Answer:
[tex]V=26.4cm^3[/tex]
Step-by-step explanation:
It is given that The cross-sectional area parallel to the bases of the two figures above is the same at every level, therefore the area of the triangle will be equal to the area of cone, thus using Pythagoras theorem, we have
[tex](4.8)^2+(base)^2=(6.0)^2[/tex]
[tex](base)^2=36-23.04[/tex]
[tex]Base=3.6cm[/tex]
Now, area of triangle is given as:
[tex]A=\frac{1}{2}{\times}3.6{\times}4.8[/tex]
[tex]A=8.64cm^2[/tex]
Also, Volume of cone is given as:
[tex]V=\frac{1}{3}{\pi}r^2h[/tex]
Using the given property, we have
[tex]V=\frac{1}{3}{\times}8.64{\times}9.2[/tex]
[tex]V=26.4cm^3[/tex]
Thus, the volume of the cone is [tex]26.4cm^3[/tex].
