Let’s solve this step by step:

Problem Analysis

We are tasked with finding the dimensions of a rectangle where:

• The length is 20 meters shorter than five times the width.
• The area of the rectangle is $60 \, \text{m}^2$.

Define Variables:

Let:

• $w$: Width of the rectangle (in meters)
• $l$: Length of the rectangle (in meters)

Form Equations:

1. The length is related to the width by: $l = 5w - 20$

2. The area of the rectangle is given by: $l \cdot w = 60$

Substitute $l = 5w - 20$ into the area equation: $(5w - 20) \cdot w = 60$

Solve for $w$:

Expand and simplify: $5w^2 - 20w = 60$

Rearrange into standard quadratic form: $5w^2 - 20w - 60 = 0$

Divide the entire equation by 5 to simplify: $w^2 - 4w - 12 = 0$

Factorize the quadratic equation: $(w - 6)(w + 2) = 0$

Solve for $w$: $w = 6 \quad \text{or} \quad w = -2$

Since width cannot be negative: $w = 6 \, \text{meters}$

Find $l$:

Substitute $w = 6$ into $l = 5w - 20$: $l = 5(6) - 20 = 30 - 20 = 10 \, \text{meters}$

Verify:

Check the area: $l \cdot w = 10 \cdot 6 = 60 \, \text{m}^2$

Final Answer:

The dimensions of the rectangle are:

• Width: $6 \, \text{meters}$
• Length: $10 \, \text{meters}$

Correct Option:

The answer is $6, 10$.

Would you like further clarification or more examples?

Related Questions:

1. How do you solve other quadratic equations geometrically?
2. What are common applications of area equations in real life?
3. How would this problem change if the rectangle was a square?
4. Can you solve similar problems using graphing methods?
5. How do quadratic equations apply to physical measurements like volume?

Tip:

Always double-check units in area problems to ensure consistency!