Question

find the exact value of the perimeter, and then approximate the value to 1 decimal place.
part: 0 / 2
part 1 of 2
the perimeter of the triangle is cm.

Explanation:

Step1: Simplify the square - root terms

• For \(2\sqrt{24}\), we have \(2\sqrt{24}=2\sqrt{4\times6}=2\times2\sqrt{6} = 4\sqrt{6}\).
• For \(\sqrt{54}\), we have \(\sqrt{54}=\sqrt{9\times6}=3\sqrt{6}\).

Step2: Calculate the perimeter

The perimeter \(P\) of a triangle is the sum of its side - lengths. Given side - lengths \(a = 2\sqrt{6}\), \(b = 2\sqrt{24}\), and \(c=\sqrt{54}\).
Substitute the simplified values: \(P=2\sqrt{6}+4\sqrt{6}+3\sqrt{6}\).
Using the distributive property \(a\sqrt{6}+b\sqrt{6}+c\sqrt{6}=(a + b + c)\sqrt{6}\), where \(a = 2\), \(b = 4\), and \(c = 3\).
\(P=(2 + 4+3)\sqrt{6}=9\sqrt{6}\) \(cm\).

Answer:

\(9\sqrt{6}\)