Question

Whats the surface area of a cone with a height of "50" and a diameter of "40"

Answer 1

Answer:

4637.883499 units²

Step-by-step explanation:

Radius 'r'

40/2 = 20

Slant height 's'

sqrt(20²+50²) = 10sqrt(29)

Surface area:

(pi × r × s) + (pi × r²)

[3.14×20×10sqrt(29)] + (3.14×20²)

4637.883499 units²

Answer 2

Answer:

56844.9 units squared

Step-by-step explanation:

The surface area of a cone is denoted by: $A=\pi r^2+\frac{1}{2} \pi r^2*l$ , where r is the radius and l is the slant height. The slant height is basically the length from a point on the base circle to the top vertex of the cone.

Here, since our diameter is 50 and diameter is twice the radius, then our radius is r = 50/2 = 25.

To find the slant height, we have to use the Pythagorean Theorem:

$l^2=r^2+h^2$, where h is the height

$l^2=25^2+50^2=625+2500=3125$

$\sqrt{l^2} =\sqrt{3125}$

$l=25\sqrt{5}$

Now, plug these values of r and l into the first equation above:

$A=\pi r^2+\frac{1}{2} \pi r^2*l$

$A=\pi *25^2+\frac{1}{2} \pi *25^2*25\sqrt{5} =625\pi +\frac{15625\sqrt{5} }{2} \pi$ ≈ 56844.9 units squared