The perimeter of a rectangle is 80 m and its length is 1 m more than twice the breadth. What are the lengthsl and breadth?

Given the problem, we know the following:

1. The perimeter of the rectangle is 80 meters.
2. The length \( l \) is 1 meter more than twice the breadth \( b \).

### Step 1: Write down the formulas.
The perimeter \( P \) of a rectangle is given by:
\[
P = 2(l + b)
\]
We're given that \( P = 80 \) meters, so:
\[
2(l + b) = 80
\]
Dividing both sides by 2:
\[
l + b = 40
\]

We're also told that the length is 1 meter more than twice the breadth:
\[
l = 2b + 1
\]

### Step 2: Substitute the expression for \( l \) in the perimeter equation.
Substitute \( l = 2b + 1 \) into \( l + b = 40 \):
\[
(2b + 1) + b = 40
\]
Simplify the equation:
\[
3b + 1 = 40
\]
Subtract 1 from both sides:
\[
3b = 39
\]
Divide by 3:
\[
b = 13
\]

### Step 3: Find the length.
Now, substitute \( b = 13 \) back into the equation for \( l \):
\[
l = 2(13) + 1 = 26 + 1 = 27
\]

### Conclusion:
The length of the rectangle is 27 meters, and the breadth is 13 meters.

Would you like more details or have any questions?

Here are some related questions:
1. How would the dimensions change if the perimeter were different?
2. What if the length were three times the breadth instead of twice?
3. How can you express the area of the rectangle with these dimensions?
4. What happens if the length and breadth were reversed?
5. How does the relationship between length and breadth affect the shape of the rectangle?

**Tip:** Always double-check your substitution steps when solving systems of equations to avoid simple errors.

Learn how to solve for the dimensions of a rectangle with a perimeter of 80 meters. The problem involves using algebraic equations to find the length and breadth based on given relationships. Suitable for students in Grades 7-9.

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