Math, asked by skharshitha9948, 11 months ago

Show that the least perimeter of an isosceles triangle

Answers

Answered by akshitdhull10
2

Answer:

ABC is an isosceles triangle with sides AB=AC. O is the center of circle of radius inscribed in the triangle. AD is the altitude,

So OD=OF=OE=r

So BD = CD  ............. (1) ( altitude bisect the side)

BD = BE and CD = CF      .............. (2) ( tangents from the same point are equal in length)

From (1) and (2) 

BD=BE=CD=CF ........ (3)

Similariy AE = AF  ........... (4) (tangents from the same point)

 

Perimeter of triangle (P) = AB + BC + AC 

(P) = AE + EB + BD + CD + CF+AF

(P) = AE + AF + EB + BD + CD + CF

(P) = 2AE + 4BD    (using 3 and 4)

In Right Triangle OEA, 

AE = OETan x

=> AE = rtan x 

=> AE = r cot x     .............. (5)

In Right Triangle ABD

BD = AD*tanx 

=> BD = (AO + OD)* tan x

=>BD = rsin x + rtan x

=> BD = r secx + r tanx     ............... (6)

From 5 and 6 

=> (P) = 2r cotx + 4r(sec x  + tanx)

=> (P) = r ( 2 cotx + 4 sec x+ 4 tanx)

Differentiate w.r.t. x

we get 

dPdx = r( -2cosec2x + 4 sec x tan x + 4 sec2x)               ........ (7)

=> dPdx = r( -2sin2x + 4sin xcos2x + 4cos2x)

 

To find critical value, Put dPdx = 0

=> r( -2sin2x + 4sin xcos2x + 4cos2x) = 0

=> ( -2sin2x + 4sin xcos2x + 4cos2x) = 0

=> -2cos2x + 4sin3x + 4sin2x sin2x ×cos2x  = 0

=> -2(1-sin2x)+ 4sin3x + 4sin2x  = 0

=> -2+2sin2x+ 4sin3x + 4sin2x  = 0

=> -2+ 4sin3x + 6sin2x  = 0

=>2sin3x + 3sin2x - 1 = 0

=>(2sin3x + 2sin2x) + (sin2x - 1) = 0

=>2sin2x(sin x  + 1) + (sinx +1) (sin x - 1) = 0

=> (sin x  + 1) (2sin2x + sin x - 1) = 0

=> (sin x  + 1) = 0  or (2sin2x + sin x - 1) = 0

=> sin x  = -1  is not possible because x cant be more than 90° so  (2sin2x + sin x - 1) = 0

=>  2sin2x + 2sin x - sin x - 1 = 0

=>   2sinx( sin x + 1)  -  1( sin x + 1) = 0

=>  (2sinx-  1) ( sin x + 1) = 0

=>  (2sinx-  1) = 0 or ( sin x + 1) = 0 

=> sin x  = -1  is not possible because x cant be more than 90° so  (2sin x - 1) = 0

=>  2sinx-  1= 0 

=>  sinx = 12

=> x = 30°

at  sinx = 12 sign changes from -ve to +ve. so it is point of local minima, so at this point value ll be least.

Hence least perimeter is 

 (P) = r ( 2 cot 30° + 4 sec 30°+ 4 tan30°)

=> (P) = r ( 2 3 + 4×23+ 4 13)

=> (P) = r  6+ 8+43

=> (P) = r 183

=> (P) = 63r

Hence Proved

hopw it will help you...

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