect(c, f)
g = Segment(B, D)
h = Segment(D, C)
γ = Angle(C, A, B)
δ = Angle(A, D, B)
ε = Angle(B, C, A)
inputBox1 = Textfield(a)
text1: "Side \:\:BD \:=" + (LaTeX[g]) + ""
text2: "Side\:\:BC=" + (LaTeX[e]) + ""
text3: "$ Side\:DC=" + (LaTeX[h]) + " $"
b = 1.7320508075688772
t1: ((sqrt(3) * g) * g)
t2: t1 / 4
poly1, c_1, b_1, d_1 = Polygon(B, D, C)
i = Area(B, D, C)
text5: "$ =\frac{\sqrt{3}}{4}×" + (LaTeX[g]) + "^{2}$
$=\frac{" + (LaTeX[b]) + "×" + (LaTeX[g]) + "×" + (LaTeX[g]) + "}{4}=\frac{" + (LaTeX[t1]) + "}{4}=" + (LaTeX[t2]) + ""
text6: "$The \:Area \:calculated \:by \:the \:Computer =" + (LaTeX[i]) + "\:Sq.cm"
k = Circle(D, (0.75 * a))
p = Circle(C, (0.75 * a))