3D Pythagoras and trigonometry Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE 3D Pythagoras and trigonometry questions. See exactly how to solve problems on 3D Pythagoras, space diagonal of a cuboid, working in a 2D sub-triangle, base diagonal.

3D Pythagorasspace diagonal of a cuboidworking in a 2D sub-trianglebase diagonalapplied 3D problempyramids
GCSE Higher70 questionsStep-by-step solutions
Question 1
2 markseasy
The diagram shows a cuboid ABCDEFGHABCDEFGH with horizontal base ABCDABCD and top face EFGHEFGH, where EE is directly above AA, FF is directly above BB, GG is directly above CC and HH is directly above DD. AB=3AB = 3 cm, BC=4BC = 4 cm and AE=12AE = 12 cm. Work out the length of the diagonal AGAG.

Worked solution

  1. State the 3D version of Pythagoras theorem.

    d2=a2+b2+c2d^2 = a^2 + b^2 + c^2

    For a cuboid with edges aa, bb and cc, the space diagonal dd satisfies d2=a2+b2+c2d^2 = a^2 + b^2 + c^2. It comes from using Pythagoras twice: once across the base, then once in the upright triangle.

  2. Substitute the three edges of the cuboid.

    AG2=32+42+122=9+16+144=169AG^2 = 3^2 + 4^2 + 12^2 = 9 + 16 + 144 = 169

    The three edges meeting at AA are 33 cm, 44 cm and 1212 cm. Squaring and adding gives AG2=169AG^2 = 169. Underneath, this is Pythagoras used twice: across the base to get ACAC, then up the triangle ACGACG.

  3. State the length of the diagonal.

    AG=169=13 cmAG = \sqrt{169} = 13\text{ cm}

    The diagonal AGAG is 1313 cm.

Answer
AG=13 cmAG = 13\text{ cm}
Question 2
2 markseasy
The diagram shows a cuboid ABCDEFGHABCDEFGH with horizontal base ABCDABCD and top face EFGHEFGH, where EE is directly above AA, FF is directly above BB, GG is directly above CC and HH is directly above DD. AB=9AB = 9 cm, BC=12BC = 12 cm and AE=20AE = 20 cm. Work out the length of the diagonal AGAG.

Worked solution

  1. State the 3D version of Pythagoras theorem.

    d2=a2+b2+c2d^2 = a^2 + b^2 + c^2

    For a cuboid with edges aa, bb and cc, the space diagonal dd satisfies d2=a2+b2+c2d^2 = a^2 + b^2 + c^2. It comes from using Pythagoras twice: once across the base, then once in the upright triangle.

  2. Substitute the three edges of the cuboid.

    AG2=92+122+202=81+144+400=625AG^2 = 9^2 + 12^2 + 20^2 = 81 + 144 + 400 = 625

    The three edges meeting at AA are 99 cm, 1212 cm and 2020 cm. Squaring and adding gives AG2=625AG^2 = 625. Underneath, this is Pythagoras used twice: across the base to get ACAC, then up the triangle ACGACG.

  3. State the length of the diagonal.

    AG=625=25 cmAG = \sqrt{625} = 25\text{ cm}

    The diagonal AGAG is 2525 cm.

Answer
AG=25 cmAG = 25\text{ cm}
Question 3
2 markseasy
The diagram shows a cuboid ABCDEFGHABCDEFGH with horizontal base ABCDABCD and top face EFGHEFGH, where EE is directly above AA, FF is directly above BB, GG is directly above CC and HH is directly above DD. AB=2AB = 2 cm, BC=3BC = 3 cm and AE=6AE = 6 cm. Work out the length of the diagonal AGAG.

Worked solution

  1. State the 3D version of Pythagoras theorem.

    d2=a2+b2+c2d^2 = a^2 + b^2 + c^2

    For a cuboid with edges aa, bb and cc, the space diagonal dd satisfies d2=a2+b2+c2d^2 = a^2 + b^2 + c^2. It comes from using Pythagoras twice: once across the base, then once in the upright triangle.

  2. Substitute the three edges of the cuboid.

    AG2=22+32+62=4+9+36=49AG^2 = 2^2 + 3^2 + 6^2 = 4 + 9 + 36 = 49

    The three edges meeting at AA are 22 cm, 33 cm and 66 cm. Squaring and adding gives AG2=49AG^2 = 49. Underneath, this is Pythagoras used twice: across the base to get ACAC, then up the triangle ACGACG.

  3. State the length of the diagonal.

    AG=49=7 cmAG = \sqrt{49} = 7\text{ cm}

    The diagonal AGAG is 77 cm.

Answer
AG=7 cmAG = 7\text{ cm}
Question 4
2 markseasy
The diagram shows a cuboid ABCDEFGHABCDEFGH with horizontal base ABCDABCD and top face EFGHEFGH, where EE is directly above AA, FF is directly above BB, GG is directly above CC and HH is directly above DD. AB=6AB = 6 cm, BC=6BC = 6 cm and AE=7AE = 7 cm. Work out the length of the diagonal AGAG.

Worked solution

  1. State the 3D version of Pythagoras theorem.

    d2=a2+b2+c2d^2 = a^2 + b^2 + c^2

    For a cuboid with edges aa, bb and cc, the space diagonal dd satisfies d2=a2+b2+c2d^2 = a^2 + b^2 + c^2. It comes from using Pythagoras twice: once across the base, then once in the upright triangle.

  2. Substitute the three edges of the cuboid.

    AG2=62+62+72=36+36+49=121AG^2 = 6^2 + 6^2 + 7^2 = 36 + 36 + 49 = 121

    The three edges meeting at AA are 66 cm, 66 cm and 77 cm. Squaring and adding gives AG2=121AG^2 = 121. Underneath, this is Pythagoras used twice: across the base to get ACAC, then up the triangle ACGACG.

  3. State the length of the diagonal.

    AG=121=11 cmAG = \sqrt{121} = 11\text{ cm}

    The diagonal AGAG is 1111 cm.

Answer
AG=11 cmAG = 11\text{ cm}
Question 5
1 markeasy
The diagram shows a cuboid ABCDEFGHABCDEFGH with horizontal base ABCDABCD and top face EFGHEFGH, where EE is directly above AA, FF is directly above BB, GG is directly above CC and HH is directly above DD. AB=6AB = 6 cm, BC=8BC = 8 cm and AE=5AE = 5 cm. Work out the length of the diagonal ACAC of the base.

Worked solution

  1. Pick out the right-angled triangle you need.

    AC2=AB2+BC2AC^2 = AB^2 + BC^2

    The diagonal ACAC lies entirely in the base, so this is an ordinary 2D Pythagoras problem in the rectangle ABCDABCD. The height of the cuboid is not needed at all.

  2. Substitute the two base edges.

    AC2=62+82=36+64=100AC^2 = 6^2 + 8^2 = 36 + 64 = 100

    AB=6AB = 6 cm and BC=8BC = 8 cm are the two shorter sides of triangle ABCABC, which is right-angled at BB.

  3. State the length of the base diagonal.

    AC=100=10 cmAC = \sqrt{100} = 10\text{ cm}

    The base diagonal ACAC is 1010 cm.

Answer
AC=10 cmAC = 10\text{ cm}

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