In Fig. 9, is shown a sector OAP of a circle with centre O, containing \[\angle \theta .\text{ }AB\] is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter of shaded region is \[r\left[ \tan \,\theta +\sec \,\theta +\frac{\pi \theta }{180}-1 \right]\] |
Answer:
Given, the radius of circle with centre O is r. \[\angle POA=\theta \]. then, length of the arc \[\overset\frown{PA}=\frac{2\pi r\theta }{360{}^\circ }=\frac{\pi r\theta }{180{}^\circ }\] And \[\tan \theta =\frac{AB}{r}\] \[AB=r\,\tan \theta \] And \[\sec \theta =\frac{OB}{r}\] \[OB=r\,\sec \theta \] Now, \[PB=OB-OP\] \[=r\,\sec \,\theta -r\] \[\therefore \] Perimeter of shaded region \[=AB+PB+\overset\frown{PA}\]. \[=r\,\tan \,\theta +r\,\sec \,\theta -r+\frac{\pi r\theta }{180{}^\circ }\] \[=r\left[ \tan \theta +\sec \theta +\frac{\pi \theta }{180}-1 \right]\] Hence Proved.
You need to login to perform this action.
You will be redirected in
3 sec