Question

The height of a cone and the radius of its base are respectively 9 and 3 cm the cone is cut by a plane parallel to its base is divided into two parts volume of frustum of cone is 44 cm cube and the radius of upper circular surface of frustum is

Answer 1

Hey mate !

_________________

Let DO' = r cm and OO'= h cm.

As ∆ADO' and ∆ABO are similar

$\therefore {\mathsf{\frac{AO'}{AO} = \frac{DO'}{BO}}}$


$\mathsf{ \implies \frac{9 - h}{h} = \frac{r}{3} } \\ \\ \mathsf{ \implies 9 - h = 3r} \\ \\ \mathsf{ \implies h = 9 - 3r}$


Volume of Frustum

$\mathsf{ \implies \frac{1}{3}\pi h \: (r {}^{2} \: _1 + r {}^{2} \: _2 +r _1r _2) } \\ \\ \mathsf{ \implies 44 = \frac{1}{3} \times \frac{22}{7} \times (9 - 3r)(3 {}^{2} + 3r + r {}^{2} ) } \\ \\ \mathsf{ \implies \frac{44 \times 7}{22} = 3 { }^{3} - r {}^{3} } \\ \\ \mathsf{ \implies 14 = 27 - r {}^{3} } \\ \\ \mathsf{ \implies r {}^{3} = 27 - 14 } \\ \\ \mathsf{ \implies r {}^{3} = 13 } \\ \\ \mathsf{ \implies r = \sqrt[3]{13} } \\ \\ \bold{hence} \\ \\ \mathsf{ \therefore r = \sqrt[3]{13} }$


___________________

Thanks for the question !

Answer 2

hope you may like it

hope you may like it plzz mark me as brainlisit