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 $7\sqrt{54}$.
For $2\sqrt{24}=2\sqrt{4\times6}=2\times2\sqrt{6}=4\sqrt{6}$.
For $7\sqrt{54}=7\sqrt{9\times6}=7\times3\sqrt{6}=21\sqrt{6}$.

Step2: Calculate 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. $28\sqrt{6}$