Sphere in Cone

Take a cross-section. We now have a circle inscribed in an isosceles triangle with base 12 and legs 10. The area of this triangle is $6\times 8=48$ . We then use the triangle area formula $A=is$ , where $A$ is the area, $i$ is the radius of the inscribed circle, and $s$ is the semiperimeter (half of the perimeter). We want to find $i$ . $A=is$ $48=i(6+10)$ $48=16i$ $i=3$ So we know that the radius of the sphere is $3$ . From there, we can calculate the volume. $V=\frac{4}{3}\pi r^3$ $V=\frac{4}{3}\pi 3^3$ $V=\frac{4}{3}27\pi$ $V=\boxed{36\pi}$

cone radius = m = 6

cone height = h = 8

cone hypotenuse = s = 10 (sqrt(6^2 + 8^2)

sphere radius = r = ?

Then

$\frac { m }{ s } =\frac { r }{ h-r } \quad \Leftrightarrow \quad \frac { 6 }{ 10 } =\frac { r }{ 8-r } \\ \\ \Rightarrow \quad 6(8-r)\quad =\quad 10(r)\\ \Rightarrow \quad 48-6r\quad =\quad 10r\\ \Rightarrow \quad 16r\quad =\quad 48\\ \Rightarrow \quad r\quad \quad \quad =\quad 3$

We found the sphere radius, now lets calculate the volume

$V\quad =\quad \frac { 4 }{ 3 } \pi { r }^{ 3 }\\ V\quad =\quad \frac { 4 }{ 3 } \pi .{ 3 }^{ 3 }\quad =\quad 4.{ 3 }^{ 2 }\pi \quad =\quad 36\pi$

Area of triangle is=1/2 * 12 * 8= 2 ( 1/2 * 6 * r + 1/2 * 4 * r + 1/2 * 6 * r), => r = 3, volume of circle = 4/3 * pi * (r^3)=36pi

From similar triangles, r/(h-r) is to 6/10 where h=8. Solving for r, r=3. Volume of sphere = 4/3 pi r^3 = 4/3 pi 3^3 = 36*pi

Siple solution ...

Base radius is given 8 , so sphear will at 4 Now according to simple ratio ....... 8/6 = 4/r

So r=3 and put r in equation 4/3 (pie) r^3 = 4/3 (pie) 3^3 =4 (pie) 9 =36 * (pie)

by taking problem in 2D; by triangle side ratio ; r == 8 6/16 ==3; V==4/3 pi 3 3*3; V==36pi --------------Ans

we just have to find the inradius of an isosceles triangle whose non-equal side is 12 and altitude is 8. therefore, equal sides are 10. (using pythagoras theorem) then use pythagoras to find inradius.

i juz guessed it out.. cone radius=6 as sphere iz inscribed in it .. its radius may be half of it .. touching all its edges .. so therfore.. the volume of sphere iz= \frac { 4 }{ 3 } \times \quad \pi \quad \times \quad 3\quad \times \quad 3\quad \times \quad 3

36\quad \pi

If radious of the sphere is r then (8-r) x sin(tan[inverse] 6/8) = r.........solving this we get r=3.....and the volume is 36pi

We know that volume of sphere is 4/3 pia r3. since radius is 3 substituting in above,we get 36 pia as answer. K.K.GARG.India