Answer:
volume: 49280 cm³ and curved S.A = 5826.64 cm²
Explanation:
using similarities:
[tex]\sf \frac{45 + h2}{28} = \frac{h2}{7}[/tex]
[tex]\sf 315 + 7h2 = 28h2[/tex]
[tex]\sf 315 = 28h2-7h2[/tex]
[tex]\sf 315 = 21h2[/tex]
[tex]\sf 15 = h2[/tex]
total height of the cone: 45 + 15 = 60 cm
solve for volume:
[tex]\sf volume \ of \ cone \ = \frac{1}{3} \pi r^2 h[/tex]
[tex]\sf volume \ of \ cone \ = \frac{1}{3} \pi (28)^2 (60)[/tex]
[tex]\sf volume \ of \ cone \ = 15680\ *\frac{22}{7}[/tex]
[tex]\sf volume \ of \ cone \ = 49280 \ cm^3[/tex]
First find the slant height using Pythagoras theorem -
[tex]\sf l^2 = r^2 + h^2[/tex]
[tex]\sf l =\sqrt{ r^2 + h^2}[/tex]
[tex]\sf l =\sqrt{ (28)^2 + (60)^2}[/tex]
[tex]\sf l =66.212 \ cm[/tex]
solve for curved surface area:
[tex]\sf curved \ surface \ area = \pi rl[/tex]
[tex]\sf curved \ surface \ area = \pi (28)(66.2)[/tex]
[tex]\sf curved \ surface \ area =5826.64 \ cm^2[/tex]