3D Pythagoras and trigonometry

Higher only

The idea

Solve 3D problems by finding right-angled triangles inside the shape. The diagonal of a cuboid is $\sqrt{a^2 + b^2 + c^2}$ .

Find the triangle that contains the length or angle you need.
Use Pythagoras on the base first to get a diagonal.
Use that diagonal in an upright triangle with Pythagoras or trigonometry.

Watch out The angle between a line and a plane is measured to the line's projection on the plane, its 'shadow' straight below it.

See it worked

Worked exampleNon-calculator3 marks

A cuboid measures 3 cm by 4 cm by 12 cm.

Work out the length of the longest straight rod that fits inside the cuboid.

Diagonal of the 3 by 4 base. $\sqrt{3^2 + 4^2} = 5$
Then with the height 12. $\sqrt{5^2 + 12^2} = \sqrt{169} = 13$

Answer 13 cm

Your turn

Question 1Non-calculator3 marks

A cuboid has side lengths 3 cm, 4 cm and 12 cm. Work out the length of its space diagonal, joining two opposite vertices through the interior.

Question 2Calculator allowed5 marks

A cuboid has horizontal base 6 cm by 8 cm and vertical height 24 cm.

(a) Find its space diagonal.

(b) Find the angle between the space diagonal and the base, to 1 decimal place.

Question 3Calculator allowed4 marks

A right pyramid has a rectangular base 10 cm by 16 cm. Its apex is vertically above the centre of the base. Its volume is 640 cm³. Work out its total surface area, including the base. Give your answer to 3 significant figures.

Volume of a pyramid $= \frac{1}{3} \times$ area of base $\times$ perpendicular height

You must show your working.

In past papers

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