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
a) To find the base radius of a cone made from a semi-circle of radius 12 cm, we need to use the formula:
\( \text{Base Radius} = \frac{\text{Semi-Circle Radius}}{2} \)
Substituting the given radius, we get:
\( \text{Base Radius} = \frac{12 \, \text{cm}}{2} = 6 \, \text{cm} \)
Therefore, the base radius of the cone is 6 cm.
b) To find the base radius of a cone made from a sector of a circle with radius 6 cm and an angle of 144 degrees, we need to use the formula:
\( \text{Base Radius} = \frac{\text{Circle Radius} \times \text{Angle}}{360} \)
Substituting the given radius and angle, we get:
\( \text{Base Radius} = \frac{6 \, \text{cm} \times 144^\circ}{360} \)
Simplifying the expression, we have:
\( \text{Base Radius} = \frac{864 \, \text{cm} \cdot \text{deg}}{360} \)
\( \text{Base Radius} = 2.4 \, \text{cm} \cdot \text{deg} \)
Therefore, the base radius of the cone is 2.4 cm.
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.