Q. 274.0( 2 Votes )

Prove that the le

Answer :


Let ABC be the isosceles triangle with AB = AC and the circle be inscribed in it of radius r and centre O


The centre of incircle is intersections of altitudes and in isosceles triangle the altitude from vertex A is also the bisector hence AF bisects BC at F hence BF = FC


Now we require some parameter based on which the perimeter of triangle will change


Let that parameter be the angle θ as shown


We have to find a θ such that the perimeter of ΔABC is minimum and also the minimum perimeter


let P be the perimeter


P = AB + BC + AC


P = 2AB + BC


From figure BC = BF + FC but BF = FC hence BC = 2BF


P = 2AB + 2BF


Consider ΔABF



Hence


P = 2AB + 2(ABsinθ)


P = 2AB(1 + sinθ) …(a)


Again consider ΔABF



But from figure AF = AO + OF and OF is radius r hence AF = AO + r



Now consider ΔADO



But OD is the radius r hence


Put in (b)




Put this value of AB in (a)




We know that sin2θ = 2sinθcosθ



Now we have found P in terms of theta, to find minimum value of P we have to first find the critical points where


Differentiate P with respect to θ and equate to 0


Differentiating using the rule which says that


Here u = (1 + sinθ)2 and v = sin2θ




As θ is the angle of triangle shown it has to be greater than 0 and less than 90° hence sin2θ is not 0


Hence


0 = 2(1 + sinθ)cosθsin2θ – 2(1 + sinθ)2cos2θ


cosθsin2θ = (1 + sinθ)cos2θ


Since sin2θ = 2sinθcosθ and cos2θ = 1 – 2sin2θ


2sinθcos2θ = (1 + sinθ)(1 – 2sin2θ)


Since sin2θ + cos2θ = 1


2sinθ(1 – sin2θ) = (1 + sinθ)(1 – 2sin2θ)


2sinθ(1 + sinθ)(1 – sinθ) = (1 + sinθ)(1 – 2sin2θ)


2sinθ(1 – sinθ) = 1 – 2sin2θ


2sinθ – 2sin2θ = 1 – 2sin2θ


2sinθ = 1


sinθ = 1/2


θ = 30°


Consider θ = 30° on number line we start with negative from left and for each point on which we got we change the sign. Here we have only one point 30° on which the sign for changes from minus to plus hence 30° is point of minima



Put θ = 30° in P to get the minimum perimeter







P = 6√3r


Hence proved the minimum perimeter is 6√3r


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