2 Answers

Arun Kumar Khattri · Answer 1 · 2021-12-28T03:00:48+0000

The surface area of the right cone with slanted height of 13 and radius of 4 is 221.06 square units.

What is the surface area of the right cone below?

Surface area of a right cone

$=\pi \: r(r + \sqrt{ {h}^(2) + {r}^(2) })$

$=3.14 *4 (4 + \sqrt{ {13}^(2) + {4}^(2) })$

$=12.56 (4 + \sqrt{ {169} + {16} })$

$=12.56 (4 + \sqrt{ {185} })$

=12.56 (4 + 13.60 )

= 12.56(17.60)

= 221.06

Therefore, surface area equals 221.06 square units.

Mark Fowler · Answer 2 · 2022-01-01T06:38:04+0000

Answer:

213.69 units

Explanation:

We have to first find the height of the cone.

We can use Pythagoras rule because the slant height, height and radius of a cone all form a right angled triangle:

h^2 = a^2 + b^2

where h = hypotenuse

a and b = the other two sides of the triangle

The radius is 4 units and the slant height is 13 units:

$13^2 = a^2 + 4^2\\\\169 = a^2 + 16\\\\a^2 = 169 - 16 = 153\\\\a = √(153)\\\\a = 12.37units$

The height of the cone is 12.37 units.

The surface area of a cone is given as:

SA =
$\pi r (r + √(h^2 + r^2) )$

$SA = \pi * 4(4 + √(12.37^2 + 4^2) )\\\\SA = 12.57(4 + √(13^2) )\\SA = 12.57 (4 + 13)\\\\SA = 12.57(17)\\\\SA = 213.69 units$

The surface area of the cone is 213.69 units.

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