A rectangular city park measures 30 m by 20 m. At each corner a quarter-circle flower bed of radius 7 m is laid out. A fence follows the boundary of the remaining area as a running track. How long is the track?

Step-by-step solution

1. 1

   Find the perimeter before the corners are cut.

   $P = 2(30 + 20) = 100\ \text{m}$

   This is the answer to a different question — the perimeter of the whole park — and it is option A, the trap for anyone who stops here.

2. 2

   See what each corner removes and adds. A quarter circle of radius $7$ at a corner eats $7\ \text{m}$ along each of the two sides meeting there, so it removes $7 + 7 = 14\ \text{m}$ of straight fence. In its place comes the quarter arc:

   $L_{\text{arc}} = \frac14 \cdot 2\pi r = \frac{2\pi(7)}{4} = 3.5\pi\ \text{m}$

3. 3

   Evaluate the arc. School convention for $r = 7$ uses $\pi = \tfrac{22}{7}$, which makes the arithmetic exact:

   $3.5\pi = 3.5 \times \frac{22}{7} = 11\ \text{m}$

   (With the true $\pi$ the arc is $10.9956\ \text{m}$ — the difference will turn out to be under $2\ \text{cm}$ in the final answer.)

4. 4

   Net the change at one corner, then at four. Each corner loses $14$ m and gains $11$ m, a net loss of $3$ m. Over four corners:

   $4 \times 3 = 12\ \text{m lost}$

   The corners fit because $7 + 7 = 14 \le 20$, the shorter side.

5. 5

   Compute the track length.

   $100 - 12 = 88\ \text{m}$

   so the answer is B.

6. 6

   Check by adding the pieces directly. Straight portions: $100 - 4(14) = 44\ \text{m}$. Four arcs: $4 \times 11 = 44\ \text{m}$. Total $88\ \text{m}$, matching. Using the true $\pi$ instead gives $44 + 4(10.99557) = 87.982\ \text{m}$, which still rounds to $88$ — so the choice of $\pi$ does not change the multiple-choice answer here.