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📅 May 5, 2026 👤 scraface

May 05, 2026

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📌 Solution: The lateral surface area of a cylinder is $ 2\pi r h $. The area of a circle with radius $ 2r $ is $ \pi (2r)^2 = 4\pi r^2 $. Setting them equal: $ 2\pi r h = 4\pi r^2 $. Cancel $ 2\pi r $ to get $ h = 2r $. Thus, the ratio $ \frac{h}{r} = \boxed{2} $.

📌 Question: A circle is inscribed in a square with side length $ s $. What is the ratio of the circle’s area to the square’s area, expressed in terms of $ \pi $?

📌 Solution: The circle’s diameter equals the square’s side, so radius $ r = \frac{s}{2} $. Circle area: $ \pi \left(\frac{s}{2}\right)^2 = \frac{\pi s^2}{4} $. Square area: $ s^2 $. Ratio: $ \frac{\frac{\pi s^2}{4}}{s^2} = \frac{\pi}{4} $. The ratio is $ \boxed{\dfrac{\pi}{4}} $.

📌 Question: An equilateral triangle’s side is reduced by 3 cm, decreasing its area by $ 15\sqrt{3} \, \text{cm}^2 $. What was the original side length?

📌 Solution: Let original side be $ s $. Original area: $ \frac{\sqrt{3}}{4} s^2 $. New side $ s - 3 $, new area: $ \frac{\sqrt{3}}{4} (s - 3)^2 $. The difference: $ \frac{\sqrt{3}}{4} [s^2 - (s - 3)^2] = 15\sqrt{3} $. Simplify: $ \frac{\sqrt{3}}{4} (6s - 9) = 15\sqrt{3} $. Cancel $ \sqrt{3} $ and solve $ \frac{6s - 9}{4} = 15 $, leading to $ 6s - 9 = 60 $, so $ s = \frac{69}{6} = 11.5 $. Original side length is $ \boxed{11.5} \, \text{cm} $.

📌 Question: A quantum dot (modeled as a sphere) has radius $ r $. If its surface area equals the area of a circle with radius $ \sqrt{2}r $, find $ r $ in terms of the circle’s radius.

📌 Solution: Sphere surface area: $ 4\pi r^2 $. Circle area: $ \pi (\sqrt{2}r)^2 = 2\pi r^2 $. Setting equal: $ 4\pi r^2 = 2\pi r^2 $. This implies $ 4 = 2 $, a contradiction. Thus, no solution exists unless the circle’s radius is adjusted. However, if the problem states equivalence, the only possibility is $ r = 0 $, which is trivial. Rechecking the question reveals a misstatement; assuming the circle’s radius is $ R $, then $ 4\pi r^2 = \pi R^2 \Rightarrow R = 2r $. The original question’s setup is inconsistent, but if forced, $ r = \frac{R}{2} $, so $ \boxed{r = \dfrac{R}{2}} $.

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