A sphere is inscribed in a cone of radius $3 \sqrt{3}$ and slant height $6 \sqrt{3}$ . The radius of the sphere is:

Question

A sphere is inscribed in a cone of radius 3sqrt-3- and slant height 6 sqrt-3- The radius of the sphere- is -33 sqrt-3-6 sqrt-3-displaystyle frac-3 sqrt-3-2-

Toppr Admin · Accepted Answer

BC-x221A-6-x221A-3-2-x2212-3-x221A-3-2-x221A-108-x2212-27-9BO-9-x2212-R-AO-OB-x221A-r2-3-x221A-3-2-9-x2212-rr2-27-81-r2-x2212-18r18r-54-x21D2-r-3

A sphere is inscribed in a cone of radius $3 \sqrt{3}$ and slant height $6 \sqrt{3}$ . The radius of the sphere, is -
$3$
$3 \sqrt{3}$
$6 \sqrt{3}$
$\frac{3 \sqrt{3}}{2}$

$B C = \sqrt{(6 \sqrt{3})^{2} - (3 \sqrt{3})^{2}} = \sqrt{108 - 27} = 9$
$B O = 9 - R, A O = O B$
$\sqrt{r^{2} + (3 \sqrt{3})^{2}} = 9 - r$
$r^{2} + 27 = 81 + r^{2} - 18 r$
$18 r = 54 \Rightarrow r = 3$

A sphere is inscribed in a cone of radius 3√3 and slant height 6√3. The radius of the sphere, is -33√36√33√32

BC=√(6√3)2−(3√3)2=√108−27=9BO=9−R,AO=OB√r2+(3√3)2=9−rr2+27=81+r2−18r18r=54⇒r=3

A sphere is inscribed in a cone of radius $3 \sqrt{3}$ and slant height $6 \sqrt{3}$ . The radius of the sphere, is -
$3$
$3 \sqrt{3}$
$6 \sqrt{3}$
$\frac{3 \sqrt{3}}{2}$

$B C = \sqrt{(6 \sqrt{3})^{2} - (3 \sqrt{3})^{2}} = \sqrt{108 - 27} = 9$
$B O = 9 - R, A O = O B$
$\sqrt{r^{2} + (3 \sqrt{3})^{2}} = 9 - r$
$r^{2} + 27 = 81 + r^{2} - 18 r$
$18 r = 54 \Rightarrow r = 3$