# How do you find the perimeter and area of a rectangle with width of 2sqrt7-2sqrt5 and length of 3sqrt7+3sqrt5?

May 18, 2017

Perimeter $P = 10 \sqrt{7} + 2 \sqrt{5}$

Area $A = 12$

#### Explanation:

The perimeter $P$ of a rectangle is the sum of all side lengths.

The formula is $P = 2 \left(l + w\right)$, where $l$ is the length and $w$ is the width:

$R i g h t a r r o w P = 2 \left(\left(3 \sqrt{7} + 3 \sqrt{5}\right) + \left(2 \sqrt{7} - 2 \sqrt{5}\right)\right)$

$R i g h t a r r o w P = 2 \left(3 \sqrt{7} + 2 \sqrt{7} + 3 \sqrt{5} - 2 \sqrt{5}\right)$

$R i g h t a r r o w P = 2 \left(5 \sqrt{7} + \sqrt{5}\right)$

$\therefore P = 10 \sqrt{7} + 2 \sqrt{5}$

The area $A$ of a rectangle is the product of the length and the width.

The formula is $A = l w$:

$R i g h t a r r o w A = \left(3 \sqrt{7} + 3 \sqrt{5}\right) \times \left(2 \sqrt{7} - 2 \sqrt{5}\right)$

$R i g h t a r r o w A = 3 \sqrt{7} \times 2 \sqrt{7} + 3 \sqrt{7} \times \left(- 2 \sqrt{5}\right) + 3 \sqrt{5} \times 2 \sqrt{7} + 3 \sqrt{5} \times \left(- 2 \sqrt{5}\right)$

$R i g h t a r r o w A = 6 \times 7 - 6 \sqrt{35} + 6 \sqrt{35} - 6 \times 5$

$R i g h t a r r o w A = 42 - 30$

$\therefore A = 12$

Therefore, the perimeter is $10 \sqrt{7} + 2 \sqrt{5}$ and the area is $12$.

May 18, 2017

Perimeter of rectangle is $2 \left(5 \sqrt{7} + \sqrt{5}\right)$ unit.

Area of rectangle is $12$ sq.unit.

#### Explanation:

Length of rectangle is $l = 3 \sqrt{7} + 3 \sqrt{5}$
Width of rectangle is $w = 2 \sqrt{7} - 2 \sqrt{5}$

Perimeter of rectangle is $P = 2 l + 2 w = 2 \left(3 \sqrt{7} + 3 \sqrt{5}\right) + 2 \left(2 \sqrt{7} - 2 \sqrt{5}\right)$

$P = \sqrt{7} \left(6 + 4\right) + \sqrt{5} \left(6 - 4\right) = 10 \sqrt{7} + 2 \sqrt{5} = 2 \left(5 \sqrt{7} + \sqrt{5}\right)$unit

Area of rectangle is $A = l \cdot w = \left(3 \sqrt{7} + 3 \sqrt{5}\right) \cdot \left(2 \sqrt{7} - 2 \sqrt{5}\right)$

$A = 6 \cdot 7 - \cancel{6 \cdot \sqrt{35}} + \cancel{6 \sqrt{35}} - 6 \cdot 5 = 42 - 30 = 12$ sq.unit.

Perimeter of rectangle is $2 \left(5 \sqrt{7} + \sqrt{5}\right)$ unit
Area of rectangle is $12$ sq.unit. [Ans]

May 18, 2017

12

#### Explanation:

we know area of rectangle = length*width sq unit.

So, $\textcolor{g r e e n}{A r e a}$ = $\left[3 \sqrt{7} + 3 \sqrt{5}\right] \times \left[2 \sqrt{7} - 2 \sqrt{5}\right]$

$\Rightarrow 3 \left[\sqrt{7} + \sqrt{5}\right] \times 2 \left[\sqrt{7} - \sqrt{5}\right]$

$\Rightarrow 3 \times 2 \times \left[{\left(\sqrt{7}\right)}^{2} - {\left(\sqrt{5}\right)}^{2}\right]$

$\Rightarrow 3 \times 2 \times \left[7 - 5\right] = 3 \times 2 \times 2 = 12$ sq unit

$\textcolor{red}{p e r i m e t e r}$= 2*[length+width]unit

$\Rightarrow 2 \left[\left\{3 \sqrt{7} + 3 \sqrt{5}\right\} + \left\{2 \sqrt{7} - 2 \sqrt{5}\right\}\right]$

$\Rightarrow 2 \left[3 \sqrt{7} + 3 \sqrt{5} + 2 \sqrt{7} - 2 \sqrt{5}\right]$

$\Rightarrow 2 \left[5 \sqrt{7} + \sqrt{5}\right]$

May 18, 2017

Areacolor(blue)(=12 color(white)(a)square units,color(blue)(P=2(5sqrt7+sqrt5)

#### Explanation:

$W = 2 \sqrt{7} - 2 \sqrt{5}$

$L = 3 \sqrt{7} + 3 \sqrt{5}$

perimeter of a rectangle$= P = L + W + L + W$

$P = 2 L + 2 W = 2 \left(L + W\right)$

$P = 2 \left[\left(3 \sqrt{7} + 3 \sqrt{5}\right) + \left(2 \sqrt{7} - 2 \sqrt{5}\right)\right]$

$P = 2 \left[3 \sqrt{7} + 3 \sqrt{5} + 2 \sqrt{7} - 2 \sqrt{5}\right]$

$P = 2 \left[5 \sqrt{7} + \sqrt{5}\right]$

Area$= A = W \times L$

$A = \left(2 \sqrt{7} - 2 \sqrt{5}\right) \times \left(3 \sqrt{7} + 3 \sqrt{5}\right)$

$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$2 \sqrt{7} - 2 \sqrt{5}$
$\textcolor{w h i t e}{a a a a a a a a a a}$$\times \underline{3 \sqrt{7} + 3 \sqrt{5}}$
$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$6 \times 7 - 6 \sqrt{5} \sqrt{7}$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a}$$6 \sqrt{5} \sqrt{7} - 6 \times 5$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$overline(42-30color(blue)(=12 square units