Question

the width of a rectangle is \\(sqrt3{216^2}\\) ft and the length is \\(64^{\frac{2}{3}}\\) ft. what is the area of the rectangle?
answer
\boxed{} ft²

Explanation:

Step1: Simplify the width

Use the property $\sqrt[n]{a^m}=a^{\frac{m}{n}}$. For $\sqrt[3]{216^{2}}$, since $216 = 6^3$, then $\sqrt[3]{216^{2}}=(6^{3})^{\frac{2}{3}}$.
By the power - of - a - power rule $(a^{m})^{n}=a^{mn}$, so $(6^{3})^{\frac{2}{3}}=6^{3\times\frac{2}{3}} = 6^{2}=36$.

Step2: Simplify the length

Use the property $a^{\frac{m}{n}}=\sqrt[n]{a^{m}}$. For $64^{\frac{2}{3}}$, since $64 = 4^3$, then $64^{\frac{2}{3}}=(4^{3})^{\frac{2}{3}}$.
By the power - of - a - power rule $(a^{m})^{n}=a^{mn}$, so $(4^{3})^{\frac{2}{3}}=4^{3\times\frac{2}{3}}=4^{2} = 16$.

Step3: Calculate the area of the rectangle

The area formula of a rectangle is $A = lw$ (where $l$ is the length and $w$ is the width).
Substitute $l = 16$ and $w = 36$ into the formula, then $A=36\times16$.
$36\times16=(30 + 6)\times16=30\times16+6\times16=480+96 = 576$.

Answer:

$576$