O-LevelMathematicsMensuration4024/21 · Oct/Nov 2021 · Q1012 Marks

10 (a) A cuboid measures 6.2 cm by 4.8 cm by 2.5 cm. Each measurement is given correct to the nearest millimetre. Calculate the upper bound of the surface area of the cuboid. cm² (b) [Volume of a pyramid = 1/3 × base area × height] [Figure diagram of a square-based pyramid ABCDE] The diagram shows a square-based pyramid ABCDE. Vertex E is vertically above X, the centre of the square base. The height of the pyramid, EX, is 17 cm. EC = 19 cm. (i) Show that the length of the base is 12 cm. (ii) Calculate the volume of the pyramid. cm³ (iii) Calculate angle CBE. Angle CBE =

(a)A cuboid measures 6.2 cm by 4.8 cm by 2.5 cm. Each measurement is given correct to the nearest millimetre. Calculate the upper bound of the surface area of the cuboid.[3]
(b)(i)Show that the length of the base is 12 cm.[2]
(b)(ii)Calculate the volume of the pyramid.[2]
(b)(iii)Calculate angle CBE.[3]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.

[3] (a) 117.235
[3] B1 for 6.25, 4.85 and 2.55 seen or for 62.5, 48.5 and 25.5 seen
[3] M1 for (their 6.25 × their 4.85 + their 6.25 × their 2.55 + their 2.55 × their 4.85) [× 2] or for (their 62.5 × their 48.5 + their 62.5 × their 25.5 + their 25.5 × their 48.5) [× 2]
[2] (b)(i) XC² = 19²-17² oe
[2] M2 M1 for 19² = XC² + 17²
[2] their XC² + their XC² or d² + d² = (2 × their XC)²
[2] M1 Dependent on use of Pythagoras for XC
[2] √144 = 12
[2] A1
[2] (b)(ii) 816
[2] M1 for 1/3 × 12² × 17 oe
[3] (b)(iii) 71.5 to 71.6
[3] M2 for cos[ ] = 6/19 oe or for cos[ ] = (19²+12²-19²) / (2×19×12) oe
[3] Or M1 for 19² = 12² + 19² – 2×12×19×cos[]

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