Boundary tap in a quadrilateral plot
A ward in Kavre has a quadrilateral community plot ABCD. Survey measurements: AB = 48 m, BC = 20 m, AC = 52 m, CD = 30 m, DA = 60 m. The ward chair claims that because 48, 20 and 52 form a Pythagorean triple, angle ABC is a right angle, so the plot is easy to divide. A donor will pay compensation for the triangular part ABC at NPR 1,200 per square metre, and for triangle ACD at NPR 900 per square metre. The ward also wants to place a public tap inside triangle ABC at a point equidistant from the three corners A, B and C, so that no household is favoured. A ward member says this is always possible inside any triangle. The engineer is asked to check the right angle, compute the two areas and compensation, and advise whether an interior equidistant point exists.
1. Verify whether angle ABC is a right angle, and compute the area and compensation for triangle ABC.
Read the worked answer
Check the converse of Pythagoras:
AB² + BC² = 48² + 20² = 2,304 + 400 = 2,704.
AC² = 52² = 2,704.
Since AB² + BC² = AC², angle ABC is 90°. Therefore AB and BC are perpendicular legs.
Area of ABC = ½ × 48 × 20 = 480 m².
Compensation = 480 × 1,200 = NPR 576,000.
2. Compute the area of triangle ACD and the total compensation for both triangles.
Read the worked answer
Triangle ACD has sides 52 m, 30 m and 60 m.
Semi-perimeter s = (52 + 30 + 60) ÷ 2 = 71 m.
Area = sqrt[71(71 − 52)(71 − 30)(71 − 60)]
= sqrt(71 × 19 × 41 × 11) = sqrt(608,399) ≈ 780.0 m².
Compensation for ACD = 780 × 900 = NPR 702,000.
Total compensation = 576,000 + 702,000 = NPR 1,278,000.
The near-integer area is a useful check, but the compensation should be based on the unrounded area unless the contract says otherwise.
3. Can the ward place the tap strictly inside triangle ABC and equidistant from A, B and C? Advise the engineer.
Read the worked answer
A point equidistant from A, B and C is the circumcentre. In a right triangle, the circumcentre is the midpoint of the hypotenuse. Here the hypotenuse is AC = 52 m, so the midpoint is 26 m from A, B and C.
However, that midpoint lies on AC, which is the boundary between triangles ABC and ACD, not strictly inside triangle ABC. So an interior point equidistant from all three corners does not exist here. Placing the tap there would put it on a shared boundary and could revive the dispute.
If the real rule is equidistant from the three sides, use the incenter instead. If a boundary location is acceptable, the midpoint works, but the ward should get written consent from both sides.