Surds and exact values – Year 10 GCSE maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. Area \(=\pi\times25=\) \(25\pi\) cm² (1). Circumference \(=\) \(10\pi\) cm (1) Halve the diameter to get the radius 5 cm, then use \(A=\pi r^2\) and \(C=\pi d\). The area is \(25\pi\) cm² and the circumference is \(10\pi\) cm.
   2. \(\dfrac{60}{360}\times\pi\times36=\) \(6\pi\) cm² (1 for method, 1 for the answer) The sector is \(\tfrac{60}{360}=\tfrac16\) of the circle of area \(36\pi\). The area of the sector is \(6\pi\) cm².

2. Answer to Question 2 (4 marks)

   1. \(\sqrt{25+100}=\sqrt{125}\) (1) \(=\) \(5\sqrt5\) cm (1) Use Pythagoras for the hypotenuse, then simplify with the largest square factor: \(\sqrt{125}=\sqrt{25\times5}=5\sqrt5\).
   2. \(\tfrac12\times5\times10=\) 25 cm² (1) The two shorter sides are perpendicular, so they are the base and the height. The area is 25 cm².
   3. \(5+10+5\sqrt5=\) \(15+5\sqrt5\) (1) Add the three sides. The perimeter is \(15+5\sqrt5\) cm, so \(k=5\).

   Check (a) with a calculator: \(5\sqrt5\approx11.18\), and \(\sqrt{125}\approx11.18\).

3. Answer to Question 3 (4 marks)

   1. \(5\sqrt3+3\sqrt3=\) \(8\sqrt3\) (1 for each simplified surd, 1 for the sum) Simplify each surd: \(\sqrt{75}=\sqrt{25\times3}=5\sqrt3\) and \(\sqrt{27}=\sqrt{9\times3}=3\sqrt3\). Then add the like surds.
   2. \(16-3=\) 13 (1) This is a difference of two squares: \(4^2-(\sqrt3)^2=16-3\).
   3. \(4+4\sqrt5+5=\) \(9+4\sqrt5\) (1) Expand \((2+\sqrt5)(2+\sqrt5)\): there are two middle terms of \(2\sqrt5\), and \(\sqrt5\times\sqrt5=5\).

   Check (b): the product has no surd because the middle terms \(-4\sqrt3\) and \(+4\sqrt3\) cancel. A common error in (c) is to write \((2+\sqrt5)^2=4+5=9\), forgetting the middle terms.

4. Answer to Question 4 (4 marks)

   1. \(\dfrac{14\sqrt7}{7}=\) \(2\sqrt7\) (1, 1) Multiply the top and bottom by \(\sqrt7\), then simplify \(\tfrac{14}{7}=2\). The answer is \(2\sqrt7\).
   2. \(\dfrac{6(3+\sqrt3)}{9-3}\) (1) \(=\) \(3+\sqrt3\) (1) Multiply the top and bottom by the conjugate \(3+\sqrt3\). The bottom is \(9-3=6\), and \(\tfrac{6}{6}=1\). The answer is \(3+\sqrt3\), so \(a=3\) and \(b=1\).

   Check (b) with a calculator: \(6\div(3-1.732)\approx4.732\) and \(3+\sqrt3\approx4.732\). Using \(3-\sqrt3\) instead of the conjugate leaves a surd in the denominator.

5. Answer to Question 5 (4 marks)

   1. \(\sqrt{25+144}=\) 13 cm (1) The radius, the height and the slant height form a right-angled triangle. The slant height is 13 cm.
   2. \(\tfrac13\pi\times25\times12=\) \(100\pi\) cm³ (1) Use \(V=\tfrac13\pi r^2h\) with \(r=5\) and \(h=12\). The volume is \(100\pi\) cm³.
   3. \(\pi\times5\times13=\) \(65\pi\) cm² (1) Use \(\pi rl\) with the slant height \(l=13\). The curved surface area is \(65\pi\) cm².
   4. \(65\pi+25\pi=\) \(90\pi\) cm² (1) Add the circular base, \(\pi\times5^2=25\pi\). The total surface area is \(90\pi\) cm².

Total : 20 marks