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A Solid Cone of Base Radius 10 Cm is Cut into Two Parts Through the Mid-point of Its Height, by a Plane Parallel to Its Base. Find the Ratio in the Volumes of Two Parts of the Cone. - Mathematics

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प्रश्न

A solid cone of base radius 10 cm is cut into two part through the mid-point of its height, by a plane parallel to its base. Find the ratio in the volumes of two parts of the cone.

थोडक्यात उत्तर

उत्तर

Let the height of the cone be H.
Now, the cone is divided into two parts by the parallel plane
∴ OC = CAH2
Now, In ∆OCD and OAB
∠OCD = OAB          (Corresponding angles)
∠ODC = OBA          (Corresponding angles)
By AA-similarity criterion ∆OCD ∼ ∆OAB

\[\therefore \frac{CD}{AB} = \frac{OC}{OA}\]

\[ \Rightarrow \frac{CD}{10} = \frac{H}{2 \times H}\]

\[ \Rightarrow CD = 5 cm\]

\[\frac{\text { Volume of first part }}{\text { Volume of second part }} = \frac{\frac{1}{3}\pi \left( CD \right)^2 \left( OC \right)}{\frac{1}{3}\pi CA\left[ \left( AB \right)^2 + \left( AB \right)\left( CD \right) + {CD}^2 \right]}\]

\[ = \frac{\left( 5 \right)^2}{\left[ \left( 10 \right)^2 + \left( 10 \right)\left( 5 \right) + 5^2 \right]}\]

\[ = \frac{25}{100 + 50 + 25}\]

\[ = \frac{25}{175}\]

\[ = \frac{1}{7}\]

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पाठ 14: Surface Areas and Volumes - Exercise 14.3 [पृष्ठ ७९]

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आरडी शर्मा Mathematics [English] Class 10
पाठ 14 Surface Areas and Volumes
Exercise 14.3 | Q 18 | पृष्ठ ७९

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