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# Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius 10 cm is square of side 10√2 cm.

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## Let ABCD be a rectangle inscribed in a circle of radius 10 cm with centre at O, then DB=20 cm.Let ∠OBA=θ,(0<θ<π2) and θ is in radian.Then, AB=20cosθ AD=20sinθLet p be the perimeter of the rectangle ABCD, thenp=2AB+2AD=2(20cosθ+20sinθ)p=40(cosθ+sinθ) ...(i)Differentiating (i) w.r.t. θ, we getdpdθ=40(−sinθ+cosθ) ...(ii)Again differentiating, we getd2pdθ2=−40(cosθ+sinθ) ...(iii)Now, for maximum value dpdθ=0⇒40(−sinθ+cosθ)=0⇒cosθ=sinθ⇒tanθ=1θ=tan−1(1)⇒θ=π4Also, (d2pdθ2θ=π/4)=−40(1√2+1√2)=−40×2√2=−40√2<0(−ve)⇒p is maximum when θ=π4.Therefore, p is maximum whenAB=20cosπ4=20×1√2=10√2AD=20sinπ4=20×1√2=10√2i.e., when adjacent sides are equal and each of them is 10√2 cm.That is, a rectangle which can be inscribed in a circle of radius 10 cm is a square of side 10√2 cm.   Suggest Corrections  0      Similar questions
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