Question

The length of a rectangle is 4cm more than the width. If the length were decreased 5cm and the width decreased 2cm. The perimeter would be 10cm. Find the dimensions of the original rectangle

Answer 1

Let's denote the width of the rectangle as \( w \) cm. According to the problem, the length \( l \) is given as:

\[
l = w + 4
\]

If the length is decreased by 5 cm and the width is decreased by 2 cm, the new dimensions would be:

\[
\text{New Length} = l - 5 = (w + 4) - 5 = w - 1
\]
\[
\text{New Width} = w - 2
\]

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

\[
P = 2(\text{Length} + \text{Width})
\]

According to the problem, if we decrease the dimensions as stated, the perimeter becomes 10 cm. Therefore, we can set up the equation:

\[
2((w - 1) + (w - 2)) = 10
\]

Now, let's simplify the equation:

\[
2(2w - 3) = 10
\]

Dividing both sides by 2 gives:

\[
2w - 3 = 5
\]

Adding 3 to both sides results in:

\[
2w = 8
\]

Dividing both sides by 2 gives:

\[
w = 4
\]

Now that we have the width, we can find the length:

\[
l = w + 4 = 4 + 4 = 8
\]

Thus, the dimensions of the original rectangle are:

\[
\text{Width} = 4 \text{ cm}, \quad \text{Length} = 8 \text{ cm}
\]

To verify, let's check the new dimensions after the decreases:

1. New Length: \( 8 - 5 = 3 \) cm
2. New Width: \( 4 - 2 = 2 \) cm

Calculating the perimeter with these new dimensions:

\[
P = 2(3 + 2) = 2 \times 5 = 10 \text{ cm}
\]

This confirms our solution is correct.

**Final answer:** The dimensions of the original rectangle are 4 cm (width) and 8 cm (length).