A rectangle has a width of 2(sqrt 5) cm and an area of 50 cm^2. Find the length of the rectangle.

Question:

A rectangle has a width of {eq}2 \sqrt{5} \, \mathrm{cm} {/eq} and an area of {eq}50 \, \mathrm{cm^2} {/eq}. Find the length of the rectangle.

Rectangular Area:

The area of the rectangle is the product of the length and the width. Wherein the length is the longer side of the rectangle while the width is the shorter side of the rectangle.

Answer and Explanation:

The width of the rectangle is {eq}2 \sqrt 5 {/eq} with an area of {eq}\rm 50 \ cm^2 {/eq}. To determine the length, use the given information for the formula of the rectangular area and solve for the length:

{eq}\begin{align} A &= lw \\ 50 &= l(2 \sqrt 5) \end{align} {/eq}

Dividing both sides by {eq}2 \sqrt 5 {/eq} and rationalizing:

{eq}\begin{align} \dfrac{50}{2 \sqrt 5} &= \dfrac {l(2 \sqrt 5)}{2 \sqrt 5} \\ l &= \left (\dfrac{50}{2 \sqrt 5} \right) \times \left (\dfrac {\sqrt 5}{\sqrt 5} \right) \\ l &= \dfrac{50 \sqrt 5 }{(2)(5)} \\ l &= \dfrac{50 \sqrt 5 }{10} \\ l &= 5 \sqrt 5 \end{align} {/eq}

The length of the rectangle is {eq}5 \sqrt 5 \rm \ cm. {/eq}