The length of a rectangle is 10 feet more than twice its width. The perimeter of

the rectangle is 170 feet. Find the dimensions of the rectangle. ANS: Length= 60 feet, Width= 25 feet

I already know the answer, its in my review sheet. If someone can explain to me how you get that answer, I will appreciate it! Thanks.

I have shown you how to set up 2 of these kind of problems.

Why don't you show me your own attempts at one of them?

The length of a rectangle is the perimeter of the. The length of a rectangle with the perimeter of the rectangle is 126 feet

To find the dimensions of the rectangle, let's assign variables. Let's say the width of the rectangle is represented by "w" feet. According to the problem, the length of the rectangle is 10 feet more than twice its width. So the length would be 2w + 10 feet.

Now, we have the width and length of the rectangle. To find the perimeter, we add up all four sides. In a rectangle, the opposite sides are equal in length. So the perimeter would be:

Perimeter = 2(length + width)

Substituting the expressions for width and length, we get:

170 = 2((2w + 10) + w)

Now we can simplify the equation:

170 = 2(3w + 10)

Distribute the 2:

170 = 6w + 20

Subtract 20 from both sides:

150 = 6w

Divide both sides by 6:

25 = w

So the width of the rectangle is 25 feet. Now we can substitute this value back into the expression for the length:

Length = 2w + 10 = 2(25) + 10 = 50 + 10 = 60

Therefore, the dimensions of the rectangle are: Width = 25 feet, Length = 60 feet.