The radius of the base of a hollow cone is 8, | vedclass

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(A) Let the radius of the base of the cone be $R = 8\, cm$ and the height be $H = 15\, cm$.The slant height $L$ of the cone is given by $L = \sqrt{R^2

Vedclass · Accepted Answer

$5:3$

Vedclass · Answer

(A) Let the radius of the base of the cone be $R = 8\, cm$ and the height be $H = 15\, cm$.The slant height $L$ of the cone is given by $L = \sqrt{R^2 + H^2} = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\, cm$.Let $r$ be the radius of the largest sphere that can be inscribed in the cone.The formula for the radius of the inscribed sphere in a cone is $r = \frac{R \cdot H}{R + L}$.Substituting the values: $r = \frac{8 \cdot 15}{8 + 17} = \frac{120}{25} = 4.8\, cm$.We need to find the ratio of the radius of the base of the cone $(R)$ to the radius of the sphere $(r)$:Ratio $= R : r = 8 : 4.8$.To simplify,multiply both sides by $10$: $80 : 48$.Divide by $16$: $80/16 : 48/16 = 5 : 3$.Thus,the ratio is $5:3$.

The radius of the base of a hollow cone is $8\, cm$ and its height is $15\, cm$. $A$ sphere of the largest possible radius is placed inside the cone. What is the ratio of the radius of the base of the cone to the radius of the sphere?

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