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| 1. |
A chord of a circle of a radius 15 subtends of 60° at the centre. Find the areas |
| Answer» r = 15 cm,\xa0θ = 60oArea of the minor sector = {tex}\\frac\\theta{360^\\circ}\\mathrm{πr}^2\\;=\\;\\frac{\\displaystyle60^\\circ}{\\displaystyle360^\\circ}\\times3.14\\;\\times15\\times15{/tex} = 117.75 cm2In {tex}\\triangle{/tex}AOB, draw OM\xa0{tex}\\perp{/tex} ABIn right triangle OMA and OMB,OA = OB .........Radii of the same circleOM = OM .........Common side{tex}\\therefore{/tex}\xa0{tex}\\triangle{/tex}OMA\xa0{tex}\\cong{/tex}{tex}\\triangle{/tex}OMB .........RHS congruence criterion{tex}\\therefore{/tex}\xa0AM = BM .......CPCT{tex}\\Rightarrow{/tex}\xa0AM = BM = {tex}\\frac 12{/tex}AB{tex}\\angle{/tex}AOM = {tex}\\angle{/tex}BOM .......CPCT{tex}\\Rightarrow{/tex}\xa0{tex}\\angle{/tex}AOM = {tex}\\angle{/tex}BOM = {tex}\\frac 12{/tex}{tex}\\angle{/tex}AOB =\xa0{tex}\\frac 12{/tex}\xa0{tex}\\times{/tex} 60o = 30o{tex}\\therefore{/tex}\xa0In right triangle OMA, cos30o = {tex}\\frac {OM}{OA}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{\\sqrt3}2{/tex}= {tex}\\frac {OM}{15}{/tex}{tex}\\Rightarrow{/tex}\xa0OM = {tex}\\frac{15\\sqrt3}2{/tex}cmsin30o = {tex}\\frac {AM}{OA}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac 12{/tex}= {tex}\\frac {AM}{15}{/tex}{tex}\\Rightarrow{/tex}\xa0AM = {tex}\\frac{15}2{/tex}cm{tex}\\Rightarrow{/tex}\xa0AB = 15 cm{tex}\\therefore{/tex}\xa0Area of {tex}\\triangle{/tex}AOB =\xa0{tex}\\frac 12{/tex}\xa0{tex}\\times{/tex} AB\xa0{tex}\\times{/tex} OM= {tex}\\frac 12{/tex}\xa0{tex}\\times{/tex} 15\xa0{tex}\\times{/tex}{tex}\\frac{15\\sqrt3}2{/tex} = {tex}\\frac{225\\sqrt3}4{/tex}= {tex}\\frac {225 × 1.73}4{/tex} = 97.3125 cm2{tex}\\therefore{/tex}\xa0Area of the corresponding minor segment of the circle = Area of minor sector - Area of {tex}\\triangle{/tex}AOB= 117.75 - 97.3125 = 20.4375 cm2and, area of the corresponding major segment of the circle = {tex}\\pi{/tex}r2 - area of the corresponding minor segment of the circle= 3.14\xa0{tex}\\times{/tex} 15\xa0{tex}\\times{/tex} 15 - 20.4375= 706.5 - 20.4375 = 686.0625 cm2 | |