Question

find the dimensions and the area of a rectangle with perimeter of 40 inches, such that its length is one inch less than twice the width

Answer 1

length = 13 in, width = 7 in , area = 91 in²

Explanation:

let width be w then length = 2w - 1

perimeter(P) of rectangle is calculated as

P = 2 × length + 2× width

= 2(2w - 1) + 2w

= 4w - 2 + 2w

= 6w - 2

given P = 40 , then

6w - 2 = 40 ( add 2 to both sides )

6w = 42 ( divide both sides by 6 )

w = 7

then

width = 7 in and length = 2w - 1 = 2(7) - 1 = 14 - 1 = 13 in

the area (A) of a rectangle is calculated as

A = length × width = 13 × 7 = 91 in²