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A Toy is in the Form of a Hemisphere Surmounted by a Right Circular Cone of the Same Base Radius as that of the Hemisphere. If the Radius of the Base of the Cone is 21 Cm and Its Volume is `2/3` of the Volume of Hemisphere, Calculate the Height of the Cone and the Surface Area of the Toy. `(Use Pi = 22/7)` - Mathematics

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Question

A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere. If the radius of the base of the cone is 21 cm and its volume is `2/3` of the volume of hemisphere, calculate the height of the cone and the surface area of the toy.
 `(use pi = 22/7)`

Solution

Let the height of the conical part be h.

Radius of the cone = Radius of the hemisphere = r = 21 cm

The toy can be diagrammatically represented as

 

Volume of the cone = `1/3pir^2h`

Volume of the hemisphere = `2/3pir^3`

According to given information:

Volume of the cone `=2/3`× Volume of the hemisphere

`therefore 1/3pir^2h=2/3xx2/3pir^3`

⇒`h=(2/3xx2/3pir^3)/(1/3pir^2)`

⇒`h=4/3r`

`therefore h=4/3xx21cm =28 cm`

Thus, surface area of the toy = Curved surface area of cone + Curved surface area of hemisphere

= πrl + 2πr2

`=pirsqrt(h^2+r^2)+2pir^2`

`=pir(sqrt(h^2+r^2)+2r)`

`=22/7xx21cm(sqrt((28cm)^2+(21cm)^2)+2xx21cm)`

`=66(sqrt(784+441)+42)cm^2`

`66(sqrt(1225)+42)cm^2`

= 66(35+42) cm2

= 66 x 77 cm2

= 5082 cm2

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Chapter 14: Surface Areas and Volumes - Exercise 14.2 [Page 62]

APPEARS IN

RD Sharma Mathematics [English] Class 10
Chapter 14 Surface Areas and Volumes
Exercise 14.2 | Q 26 | Page 62

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