Question

A cone, a hemisphere and a cylinder stand on equal bases and have the same
height. Show that their volumes are in the ratio 1:2:3.

Class 10.
Chapter Surface Areas and Volumes.​

Answer 1

Step-by-step explanation:

Given.

A cone, a hemisphere, and a cylinder stand on equal bases.

Hence radius of base of cone = radius of hemisphere = radius of cylinder=r

And,

They also have same height =h

Hight of cone =h

Hight of cylinder=h

Hight of Hemisphere r=h (r is the radius of sphere)

Now,

Volume of cone V

C

=

3

1

πr

2

h...….(1)

Volume of hemisphere V

H

=

3

2

πr

2

×r

=

3

2

πr

2

×h

V

H

=

3

2

πr

2

h…….(2)

Volume of cylinder V

c

=πr

2

h...…..(3)

From equation 1,2 and 3

V

C

:V

H

:V

c

=

3

1

πr

2

h:

3

2

πr

2

h:πr

2

h

V

C

:V

H

:V

c

=

3

1

:

3

2

:1

V

C

:V

H

:V

c

=1:2:3

Answer 2

Hlo...ur goutami...i am..ankit...xd...

SolutioN :-

Let the speed of the stream be x km/h.

so, speed of boat in downstream is ( 11 + x )

and also speed of boat in upstream is ( 11 - x )

Distance cover by boat is 12 km.

The speed of boat in still water is 11 km/h

\boxed{\tt \bullet \: \: \: Distance = Speed \times Time.}

∙Distance=Speed×Time.

\begin{gathered} \tt \rightarrow \dfrac{12}{11 - x} + \dfrac{12}{11 + x} = 2 \times \dfrac{3}{4} \\ \\ \end{gathered}

→

11−x

12

+

11+x

12

=2×

4

3

\begin{gathered} \tt \rightarrow \dfrac{12}{11 - x} + \dfrac{12}{11 + x} = \dfrac{11}{4} \\ \\ \end{gathered}

→

11−x

12

+

11+x

12

=

4

11

\begin{gathered} \boxed{\tt \bullet \: \: \: {a}^{2} + {b}^{2} = (a + b)(a - b)} \\ \\ \end{gathered}

∙a

2

+b

2

=(a+b)(a−b)