Question

At the centre of a city's municipal park there is a large circular pool. A fish is released in the water at the edge of the pool. The fish swims north for 300 feet before it hits the edge of the pool. It then turns east and swims for 400 feet before hitting the edge again. What is the area of the pool?

• A. \(62500\pi\)
• B. \(125000\pi\)
• C. \(250000\pi\)
• D. \(500000\pi\)
• E. Cannot be answered from the given data.

Hint:

Model straight swims as chords; use the circle equation at chord endpoints and subtract to eliminate \(R^2\). Differences along perpendicular directions often yield a right triangle whose legs determine \(R\).

Answer 1

The Correct Option is A

Step 1: Let the circle be \((x-h)^2+(y-k)^2=R^2\). After release at point \(P\) on the edge, the fish swims north to point \(Q=(x_0,y_0)\) on the edge with vertical travel \(300\). Thus \(P=(x_0,y_0-300)\) and \[ (x_0-h)^2+(y_0-k)^2=R^2, (x_0-h)^2+(y_0-300-k)^2=R^2. \] Subtracting gives \(y_0-k=150\).
Step 2: From \(Q\), the fish swims east \(400\) to point \(R=(x_0+400,y_0)\) on the edge: \[ (x_0-h)^2+(y_0-k)^2=R^2, (x_0+400-h)^2+(y_0-k)^2=R^2. \] Subtracting gives \(x_0-h=-200\).
Step 3: Hence \[ R^2=(x_0-h)^2+(y_0-k)^2=(-200)^2+150^2=40000+22500=62500. \] Therefore, the pool's area is \(\pi R^2=62500\pi\).