Look at the picture.
The triangles ABC and DEC are similar. Therefore the sides are in proportion:
[tex]\dfrac{AC}{DC}=\dfrac{AB}{DE}[/tex]
AC = 4cm + 16cm = 20cm
DC = 16cm
AB = 15cm
DE = r
Substitute:
[tex]\dfrac{20}{16}=\dfrac{15}{r}\to\dfrac{5}{4}=\dfrac{15}{r}[/tex] cross multiply
[tex]5r=(4)(15)[/tex]
[tex]5r=60[/tex] divide both sides by 5
[tex]r=12\ cm[/tex]
The volume of a larger cone:
[tex]V_l=\dfrac{1}{3}\pi\cdot15^2\cdot20=\dfrac{1}{\not3_1}\pi\cdot15\!\!\!\!\diagup^5\cdot15\cdot20=1500\pi\ cm^3[/tex]
The volume of a smaller cone:
[tex]V_s=\dfrac{1}{3}\pi\cdot12^2\cdot16=\dfrac{1}{\not3_1}\pi\cdot12\!\!\!\!\!\diagup^4\cdot12\cdot16=768\pi\ cm^3[/tex]
The volume of the frustum:
[tex]V=V_l-V_s\to V=1500\pi-768\pi732\pi\ cm^3[/tex]
Answer: 732π cm³.
