Question

Find the base radius of a cone made from:
a) a semi circle of radius 12cm
b) a sector of a circle of radius 6cm and angle 144degree

Answer 1

a) The diameter of the semi circle is equal to the slant height of the cone, as shown below:

[asy]
size(100);
draw(Circle((0,0),1));
draw((-1,0)--(1,0));
draw((0,1)--(0,0));
draw((0,0)--(0.8,-0.6));
label("$12$",(0.5,0),S);
label("$r$",(0.4,-0.3),S);
[/asy]

Using the Pythagorean theorem, we have:

$r^2 = 12^2 - \left(\frac{12}{2}\right)^2 = 144 - 36 = 108$

Therefore, the base radius of the cone is $\sqrt{108} = \boxed{6\sqrt{3}}$ cm.

b) The radius of the sector is equal to the slant height of the cone, as shown below:

[asy]
size(100);
draw(Circle((0,0),1));
draw((-1,0)--(1,0));
draw((0,1)--(0,0));
draw((0,0)--(0.5,-0.866));
label("$6$",(0.25,0),S);
label("$\theta$",(-0.2,0.12),N);
draw((0,0)--(0.8,0));
label("$r$",(0.4,0),S);
[/asy]

Using the formula for the length of an arc of a sector, we have:

$\frac{\theta}{360^\circ} \cdot 2\pi \cdot 6 = r$

Substituting $\theta = 144^\circ$, we have:

$\frac{144}{360} \cdot 2\pi \cdot 6 = r$

$r = \boxed{3\pi}$ cm.