Question

Obtain a quadratic equation involving $R$ and $r$ from the above information.

Answer 1

The areas of the two circles are given by \( \pi R^2 \) and \( \pi r^2 \) for the larger and smaller circle, respectively. We know that:

\[ \pi R^2 + \pi r^2 = 130 \]

Simplifying:

\[ R^2 + r^2 = \frac{130}{\pi} \approx \frac{130}{3.14} = 41.4 \]

Thus, the first equation is:

\[ R^2 + r^2 = 41.4 \, \text{(Eq. 1)} \]

Answer 2

Write a quadratic equation involving only \( r \).

The distance between the centers of the circles is 14 m. Using the Pythagorean theorem for the distance between the centers:

\[ R + r = 14 \]

Squaring both sides:

\[ (R + r)^2 = 14^2 \]

Expanding:

\[ R^2 + 2Rr + r^2 = 196 \]

Substitute \( R^2 + r^2 = 41.4 \) from Eq. 1:

\[ 41.4 + 2Rr = 196 \]

Solving for \( Rr \):

\[ 2Rr = 196 - 41.4 = 154.6 \]

Thus:

\[ Rr = 77.3 \]

Answer 3

Find the radius \( r \) and the corresponding area irrigated.

Now, substitute \( R = 14 - r \) into the equation \( Rr = 77.3 \):

\[ (14 - r)r = 77.3 \]

Expanding:

\[ 14r - r^2 = 77.3 \]

Rearranging:

\[ r^2 - 14r + 77.3 = 0 \]

Solving this quadratic equation using the quadratic formula:

\[ r = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(1)(77.3)}}{2(1)} \] \[ r = \frac{14 \pm \sqrt{196 - 309.2}}{2} \approx \frac{14 \pm \sqrt{-113.2}}{2} \]

Since the discriminant is negative, there is no real solution, suggesting an error in the setup or interpretation of the problem. The formula setup and the initial assumptions need to be checked carefully.