A right-angled cone (with base radius 5 cm and height 12 cm), as shown in the figure below, is rolled on the ground keeping the point P fixed until the point Q (at the base of the cone, as shown) touches the ground again. By what angle (in radians) about P does the cone travel?
To find the angle about point P by which the cone travels, we need to analyze the rolling motion of the cone.
\(l = \sqrt{r^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, \text{cm}\)
\(C = 2\pi r = 2\pi \times 5 = 10\pi \, \text{cm}\)
\(\text{Arc length} = 10\pi\)
\(\text{Circumference of circle with radius } l = 2\pi \times l = 2\pi \times 13 = 26\pi\)
The angle in radians is:
\(\theta = \frac{\text{Arc length}}{\text{circumference with radius} \, l} = \frac{10\pi}{26\pi} = \frac{10}{26} = \frac{5}{13}\)
Therefore, the angle by which the cone travels about point P is \(\frac{10\pi}{13}\) radians.
Thus, the correct answer is \(\frac{10\pi}{13}\).
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$^2$.
