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 square - roots

Simplify \(2\sqrt{24}\) and \(7\sqrt{54}\).
Since \(24 = 4\times6\), then \(2\sqrt{24}=2\sqrt{4\times6}=2\times2\sqrt{6} = 4\sqrt{6}\).
Since \(54=9\times6\), then \(7\sqrt{54}=7\sqrt{9\times6}=7\times3\sqrt{6}=21\sqrt{6}\).

Step2: Calculate the perimeter

The perimeter \(P\) of a triangle with side lengths \(a = 3\sqrt{6}\), \(b = 4\sqrt{6}\), and \(c = 21\sqrt{6}\) is \(P=a + b + c\).
\(P=3\sqrt{6}+4\sqrt{6}+21\sqrt{6}=(3 + 4+21)\sqrt{6}=28\sqrt{6}\).

Answer:

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