The lateral surface area of a right circular cone is three times the volume of that cone. If the height and the radius of the cone are h units and r u - Quantitative Aptitude MCQ WBCS Mains 2026 | ExamTest.live

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The lateral surface area of a right circular cone is three times the volume of that cone. If the height and the radius of the cone are h units and r units respectively, then \frac{1}{h ^{2}} + \frac{1}{r ^{2}} is equal to:

Solution & Step-by-step Explanation

Let S be the lateral surface area and V be the volume of the cone. S = π r l = π r r^{2} + h^{2} V = \frac{1}{3} π r^{2} h According to the question, S = 3 V : π r r^{2} + h^{2} = 3 \times (\frac{1}{3} π r^{2} h) π r r^{2} + h^{2} = π r^{2} h Dividing both sides by π r : r^{2} + h^{2} = r h Squaring both sides: r^{2} + h^{2} = r^{2} h^{2} Divide the entire equation by r^{2} h^{2} : \frac{r ^{2}}{r ^{2} h ^{2}} + \frac{h ^{2}}{r ^{2} h ^{2}} = \frac{r ^{2} h ^{2}}{r ^{2} h ^{2}} \frac{1}{h ^{2}} + \frac{1}{r ^{2}} = 1

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The lateral surface area of a right circular cone is three times the volume of that cone. If the height and the radius of the cone are h units and r units respectively, then \frac{1}{h ^{2}} + \frac{1}{r ^{2}} is equal to:

1

3

9

\frac{1}{9}

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