Splitting a Field: An Ambiguous Fence and a Disputed Area
A surveyor has measured a four-sided field ABCD. Walking the boundary, the sides are AB = 42 m, BC = 58 m, CD = 35 m and DA = 47 m, and the diagonal AC is 65 m. The owner has already bought a 60 m roll of fencing and wants a new straight fence from corner C to a point P on the side AB, with CP = 60 m, to split the field into two paddocks. A neighbour disputes the field's area, claiming it is only 1 800 m² and that the boundary should move in his favour. The land registry requires the area to the nearest square metre. The surveyor's notes record that the ground rises steadily at about 6° along the line AC, and that all lengths were measured with a tape held along the ground. Nobody has checked whether ABCD is convex, and the tape was stretched taut for every measurement.
1. Find the angle ABC and the total area of the field.
Read the worked answer
Split the quadrilateral along AC into triangles ABC and ACD.
Triangle ABC (sides 42, 58, 65), cosine rule at B:
(\cos B = \dfrac{42^2+58^2-65^2}{2(42)(58)} = \dfrac{1764+3364-4225}{4872} = \dfrac{903}{4872} = 0.18535)
(B = 79.3^\circ).
Area (= \tfrac12(42)(58)\sin 79.3^\circ = 1218 \times 0.9827 = 1196.9\ \text{m}^2).
Triangle ACD (sides 65, 35, 47), Heron: (s = \tfrac{65+35+47}{2} = 73.5)
Area (= \sqrt{73.5(73.5-65)(73.5-35)(73.5-47)} = \sqrt{73.5 \times 8.5 \times 38.5 \times 26.5})
(= \sqrt{637,401} = 798.4\ \text{m}^2).
Total area = 1196.9 + 798.4 = 1 995 m² (to the nearest square metre).
Both triangles satisfy the triangle inequality (42 + 58 > 65 and 35 + 47 > 65), so the quadrilateral closes and the data are at least self-consistent.
2. Find the distance AP for the new fence, and explain why only one position of P is valid.
Read the worked answer
First get the angle at A in triangle ABC:
(\cos A = \dfrac{42^2+65^2-58^2}{2(42)(65)} = \dfrac{2625}{5460} = 0.48077 \Rightarrow A = 61.3^\circ).
In triangle ACP: AC = 65 is opposite angle P, and CP = 60 is opposite angle A. Sine rule:
(\sin P = \dfrac{65\sin 61.3^\circ}{60} = \dfrac{65 \times 0.8770}{60} = 0.9501)
(P = 71.8^\circ) or (P = 180 - 71.8 = 108.2^\circ).
This is the ambiguous SSA case, so both must be tested against the constraint that P lies on the segment AB (AP ≤ 42 m):
- (P = 71.8^\circ): angle ACP = 180 − 61.3 − 71.8 = 46.9°, so (AP = \dfrac{60\sin 46.9^\circ}{\sin 61.3^\circ} = 49.9) m. That is longer than AB, so P would sit beyond B on the extension of AB - not on the side. Reject.
- (P = 108.2^\circ): angle ACP = 180 − 61.3 − 108.2 = 10.5°, so (AP = \dfrac{60\sin 10.5^\circ}{\sin 61.3^\circ} = 12.5) m. Valid.
So AP = 12.5 m from A, and the position is unique. The sine rule alone cannot decide this; the length constraint does.
3. Assess the neighbour's claim of 1 800 m², and comment on how the 6° slope affects the answer.
Read the worked answer
The measured boundary gives 1 995 m², so the neighbour's 1 800 m² is not supported - the gap is about 195 m², roughly 10%. The most sensitive measurement is the diagonal AC, since it controls both triangles; re-measuring AC (or the angle between AB and AD) is the cheapest way to test whether the figure is wrong, and a small error there could move the area by tens of square metres.
Slope: the tape followed the ground, so the recorded 42, 58, 35, 47 and 65 are slope distances. If the field is a uniform 6° slope, horizontal distances are the slope distances times cos 6° = 0.9945, so the plan area is smaller than the surface area by a factor of about ((0.9945)^2). Treating the 1 995 m² as a surface area, the plan area would be about (1995 \times 0.989 = 1,973\ \text{m}^2); treating it as plan area, the surface area is (1995/\cos 6^\circ = 2,006\ \text{m}^2). Either way the correction is around 1%, not 10%.
Judgement: reject the claim, but state explicitly in the report whether 1 995 m² is a plan or surface area, because the registry charges on plan area and the difference, though small, is a legitimate challenge.