A solid is in the form of the cone mounted on the hemisphere is such a way that the center - Maths - Surface Areas and Volumes

Question

A solid is in the form of the cone mounted on the hemisphere is
such a way that the center of the base of the cone just coincide
with the center of the base of the hemisphere Slant hieght of the
cone is L and radius of the base of the cone is 1/2 r ,where r is
the radius of the hemisphere Prove that the total surface area of
the solid is pi / 4 11r + 2l r square units

Gursheen kaur. · Accepted Answer

Given: Radius of base of cone = 1/2 r

and slant height of cone = l

To prove: TSA of solid = $\frac{π}{4} [11 r + 2 l] r$
Proof:

As we know that,

SA of cone = $π r (l + r) = π \frac{r}{2} (l + \frac{r}{2}) = \frac{π r}{2} (l + \frac{r}{2})$ .....................(i)

Now, the cone is mounted on hemisphere , which implies SA of hemisphere is half of sphere

SA of hemisphere= $3 π r^{2}$ .......................(ii)

Now, to get the new SA of cone and new SA of hemisphere, subtract the area of base of cone twice because once the cone is placed at top ,it will remove the SA from cone as well as hemisphere.

From (i) and (ii),

New SA of cone = $\frac{π r}{2} (l + \frac{r}{2}) - \frac{π r^{2}}{4} = \frac{π r l}{2} + \frac{π r^{2}}{4} - \frac{π r^{2}}{4} = \frac{π r l}{2}$ ...............(iii)

and

New SA of hemisphere= $3 π r^{2} - \frac{π r^{2}}{4} = π r^{2} (3 - \frac{1}{4}) = π r^{2} (\frac{12 - 1}{4}) = π r^{2} (\frac{11}{4})$ ..............(iv)

therefore,

Total SA of solid=SA of cone +SA of hemisphere

$= \frac{π r l}{2} + π r^{2} (\frac{11}{4}) = \frac{π r}{2} (l + \frac{11}{2} r) = \frac{π r}{2} (\frac{2 l + 11 r}{2}) = \frac{π r}{4} (2 l + 11 r)$
Hence proved.