Question

The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be 1 27 th of the volume of the given cone, at what height above the base is the section made

Answer 1

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♠ ғȏяṃȗʟѧ ȗśєԀ :



ṿȏʟȗṃє ȏғ ċȏṅє =

$\sf{\frac{1}{3} \times \pi \times {radius}^{2} \times height}$





♠ ℓєт тнє яα∂ιυѕ σf тнє σяιgιиαℓ ¢σиє вє "R", яα∂ιυѕ σf ѕмαℓℓєя ¢σиє вє "r" αи∂ нєιgнт σf тнє ѕмαℓℓєя ¢σиє вє "h". αℓѕσ, gινєи нєιgнт σf тнє σяιgιиαℓ ¢σиє ( "H" ) ιѕ 30 ¢м.





♠ Now,

°•° ∆OCD ≈ ∆OAB


=> $\sf{\frac{r}{R} = \frac{h}{H}}$

=> $\sf{\frac{r}{R} = \frac{h}{30}}$

=> $\sf{r = \frac{hR}{30}...eq i}$





♠ Now, it is given that, volume of smaller cone is 1/27 times the volume of original cone.



=> Volume of original cone = 27 (volume of new cone)


=> $\sf{\frac{1}{3} \times \pi \times {R}^{2} \times H =27(\frac{1}{3} \times \pi \times {r}^{2} \times h)}$


=> $\sf{\frac{1}{3} \times \pi \times {R}^{2} \times 30 =27(\frac{1}{3} \times \pi \times {r}^{2} \times h)}$


=> $\sf{{R}^{2} \times 30 =27( {r}^{2} \times h)}$


=>$\sf{{R}^{2} \times \frac {30}{27} = {r}^{2} \times h}$


=> $\sf{{r}^{2} \times h ={R}^{2} \times \frac {30}{27}}$


=> $\sf{{(\frac{hR}{30})}^{2} \times h= {R}^{2} \times \frac {30}{27}}$...(using i)


=> $\sf{\frac{ {h}^{2} {R}^{2} h }{900} = {R}^{2} \times \frac {30}{27}}$


=> $\sf{\frac{ {h}^{3} }{900} = \frac{10}{9}}$


=> $\sf{ {h}^{3} = \frac{10 \times 900}{9} }$


=> $\sf{ {h}^{3} = 1000}$


=> $\sf{h = \sqrt[3]{1000} }$


=> $\sf{h = 10 \: cm}$



•°• Height of smaller cone = 10 cm





♠ •°• In the figure,

Total height, OA = 30 cm

Height of small cone, OC = 10 cm



Therefore,

Height at which the section is made, AC,

= OA - OC

= 30 - 10

= 20 cm.





♥ •°• The section has been made at a height of 20 cm above the base. ♥
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ⓣⓗⓐⓝⓚ ⓨⓞⓤ.. (^_-)

Answer 2

Answer:

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