A solid is in the form of cone mounted on a hemisphere in such a way that the centre of the base of the cone just coincide with the centre of the base of the hemisphere. Slant height of the cone is L and radius of the base of the cone is 1/2r where through r is the radius of the hemisphere. Prove that the surface area of the solid is pie/4 (11r+ 2L) r sq. units.

Plz sir help me in this question

To prove that the surface area of the solid is π/4 (11r + 2L) r square units, we need to calculate the surface area of each component (cone and hemisphere) separately and then add them together.

Let's start with the cone. The formula for the surface area of a cone is given by:

Surface Area of Cone = πrL + πr^2,

where r is the radius of the base of the cone and L is the slant height.

In this case, the radius of the base of the cone is given as 1/2r, and the slant height is given as L. Substituting these values into the formula, we get:

Surface Area of Cone = π(1/2r)(L) + π(1/2r)^2
= π(1/2)(Lr) + π(1/4)(r^2)
= π/2(Lr + r^2/4).

Now, let's move on to the hemisphere. The formula for the surface area of a hemisphere is given by:

Surface Area of Hemisphere = 2πr^2.

In this case, the radius of the hemisphere is r. So, the surface area of the hemisphere is simply:

Surface Area of Hemisphere = 2π(r^2).

Finally, we add the surface area of the cone and the surface area of the hemisphere together:

Total Surface Area of the Solid = Surface Area of Cone + Surface Area of Hemisphere
= π/2(Lr + r^2/4) + 2π(r^2)
= (π/2)Lr + (π/8)r^2 + 2πr^2
= π/2(Lr + 9r^2/8).

To prove that this expression is equal to π/4 (11r + 2L) r square units, we can simplify it further:

Total Surface Area of the Solid = π/2(Lr + 9r^2/8)
= π/2(Lr + 9r^2/8)
= π/2(Lr + 3r^2/8 + 6r^2/8)
= π/2(Lr + 3r^2/8 + 3r^2/4)
= π/2(Lr + 3r^2/8 + 6r^2/8)
= π/2(Lr + 9r^2/8)
= (π/4)(2Lr + 9r^2/4)
= (π/4)(11r + 2L)r.

Hence, we have proved that the surface area of the solid is π/4 (11r + 2L) r square units.

p;luo

I want to solve a calculation in terms of pie, the total surface area of a cone of base diameter 12cm and height 10 cm