Question

plz solve my sum
I give you marks ​

Answer 1

Answer:

1) π times radius times slant height

2) If the radius of a sphere is increased from 'r' to '2r', then the ratio of the volumes of the two sphere is 1:8.

3) $2$π$r^2$ is the Curved Surface Area of a Hemisphere.

4) The radius of this sphere is $3\sqrt[3]{7} cm$

5) 270 number of persons can be accommodated in the dining Hall.

Step-by-step explanation:

1) The expression for the Curved Surface Area of a right circular cone is πrl where r is the radius of the base and l is the slant height.

2) If the radius of a sphere is increased from 'r' to '2r', then the ratio of the volumes of the two sphere is 1:8.

Volume of the first sphere is $\frac{4}{3}$π$r^3$

And Volume of the first sphere is $\frac{4}{3}$π$(2r)^3 =$  $\frac{4}{3}$π . $8r^3$

3) The Curved Surface Area of a Solid Sphere is $4$π$r^2$

Therefore, Curved Surface Area of a Hemisphere is $\frac{1}{2}$ . $4$π$r^2$ $=2$π$r^2$

$2$π$r^2$ is the Curved Surface Area of a Hemisphere.

4) Let's assume that, the radius is r.  Then the volume of the sphere is $\frac{4}{3}$π$r^3$

According to the problem:

$\frac{4}{3}$π$r^3$ = 5544

∴ $r^3 = 189 = (3\sqrt[3]{7} )^3$

∴ $r = 3\sqrt[3]{7} cm$

The radius of this sphere is $3\sqrt[3]{7} cm$

5) Volume of the dinning hall = 20 x 15 x 4.5 $m^3$ = 15 x 90 $m^3$

Each person requires 5 $m^3$ of Air.

Therefore, $\frac{15 . 90}{5}$ $= 270$ number of persons can be accommodated in the dining Hall.

Answer 2

Answer:

Answer:

1) π times radius times slant height

2) If the radius of a sphere is increased from 'r' to '2r', then the ratio of the volumes of the two sphere is 1:8.

3) 22 πr^2r

2

is the Curved Surface Area of a Hemisphere.

4) The radius of this sphere is 3\sqrt[3]{7} cm3

3

7

cm

5) 270 number of persons can be accommodated in the dining Hall.

Step-by-step explanation:

1) The expression for the Curved Surface Area of a right circular cone is πrl where r is the radius of the base and l is the slant height.

2) If the radius of a sphere is increased from 'r' to '2r', then the ratio of the volumes of the two sphere is 1:8.

Volume of the first sphere is \frac{4}{3}

3

4

πr^3r

3

And Volume of the first sphere is \frac{4}{3}

3

4

π(2r)^3 =(2r)

3

= \frac{4}{3}

3

4

π . 8r^38r

3

3) The Curved Surface Area of a Solid Sphere is 44 πr^2r

2

Therefore, Curved Surface Area of a Hemisphere is \frac{1}{2}

2

1

. 44 πr^2r

2

=2=2 πr^2r

2

22 πr^2r

2

is the Curved Surface Area of a Hemisphere.

4) Let's assume that, the radius is r. Then the volume of the sphere is \frac{4}{3}

3

4

πr^3r

3

According to the problem:

\frac{4}{3}

3

4

πr^3r

3

= 5544

∴ r^3 = 189 = (3\sqrt[3]{7} )^3r

3

=189=(3

3

7

)

3

∴ r = 3\sqrt[3]{7} cmr=3

3

7

cm

The radius of this sphere is 3\sqrt[3]{7} cm3

3

7

cm

5) Volume of the dinning hall = 20 x 15 x 4.5 m^3m

3

= 15 x 90 m^3m

3

Each person requires 5 m^3m

3