Shape and Space 9

4 marks · Normal

The diagram shows a rectangular box. The base AB = 9 cm, BC = 5 cm, and the height is BE = 12 cm. Point F is directly above C.

(a)

Find the length AF.

[2 marks]

(b)

Find the angle that AF makes with the base ABCD.

[2 marks]

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(a) 15 cm

(b) 53.1° degrees

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · Identifies right triangle and applies 3D Pythagoras: or equivalent route.

A1 (a) · 1 mark · Finds cm (3 sf). (follow-through)

M2 (b) · 1 mark · Uses inverse cosine or tangent to determine angle with base.

A2 (b) · 1 mark · Correct angle = 53.1° (1 dp). (follow-through)

Worked solution

a) The 3D diagonal AF can be found using Pythagoras’ theorem:

b) The angle AF makes with the base can be found by considering the right triangle with vertical and horizontal projections: