Question 1044876
In a right circular cone, the base radius, {{{r}}} is perpendicular to the height, {{{h}}} .
Radius, height, and slant height form a right triangle,
and the slant height can be calculated using the Pythagorean theorem as
{{{sqrt(h^2+r^2)}}} .
So, when given {{{r}}} and {{{h}}} , the lateral surface area of a right circular cone can be calculated as
{{{pi*r*sqrt(h^2+r^2)}}} .
With {{{h=24}}} and {{{r=10}}} ,
the slant height is {{{sqrt(24^2+10^2)=26}}} ,
and the lateral surface area is {{{pi*10*26=260pi}}} .
 
I assume that your composite figure is a cone with a hemisphere taken out of the base, so that the cross section looks like this:
{{{drawing(240,280,-12,12,-25,3,
green(arrow(0,0,-8,0)),green(arrow(-8,0,0,0)),
green(triangle(10,0,0,0,0,-24)),
green(rectangle(0,0,1,-1)),
locate(-4.5,0,green(8)),locate(4.5,2.5,green(10)),
locate(0.3,-12,green(h=24)),
green(arrow(0,1,10,1)),green(arrow(10,1,0,1)),
line(-10,0,-8,0),line(8,0,10,0),
line(-10,0,0,-24),line(10,0,0,-24),
arc(0,0,16,16,0,180),locate(5.5,-11,26)
)}}} , and only a ring is left of the base of the cone,like this: {{{drawing(240,240,-12,12,-12,12,circle(0,0,10),
circle(0,0,8),locate(-3.5,4,circular),locate(-2,2,hole),
green(arrow(0,0,8,0)),green(arrow(8,0,0,0)),
locate(3.5,0,green(8)),locate(-5.5,0,green(10)),
green(arrow(-2,0,-10,0)),green(arrow(-8,0,0,0)),
locate(-1.5,-8,ring)
)}}} .
 
The surface area of that ring that is left over of the base of the cone
is the area of a circle of radius {{{10}}}
minus the area of a circle of radius {{{8}}} ;
{{{pi*10^2-pi*8^2=100pi-64pi=36pi}}} .
 
The surface area of a sphere of radius {{{r=8}}} is {{{4*pi*8^2}}} ,
so the surface of a hemisphere of radius {{{8}}} is
{{{2*pi*8^2=2*pi*64=128pi}}} ,
 
The surface area of your composite figure is mad of three parts:
lateral surface area of the cone ={{{260pi}}} ,
area of ring on the cone base ={{{36pi}}} , and
area of hemisphere ={{{128pi}}} .
Total surface area ={{{260pi+36pi+128pi=highlight(424pi)}}} .