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G20Pythagoras' theorem (incl. 3D)

Edexcel GCSE Maths (1MA1), Higher tier · Geometry and measures

Practise Pythagoras' theorem (incl. 3D). 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.

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Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy3 marks
The diagram shows a right-angled triangle.
Diagram NOT accurately drawn
[object Object]
Work out the value of \(x\).[3]
Show the answer and mark scheme
Answer: \(x = 15\)
  • M1 for \(9^{2} + 12^{2}\) (\(= 225\))
  • M1 for \(\sqrt{225}\)
  • A1 for 15 cao

Worked solution: \(x^{2} = 9^{2} + 12^{2} = 81 + 144 = 225\), so \(x = \sqrt{225} = 15\).

Question 2Medium3 marks
\(XYZ\) is a right-angled triangle.
Diagram NOT accurately drawn
[object Object]
Work out the length of \(YZ\).[3]
Show the answer and mark scheme
Answer: \(8\) cm
  • M1 for \(10^2 - 6^2\) (= 64)
  • M1 for \(\sqrt{64}\)
  • A1 for \(8\)

Worked solution: \(YZ^2 = 10^2 - 6^2 = 64\)
\(YZ = \sqrt{64} = 8\) cm

Question 3Hard4 marks
A closed umbrella is 1.2 m long. Model it as a straight line segment.
Harry wants to pack the umbrella in a box in the shape of a cuboid measuring 80 cm by 60 cm by 70 cm.
Will the umbrella fit in the box?
You must show your working.[4]
Show the answer and mark scheme
Answer: Yes: the longest diagonal is \(\sqrt{80^2 + 60^2 + 70^2} = \sqrt{14\,900} = 122.1\) cm, which is more than 120 cm.
  • P1 for the diagonal of a face, e.g. \(\sqrt{80^2 + 60^2} = 100\)
  • P1 for \(\sqrt{100^2 + 70^2}\) or \(\sqrt{80^2 + 60^2 + 70^2}\)
  • A1 for 122.06... (or \(\sqrt{14\,900}\))
  • C1 for 'yes' with a comparison of 122.1 cm and 120 cm (1.2 m)

Worked solution: Base diagonal \(= \sqrt{80^2 + 60^2} = \sqrt{10\,000} = 100\) cm.
Space diagonal \(= \sqrt{100^2 + 70^2} = \sqrt{14\,900} = 122.06\ldots\) cm.
1.2 m = 120 cm < 122.06 cm, so the umbrella fits along the space diagonal.

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