Question

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Answer 1

Question :-

The radius of the base of the right circular cone is r . It is cut by a plane parallel to the base at a height h from the base. The slant height of the frustum is $\rm{\sqrt{h^2+\dfrac{r^2}{9}}}$ . Show that the volume of the frustum is $\rm{\dfrac{19 \pi r^2 h }{27}}$.

Figure :-

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• AO = BD = h is the height of frustum formed
• OC = r is the radius of larger base
• AB is the radius of smaller base
• BC is the slant height of frustum , given by  $\rm{BC=\sqrt{h^2+\dfrac{r^2}{9}}}$

Formula to be used :-

• Volume of frustum

$\boxed{\rm{volume\;of\;frustum=\frac{1}{3}\pi h({r_1}^2+{r_2}^2+r_1r_2)}}$

(where h is the height of frustum , r₁ and r₂ are the radius of two circular bases of frustum)

Solution :-

Using pythagoras theorem Finding AB

By pythagoras theorem in Δ BCD

$\implies\rm{BC^2=BD^2 + CD^2}\\\\\implies\rm{\left(\sqrt{h^2+\dfrac{r^2}{9}}\right)^2=h^2+CD^2}\\\\\implies\rm{h^2+\dfrac{r^2}{9}=h^2+CD^2}\\\\\implies\red{\boxed{\rm{CD=\dfrac{r}{3}}}}$

As we can see in the figure

$\implies\rm{AB=OC-CD}\\\\\implies\rm{AB=r-\dfrac{r}{3}}\\\\\implies\red{\boxed{\rm{AB=\dfrac{2r}{3}}}}$

so, the radius of smaller base of frustum is 2r / 3.

Calculating the Volume of frustum formed

$\implies\rm{volume\:of\:frustum=\dfrac{1}{3}\pi h\left ((r)^2+{(\dfrac{2r}{3})}^2+(r)(\dfrac{2r}{3})\right)}\\\\\implies\rm{volume\:of\:frustum=\dfrac{1}{3}\pi h \left(r^2+\dfrac{4r^2}{9}+\dfrac{2r^2}{3}\right)}\\\\ \implies\rm{volume\:of\:frustum=\dfrac{1}{3}\pi h \left( \dfrac{9r^2+4r^2+6r^2}{9}\right)}\\\\\implies\red{\boxed{\boxed{\rm{volume\:of\:frustum=\dfrac{19\pi r^2h}{27}}}}}$

Proved .