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0580/33 Oct/Nov 2024, Question 7: Pythagoras' Theorem

Cambridge IGCSE Mathematics 0580 · Paper 33 (Core) · Calculator · 12 marks

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The question, in short

Name a 3D solid from its picture, measure and classify an angle, use circle geometry with AB = 11 cm and BC = 5 cm, and find a sphere's radius from surface area 250 cm².

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

Part (a) · 1 mark

State the mathematical name of the solid shown, which has two circular ends joined by a curved surface.

  1. 1Two circular ends and a curved surface: cylinder

AnswerCylinder

Part (bi) · 1 mark

Use a protractor to find the size of the marked angle x.

  1. 1Place the protractor on the vertex, zero on one arm, read the other arm
  2. 2x = 137°

Answer137 (accept 135 to 139)

Part (bii) · 1 mark

State the mathematical classification of the angle x shown in the diagram.

  1. 1Between 90° and 180°: obtuse

AnswerObtuse

Part (ci) · 1 mark

Explain, using a circle theorem, why angle ACB equals 90° when AB passes through centre O.

  1. 1AB is a diameter, and the angle in a semicircle is 90°

AnswerAngle in a semicircle is 90°

Part (cii) · 2 marks

Work out the circumference of the circle using the given diameter information.

  1. 1AB passes through O, so AB is a diameter: d = 11 cm.
  2. 2Circumference = π × d = π × 11
  3. 3= 34.557… ≈ 34.6 cm

Answer34.6 or 34.55 to 34.562

Part (ciii) · 3 marks

Demonstrate that the length AC rounds to 9.8 cm at 2 significant figures.

  1. 1Pythagoras: AC² + 5² = 11²
  2. 2AC² = 121 − 25 = 96
  3. 3AC = √96 = 9.79… = 9.8 cm to 2 s.f.

Answer11² − 5² = AC², √96 = 9.79… or 9.80

Part (d) · 3 marks

Find the radius of a sphere whose surface area is 250 cm², using A = 4πr².

  1. 14πr² = 250
  2. 2r² = 250 ÷ 4π = 19.89…
  3. 3r = √19.89… = 4.460… ≈ 4.46 cm

Answer4.46 or 4.460 …

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Also searched as: 0580/33/O/N/24 · 0580_w24_qp_33 · 0580 w24 33 · question 7.
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