Question

Plz answer 22 and 23 question​

Answer 1

22)R=28cm

r=21cm

let the height of frustum be h cm

volume held by bucket=28.490 litres=28490cm^3

also,

volume of frustum=

pi(R^2+Rr+r^2)h/3

=>28490=22/7(28^2+28×21+21^2)h/3

h=15 cm.

23)Given, the cone and hemisphere base have equal radius.

Diameter of hemisphere = 7 cm

∴ radius of hemisphere=7/2=3.5cm

and radius of cone = 3.5 cm

Let H be the height of toy.

H  = h + r , Where h = height of cone

H = h + 3.5

Now, volume of toy = volume of cone + volume of hemisphere

231=1/3pir^2h+2/3pir^3

$231 =(\pi{r}^{2}) \div 3(h + 2r)$

put all the values,we get h as 11cm

height of the toy is h+r=11+3.5=14.5cm