The length of a rectangle is 1 foot more than twice the width the perimeter is 44 feet find the dimensions of the rectangle

The length of rectangle is 15 feet and width of rectangle is 7 feet

Solution:

Let "a" be the width of rectangle

Given that, length of rectangle is 1 foot more than twice the width

Therefore, we get

length = 1 + twice the width

length = 1 + 2a eqn 1

Given that perimeter is 44 feet

The perimeter of rectangle is given as:

Perimeter = 2(length + width)

44 = 2(1 + 2a + a)

44 = 2(1+3a)

22 = 1 + 3a

3a = 22 - 1

3a = 21

a = 7

Thus from eqn 1,

length = 1 + 2(7) = 1 + 14 = 15

Thus length of rectangle is 15 feet and width of rectangle is 7 feet

a

step-by-step explanation:

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