# How do you find the area of a rectangle with sides 6a^2b^4 and 3ab^2?

Oct 18, 2016

$A = 18 {a}^{3} {b}^{6}$

#### Explanation:

The area of a rectangle is found by multiplying the length by the breadth.

$A = l \times b$

If the sides are $6 {a}^{2} {b}^{4} \mathmr{and} 3 a {b}^{2}$, then the area is the product if these sides.

$A = 6 {a}^{2} {b}^{4} \times 3 a {b}^{2}$

Just multiply the numbers and add the indices of like bases

$A = 18 {a}^{3} {b}^{6}$

Do not be put off by the question! The lengths of the sides of the rectangle are given using variables and powers.

Do the same as you would with numbers - just multiply the terms.

Oct 18, 2016

$A = 18 {a}^{3} {b}^{6}$

#### Explanation:

In this exercise we will using the product of powers with same base that's :

${\textcolor{red}{b}}^{4} \cdot {\textcolor{red}{b}}^{2} = {\textcolor{red}{b}}^{\textcolor{b l u e}{4 + 2}}$

Knowing the area of rectangle that's :
${A}_{r e c} = L \cdot W$

where $L$ is the length of a rectangle and $W$ is its width
Here,
$L = 6 {a}^{2} {b}^{4} \mathmr{and} W = 3 a {b}^{2}$

$A = \left(6 {a}^{2} {b}^{4}\right) \cdot \left(3 a {b}^{2}\right)$
$A = \left(6 \cdot 3\right) \left({a}^{2} \cdot a\right) \left({b}^{4} \cdot {b}^{2}\right)$

$A = 18 {a}^{\textcolor{b l u e}{2 + 1}} {b}^{\textcolor{b l u e}{4 + 2}}$

$A = 18 {a}^{3} {b}^{6}$