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.
Answers
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:
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