Question

A cylindrical bucket of diameter 28 cm and height 20 cm was full of sand. When the sand in the bucket was poured on the ground, the sand got converted into a shape of a cone. If the height of the cone was 14 cm, what was the base area of the cone ?

Answer 1

Answer:

2640cm²

Step-by-step explanation:

diameter of cylinder is 28cm

radius R=14cm

height H=20cm

let radius of cone=r

height, h=14cm

volume of cylinder=volume of cone

πR²H=1/3πr²h

3(14)²x20=r²x14

r²=3x14x20

r²=3x7x2x5x2²

r=2√210cm

base area=πr²

=22/7x4x210

=22x4x30=120x22=2640cm²

Answer 2

Answer:

For cylindrical bucket,

r =radius =

$\frac{diameter}{2} = \frac{28}{2} = 14cm \:$

h=height =20cm.

For cone of sabd poured on the ground,

R=radius of base of the cone of sand =?

H= height of the cone of sand =14cn

To find :Base area of the cone

Formula:

$volume \: of \: a \: cylinder \: = \pi \: {r}^{2} h \\ \\ volume \: of \: a \: cone \: = \frac{1}{3} \pi \: {r}^{2} h \\ \\ base \: area \: of \: th \: cone \: = \pi {r}^{2}$

Volume of a cone =1/3πR^2H

base area of the cone =πR^2

Solution :

A cylinder bucket full of sand is poured on the ground in the shape of a cone.

Volume of cone of sand formed

=Volume of cylindrical bucket full of sand.

1/3πR^2H = πr2h

πR^2=3*πr^2h/H

base area of the cone =3*π*r^2h/H

base area of the cone

$\\ 3 \times \frac{22}{7} \times \frac{14 \times 14 \times 20}{14} \\ = 66 \times 2 \times 20 \\ = 66 \times 40 \\ = 2640sq .cm$

the Base area of the cone =2640sq. cm