A Triangular Park: A Sprinkler Rule, Two Candidate Sites and a Bench on the Path
A panchayat plans a small triangular park on a plot drawn to scale on graph paper, where 1 unit represents 10 m. The three corner pillars are fixed at A(2, 3), B(8, 3) and C(8, 11). The gardener wants a sprinkler that waters the three corners equally and proposes placing it at the midpoint of the longest side, arguing that "the midpoint of the longest side is always the same distance from all three corners". A second proposal, chosen by eye, puts the sprinkler at (5, 6.5). The works inspector must approve exactly one site. Pipe will be laid in three straight lines from the sprinkler to the three corners and costs ₹120 per metre. There is also a straight footpath planned from A to C, with a bench to be fixed on the path one-quarter of the way from A to C, measured from A. The inspector also wants confirmation that the corner at B is a right angle before accepting the gardener's rule.
1. Show that the triangle is right-angled at B and give all three sides in metres. Then say whether the gardener's rule is valid in general.
Read the worked answer
Use the distance formula, or equivalently the side lengths in units:
- AB² = (8 − 2)² + (3 − 3)² = 36, so AB = 6 units = 60 m
- BC² = (8 − 8)² + (11 − 3)² = 64, so BC = 8 units = 80 m
- AC² = (8 − 2)² + (11 − 3)² = 36 + 64 = 100, so AC = 10 units = 100 m
Since AB² + BC² = 36 + 64 = 100 = AC², the converse of Pythagoras gives a right angle at B, and the longest side is the hypotenuse AC.
The rule as stated is false in general. The midpoint of a side is equidistant from all three vertices only when that side is the hypotenuse of a right triangle - the midpoint of the hypotenuse is the circumcentre. For a scalene triangle with no right angle, the midpoint of the longest side is nearer to the two endpoints of that side than to the third vertex. The gardener has generalised from a special case, so the inspector should approve the position on a correct argument, not on the gardener's reasoning.
2. Test both proposed sprinkler sites and recommend one. Include the pipe cost of the recommended site.
Read the worked answer
Site 1, midpoint of AC: midpoint = ((2 + 8)/2, (3 + 11)/2) = (5, 7). Distances: to A, √(3² + 4²) = √25 = 5 units; to C, √(3² + 4²) = 5 units; to B, √(3² + 4²) = 5 units. All three are 5 units = 50 m. Pipe = 3 × 50 = 150 m, cost = 150 × 120 = ₹18,000. The corner is watered equally.
Site 2, (5, 6.5): to A, √(3² + 3.5²) = √21.25 ≈ 4.61 units = 46.1 m; to C, √(3² + 4.5²) = √29.25 ≈ 5.41 units = 54.1 m; to B, √(3² + 3.5²) ≈ 4.61 units = 46.1 m. Pipe ≈ 146.3 m, cost ≈ ₹17,556 - about ₹444 cheaper, but the distances differ by roughly 8 m, so the far corner is under-watered.
Recommendation: Site 1, (5, 7). A saving of ₹444 does not justify a sprinkler that waters one corner at 4.61 units and another at 5.41 units; the specification "waters the three corners equally" is the binding requirement, and cost is secondary.
3. Find the coordinates of the bench, which lies on AC one-quarter of the way from A to C.
Read the worked answer
The bench divides AC in the ratio 1 : 3 (from A towards C), so use the section formula with m : n = 1 : 3:
x = (1·8 + 3·2)/(1 + 3) = (8 + 6)/4 = 14/4 = 3.5 y = (1·11 + 3·3)/(1 + 3) = (11 + 9)/4 = 20/4 = 5
Bench at (3.5, 5) in graph units, i.e. (35 m, 50 m) on the ground.
Check by a second route: "one-quarter of the way from A to C" means A + ¼(C − A) = (2, 3) + ¼(6, 8) = (2 + 1.5, 3 + 2) = (3.5, 5). The two methods agree. Note the ratio is 1 : 3, not 1 : 4 - a common slip is to divide the coordinates by 4 without weighting, which would give (2.5, 3.5) and place the bench wrongly.