Question

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

Answer 1

The areas of the larger and smaller circles are \( \pi R^2 \) and \( \pi r^2 \), respectively. The sum of these areas is \( 130 \):

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

Dividing by \( \pi \) and approximating \( \pi \approx 3.14 \):

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

This yields the equation:

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

Answer 2

Formulate a quadratic equation solely in terms of \( r \).

The distance between the centers of the circles measures 14 m. Applying the Pythagorean theorem to the distance between the centers yields:

\[ R + r = 14 \]

Squaring both sides of the equation:

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

Expanding the squared term:

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

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

\[ 41.4 + 2Rr = 196 \]

Isolate and solve for the product \( Rr \):

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

Consequently:

\[ Rr = 77.3 \]

Answer 3

Determine the radius \( r \) and the associated irrigated area.

Substitute \( R = 14 - r \) into the equation \( Rr = 77.3 \):

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

Expansion yields:

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

Rearranging the terms gives:

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

Applying the quadratic formula to solve this equation:

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

The negative discriminant indicates no real solution exists, suggesting an error in the problem's formulation or interpretation. A thorough review of the formula and initial assumptions is required.