Question

persevere if the area of a rectangle is 144\sqrt{5} square inches, what are possible dimensions of the rectangle? explain your reasoning. because select choice \sqrt{12}× select choice \sqrt{15}=144\sqrt{5}, select choice \sqrt{12} in. is a possible length and select choice \sqrt{15} in. is a possible width.

Explanation:

Step1: Recall area formula for rectangle

The area of a rectangle is $A = l\times w$, where $l$ is the length and $w$ is the width. Here $A = 144\sqrt{5}$. We want to find $l$ and $w$ such that $l\times w=144\sqrt{5}$. Given the form $a\sqrt{12}\times b\sqrt{15}$, we first simplify $\sqrt{12}\times\sqrt{15}$.

Step2: Simplify square - root product

We know that $\sqrt{12}\times\sqrt{15}=\sqrt{12\times15}=\sqrt{180}=\sqrt{36\times5}=6\sqrt{5}$.

Step3: Solve for $a$ and $b$

We have the equation $(a\sqrt{12})\times(b\sqrt{15})=144\sqrt{5}$. Since $\sqrt{12}\times\sqrt{15}=6\sqrt{5}$, the equation becomes $ab\times6\sqrt{5}=144\sqrt{5}$. Divide both sides of the equation $ab\times6\sqrt{5}=144\sqrt{5}$ by $6\sqrt{5}$, we get $ab = 24$. We can choose $a = 12$ and $b = 2$.
So $12\sqrt{12}$ inches is a possible length and $2\sqrt{15}$ inches is a possible width.

Answer:

$12\sqrt{12}$; $2\sqrt{15}$; $12\sqrt{12}$; $2\sqrt{15}$