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| 1. |
Prove the perimetèr of the shaded region is r(tanø+secø+pieø/180-1) |
| Answer» Given, Radius\xa0= r{tex} \\triangle{/tex}AOB is a right triangle.In the right {tex} \\triangle{/tex}OAB,{tex} \\tan \\theta = \\frac { \\mathrm { AB } } { \\mathrm { OA } }{/tex}{tex} \\Rightarrow A B = O A \\times \\tan \\theta{/tex}{tex} = r \\tan \\theta{/tex}Now, area of {tex} \\triangle{/tex}OAB ={tex}\\frac12 b \\times h{/tex}={tex} \\frac { 1 } { 2 } \\mathrm { OA } \\times \\mathrm { AB }{/tex}{tex} = \\frac { 1 } { 2 } \\times r \\times r \\tan \\theta{/tex}{tex} = \\frac { 1 } { 2 } r ^ { 2 } \\tan \\theta{/tex}and area of sector OAC ={tex} \\pi r ^ { 2 } \\times \\frac { \\theta } { 360 ^ { \\circ } }{/tex}\xa0length of arc AC =\xa0{tex} 2 \\pi r \\times \\frac { \\theta } { 360 ^ { \\circ } }{/tex}{tex} = \\frac { 2 \\pi r \\theta } { 360 ^ { \\circ } } = \\frac { \\pi r \\theta } { 180 ^ { \\circ } }{/tex}{tex} \\therefore{/tex}Perimeter of the shaded region= Arc AC + AB + BC{tex} = \\frac { \\pi r\\theta } { 180 } + r \\tan \\theta + ( \\mathrm { OB } - \\mathrm { OC } ){/tex}{tex} = \\frac { \\pi r\\theta } { 180 } + r \\tan \\theta + ( r \\sec \\theta - r ){/tex}{tex} = r \\left( \\frac { \\pi \\theta } { 180 } + \\tan \\theta + \\sec \\theta - 1 \\right){/tex}{tex} = r \\left( \\tan \\theta + \\sec \\theta + \\frac { \\pi \\theta } { 180 } - 1 \\right){/tex} | |