## Basic College Mathematics (9th Edition)

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.
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