# Using Pythagoras in 3D problems

In this worksheet, students must use Pythagoras' Theorem to calculate lengths in 3D objects.

Key stage:  KS 4

Curriculum topic:  Geometry and Measures

Difficulty level:

### QUESTION 1 of 10

Pythagoras' Theorem can be applied on any right-angled triangle, even when it is part of a 3 dimensional shape.

Follow this example to see how it is used to find lengths in 3D shapes.

Example

ABCDEFGH is a cuboid (not drawn to scale).

AB = a = 5 cm,
BC = b = 5 cm,
CH = c = 10 cm.

Find the lengths DB and EB in cm to 2 decimal places.

To find DB

Look at Triangle DAB, which is right-angled at A.

By Pythagoras' Theorem,

DB2 = a2 + b2 = 52 + 52 = 25 + 25 = 50

DB = √50 = 7.07 cm

To find EB

Look at Triangle EDB, which is right-angled at D.

By Pythagoras' Theorem,

EB2 = ED2 + DB2 = 102 + 50 = 100 + 50 = 150

(Notice how we use the previous fact that DB2 = 50, no need to use the rounded off DB = 7.07.)

EB = √150 = 12.25 cm

ABCDEFGH is a cuboid (not drawn to scale).

AB = a = 4 cm,
BC = b =  3 cm,
CH = c = 12 cm.

Find the length DB in cm.

ABCDEFGH is a cuboid (not drawn to scale).

AB = a = 4 cm,
BC = b =  3 cm,
CH = c = 12 cm.

Find the length EB in cm.

PQRST is a right square-based pyramid with vertex P directly above the intersection of the base diagonals, M.

QR = a = 7 cm,
PQ = b = 26 cm.

Find the length QS in cm to 2 decimal places.

PQRST is a right square-based pyramid with vertex P directly above the intersection of the base diagonals, M.

QR = a = 7 cm,
PQ = b = 26 cm.

Find the length QM in cm to 2 decimal places.

PQRST is a right square-based pyramid with vertex P directly above the intersection of the base diagonals, M.

QR = a = 7 cm,
PQ = b = 26 cm.

Find the length PM in cm to 2 decimal places.

ABCDEFGH is a cuboid (not drawn to scale).

AB = a = 4 cm,
BC = b = 9 cm,
CH = c = 10 cm.

Find the length DB in cm.

ABCDEFGH is a cuboid (not drawn to scale).

AB = a = 4 cm,
BC = b = 9 cm,
CH = c = 10 cm.

Find the length EB in cm.

PQRST is a right square-based pyramid with vertex P directly above the intersection of the base diagonals, M.

QR = a = 8 cm,
PQ = b = 20 cm.

Find the length QS in cm to 2 decimal places.

PQRST is a right square-based pyramid with vertex P directly above the intersection of the base diagonals, M.

QR = a = 8 cm,
PQ = b = 20 cm.

Find the length QM in cm to 2 decimal places.

PQRST is a right square-based pyramid with vertex P directly above the intersection of the base diagonals, M.

QR = a = 8 cm,
PQ = b = 20 cm.

Find the length PM in cm to 2 decimal places.

• Question 1

ABCDEFGH is a cuboid (not drawn to scale).

AB = a = 4 cm,
BC = b =  3 cm,
CH = c = 12 cm.

Find the length DB in cm.

5
EDDIE SAYS
ΔDAB is a 3, 4, 5 Pythagorean Triad.
• Question 2

ABCDEFGH is a cuboid (not drawn to scale).

AB = a = 4 cm,
BC = b =  3 cm,
CH = c = 12 cm.

Find the length EB in cm.

13
EDDIE SAYS
ΔDAB is a 3, 4, 5 Pythagorean Triad.
ΔEDB is a 5, 12, 13 Pythagorean Triad.
• Question 3

PQRST is a right square-based pyramid with vertex P directly above the intersection of the base diagonals, M.

QR = a = 7 cm,
PQ = b = 26 cm.

Find the length QS in cm to 2 decimal places.

9.90
EDDIE SAYS

Look at ΔQRS, which is right-angled at R.

By Pythagoras' Theorem,

QS2 = a2 + a2 = 72 + 72 = 49 + 49 = 98

QS = √98 = 9.889... = 9.90 cm

• Question 4

PQRST is a right square-based pyramid with vertex P directly above the intersection of the base diagonals, M.

QR = a = 7 cm,
PQ = b = 26 cm.

Find the length QM in cm to 2 decimal places.

4.95
EDDIE SAYS
This is half of QS, which we found in the previous question.
• Question 5

PQRST is a right square-based pyramid with vertex P directly above the intersection of the base diagonals, M.

QR = a = 7 cm,
PQ = b = 26 cm.

Find the length PM in cm to 2 decimal places.

25.52
EDDIE SAYS

Look at ΔPMQ, which is right-angled at M.

By Pythagoras' Theorem,

PM2 = PQ2 - QM2 = 262 - 4.952 = 676 - 24.5 = 651.5

PM = √651.5 = 25.5244... = 25.52 cm

• Question 6

ABCDEFGH is a cuboid (not drawn to scale).

AB = a = 4 cm,
BC = b = 9 cm,
CH = c = 10 cm.

Find the length DB in cm.

9.85
EDDIE SAYS

Look at ΔDAB, which is right-angled at A.

By Pythagoras' Theorem,

DB2 = a2 + b2 = 42 + 92 = 16 + 81 = 97

DB = √97 = 9.8488... = 9.85 cm

• Question 7

ABCDEFGH is a cuboid (not drawn to scale).

AB = a = 4 cm,
BC = b = 9 cm,
CH = c = 10 cm.

Find the length EB in cm.

14.04
EDDIE SAYS

Look at ΔEDB, which is right-angled at D.

By Pythagoras' Theorem, and remembering that DB2 =97 from previous question,

EB2 = ED2 + DB2 = 102 + 97 = 100 + 97 = 197

EB = √197 = 14.0356... = 140.04 cm

• Question 8

PQRST is a right square-based pyramid with vertex P directly above the intersection of the base diagonals, M.

QR = a = 8 cm,
PQ = b = 20 cm.

Find the length QS in cm to 2 decimal places.

11.31
EDDIE SAYS

Look at ΔQRS, which is right-angled at R.

By Pythagoras' Theorem,

QS2 = a2 + a2 = 82 + 82 = 64 + 64 = 128

QS = √128 = 11.3137... = 11.31 cm

• Question 9

PQRST is a right square-based pyramid with vertex P directly above the intersection of the base diagonals, M.

QR = a = 8 cm,
PQ = b = 20 cm.

Find the length QM in cm to 2 decimal places.

5.66
EDDIE SAYS
This is half of QS, which we found in the previous question.
• Question 10

PQRST is a right square-based pyramid with vertex P directly above the intersection of the base diagonals, M.

QR = a = 8 cm,
PQ = b = 20 cm.

Find the length PM in cm to 2 decimal places.

19.18
EDDIE SAYS

Look at ΔPMQ, which is right-angled at M.

By Pythagoras' Theorem,

PM2 = PQ2 - QM2 = 202 - 5.662 = 400 - 32 = 368

PM = √368 = 19.183... = 19.18 cm

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