3D pythagoras / trig. Quiz (Grade 7)

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Score: 0 / 10

3D pythagoras / trig. — Notes & Worked Examples Show / Hide

Explanation:

To solve 3D Pythagoras and trigonometry problems, identify and sketch the key 2D right-angled triangle inside the 3D shape. For a cuboid with dimensions l, w, and h, the floor diagonal is √(l² + w²) and the 3D space diagonal can be found in one step using d = √(l² + w² + h²).

The angle between a line (such as a space diagonal or slant edge) and a horizontal plane (the base) is always the angle between the line and its projection (shadow) on the base directly underneath it.

Example 1: A cuboid ABCD.EFGH measures 12 cm by 9 cm by 8 cm (vertical height). Find space diagonal AG and its angle with the base ABCD (1 d.p.).

Solution:

Base diagonal AC = √(12² + 9²) = √225 = 15 cm

Space diagonal AG = √(15² + 8²) = √289 = 17 cm

In ΔACG: tan(∠GAC) = 815 ⇒ ∠GAC = 28.072...°

Answer: AG = 17 cm, angle = 28.1°

Example 2: A square-based pyramid has base side 10 cm and slant edges 13 cm. Find its vertical height h (3 s.f.).

Solution:

Base diagonal = √(10² + 10²) = √200 cm

Half diagonal (corner to M) = √2002 = √50 cm

h = √(13² − (√50)²) = √(169 − 50) = √119

Answer: h = 10.9 cm (3 s.f.)

One Page Guide (PDF) 3D pythagoras / trig. Worksheets

Question 1

[2 marks]

The diagram shows a cuboid of length 12 cm, width 3 cm, and height 4 cm.
Calculate the exact length of the space diagonal AG.

Worked Solution & Answer

AG = √(12² + 3² + 4²) = √(169) = 13 cm B1

Answer: 13 cm cm

Question 2

[2 marks]

The diagram shows a cuboid of length 24 cm, width 8 cm, and height 6 cm.
Calculate the exact length of the space diagonal AG.

Worked Solution & Answer

AG = √(24² + 8² + 6²) = √(676) = 26 cm B1

Answer: 26 cm cm

Question 3 IGCSE Jun 2018 2H Q19

[3 marks]

The diagram shows a triangular prism.

AF = 10 cm, AB = 24 cm and BC = 8 cm.
Angle FAB = angle ADC = angle BCD = 90°
Work out the size of the angle between the line BE and the plane ABCD.
Give your answer correct to 1 decimal place.

Worked Solution & Answer

BD² = 24² + 8² = 576 + 64 = 640 ⇒ BD = √640 = 25.298… cm M2
tan(∠DBE) = ED / BD = 10 / √640 = 0.39528… ⇒ ∠DBE = 21.568…° = 21.6° (1 d.p.) A1

Answer: 21.6 °

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Question 4 AQA Nov 2019 3H Q19

[3 marks]

Here is a cuboid.
DH = 8 cm
DB = 13 cm

(a) Work out the size of angle DBH. [2 marks]

Answer degrees

(b) Using your answer to part (a), work out the size of angle ECG. [1 mark]

Answer degrees

Worked Solution & Answer

(a)

Triangle BDH is right-angled at D, because DH is vertical and BD is in the horizontal base. M1
\tan(∠ DBH) = DHDB = 813.
∠ DBH = \arctan(813) ≈ 31.6°. A1

Answer (a): Answer 31.6 degrees

(b)

By symmetry of the cuboid, right triangle ECG has base diagonal EC = DB = 13 cm and vertical height CG = DH = 8 cm, so ∠ ECG = ∠ DBH = 31.6°. B1

Answer (b): Answer 31.6 degrees

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Question 5

[4 marks]

ABCDEFGH is a cuboid.

AB = 7.3 cm
CH = 8.1 cm
Angle BCA = 48°
Find the size of the angle between AH and the plane ABCD.
Give your answer correct to 1 decimal place.

Worked Solution & Answer

In right-angled triangle ABC: sin 48° = 7.3 / AC ⇒ AC = 7.3 / sin 48° = 9.8231 cm M3
In right-angled triangle ACH: tan(CAH) = CH / AC = 8.1 / 9.8231 = 0.82459 ⇒ angle CAH = 39.5°. A1

Answer: 39.5

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Question 6 IGCSE Jan 2023 1HR Q21

[4 marks]

The diagram shows a triangular prism, ABCDEF, with a rectangular base ABCD

AB = 6 cm DE = 2.2 cm angle DAE = 18° angle ADE = 90°
Work out the angle that BE makes with the plane ABCD
Give your answer correct to one decimal place.

Worked Solution & Answer

In △ ADE: AD = 2.2\tan 18° = 6.7709 cm M2
In base ABCD: BD = √(6² + 6.7709²) = 9.0468 cm M1
Angle DBE = \tan^-1(2.29.0468) = 13.668° = 13.7° (1 d.p.). A1

Answer: 13.7 °

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Question 7 IGCSE Jan 2020 1H Q19

[4 marks]

The diagram shows a cuboid ABCDEFGH.
EH = 9 cm, HG = 5 cm and GB = 6 cm.

Work out the size of the angle between AH and the plane EFGH.
Give your answer correct to 3 significant figures.

Worked Solution & Answer

FH = √(9² + 5²) = √106 M3
tan(AHF) = AF / FH = 6 / √106 ⇒ angle AHF = tan⁻¹(6 / √106) = 30.23... = 30.2° A1

Answer: 30.2 °

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Question 8

[4 marks]

ABCDEFGH is a cuboid.
Angle GEF = 52°
Angle BHG = 27°
EF = 15 cm

Work out the size of angle GEB.
Give your answer to the nearest degree.

Worked Solution & Answer

In right-triangle EFG: cos(52°) = EF / EG ⇒ EG = 15 ÷ cos(52°) ≈ 24.364 cm M2
In right-triangle BGH: GH = EF = 15 cm. tan(27°) = BG / GH ⇒ BG = 15 × tan(27°) ≈ 7.643 cm M1
In right-triangle BGE (right angle at G): tan(GEB) = BG / EG = 7.643 ÷ 24.364 ≈ 0.3137 ⇒ angle GEB = tan⁻¹(0.3137) ≈ 17.41° ⇒ 17°. A1

Answer: 17 °

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Question 9

[5 marks]

Here is a prism ABCDSPQR.

The base ABCD of the prism is a square of side 14 cm
T is the point on BC such that BT : TC = 4 : 3
The cross section of the prism is in the shape of a trapezium of area 147 cm²
CR = 12 cm
Find the size of the angle between the line ST and the base ABCD.
Give your answer correct to 1 decimal place.

Worked Solution & Answer

(The area of a trapezium is ½(a + b) × height)
TC = (3/7) × 14 = 6 cm M2
DT = √(14² + 6²) = √232 ≈ 15.2315 cm M2
Area of trapezium DC R S = ½(SD + 12) × 14 = 147 ⇒ SD + 12 = 21 ⇒ SD = 9 cm. tan(∠STD) = SD / DT = 9 / √232 ⇒ ∠STD = tan⁻¹(9 / √232) = 30.577... → 30.6° A1

Answer: 30.6

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Question 10 IGCSE Jan 2021 2H Q17

[7 marks]

The diagram shows a solid prism ABCDEFGH.

(a) Work out the total surface area of the prism. [4 marks]

cm²

(b) Calculate the size of the angle between AF and the plane ADEH.
Give your answer correct to one decimal place. [3 marks]

Worked Solution & Answer

(a)

(The area of a trapezium is ½(a + b) × height)
Area of trapezium ABCD = 0.5 × (37 + 28) × 20 = 650 cm² M1
Horizontal offset = (37 – 28)/2 = 4.5 cm M1
Slanted edge AB = √(20² + 4.5²) = √420.25 = 20.5 cm M1
Total surface area = 2(650) + 2(20.5 × 24) + 37 × 24 + 28 × 24 = 1300 + 984 + 888 + 672 = 3844 cm² A1

Answer (a): 3844 cm²

(b)

Projection of F onto HE is 4.5 cm from E, so distance along AD/HE is 37 – 4.5 = 32.5 cm M2
Horizontal distance from A to point below F = √(32.5² + 24²) = √1632.25 = 40.401... cm. tan θ = 20 / 40.401... => θ = 26.337... ≈ 26.3° A1

Answer (b): 26.3 °

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Score: 0 / 10 (0 / 38 marks)

3D pythagoras / trig. Worksheets Next: Histograms →

Grade 7 Quizzes

19 Topics

7A - Surds	7B - Bounds
7C - Proportion (harder)	7D - Algebraic fractions
7E - The quadratic formula	7F - Factorising harder quadratics
7G - Trigonometric & exp. Graphs	7H - Rearranging (harder)
7J - Functions (harder)	7K - Iteration
7L - Area of any triangle	7M - The sine rule
7N - The cosine rule	7P - Congruent triangles
7Q - 3D pythagoras / trig.	7R - Histograms
7S - Conditional probability	7T - Exact trig. values (harder)
7U - Area of a segment