# How do you find the area of a rectangular object which has a length of 5 square root of 8 inches and a width of 2 square root of 5 inches?

Oct 4, 2016

$20 \sqrt{10} \text{ square inches}$

#### Explanation:

The area ( A) of a rectangle is $\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{A = l \times b} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where l is the length and b, the width.

here $l = 5 \sqrt{8} \text{ and } b = 2 \sqrt{5}$

$\Rightarrow A = 5 \sqrt{8} \times 2 \sqrt{5}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\sqrt{a} \times \sqrt{b} \Leftrightarrow \sqrt{a b}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\Rightarrow A = \textcolor{b l u e}{5} \times \textcolor{red}{\sqrt{8}} \times \textcolor{b l u e}{2} \times \textcolor{red}{\sqrt{5}}$

$= \textcolor{b l u e}{10} \times \textcolor{red}{\sqrt{40}}$

Now $\sqrt{40} = \sqrt{4 \times 10} = \sqrt{4} \times \sqrt{10} = 2 \times \sqrt{10}$

$\Rightarrow A = \textcolor{b l u e}{10} \times \textcolor{red}{\sqrt{40}} = 10 \times 2 \times \sqrt{10}$

$\Rightarrow A = 20 \times \sqrt{10} = 20 \sqrt{10} \text{ square inches}$