An allotment holder is planning a rectangular plot of width \(8\) m and unknown length \(x\) m, with a semicircular herb bed of diameter \(8\) m attached al...
Question 1 Report
An allotment holder is planning a rectangular plot of width \(8\) m and unknown length \(x\) m, with a semicircular herb bed of diameter \(8\) m attached along one short edge, as shown. The total area enclosed is \(100\) m\(^{2}\).
- Form an equation in \(x\) and solve it, giving \(x\) correct to 3 significant figures. (3)
- Work out the total perimeter of the plot, correct to 3 significant figures. (3)
- Fencing costs \(\pounds6\) per metre. The allotment holder has budgeted \(\pounds150\). State, with a reason, whether this is enough. (2)
Answer Details
(a) The total area is the rectangle plus the semicircular herb bed. The rectangle has area \(8x\), and the semicircle, of diameter \(8\) m so radius \(4\) m, has area \(\dfrac{1}{2}\pi(4)^{2} = 8\pi\). Setting the total equal to \(100\) m\(^{2}\) gives \(8x + 8\pi = 100\), so \(8x = 100-8\pi\), giving \(x = \dfrac{100-8\pi}{8} = 9.36\) m, correct to 3 significant figures. [3 marks]
(b) The perimeter is made up of the two long sides, each of length \(x\), the one remaining short side of length \(8\) m, and the curved semicircular arc that replaces the other short side, of length \(\pi \times 4\). Using the unrounded value \(x = 9.3584...\): perimeter \(= 2(9.3584...) + 8 + 4\pi = 39.3\) m, correct to 3 significant figures. [3 marks]
(c) The fencing cost, using the unrounded perimeter \(39.283...\) m, is \(39.283... \times \pounds6 = \pounds235.70\), which is more than the \(\pounds150\) budget, so the allotment holder does not have enough. [2 marks]
Using the full, unrounded value of \(x\) in part (b), rather than the value already rounded to 3 significant figures, keeps the perimeter accurate; only the final answer in each part should be rounded.
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