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 by which the cone travels about point P until point Q touches the ground again, we need to understand the geometry involved.
The cone is essentially a sector of a circle when flattened. The length of the arc of this sector is equal to the circumference of the base of the cone.
The circumference of the base of the cone is given by:
where \(r = 5 \text{ cm}\).
The length of the arc of the sector is also equal to the lateral height of the cone, which can be calculated using the Pythagorean theorem:
The slant height \(l\) is:
where \(h = 12 \text{ cm}\)
The angle subtended by the arc at the center of the sector (say \(\theta\)) is proportional to the ratio of the arc length to the full length of the circle's circumference (if it were a full circle of the cone's lateral height).
This is given by:
Thus:
This matches the given correct answer option \(\frac{10\pi}{13}\).
Therefore, the cone travels through an angle of \(\frac{10\pi}{13}\) radians about point P.
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$^2$.
