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O-LevelMathematicsMensuration4024/21 · Oct/Nov 2015 · Q1012 Marks

ABCD is a trapezium with AB parallel to DC. DC = 15 cm and AB = x cm. The perpendicular distance between AB and DC is 3 cm less than the length of AB. The area of ABCD is 75 cm². Show that x² + 12x – 195 = 0.

  1. (a)(i)ABCD is a trapezium with AB parallel to DC. DC = 15 cm and AB = x cm. The perpendicular distance between AB and DC is 3 cm less than the length of AB. The area of ABCD is 75 cm². Show that x² + 12x – 195 = 0.[2]
  2. (a)(ii)Find AB, giving your answer correct to 1 decimal place.[3]
  3. (a)(iii)AD is 0.8 cm longer than BC. Given that the perimeter of the trapezium is 38.0 cm, calculate AD.[2]
  4. (b)(i)Another trapezium, LMNO, has LM parallel to ON. The reflex angle LMN = 252°. Calculate ∠MNO.[2]
  5. (b)(ii)The ratios of the angles inside the trapezium are LÔN : LMN=1:2 and OLM : MÑO=1:k. Find k, giving your answer as a fraction in its simplest form.[3]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2015). The mark or mark type is shown in brackets.

  1. [2] Correct expansion leading to x² + 12x – 195 = 0 www
  2. M1 Or equivalent equation for area
  3. [3] 9.2 cao
  4. B2 for 9.19[8...] or 9.2[0] seen
  5. OR B1 for √12² - 4 × 1 × -195 soi
  6. And B1 for (-12 ± √their 924) / 2 oe
  7. [2] 7.3
  8. M1 for 2AD - 0.8 + 15 + their 9.2 = 38.0 oe
  9. Or 2BC + 0.8 + 15 + their 9.2 = 38.0 oe
  10. Or SC1 for answer [BC = ] 6.5
  11. [2] 72°
  12. B1 for LMN = 108° seen
  13. [3] 4/7
  14. M1 for 126 : their 72 soi or B1 for 126 seen
  15. Or SC2 for answer 7/4

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About this question

Topic
Mensuration, O-Level Mathematics (syllabus 4024). Structured question, 12 marks.
Source
O-Level Mathematics, Oct/Nov 2015 Paper 2 Variant 1, question 10.

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