A wire of length 86 cm is bent in the form of a rectangle such that its length is 7 cm more than its breadth. Find the length and the breadth of the rectangle so formed.

Answer

To solve the problem, we need to find the length and breadth of the rectangle formed by a wire of length 86 cm, given that the length is 7 cm more than the breadth.

Let's denote the breadth of the rectangle as \( b \) cm. According to the problem, the length \( l \) can be expressed as:

\[ l = b + 7 \]

The perimeter \( P \) of a rectangle is given by the formula:

\[ P = 2(l + b) \]

Since the wire is bent to form the rectangle, the perimeter is equal to the length of the wire, which is 86 cm. Therefore, we can set up the equation:

\[ 2(l + b) = 86 \]

Now, substituting the expression for \( l \) into the perimeter equation:

\[ 2((b + 7) + b) = 86 \]

Now, simplify the equation:

1. Combine like terms inside the parentheses:
\[ 2(2b + 7) = 86 \]

2. Divide both sides by 2:
\[ 2b + 7 = 43 \]

3. Subtract 7 from both sides:
\[ 2b = 43 - 7 \]
\[ 2b = 36 \]

4. Divide both sides by 2 to find \( b \):
\[ b = 18 \]

Now that we have the breadth, we can find the length using the earlier expression for \( l \):

\[ l = b + 7 \]
\[ l = 18 + 7 \]
\[ l = 25 \]

Thus, the dimensions of the rectangle are:
- Length = 25 cm
- Breadth = 18 cm

### Summary of the Solution:
- Length of the rectangle = 25 cm
- Breadth of the rectangle = 18 cm