The perimeter of a rectangle is 122 cm. The length is ten more than twice the width. What are the dimensions?
L = 2W + 10
2L + 2W = 2(2W+10) + 2W = 122
Solve for W, then L.
The subject is NOT geography.
Let's denote the width of the rectangle as "w" cm.
According to the given information, the length is ten more than twice the width, which can be expressed as: 2w + 10 cm.
The perimeter of a rectangle is defined as the sum of all four sides. For our rectangle, it can be represented as:
2(length + width)
Substituting the given expressions for length and width:
2(2w + 10 + w) = 122 cm
Simplifying the equation:
2(3w + 10) = 122
6w + 20 = 122
6w = 102
w = 17
Substituting the value of w back into the expression for length:
Length = 2w + 10 = 2(17) + 10 = 44
Therefore, the dimensions of the rectangle are 17 cm (width) and 44 cm (length).
To find the dimensions of the rectangle, we need to solve the given information using algebra. Let's denote the width of the rectangle as "w" and the length as "l".
1) We are given that the perimeter of the rectangle is 122 cm. The formula for the perimeter of a rectangle is P = 2(l + w), where P is the perimeter, l is the length, and w is the width. So we can write the equation as:
122 = 2(l + w)
2) The problem also states that the length is ten more than twice the width. We can express this as:
l = 2w + 10
Now we have two equations with two variables (l and w). Let's solve them simultaneously:
Substitute the value of l from the second equation into the first equation:
122 = 2((2w + 10) + w)
Simplify and solve for w:
122 = 2(3w + 10)
122 = 6w + 20
6w = 122 - 20
6w = 102
w = 102 / 6
w = 17
Now, substitute the value of w back into the second equation to find the length l:
l = 2w + 10
l = 2(17) + 10
l = 34 + 10
l = 44
So, the dimensions of the rectangle are width = 17 cm and length = 44 cm.