Question

a right triangle whose sides are 6cm and 8 cm ( other than hypotenuse ) is made to revolve about its hypotenuse. find the volume and surface area of the double cone so formed.

Answer 1

The double cone so formed the right angled triangle ABC 

Hypotenuse AC = underoot 6 square + 8 square  

= underoot 36 + 64 

= underoot 100 

= 10 cm  

 

Area of ΔABC = ½ x AB x AC 

½ x AC x OB = ½ x 6 x 8 

½ x 10 x OB = 24 

OB = 4.8 cm  

Volume of double cone = volume of cone 1 + volume  of cone 2  

= 1/3 π r square h1 + 1/3π r square h2 

= 1/3π r square (h1+h2) 

= 1/3π r square (OA+OC) 

= 1/3 x 22/7 x (4.8) square x 10 

= 241.3 cm cube 

Surface area of double cone = surface area of cone 1 + surface area of cone 2 

π rl1 + π rl2 

π r (6+8) 

= 22/7 x 4.8 x 14 

= 211.2 cm square

Answer 2

Answer:

The double cone so formed the right angled triangle ABC 

Hypotenuse AC = underoot 6 square + 8 square  

= underoot 36 + 64 

= underoot 100 

= 10 cm  

 

Area of ΔABC = ½ x AB x AC 

½ x AC x OB = ½ x 6 x 8 

½ x 10 x OB = 24 

OB = 4.8 cm  

Volume of double cone = volume of cone 1 + volume  of cone 2  

= 1/3 π r square h1 + 1/3π r square h2 

= 1/3π r square (h1+h2) 

= 1/3π r square (OA+OC) 

= 1/3 x 22/7 x (4.8) square x 10 

= 241.3 cm cube 

Surface area of double cone = surface area of cone 1 + surface area of cone 2 

π rl1 + π rl2 

π r (6+8) 

= 22/7 x 4.8 x 14 

= 211.2 cm square