Question

the area of a rectangle is $\sqrt{12}$, with a length of $\sqrt{5}$. the width is______. (keep the answer in its simplest form.)

Explanation:

Step1: Recall the formula for the area of a rectangle

The area formula of a rectangle is \(A = l\times w\), where \(A\) is the area, \(l\) is the length, and \(w\) is the width. We can solve for \(w\) by \(w=\frac{A}{l}\). Given \(A = \sqrt{12}\) and \(l=\sqrt{5}\), so \(w=\frac{\sqrt{12}}{\sqrt{5}}\).

Step2: Rationalize the denominator

Multiply the numerator and denominator by \(\sqrt{5}\). We get \(w=\frac{\sqrt{12}\times\sqrt{5}}{\sqrt{5}\times\sqrt{5}}\).
Since \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\) (\(a\geq0,b\geq0\)), then \(\sqrt{12}\times\sqrt{5}=\sqrt{60}\), and \(\sqrt{5}\times\sqrt{5} = 5\). So \(w=\frac{\sqrt{60}}{5}\).

Step3: Simplify the square - root

Factor \(60\): \(60=4\times15\). Then \(\sqrt{60}=\sqrt{4\times15}\). Using the property \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (\(a\geq0,b\geq0\)), \(\sqrt{4\times15}=\sqrt{4}\times\sqrt{15}=2\sqrt{15}\).

Answer:

\(\frac{2\sqrt{15}}{5}\)