Question 1 [3 marks]
AG² = AD² + DC² + AE² = 2² + 6² + 3² = 49, so AG = 7.00 cm (3 s.f.).
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AG² = AD² + DC² + AE² = 2² + 6² + 3² = 49, so AG = 7.00 cm (3 s.f.).
DC² = AG² - AD² - AE² = 7² - 2² - 3² = 36, so DC = 6.00 cm (3 s.f.).
AD² = AG² - DC² - AE² = 49 - 36 - 9 = 4, so AD = 2.00 cm (3 s.f.).
AE² = AG² - AD² - DC² = 49 - 4 - 36 = 9, so AE = 3.00 cm (3 s.f.).
AG² = AD² + DC² + AE² = 5² + 8² + 4² = 105, so AG = √105 = 10.2 cm (3 s.f.).
In right-angled triangle FCD, FC = CD tan 30°. In the base, AC² = AD² + CD². Triangle AFC is right-angled at C, so tan(angle AFC) = ACFC. Hence angle AFC = 71.7° (1 d.p.).
FC = 5 tan 25°, AC = √12² + 5² = 13. In right-angled triangle AFC, AF² = AC² + FC², so AF = 13.2 cm (1 d.p.).
In right-angled triangle FCD, FC = CD tan 40°, so FC = 7.6 cm (1 d.p.).
FC = 6 tan 35°, AC = √11² + 6² = √157. In right-angled triangle AFC, tan(angle FAC) = FCAC, so angle FAC = 18.5° (1 d.p.).
FC = 8 tan 28°, AC = √6² + 8² = 10. Then angle AFC = 67.0° (1 d.p.).
Let O be the centre of the base. AO = 1√2 × AB and AE² = AO² + 10². Then cos(angle EAC) = AB²AC · AE with AC = AB√2, giving angle EAC = 70.5° (1 d.p.).
With O the centre of the base, AO = AB√2 and EA² = AO² + 8², so EA = 9.1 cm (1 d.p.).
Angle EAC = 61.2° (1 d.p.).
EA² = AB²2 + 11² = 812 + 121, so EA = 12.7 cm (1 d.p.).
Angle EAC = 60.5° (1 d.p.).
Let h be the vertical height from P to the base centre. Then h = AB√2 tan 65°. Volume = 13 × AB² × h = 1710 cm³ (3 s.f.).
h = AB√2 tan 55° with AB = 12, so h = 12.1 cm (3 s.f.).
Volume = 281 cm³ (3 s.f.).
h = 8√2 tan 48° = 6.28 cm (3 s.f.).
Volume = 1040 cm³ (3 s.f.).
AG² = 6² + 7² + 9² = 166, so AG = 12.9 cm (3 s.f.).
FC = 10 tan 22°, AC = √7² + 10² = √149. Then angle AFC = 71.7° (1 d.p.).
EA² = AB²2 + 6² = 812 + 36, so EA = 8.7 cm (1 d.p.).
h = 10√2 tan 45° = 5√2. Volume = 13 × 100 × 5√2 = 236 cm³ (3 s.f.).
AE² = AG² - AD² - DC² = 12.7² - 5² - 8² = 161.29 - 25 - 64 = 72.29, so AE = 8.50 cm (3 s.f.).
FC = 8 tan 32°, so FC = 5.0 cm (1 d.p.).
Angle EAC = 72.2° (1 d.p.).
h = 11√2 tan 52°, volume = 13 × 11² × h = 402 cm³ (3 s.f.).
DC² = AG² - AD² - AE² = 13² - 3² - 4² = 169 - 9 - 16 = 144, so DC = 12.0 cm (3 s.f.).
FC = 7 tan 45° = 7, AC = √8² + 7² = √113. Then tan(angle FAC) = FCAC, so angle FAC = 33.4° (1 d.p.).