Answer:

Решение

проверено

Проверено Топпр

Двойной конус, образованный вращением прямоугольного треугольника ABC вокруг его гипотенузы, показан на рисунке.

Гипотенуза, АС =

3

2

+4

2

знак равно

9+16

Solution

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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

= 31 πr

2 h 1+ 31

πr 2h 2

= 31 πr 2 [h 1

+h

2

]=

3

1

πr

2

[DA+DC]

= 31

×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

Answer:

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Explanation:

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