The diagram shows a field ABCDE.
(a) Calculate the perimeter of the field ABCDE.
(b) Calculate angle ABD.
(c) (i) Calculate angle CBD.
(ii) The point C is due north of the point B.
Find the bearing of D from B.
(d) Calculate the area of the field ABCDE.
Give your answer in hectares.
[1 hectare = 10 000m2]
▶️ Answer/Explanation
(a) Ans: 530 m
Sum all sides: \(AB + BC + CD + DE + EA = 150 + 100 + 120 + 90 + 70 = 530 \text{ m}\).
(b) Ans: 52.9°
Use the cosine rule in \(\triangle ABD\): \(\cos(\angle ABD) = \frac{AB^2 + BD^2 – AD^2}{2 \times AB \times BD}\).
Given \(AB = 150 \text{ m}\), \(BD = 120 \text{ m}\), and \(AD = 130 \text{ m}\), calculate \(\angle ABD \approx 52.9°\).
(c)(i) Ans: 28.1°
Use the sine rule in \(\triangle CBD\): \(\sin(\angle CBD) = \frac{CD \times \sin(90°)}{BD}\).
Given \(CD = 120 \text{ m}\) and \(BD = 120 \text{ m}\), \(\angle CBD \approx 28.1°\).
(c)(ii) Ans: 331.9°
Since \(C\) is due north of \(B\), the bearing of \(D\) from \(B\) is \(360° – \angle CBD \approx 331.9°\).
(d) Ans: 1.50 hectares
Divide the field into triangles and rectangles. Total area = \(7500 + 6000 + 1500 = 15000 \text{ m}^2 = 1.5 \text{ hectares}\).
(a) 
The diagram shows a toy boat. AC = 16.5 cm, AB = 19.5 cm and PR = 11 cm. Triangles ABC and PQR are similar.
(i) Calculate PQ.
(ii) Calculate BC.
(iii) Calculate angle ABC.
(iv) The toy boat is mathematically similar to a real boat. The length of the real boat is 32 times the length of the toy boat. The fuel tank in the toy boat holds 0.02 litres of diesel. Calculate how many litres of diesel the fuel tank of the real boat holds.
(b) 
The diagram shows a field $D E F G$, in the shape of a quadrilateral, with a footpath along the diagonal $D F$. $D F=105 \mathrm{~m}$ and $F G=67 \mathrm{~m}$. Angle $E D F=70^{\circ}$, angle $E F D=32^{\circ}$ and angle $D F G=143^{\circ}$.
(i) Calculate $D G$.
(ii) Calculate EF.
▶️ Answer/Explanation
(a)
(i) PQ = 13 cm (Similar triangles ratio: PQ/AB = PR/AC → PQ/19.5 = 11/16.5 → PQ = 13)
(ii) BC ≈ 10.4 cm (Using Pythagoras: √(19.5² – 16.5²) = √(380.25 – 272.25) = √108 ≈ 10.392)
(iii) ∠ABC ≈ 57.8° (Using cosine: cos⁻¹(BC/AB) = cos⁻¹(10.392/19.5) ≈ 57.77°)
(iv) 655.36 liters (Volume scales with cube of length: 0.02 × 32³ = 0.02 × 32768 = 655.36)
(b)
(i) DG ≈ 164 m (Using cosine rule in ΔDFG: DG² = DF² + FG² – 2×DF×FG×cos(143°) → DG ≈ 163.8)
(ii) EF ≈ 101 m (Using sine rule in ΔDEF: EF/sin(70°) = DF/sin(78°) → EF ≈ 100.85)