Answer
a)
$P$ = 10 ft
$A$ = 6 ft$^{2}$
b)
$P$ = 20 ft
$A$ = 24 ft$^{2}$
c)
The perimeter is twice as large as the original.
The area is four times larger than the original.
Work Step by Step
Area of a rectangle: $A$ = $l$$w$
Perimeter of a rectangle: $P$ = 2($l$ + $w$)
a)
$A$ = $l$$w$
$A$ = (3 ft)(2 ft)
$A$ = 6 ft$^{2}$
$P$ = 2($l$ + $w$)
$P$ = 2(3 ft + 2 ft)
$P$ = 2(5 ft)
$P$ = 10 ft
b)
$A$ = $l$$w$
$A$ = (6 ft)(4 ft)
$A$ = 24 ft$^{2}$
$P$ = 2($l$ + $w$)
$P$ = 2(6 ft + 4 ft)
$P$ = 2(10 ft)
$P$ = 20 ft
