Resurfacing a Stadium-Shaped Running Track
A club is resurfacing a stadium-shaped running track. The inner boundary is formed by a rectangle of length 40 m with a semicircle of radius 18 m attached at each end. The track has a constant width of 1.2 m, so the outer boundary is a larger stadium shape with semicircular radius 19.2 m and the same straight length. The surface is to receive two coats of paint. One litre covers 5 m² per coat, but the contractor adds 10% extra paint for absorption and edge loss. Paint is sold only in 20-litre tins. The club also needs a low fence along the outer curved sections only, not along the two straight sides. A committee member estimates the track area as “about 200 m²” and the curved fencing as “about 120 m”. The project manager wants exact expressions before deciding how many tins to order and whether the estimates are safe.
1. Calculate the exact track area and compare it with the committee’s 200 m² estimate.
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The area of a stadium shape is the rectangle area plus one full circle area, because the two semicircles make a circle.
Inner stadium: [A_{\text{in}}=40(2\cdot18)+\pi(18)^2=1440+324\pi.]
Outer stadium: [A_{\text{out}}=40(2\cdot19.2)+\pi(19.2)^2=1536+368.64\pi.]
Track area: [A=A_{\text{out}}-A_{\text{in}}=96+44.64\pi.]
Numerically, [A\approx96+140.24=236.24\ \text{m}^2.]
The committee estimate of 200 m² is too low by about 36 m², roughly 15%. It is not safe for ordering materials.
2. Determine the number of 20-litre tins needed for two coats, including 10% extra.
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One coat needs [\frac{96+44.64\pi}{5}=19.2+8.928\pi\approx47.25\ \text{L}.]
Two coats need about (94.50) L. Adding 10% gives [94.50\times1.10\approx103.95\ \text{L}.]
Each tin contains 20 L, so [\frac{103.95}{20}\approx5.20.]
Paint is sold in whole tins, so the club must buy 6 tins. Buying only 5 tins would provide 100 L, which is below the required amount even before any further wastage.
3. Calculate the length of fencing needed along the outer curved sections only and assess the “about 120 m” estimate.
Read the worked answer
The outer curved sections are the two semicircles at the ends, which together form one full circle of radius 19.2 m. Their total length is [C=2\pi(19.2)=38.4\pi\approx120.64\ \text{m}.]
The straight sides are excluded, so this is the required fencing length. The estimate “about 120 m” is close but slightly low; using 120 m would leave the fence about 0.64 m short. For ordering, the club should use at least 121 m or the exact value (38.4\pi) m.