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CBSE Class 12-science Answered

Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is 6 r square root of 3
Asked by manasd | 05 Nov, 2016, 10:14: PM
answered-by-expert Expert Answer
begin mathsize 11px style AD space is space the space altitude space of space the space isosceles space triangle space ABC. Let space apostrophe straight r apostrophe space be space the space radius space of space the space inscribed space circle. So space OD equals OE equals OF equals straight r comma space where space straight O space is space the space center space of space the space inscribed space circle. straight D space is space the space foot space of space perpendicular space on space BC. space AB space and space AC space are space the space equal space sides. BD equals DC space left parenthesis foot space of space perpendicular space bisects space the space side space BC right parenthesis... left parenthesis 1 right parenthesis BD equals BE space and space CD equals CF space left parenthesis lengths space of space tangents space drawn space from space external space point space are space equal space right parenthesis... left parenthesis 2 right parenthesis From space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis comma space BD equals BE equals DC equals CF.... left parenthesis 3 right parenthesis Similarly space AE equals AF space left parenthesis lengths space of space tangents space from space point space straight A space to space the space circle right parenthesis... left parenthesis 4 right parenthesis Perimeter space of space the space triangle equals AB plus BC plus AC equals AE plus BE plus BD plus DC plus CF plus AF equals stack AE plus AF with underbrace below plus stack BE plus BD plus DC plus CF with underbrace below equals 2 AE plus 4 BE space left parenthesis Using space left parenthesis 3 right parenthesis space and space left parenthesis 4 right parenthesis right parenthesis In space right space triangle space OEA comma space AE equals OE over tanx equals straight r over tanx.... left parenthesis 5 right parenthesis In space right space triangle space ABD comma space BD equals ADtanx rightwards double arrow BD equals ADtanx equals open parentheses AO plus OD close parentheses tanx equals open parentheses straight r over sinx plus straight r close parentheses tanx.... left parenthesis 6 right parenthesis So space perimeter equals 2 AE plus 4 BE equals fraction numerator 2 straight r over denominator tanx end fraction plus 4 open parentheses straight r over sinx plus straight r close parentheses tanx rightwards double arrow straight P left parenthesis straight x right parenthesis equals straight r open parentheses 2 cotx plus 4 secx plus 4 tanx close parentheses To space find space extrema comma space equate space the space first space derivative space to space zero. fraction numerator dP left parenthesis straight x right parenthesis over denominator dx end fraction equals straight r open parentheses negative 2 cosec squared straight x plus 4 secxtanx plus 4 sec squared straight x close parentheses equals 0 rightwards double arrow straight r open parentheses negative fraction numerator 2 over denominator sin squared straight x end fraction plus fraction numerator 4 sinx over denominator cos squared straight x end fraction plus fraction numerator 4 over denominator cos squared straight x end fraction close parentheses equals 0 rightwards double arrow straight r open parentheses fraction numerator negative 2 cos squared straight x plus 4 sin cubed straight x plus 4 sin squared straight x over denominator sin squared xcos squared straight x end fraction close parentheses equals 0 rightwards double arrow negative 2 open parentheses 1 minus sin squared straight x close parentheses plus 4 sin cubed straight x plus 4 sin squared straight x equals 0 rightwards double arrow 2 sin cubed straight x plus 3 sin squared straight x minus 1 equals 0 By space inspection comma space sinx equals negative 1 space is space straight a space solution. Hence comma space sinx plus 1 space is space straight a space factor space of space 2 sin cubed straight x plus 3 sin squared straight x minus 1 So space 2 sin cubed straight x plus 3 sin squared straight x minus 1 equals 0 rightwards double arrow open parentheses sinx plus 1 close parentheses open parentheses 2 sin squared straight x plus sinx minus 1 close parentheses equals 0 sinx space can space not space be space minus 1 space because space apostrophe straight x apostrophe space can space not space be space more space than space 90 degree So space 2 sin squared straight x plus sinx minus 1 equals 0 rightwards double arrow open parentheses 2 sinx minus 1 close parentheses open parentheses sinx plus 1 close parentheses equals 0 Again space sinx space can space not space be space minus 1. space So space 2 sinx minus 1 equals 0 rightwards double arrow sinx equals 1 half rightwards double arrow straight x equals 30 degree At space sinx equals 1 half comma space the space first space derivative space changes space sign space from space minus ve space to space plus ve. So space it space is space straight a space point space of space minima space for space straight P left parenthesis straight x right parenthesis. Hence comma space least space perimeter equals right enclose straight P left parenthesis straight x right parenthesis end enclose subscript straight x equals 30 degree end subscript equals straight r open parentheses 2 cot 30 degree plus 4 sec 30 degree plus 4 tan 30 degree close parentheses equals straight r open parentheses 2 square root of 3 plus 4 cross times fraction numerator 2 over denominator square root of 3 end fraction plus 4 cross times fraction numerator 1 over denominator square root of 3 end fraction close parentheses equals straight r open parentheses fraction numerator 18 over denominator square root of 3 end fraction close parentheses equals 6 square root of 3 straight r space left parenthesis Proved right parenthesis end style
Answered by Rebecca Fernandes | 06 Nov, 2016, 10:15: AM

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