What is the surface area of the geometric solid produced by the 30-60-90 triangle below when it is rotated 360 degrees about the axis RD? Support your answer.

The surface area of the cone is 9.42cm²

Explanation:

When a triangle with the interior angle as 30°-60°-90° is rotated then a cone is formed.

So, we need to find the surface area of the cone.

Surface area of a cone is πr ² + πrl

where,

r = radius of the cone

l = lateral height of the cone

tan (30°) =

So,

BC = 1

AB = √3

AC = ?

(AC)² = (BC)² + (AB)²

(AC)² = (1)² + (√3)²

AC = 2

Lateral height, h = 2cm

Radius = BC = 1cm

Thus,

surface area of the cone = πr ² + πrl

Therefore, surface area of the cone is 9.42cm²

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

step-by-step explanation: