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Thus, the length of the shortest altitude is \(\boxed{\frac{56}{5}}\) units.
An equilateral triangle has an area of \(36\sqrt{3}\) square centimeters. If each side of the triangle is decreased by 4 cm, by how many square centimeters is the area reduced?
The area \(A\) of an equilateral triangle with side \(s\) is:
Given \(A = 36\sqrt{3}\), solve for \(s\):
\frac{\sqrt{3}}{4} s^2 = 36\sqrt{3} \implies \frac{s^2}{4} = 36 \implies s^2 = 144 \implies s = 12 \text{ cm}
New side length after decreasing by 4 cm:
s' = 12 - 4 = 8 \text{ cm}
A' = \frac{\sqrt{3}}{4} (8)^2 = \frac{\sqrt{3}}{4} \times 64 = 16\sqrt{3}
Reduction in area:
\Delta A = 36\sqrt{3} - 16\sqrt{3} = 20\sqrt{3}