The running track at a school sports day has two semicircular ends. Each end has diameter \(70\) m. Calculate the length of one semicircular arc, to 3 signi...
Question 1 Report
The running track at a school sports day has two semicircular ends. Each end has diameter \(70\) m. Calculate the length of one semicircular arc, to 3 significant figures. (2)
Answer Details
The full circumference of a circle is \(C = \pi d\), where \(d\) is the diameter. A semicircular arc is exactly half of that circumference, so the arc length is \(\frac{1}{2}\pi d\).
With \(d = 70\) m,
\[\text{arc} = \frac{1}{2} \times \pi \times 70 = 35\pi\][1]
Evaluating, \(35\pi = 109.955...\), so the arc is \(110\) m correct to 3 significant figures. [1]
Two traps are worth naming. First, the value given is a diameter, so using \(\pi \times 70\) as if it were \(2\pi r\) would double the answer; the radius is \(35\) m and \(\frac{1}{2} \times 2\pi \times 35\) gives the same \(35\pi\). Second, the semicircular arc is the curved part only; the straight diameter across the end is not part of an arc length, though it would be included if the question asked for the perimeter of the semicircular region.
Download The App On Google Playstore
Everything you need to excel in your exams
