A-Level Mathematics October/November 2024 Q7(b): Given instead that the area of each semicircle is 50π cm², find the exact perimeter of…

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  The area of each semicircle is given as $50\pi \text{ cm}^2$. Let the radius of the semicircle be $r$. Area of semicircle = $\frac{1}{2}\pi r^2$ $\frac{1}{2}\pi r^2 = 50\pi$ $r^2 = 100$ $r = 10$ cm

  The diameter of the semicircle is $2r = 20$ cm. This diameter forms a chord of the large circle, which has a radius $R=20$ cm (from the information given in the first part of the question).

  Let the angle subtended by one of these chords at the centre O be $\theta$. We have an isosceles triangle with two sides equal to the radius of the large circle ($R=20$) and the third side being the chord (20). This means the triangle is equilateral. Therefore, the angle $\theta = \frac{\pi}{3}$ radians.

  Alternatively, using the cosine rule: $20^2 = 20^2 + 20^2 - 2(20)(20)\cos\theta$ $400 = 400 + 400 - 800\cos\theta$ $800\cos\theta = 400$ $\cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}$

  The perimeter of the metal plate consists of the major arc of the large circle and the curved arc of the two semicircles.

  The total angle subtended by the two chords at the centre is $2\theta = \frac{2\pi}{3}$. The angle for the major arc is $2\pi - 2\theta = 2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}$.

  Length of the major arc = $R \times (\text{angle}) = 20 \times \frac{4\pi}{3} = \frac{80\pi}{3}$ cm.

  Length of the curved part of one semicircle = $\pi r = \pi(10) = 10\pi$ cm. Length of the curved part of two semicircles = $2 \times 10\pi = 20\pi$ cm.

  Total perimeter of the metal plate = (Major arc length) + (Arc length of two semicircles) Total Perimeter = $\frac{80\pi}{3} + 20\pi$ Total Perimeter = $\frac{80\pi}{3} + \frac{60\pi}{3}$ Total Perimeter = $\frac{140\pi}{3}$ cm.

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  How the marks are awarded

  • B1 — Correctly setting up the area of a semicircle equation (1/2 πr² = 50π) and solving it to find the radius r = 10.
  • M1 — Using the calculated radius (r=10) to find the chord length (20) and then using this in the context of the large circle (radius R=20) to find the angle θ = π/3. This can be done by identifying the equilateral triangle or using the cosine rule.
  • DM1 — Calculating the length of the major arc. This requires finding the correct angle (2π - 2θ) and using the arc length formula s = Rα. This mark is dependent on having found a value for θ.
  • DM1 — Formulating the total perimeter by adding the arc length of the two semicircles (2 * πr) to the major arc length calculated previously. This mark is dependent on the first DM1 mark.
  • A1 — Obtaining the final, correct answer of 140π/3, presented as a single exact term.

  Common mistakes

  • Calculating the radius by using the formula for a full circle, i.e., πr² = 50π, leading to an incorrect radius.
  • Calculating the length of the minor arc (20 × 2π/3) instead of the major arc (20 × 4π/3) for the perimeter.
  • Mixing degrees and radians, for example, finding θ = 60° but then using it in the arc length formula s = rθ without converting to radians.
  • Forgetting to include all parts of the perimeter, for instance, only calculating the major arc length and forgetting to add the two semicircle arcs.

  Examiner tip: Always break down complex perimeter or area problems into their simpler geometric components, calculate each part methodically, and then combine them for the final answer.

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