Question

A right-angled triangle whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose a value of π as found appropriate.)

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Answer 1

Answer:

The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.

Hypotenuse, AC =

3

2

+4

2

=

9+16

=

25

=5 cm

Area of ΔABC=

2

1

×AB×AC

⇒

2

1

×AC×DB=

2

1

×4×3

⇒

2

1

×5×DB=6

So, DB =

5

12

=2.4 cm

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

=

3

1

πr

2

h

1

+

3

1

πr

2

h

2

=

3

1

πr

2

[h

1

+h

2

]=

3

1

πr

2

[DA+DC]

=

3

1

×3.14×2.4

2

×5

=30.14 cm

3

Answer 2

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The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.

Hypotenuse, AC =

3

2

+4

2

=

9+16

=

25

=5 cm

Area of ΔABC=

2

1

×AB×AC

⇒

2

1

×AC×DB=

2

1

×4×3

⇒

2

1

×5×DB=6

So, DB =

5

12

=2.4 cm

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

=

3

1

πr

2

h

1

+

3

1

πr

2

h

2

=

3

1

πr

2

[h

1

+h

2

]=

3

1

πr

2

[DA+DC]

=

3

1

×3.14×2.4

2

×5

=30.14 cm

3

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

=πrl

1

+πrl

2

=πr[4+3]=3.14×2.4×7

=52.75 cm

2

hope so it will help you

thank you!