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Q16. Answer D.

Explanation A pentagon is a five-sided polygon and an octagon is an eight-sided polygon. They can be divided as follows:

Q17. Answer D.

Explanation Circumference = 2πr = 2 × 3.14 × 5 = 31.4.

Q18. Answer B.

Explanation C = πd.

Q19. Answer D.

Explanation 31.4 = 3.14 × 2 × X; 31.4/3.14 = 10; 10/2 = 5.

Q20. Answer C.

Explanation Area of a circle = πr2 = 3 × 3 × 3.14 = 28.26.

Q21. Answer A.

Explanation r =1/2d; 3.14 r2 = 3.14 × 5 × 5 = 78.5.

Q22. Answer C.

Explanation C = 2πr; 25.12 = 2πr; r = 25.12/6.28 = 4.

Area = πr2 = 3.14 × 16 = 50.24.

Q23. Answer B.

Explanation C = πd; d = 21/3 = 7; r = d/2 = 3.5.

Area = πr2; r2 = 12.25; 12.25 × 3 = 36.75.

Q24. Answer D.

Explanation Volume = 4/3πr3, r3 = 27, 4/3 × 3.14 × 27 = = 113.04.

Q25. Answer C.

Explanation A hemisphere is half a sphere. So it is the same calculation as for Q24 divided by 2.

Q26. Answer C.

Explanation The shape comprises 3 identical surfaces, each of which occurs twice. So surface area = 2(2×3 + 5×3 + 2×5).

Q27. Answer A.

Explanation Area = length × breadth × height.

Q28. Answer D.

Explanation 60m = 60,000mm; v = πr2h = 3.14 × 144 × 60,000 = 27,129,600mm3 = 27,129.6m3.

Q29. Answer B.

Explanation Area = 36/360 × πr2 = 0.1 × 3.14 × 9 = 2.826cm2.

Q30. Answer D.

Explanation Arc = 72/360 × 2πr = 0.2 × 31.4 = 6.28.

Q31. Answer C.

Explanation Surface area = 2πr2 + 2πrh = 157 + 188.4.

Q32. Answer A.

Explanation The diagram illustrates well Pythagoras’ theorem so x = 9cm2 + 25cm2 = 34cm2.

Q33. Answer B.

Explanation The hypotenuse is the sloping side of the triangle. The square of its length = the sum of the square of the other two sides, so length of H = = 5.

Q34. Answer D.

Explanation x2 = 102 – 82 = 100 – 64 = 36; = 6cm.

Q35. Answer D.

Explanation A right-angled triangle can be drawn on the grid with sides 3cm and 4cm in length, so length of line x = 32 + 42 = 9 + 16 = 25; = 5.

Q36. Answer C.

Explanation If plotted the points would give a line on which a right-angled triangle can be drawn with sides 5cm and 12cm. So distance between points squared = 52 + 122 = 25 + 144 = 169; distance = = 13cm.

Q37. Answer B.

Explanation To get the circumference you need either the radius or diameter. The hypotenuse of the right-angled triangle is the length of the diameter so: c = πd; diameter squared = 42 + 62 = 16 + 36 = 52; d = = 7.21; c = 3.14 × 7.21 = 22.63.

Q38. Answer A.

Explanation 262 = 242 + 102.

Q39. Answer C.

Explanation Tan a =

So x = tan a × 8.

Q40. Answer A.

Explanation Sin a =

So sin a = 8/x.

Q41. Answer B.

Explanation To find an angle from its cosine use the cos–1 function on a calculator.

Cos a = 7/11; size of a = cosine a–1 7/11.

Q42. Answer C.

Explanation Sin 30 = o/h 0.5 = 3/h so h = 6cm.

Q43. Answer A.

Explanation Cos 30 = y/6; 0.866 = y/6; 0.866 × 6 = y.

Q44. Answer D.

Explanation Sin 66.6 = 138/x; 0.917 = 138/x; 0.917 × x = 138; x = 138/.917.

Diameter of semicircle = 150.49; circumference of semicircle = ½ πd + d = 236.269 + 150.49 = 387 (nearest whole cm).

Q45. Answer A.

Explanation Sin 1 = 90° so the triangles are right angled. To find the length of the diameter we use the sine ratio (sin angle = o/h). Sin .5 = 30; = 5.2. .5 = x/5.2; .5 × 5.2 = 2.6cm.

c = πd; c = 3.14 × 2.6 = 8.164.

Q46. Answer A.

Explanation Draw the plane on the diagram and you form a right-angled triangle of which you are given the lengths of two sides. From this you can calculate F,A,B. So tan angle FAB = 18/32.5, so tan angle = 18/32.5 = 0.55385 = 29°.

Q47. Answer D.

Explanation We need to know AC and can find it using Pythagoras’ theorem. AC2 = 32.52 + 252 = 1,056.25 + 625 = 1,681.25; = 41cm. Again using Pythagoras’ theorem we can now find AG.

AG2 = AC2 + CG2.

AG2 = 1681.25 + 324 = 2005.25; AG = = 44.78 (to two decimal places).

Q48. Answer A.

Explanation Plot the relative positions and you realize a right-angled triangle is