Answer
Perimeter $=8\sqrt {2}$ inches.
Area $=7$ square inches.
Work Step by Step
Length of the rectangle $l=\sqrt {8}+1$ inches.
Width of the rectangle $w=\sqrt {8}-1$ inches.
Perimeter:
$\Rightarrow P=2l+2w$
Substitute values.
$\Rightarrow P=2(\sqrt {8}+1)+2(\sqrt {8}-1)$.
Clear the parentheses.
$\Rightarrow P=2\sqrt {8}+2+2\sqrt {8}-2$.
Simplify.
$\Rightarrow P=4\sqrt {8}$.
Factor the radicands.
$\Rightarrow P=4\sqrt {2^2\cdot 2}$.
Simplify.
$\Rightarrow P=4\cdot 2\sqrt {2}$.
Simplify.
$\Rightarrow P=8\sqrt {2}$.
Hence, the perimeter is $8\sqrt {2}$ inches.
Area :
$\Rightarrow A=lw$
Substitute values.
$\Rightarrow A=(\sqrt {8}+1)(\sqrt {8}-1)$
Use the special formula $(A+B)(A-B)=A^2-B^2$.
$\Rightarrow A=[(\sqrt {8})^2-(1)^2]$
Clear the parentheses.
$\Rightarrow A=8-1$
Simplify.
$\Rightarrow A=7$
Hence, the area is $7$ square inches.
