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An equilateral triangle has an area of \(64\sqrt{3}\) \(\text{cm}^2\). If each side of the triangle is decreased by 4 cm, by how many square centimeters is the area decreased?

 Jun 22, 2019
 #1
avatar+4330 
+2

The area of an equilateral triangle is given by \(\frac{\sqrt{3}}{4}s^2\), where \(s\) is the side length of the equilateral triangle. Consequently, setting this formula and expression equal to \(64\sqrt{3}\), we get \(s^2=256\), and \(s=16\) centimeters. Since each side of the triangle decreases by four centimeters, the new side length of the equilateral triangle is twelve centimeters. Thus, the new area is \(36\sqrt{3}\) centimeters, and it has decreased by \(64\sqrt{3}-36\sqrt{3}=\boxed{28\sqrt{3}}\)centimeters.

 

-tertre

 Jun 22, 2019
 #2
avatar+7763 
+2

\(\text{Let }l \text{ cm be the side length of the equilateral triangle.}\\ \dfrac{\sqrt3}{4} l^2 = 64\sqrt3\\ l^2 = 256\\ l = 16\\ \text{New length} = 12\text{ cm}\\ \text{New area} = \dfrac{\sqrt3}{4} (12^2) = 36\sqrt3\text{ cm}^2\\ \text{Area decrease} = 64\sqrt3 - 36\sqrt3= 28\sqrt3\text{ cm}^2\)

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 Jun 22, 2019

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