On a circular metal plate of uniform thickness, 16 holes, each of diameter 2 cm, are made. If the plate thereby has lost one-ninth of its original weight, then what is the diameter of the plate?
24 cm
Since thickness and density are uniform, weight lost is proportional to area removed. Area of the 16 holes (each radius 1 cm) = \(16 \times \pi (1)^2 = 16\pi\ \text{cm}^2\). This equals one-ninth of the plate's area: \(16\pi = \frac{1}{9}\pi R^2 \Rightarrow R^2 = 144 \Rightarrow R = 12\ \text{cm}\). So the diameter of the plate = \(2R = 24\ \text{cm}\).
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