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the sides of a triangle measure $3sqrt{6}$, $2sqrt{24}$ and $7sqrt{54}$…

Question

the sides of a triangle measure $3sqrt{6}$, $2sqrt{24}$ and $7sqrt{54}$. what is the perimeter of the triangle?
a. $31sqrt{2}$
b. $17sqrt{6}$
c. $24sqrt{21}$
d. $28sqrt{6}$
e. $12sqrt{84}$

Explanation:

Step1: Simplify each radical term

  • Simplify \(2\sqrt{24}\):

\(2\sqrt{24}=2\sqrt{4\times6}=2\times2\sqrt{6}=4\sqrt{6}\)

  • Simplify \(7\sqrt{54}\):

\(7\sqrt{54}=7\sqrt{9\times6}=7\times3\sqrt{6}=21\sqrt{6}\)

Step2: Calculate the perimeter

The perimeter \(P = 3\sqrt{6}+2\sqrt{24}+7\sqrt{54}\)
Substitute the simplified terms:
\(P=3\sqrt{6}+4\sqrt{6}+21\sqrt{6}=(3 + 4+21)\sqrt{6}=28\sqrt{6}\)

Answer:

D. \(28\sqrt{6}\)