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

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1. Answer to Question 1

   1. \(2\pi\times5=\) \(10\pi\) cm \(=31.4\) cm Use \(C=2\pi r\) with \(r=5\). The exact answer is \(10\pi\) cm, which is about 31.4 cm.
   2. radius 6, area \(=\pi\times36=\) \(36\pi\) cm² \(=113.1\) cm² Halve the diameter to get the radius 6 cm, then use \(A=\pi r^2\). The exact area is \(36\pi\) cm².
   3. \(\tfrac12\pi\times64=\) \(32\pi\) cm² \(=100.5\) cm² A semicircle is half of a circle of area \(\pi\times8^2=64\pi\). The exact area is \(32\pi\) cm².
   4. \(\pi\times9\times10=\) \(90\pi\) cm³ \(=282.7\) cm³ Use \(V=\pi r^2h\) with \(r=3\) and \(h=10\). The exact volume is \(90\pi\) cm³. Exact answers in terms of \(\pi\) avoid rounding errors.

2. Answer to Question 2

   1. \(\sqrt{16}=4\): not a surd (rational). 16 is a square number, so its square root is a whole number.
   2. \(\sqrt{12}=3.4641\ldots\) does not end or repeat: a surd. 12 is not a square number, so \(\sqrt{12}\) is irrational. We can simplify it to \(2\sqrt3\), but it is still a surd.
   3. \(\sqrt{\tfrac94}=\tfrac32\): not a surd. Both 9 and 4 are square numbers, so the root is the fraction \(\tfrac32\).
   4. \(\sqrt7\) is a surd. \(\pi\) is irrational but is not a root, so it is not a surd. 7 is not a square number, so \(\sqrt7\) is irrational and is a surd.

3. Answer to Question 3

   1. \(\sqrt{25\times2}=\) \(5\sqrt2\) 25 is the largest square number that divides 50, and \(\sqrt{25}=5\) comes outside the root. So \(\sqrt{50}=5\sqrt2\).
   2. \(\sqrt{36\times2}=\) \(6\sqrt2\) 36 is the largest square factor of 72; using \(9\times8\) would not finish the job. So \(\sqrt{72}=6\sqrt2\).
   3. \(\sqrt{16\times3}=\) \(4\sqrt3\) 16 is the largest square factor of 48, and \(\sqrt{16}=4\). So \(\sqrt{48}=4\sqrt3\).
   4. \(\sqrt{100\times2}=\) \(10\sqrt2\) 100 is the largest square factor of 200. Check: \((10\sqrt2)^2=100\times2=200\).

4. Answer to Question 4

   1. \(8\sqrt2\) These are like surds, so we add them like \(3x+5x=8x\). The answer is \(8\sqrt2\).
   2. \(2\sqrt3+3\sqrt3=\) \(5\sqrt3\) Simplify each surd: \(\sqrt{12}=\sqrt{4\times3}=2\sqrt3\) and \(\sqrt{27}=\sqrt{9\times3}=3\sqrt3\). Now they are like surds, so the answer is \(5\sqrt3\).
   3. \(5\sqrt2-2\sqrt2=\) \(3\sqrt2\) Simplify each surd: \(\sqrt{50}=5\sqrt2\) and \(\sqrt8=2\sqrt2\). The answer is \(3\sqrt2\).
   4. \(6\sqrt5-2\sqrt5=\) \(4\sqrt5\) Simplify each surd: \(\sqrt{45}=3\sqrt5\), so \(2\sqrt{45}=6\sqrt5\), and \(\sqrt{20}=2\sqrt5\). The answer is \(4\sqrt5\).

5. Answer to Question 5

   1. \(\sqrt{36}=\) 6 Use \(\sqrt a\times\sqrt b=\sqrt{ab}\): \(3\times12=36\), which is a square number. The answer is 6.
   2. \(4-3=\) 1 (difference of two squares) This has the form \((a+b)(a-b)=a^2-b^2\), and \((\sqrt3)^2=3\), so the surds cancel. The answer is 1.
   3. \(1+2\sqrt5+5=\) \(6+2\sqrt5\) Write the square as \((1+\sqrt5)(1+\sqrt5)\). There are two middle terms of \(\sqrt5\), and \(\sqrt5\times\sqrt5=5\). The answer is \(6+2\sqrt5\).
   4. \(9-2=\) 7 This is again a difference of two squares: \(3^2-(\sqrt2)^2=9-2\). The answer is 7.

6. Answer to Question 6

   1. \(\dfrac{6\sqrt3}{3}=\) \(2\sqrt3\) Multiply the top and the bottom by \(\sqrt3\), because \(\sqrt3\times\sqrt3=3\). Then simplify \(\tfrac63=2\). The answer is \(2\sqrt3\).
   2. \(\dfrac{10\sqrt5}{5}=\) \(2\sqrt5\) Multiply the top and the bottom by \(\sqrt5\), then simplify \(\tfrac{10}{5}=2\). The answer is \(2\sqrt5\).
   3. Multiply by \(\dfrac{1-\sqrt2}{1-\sqrt2}\): \(\dfrac{1-\sqrt2}{1-2}=\) \(\sqrt2-1\) Multiply by the conjugate \(1-\sqrt2\), so the bottom becomes \(1^2-(\sqrt2)^2=-1\). Dividing by \(-1\) changes both signs. The answer is \(\sqrt2-1\).
   4. Multiply by \(\dfrac{2+\sqrt3}{2+\sqrt3}\): \(\dfrac{3(2+\sqrt3)}{4-3}=\) \(6+3\sqrt3\) Multiply by the conjugate \(2+\sqrt3\), so the bottom becomes \(4-3=1\). The answer is \(6+3\sqrt3\).

7. Answer to Question 7

   1. \(\pi\times4^2=\) \(16\pi\) m² Use \(A=\pi r^2\) with \(r=4\). The pond has area \(16\pi\) m².
   2. Outer radius \(=5\) m, area \(=25\pi\) m². The path adds 1 m to the radius, so the outer radius is \(4+1=5\) m.
   3. \(25\pi-16\pi=\) \(9\pi\) m² (\(\approx28.3\) m²). The path is the ring between the two circles, so we subtract the area of the pond from the area of the whole circle.
   4. Pond: \(2\pi\times4=\) \(8\pi\) m. Outer: \(2\pi\times5=\) \(10\pi\) m. Use \(C=2\pi r\) for each circle. The outer edge is \(2\pi\) m longer than the edge of the pond.

8. Answer to Question 8

   1. \(\sqrt{3^2+5^2}=\) \(\sqrt{34}\) m (5.83 m) The brace is the hypotenuse of a right-angled triangle with sides 3 m and 5 m. 34 has no square factor, so \(\sqrt{34}\) is already in its simplest form.
   2. \(\sqrt{25+25}=\sqrt{50}=\) \(5\sqrt2\) m The square's diagonal is the hypotenuse of a triangle with two sides of 5 m. Simplify \(\sqrt{50}=\sqrt{25\times2}=5\sqrt2\).
   3. \(5\sqrt2=\) 7.1 m Using \(\sqrt2\approx1.4142\), \(5\sqrt2\approx7.07\), which is 7.1 m to 1 decimal place.
   4. \(5\sqrt2-\sqrt{34}=7.07-5.83=\) 1.24 m longer for the square gate. Change both lengths to decimals, then subtract.

9. Answer to Question 9

   1. \(\sqrt{16+36}=\sqrt{52}=\) \(2\sqrt{13}\) m The longest side is the hypotenuse, so we use Pythagoras. Then simplify: \(\sqrt{52}=\sqrt{4\times13}=2\sqrt{13}\).
   2. \(\tfrac12\times4\times6=\) 12 m² The two shorter sides are perpendicular, so they are the base and the height. The area is 12 m².
   3. \(4+6+2\sqrt{13}=\) \(10+2\sqrt{13}\) m (17.2 m) Add the three sides. The whole numbers and the surd cannot be combined, so the exact perimeter is \(10+2\sqrt{13}\) m.
   4. \(12\times12=\) £144 Multiply the area by the price per square metre. The cloth costs £144.

10. Answer to Question 10

    1. Half the base is 3 m: \(h=\sqrt{36-9}=\sqrt{27}=\) \(3\sqrt3\) m The height cuts the equilateral triangle into two right-angled triangles with hypotenuse 6 m and base 3 m. Then \(\sqrt{27}=\sqrt{9\times3}=3\sqrt3\).
    2. \(\tfrac12\times6\times3\sqrt3=\) \(9\sqrt3\) m² Use area \(=\tfrac12\times\) base \(\times\) height with base 6 and height \(3\sqrt3\). The exact area is \(9\sqrt3\) m².
    3. \(9\sqrt3=\) 15.6 m² Using \(\sqrt3\approx1.732\), \(9\sqrt3\approx15.59\), which is 15.6 m².
    4. \(4\times9\sqrt3=\) \(36\sqrt3\) m² Multiply the exact area by 4. The total area is \(36\sqrt3\) m², about 62.4 m².

11. Answer to Question 11

    1. \(\pi\times9\times10=\) \(90\pi\) cm³ Use \(V=\pi r^2h\) with \(r^2=9\) and \(h=10\). The volume of the can is \(90\pi\) cm³.
    2. \(\tfrac13\pi\times9\times4=\) \(12\pi\) cm³ Use \(V=\tfrac13\pi r^2h\) with \(r^2=9\) and \(h=4\). The volume of the cone is \(12\pi\) cm³.
    3. \(\sqrt{3^2+4^2}=\) 5 cm The radius, the vertical height and the slant height form a right-angled triangle. The slant height is 5 cm.
    4. \(\pi rl=\pi\times3\times5=\) \(15\pi\) cm² Use the slant height \(l=5\), not the vertical height. The curved surface area is \(15\pi\) cm².

12. Answer to Question 12

    1. \(\sqrt{16+16}=\sqrt{32}=\) \(4\sqrt2\) cm The face diagonal is the hypotenuse of a right-angled triangle with two sides of 4 cm. Then \(\sqrt{32}=\sqrt{16\times2}=4\sqrt2\).
    2. \(\sqrt{(4\sqrt2)^2+4^2}=\sqrt{32+16}=\sqrt{48}=\) \(4\sqrt3\) cm The space diagonal is the hypotenuse of a right-angled triangle formed by a face diagonal and a vertical edge. Then \(\sqrt{48}=\sqrt{16\times3}=4\sqrt3\).
    3. \(4\sqrt3=\) 6.93 cm Using \(\sqrt3\approx1.7321\), \(4\sqrt3\approx6.93\) cm.
    4. \(4\sqrt3=\sqrt3\times4\) ✓ So the space diagonal is \(\sqrt3\) times the side length of 4 cm. In general, a cube of side \(a\) has space diagonal \(\sqrt{3a^2}=a\sqrt3\).

13. Answer to Question 13

    1. \(\tfrac43\pi\times216=\) \(288\pi\) cm³ Use \(V=\tfrac43\pi r^3\) with \(r^3=6^3=216\), and \(\tfrac43\times216=288\). The volume of the ball is \(288\pi\) cm³.
    2. \(4\pi\times36=\) \(144\pi\) cm² Use \(A=4\pi r^2\) with \(r^2=36\). The surface area is \(144\pi\) cm².
    3. \(\tfrac12\times288\pi=\) \(144\pi\) cm³ A hemisphere is half a sphere, so its volume is half of \(288\pi\).
    4. \(288\pi=\) 905 cm³ Using \(\pi\approx3.14159\), \(288\pi\approx904.8\). The ball has a volume of about 905 cm³.

14. Answer to Question 14

    1. \(\sqrt{12^2+1^2}=\) \(\sqrt{145}\) m The ramp is the hypotenuse of a right-angled triangle with sides 12 m and 1 m. 145 has no square factor, so \(\sqrt{145}\) is the exact length.
    2. 12.04 m \(\sqrt{145}=12.0416\ldots\), so the ramp is 12.04 m long.
    3. \(\dfrac1{12}=\) 8.3% The gradient is the rise divided by the horizontal distance: \(1\div12=0.0833\), which is about 8.3%.
    4. The gradient is exactly 1 in 12, so it just meets the guidance. A gradient of 1 in 12 is the steepest allowed, so this ramp is acceptable.

15. Answer to Question 15

    1. \(\sqrt8=2\sqrt2\), so \((2\sqrt2+\sqrt2)^2=(3\sqrt2)^2=9\times2=18\) ✓. Simplify \(\sqrt8=\sqrt{4\times2}=2\sqrt2\) first, then add the like surds before squaring.
    2. \(3\sqrt2+2\sqrt2=5\sqrt2\) ✓. Simplify each surd: \(\sqrt{18}=\sqrt{9\times2}=3\sqrt2\) and \(\sqrt8=2\sqrt2\). Then add the like surds.
    3. \(\sqrt{12}=2\sqrt3\), so \(\dfrac{4+2\sqrt3}{2}=2+\sqrt3\) ✓. Simplify \(\sqrt{12}=\sqrt{4\times3}=2\sqrt3\), then divide each term on the top by 2.
    4. Multiply top and bottom by \(\sqrt2+1\): \(\dfrac{(\sqrt2+1)^2}{2-1}=2+2\sqrt2+1=3+2\sqrt2\) ✓. The bottom becomes \((\sqrt2)^2-1^2=1\). Expanding \((\sqrt2+1)^2\) gives two middle terms of \(\sqrt2\).

16. Answer to Question 16

    1. \((3+\sqrt5)(3-\sqrt5)=9-5=\) 4 ✓ The product has the form \((a+b)(a-b)=a^2-b^2\), so the surds cancel and the area is a whole number.
    2. \(2\left[(3+\sqrt5)+(3-\sqrt5)\right]=2\times6=\) 12 cm The perimeter is twice the sum of the sides, and the \(\sqrt5\) terms cancel.
    3. \(d^2=(3+\sqrt5)^2+(3-\sqrt5)^2=14+6\sqrt5+14-6\sqrt5=28\), so \(d=\sqrt{28}=\) \(2\sqrt7\) cm Use Pythagoras. Each square has two middle terms \(\pm6\sqrt5\), which cancel. Then \(\sqrt{28}=\sqrt{4\times7}=2\sqrt7\).
    4. Sides 5.236 and 0.764: \(\sqrt{27.416+0.584}=\sqrt{28}=5.29\) cm, and \(2\sqrt7=5.29\) ✓. Both methods give 5.29 cm, so the exact answer is correct.

17. Answer to Question 17

    1. \(x=\pm\sqrt{20}=\) \(\pm2\sqrt5\) Take the square root of both sides, remembering \(\pm\). Simplify \(\sqrt{20}=\sqrt{4\times5}=2\sqrt5\).
    2. \(x+3=\pm\sqrt7\), so \(x=-3\pm\sqrt7\) Take the square root of both sides, remembering \(\pm\), then subtract 3 from both sides.
    3. \(x^2=25\), so \(x=\pm5\) Divide both sides by 2 first, then take the square root. 25 is a square number, so the solutions are whole numbers.
    4. \(x=\dfrac{6\pm\sqrt{36-8}}{2}=\dfrac{6\pm\sqrt{28}}2=\dfrac{6\pm2\sqrt7}{2}=\) \(3\pm\sqrt7\) Use the formula with \(a=1\), \(b=-6\), \(c=2\). Simplify \(\sqrt{28}=2\sqrt7\), then divide each term by 2.

18. Answer to Question 18

    1. \(2\sqrt3\), \(2\sqrt7\), \(2\sqrt{10}\) Use the largest square factors: \(\sqrt{12}=\sqrt{4\times3}\), \(\sqrt{28}=\sqrt{4\times7}\), and \(2\sqrt{10}\) is already simplified.
    2. \(12+28=40=(2\sqrt{10})^2\), so by the converse of Pythagoras the triangle is right-angled. Square each side: \((2\sqrt3)^2=12\), \((2\sqrt7)^2=28\) and \((2\sqrt{10})^2=40\).
    3. \(\tfrac12\times2\sqrt3\times2\sqrt7=2\sqrt{21}\) cm² The two shorter sides are perpendicular, so they are the base and the height. Multiply \(2\times2=4\) and \(\sqrt3\times\sqrt7=\sqrt{21}\), then halve.
    4. \(2\sqrt{21}=9.17\) cm². Using \(\sqrt{21}\approx4.5826\), the area is about 9.17 cm².

19. Answer to Question 19

    1. \(\dfrac12\times\dfrac12=\) \(\dfrac14\) Read both values from the table: \(\sin30^\circ=\tfrac12\) and \(\cos60^\circ=\tfrac12\). The exact answer is \(\tfrac14\).
    2. \(\dfrac{\sqrt3}2\times\dfrac{\sqrt3}2+\dfrac12\times\dfrac12=\dfrac34+\dfrac14=1\) ✓ Substitute the exact values from the table: \(\tfrac{\sqrt3}{2}\times\tfrac{\sqrt3}{2}=\tfrac34\) because \(\sqrt3\times\sqrt3=3\).
    3. \(6\sin60^\circ=6\times\dfrac{\sqrt3}{2}=\) \(3\sqrt3\) m (5.2 m) The ladder is the hypotenuse and the height up the wall is opposite the \(60^\circ\) angle, so we use sine. The ladder reaches \(3\sqrt3\) m up the wall.
    4. \(\dfrac1{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}=\) \(\dfrac{\sqrt3}{3}\) Multiply the top and bottom by \(\sqrt3\), so the denominator becomes 3. So \(\tan30^\circ=\tfrac{\sqrt3}{3}\).

20. Answer to Question 20

    1. \(a^2+2ab\sqrt2+2b^2\). Write the square as \((a+b\sqrt2)(a+b\sqrt2)\); the last term is \(b^2\times2=2b^2\).
    2. \(a^2+2b^2=17\) and \(2ab=12\), so \(ab=6\). Two expressions of the form \(p+q\sqrt2\) are equal only if their whole-number parts and their \(\sqrt2\) parts are equal.
    3. Try \(a=3\), \(b=2\): \(9+8=17\) ✓ and \(3\times2=6\) ✓. The pairs with \(ab=6\) are \(1\times6\), \(2\times3\), \(3\times2\) and \(6\times1\); only \(a=3\), \(b=2\) also gives \(a^2+2b^2=17\).
    4. \(\sqrt{17+12\sqrt2}=3+2\sqrt2\). Check: \(3+2.828=5.828\) and \(\sqrt{17+16.97}=\sqrt{33.97}=5.828\) ✓.