The problem states that the diameter of a circle is 6 yards. We are asked to find the ratio of the area of the original circle to the area of a new circle with a radius that is double the radius of the original circle.

1. Problem Description

The problem states that the diameter of a circle is 6 yards. We are asked to find the ratio of the area of the original circle to the area of a new circle with a radius that is double the radius of the original circle.

2. Solution Steps

First, find the radius of the original circle. Since the diameter is 6 yards, the radius $r_1$ is half of that.

$r_1 = \frac{6}{2} = 3$ yards.

The area of the original circle, $A_1$, is given by the formula:

$A_1 = \pi r_1^2 = \pi (3^2) = 9\pi$ square yards.

The radius of the new circle, $r_2$, is double the radius of the original circle:

$r_2 = 2 \times r_1 = 2 \times 3 = 6$ yards.

The area of the new circle, $A_2$, is given by the formula:

$A_2 = \pi r_2^2 = \pi (6^2) = 36\pi$ square yards.

We want to find the ratio of the area of the smaller circle (original circle) to the area of the larger circle (new circle), which is $\frac{A_1}{A_2}$.

$\frac{A_1}{A_2} = \frac{9\pi}{36\pi} = \frac{9}{36} = \frac{1}{4}$.

3. Final Answer

The ratio of the area of the smaller circle to the larger circle is $\frac{1}{4}$.