Let the width of the rectangle be $w$ cm. The length is given as 25 cm more than the width, so the length is $l = w + 25$ cm.

The formula for the perimeter $P$ of a rectangle is:

$P = 2(l + w)$

Given that the perimeter is 98 cm, we substitute the known values into the formula:

$98 = 2(w + (w + 25))$

Now, solve for $w$:

$98 = 2(2w + 25)$

$98 = 4w + 50$

$98 - 50 = 4w$

$48 = 4w$

$w = \frac{48}{4} = 12 \, \text{cm}$

Now, substitute $w = 12$ cm into the expression for the length:

$l = w + 25 = 12 + 25 = 37 \, \text{cm}$

Thus, the dimensions of the rectangle are:

• Width: 12 cm
• Length: 37 cm

Would you like more details on the steps?

Here are some related questions to practice:

1. If the width were doubled, what would the new perimeter be?
2. What is the area of the rectangle with the found dimensions?
3. How would the dimensions change if the perimeter were increased to 120 cm?
4. If the length were 10 cm less than twice the width, what would the new dimensions be?
5. How much longer is the diagonal of the rectangle compared to its width?

Tip: Always double-check your equation setup by plugging values back into the original formula to confirm correctness!