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0580/32 Oct/Nov 2023, Question 8: Pythagoras' Theorem

Cambridge IGCSE Mathematics 0580 · Paper 32 (Core) · Calculator · 10 marks

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The question, in short

Geometry and measure revision: construct a triangle from two given sides, measure a length, convert a map scale to real kilometres, find an angle in an isosceles triangle on a straight line, and verify a side of a right-angled triangle.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (ai) · 2 marks

With RS already drawn, use ruler and compasses only to build triangle RST with RT = 7 cm and ST = 4 cm, keeping all construction arcs visible.

  1. 1Draw an arc of radius 7 cm centred at R
  2. 2Draw an arc of radius 4 cm centred at S
  3. 3T is where the arcs cross; rule RT and ST, leaving the arcs

AnswerCorrect triangle with correct arcs

Part (aii) · 1 mark

Find the midpoint of RT and measure how far S is from it, giving the result in millimetres.

  1. 1Mark the midpoint of RT (3.5 cm from R) and measure from S: 52 mm

Answer53 (accept 51 to 55)

Part (b) · 2 marks

A map distance of 8.5 cm stands for a real distance at scale 1 : 50 000; work out that real distance in kilometres.

  1. 18.5 × 50 000 = 425 000 cm
  2. 2425 000 cm = 4250 m = 4.25 km

Answer4.25

Part (c) · 2 marks

On straight line ABCD, triangle BCE has BE = CE and angle DCE = 118°; determine the size of angle x at E.

  1. 1Angle ECB = 180 − 118 = 62° (angles on a straight line)
  2. 2BE = CE, so angle EBC = 62°
  3. 3x = 180 − 62 − 62 = 56

Answer56

Part (d) · 3 marks

In right-angled triangle ABC with the right angle at C, AC = 4.8 cm and AB = 8.9 cm; demonstrate that BC equals 7.5 cm to 2 significant figures.

  1. 14.8² + BC² = 8.9²
  2. 2BC² = 79.21 − 23.04 = 56.17
  3. 3BC = √56.17 = 7.49… = 7.5 cm (2 s.f.)

Answer√(8.9² − 4.8²) = 7.49… or √56.17 = 7.49…

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Also searched as: 0580/32/O/N/23 · 0580_w23_qp_32 · 0580 w23 32 · question 8.
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