Understanding the Circular Path Problem

This question asks us to find the radius of the inner circle of a circular path, given the ratio of the outer circumference to the inner circumference and the width of the path. A circular path implies two concentric circles – an inner circle and an outer circle. The region between these two circles is the path.

Setting Up the Equations

Let's denote the radius of the inner circle as \(r\) and the radius of the outer circle as \(R\).

• The circumference of the inner circle is given by the formula \(C_{inner} = 2 \pi r\).
• The circumference of the outer circle is given by the formula \(C_{outer} = 2 \pi R\).

The question states that the ratio of the outer circumference to the inner circumference is 5 ∶ 4. We can write this as an equation:

$$ \frac{C_{outer}}{C_{inner}} = \frac{2 \pi R}{2 \pi r} = \frac{5}{4} $$

Notice that \(2\pi\) cancels out, simplifying the ratio of circumferences to the ratio of radii:

$$ \frac{R}{r} = \frac{5}{4} $$

This equation tells us that the outer radius \(R\) is related to the inner radius \(r\) by \(R = \frac{5}{4}r\).

The question also provides the width of the path. The width of the circular path is the difference between the outer radius and the inner radius.

$$ \text{Width} = R - r $$

We are given that the path is 50 metres wide. So, we have a second equation:

$$ R - r = 50 $$

Solving for the Inner Radius

Now we have a system of two equations with two variables, \(R\) and \(r\):

1. \( R = \frac{5}{4}r \)
2. \( R - r = 50 \)

We can substitute the expression for \(R\) from the first equation into the second equation:

$$ \left(\frac{5}{4}r\right) - r = 50 $$

To combine the terms involving \(r\), we can find a common denominator:

$$ \frac{5}{4}r - \frac{4}{4}r = 50 $$

Subtract the fractions:

$$ \left(\frac{5-4}{4}\right)r = 50 $$

$$ \frac{1}{4}r = 50 $$

Now, solve for \(r\) by multiplying both sides of the equation by 4:

$$ r = 50 \times 4 $$

$$ r = 200 $$

The radius of the inner circle is 200 metres.

Verifying the Solution

Let's check if our answer makes sense with the given information.

• If \(r = 200\) metres, then \(R = \frac{5}{4} \times 200 = 5 \times 50 = 250\) metres.
• The path width is \(R - r = 250 - 200 = 50\) metres, which matches the given width.
• The ratio of circumferences is \(\frac{2 \pi R}{2 \pi r} = \frac{R}{r} = \frac{250}{200} = \frac{25}{20} = \frac{5}{4}\), which matches the given ratio.

The calculated inner radius of 200 metres satisfies all the conditions given in the problem.

Revision Table: Circular Path Geometry

Additional Information: Properties of Concentric Circles

Concentric circles are circles that share the same center point. In problems involving circular paths, tracks, or rings, we often deal with concentric circles.

• The distance from the center to any point on the inner circle's edge is the inner radius (\(r\)).
• The distance from the center to any point on the outer circle's edge is the outer radius (\(R\)).
• The width of the path is the constant distance between the inner and outer circles, which is \(R - r\).
• The ratio of the circumferences of two concentric circles is equal to the ratio of their radii: \(\frac{C_{outer}}{C_{inner}} = \frac{2 \pi R}{2 \pi r} = \frac{R}{r}\).
• Similarly, the ratio of the areas of two concentric circles is equal to the square of the ratio of their radii: \(\frac{A_{outer}}{A_{inner}} = \frac{\pi R^2}{\pi r^2} = \left(\frac{R}{r}\right)^2\).

These relationships are fundamental when solving geometry problems involving circular rings or paths.