A garden is in the shape of a square. The gardener grew saplings of Ashoka tree on the boundary of the garden at the distance of 1 m from each other. He wants to decorate the garden with rose plants. He chose a triangular region inside the garden to grow rose plants. In the above situation, the gardener took help from the students of class 10. They made a chart for it which looks like the given figure.
Generated by claude-sonnet-4-6 · 2026-06-15 10:29 · grounding stimulus
Model Answer
(i) Taking A as origin, the coordinates of the vertices of △PQR are:
P(4, 6), Q(3, 2), R(6, 5)
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(ii) Finding PQ and QR:
$$PQ = \sqrt{(4-3)^2 + (6-2)^2} = \sqrt{1 + 16} = \sqrt{17} \text{ m}$$
$$QR = \sqrt{(6-3)^2 + (5-2)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \text{ m}$$
OR Point dividing PR in ratio 2:1 internally:
$$x = \frac{2(6)+1(4)}{3} = \frac{16}{3}, \quad y = \frac{2(5)+1(6)}{3} = \frac{16}{3}$$
Required point = $\left(\dfrac{16}{3},\ \dfrac{16}{3}\right)$
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(iii) Finding PR:
$$PR = \sqrt{(6-4)^2 + (5-6)^2} = \sqrt{4+1} = \sqrt{5} \text{ m}$$
Since PQ = $\sqrt{17}$, QR = $3\sqrt{2}$, PR = $\sqrt{5}$ — all three sides are unequal, so △PQR is NOT an isosceles triangle.
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Source: Chapter 7 – Coordinate Geometry
Explanation
- Examiners look for: correct substitution in distance formula and section formula with working shown step by step.
- For part (i), simply state coordinates — no working needed (1 mark).
- For part (iii), you must compare at least two sides and state a conclusion clearly to earn the mark.
- Always simplify surds (e.g., √18 = 3√2) for full marks.