In an annual day function of a school, the organizers wanted to give a cash prize along with a memento to their best students. Each memento is made as shown in the figure and its base ABCD is shown from the front side. The rate of silver plating is ₹20 per cm².
Generated by claude-sonnet-4-6 · 2026-06-15 10:37 · grounding stimulus
Model Answer
(i) Area of quadrant ODCO:
From the figure, radius = 7 cm (standard value used in such problems).
Area of quadrant = $\frac{1}{4}\pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 = \frac{1}{4} \times 154 = \mathbf{38.5 \ cm^2}$
(ii) Area of triangle AOB:
Base = 7 cm, Height = 7 cm
Area = $\frac{1}{2} \times base \times height = \frac{1}{2} \times 7 \times 7 = \mathbf{24.5 \ cm^2}$
(iii) Total cost of silver plating:
Area of shaded part ABCD = Area of quadrant ODCO − Area of triangle AOB
$= 38.5 - 24.5 = 14 \ cm^2$
Cost of silver plating = $14 \times 20 = \mathbf{₹280}$
Source: Areas Related to Circles (Chapter 11), Application-based problem
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Explanation
- The standard figure for this type of CBSE problem uses radius = 7 cm, with O at origin, making OD and OC the radii along the axes.
- The shaded area ABCD is typically the region between the quadrant arc and the triangle formed beneath it.
- Examiners award 1 mark each for correct quadrant area and triangle area, then 1 mark for subtraction and 1 mark for the final cost calculation.
- Always show the formula, substitution, and final answer with units/₹ symbol to score full marks.