How can I solve this math question: "The length of a rectangle is three times more than twice the width, and the perimeter is 36 inches. What is the area of the rectangle?"

Answer 1

I think you meant the length is 3 more than twice the width.

So let L be length and w be width

L= 3+2w

Let P be perimeter

P= 2(3+2w) + 2w

Distribute

6+4w+2w=36

6+6w=36

6w=30

W=5

L=3+2w=3+2(5)=13

Area = LW=(5)(13)=65

Answer 2

I'm thinking you meant to say the length of the rectangle is 3 more than twice the width. If so, the answer is derived as follows:

Let w = width of the rectangle

Then 2w + 3 = the length (3 more than twice the width)

The perimeter of our rectangle = 2w + 2l, = 36

Substituting 2w + 3 for l we get

2w + 2(2w+3) = 36

6w + 6 = 36

6w = 30

w = 5

And l = 2w + 3 = 2(5) + 3 = 13

The area of the rectangle = l x w = 13 x 5 = 65 square inches