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  • In fig.9 is shown a sector OAP of a circle with center O, containing . AB is perpendicular to the radius OA and meets OP produced at B. Prove t

In fig.9 is shown a sector OAP of a circle with center O, containing \angle \theta. AB is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter of the shaded region is r \left [ \tan \theta + \sec \theta + \frac{\pi \theta }{180}-1 \right ]

 

 

 

 
 
 
 
 

Answers (1)

\AOB = \theta \\\\ AB \perp OA \\\\ \angle OAB = 90 \degree

area of shaded region = AB + PB + arc length AP 

Arc length AP = \frac{\theta }{360 } \times 2 \pi r = \frac{\pi \theta r }{180 } ----(1) \\\\

In right angled triangle \Delta OAB \\\\ \tan \theta = \frac{AB}{r}, AB = r \tan \theta --- (2) \\\\ \sec \theta = \frac{OB }{r} ,OB = r \sec \theta ---- (3) \\\, now , OB = OP + PB \\\\ PB = OB - OP \\\\ r \sec \theta - r \\\\ PB = r \sec \theta - r --- (4)

perimeter of shaded region = AB + PB + arc length AP 

putting values from (1) (2) (4) 

Perimeter of shaded region = 

r \tan \theta + r \sec \theta - r + \frac{\pi \theta r }{180 \degree }\\\\ r [ \tan \theta + \sec \theta -1+ \frac{\pi \theta }{180 }] \\\\ = [ \tan \theta + \sec \theta + \frac{\pi \theta }{180}-1]

Hence proved 

 

Posted by

Ravindra Pindel

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