The volume ($V$) of a right circular cone is given by the formula:
$ V = \frac{1}{3} \pi r^2 h $
where $r$ is the radius of the base and $h$ is the height of the cone.
Let the initial radius be $r$ and the initial height be $h$. The initial volume ($V_1$) is:
$ V_1 = \frac{1}{3} \pi r^2 h $
The radius is increased by 50%. The new radius ($r'$) is:
The height ($h$) remains unchanged. The new volume ($V_2$) is:
$ V_2 = \frac{1}{3} \pi (r')^2 h $
Substitute $r' = 1.5r$:
$ V_2 = \frac{1}{3} \pi (1.5r)^2 h $
$ V_2 = \frac{1}{3} \pi (2.25 r^2) h $
$ V_2 = 2.25 \times \left( \frac{1}{3} \pi r^2 h \right) $
Since $V_1 = \frac{1}{3} \pi r^2 h$, we have:
$ V_2 = 2.25 V_1 $
The increase in volume is:
$ \Delta V = V_2 - V_1 = 2.25 V_1 - V_1 = 1.25 V_1 $
The percentage increase in volume is calculated as:
$ \text{Percentage Increase} = \frac{\Delta V}{V_1} \times 100\% $
$ \text{Percentage Increase} = \frac{1.25 V_1}{V_1} \times 100\% $
$ \text{Percentage Increase} = 1.25 \times 100\% = 125\% $
Therefore, if the radius of a right circular cone is increased by 50%, its volume increases by 125%.
In the given figure, $P, Q$, and $R$ are three points on a circle of radius 10 cm with $O$ as its center, $\overline{PQ} = \overline{RQ}$, and $\angle PQR = 45^\circ$. The figure is representative.
The area of the shaded region $PQRO$ is ______________ cm$^2$.
