Governing council of a local public development authority of Dehradun decided to build an adventurous playground on the top of a hill, which will have adequate space for parking.
After survey, it was decided to build rectangular playground, with a semi-circular area allotted for parking at one end of the playground. The length and breadth of the rectangular playground are 14 units and 7 units, respectively. There are two quadrants of radius 2 units on one side for special seats.
Generated by claude-sonnet-4-6 · 2026-06-15 10:37 · grounding stimulus
Model Answer
(i) Perimeter of parking area:
The parking area is a semi-circle with diameter = 7 units → radius = 7/2 units.
Perimeter = diameter + semi-circular arc = 7 + πr = 7 + (22/7 × 7/2) = 7 + 11 = 18 units
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(ii) Total area of parking and two quadrants:
Area of semi-circle (parking) = $\frac{1}{2}\pi r^2 = \frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{4} = 19.25$ sq units
Area of two quadrants = $2 \times \frac{1}{4}\pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 4 = \frac{44}{7} \approx 6.28$ sq units
Total = 19.25 + 6.28 = 25.53 sq units
OR
Area of rectangular playground = 14 × 7 = 98 sq units
Area of parking (semi-circle) = 19.25 sq units
Ratio = 98 : 77/4 = 98 × 4 : 77 = 392 : 77 = 8 : 1.57 → 56 : 11
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(iii) Cost of fencing:
Boundary = two lengths + one breadth + semi-circular arc (the breadth at parking end is replaced by the arc)
Perimeter = 2 × 14 + 7 + πr = 28 + 7 + 11 = 46 units
Cost = 46 × ₹2 = ₹ 92
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Explanation
- Part (i): Perimeter of a semi-circle = diameter + arc (πr), not the full circumference.
- Part (ii): Two quadrants of same radius = one complete semi-circle; use $\frac{1}{2}\pi r^2$. For the OR part, keep ratio in simplest whole-number form.
- Part (iii): The fencing goes along the two long sides, one short side (the opposite end from parking), and the curved arc of the semi-circle — the shared diameter is interior, so it is not fenced. Examiners check that students don't double-count the diameter.