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?

February 22, 2026

$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 shortest altitude corresponds to the longest side, which is $c = 15$:
h_{\text{min}} = \frac{2 \times 84}{15} = \frac{168}{15} = \frac{56}{5} = 11.2
Thus, the length of the shortest altitude is $\boxed{\frac{56}{5}}$ units.
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}

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?

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