Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is 6√3r.


Let PQR is the triangle with inscribed circle of radius ‘r’, touching sides PQ at Y, QR at X and PR at Z.


OZ, OX, OY are perpendicular to the sides PR,QR,PQ.


Here PQR is an isosceles triangle with sides PQ = PR and also from the figure,


PY = PZ = x


YQ = QX = XR = RZ = y


From the figure we can see that,


Area(ΔPQR) = Area(ΔPOR) + Area(ΔPOQ) + Area(ΔQOR)


We know that area of a triangle = ×base×height











…… (1)


We know that perimeter of the triangle is Per = PQ + QR + RP


PER = (x + y) + (x + y) + 2y


PER = 2x + 4y …… (2)


From(1)





We need perimeter to be minimum and let us PER as the function of y,


We know that for maxima and minima ,






4y4 - 12y2r2 = 0


4y2(y2 - 3r2) = 0



Differentiating PER again,








>0(minima)


We got minima at y = r.


Let’s find the value of x,



x = r


PER = 2(r) + 4(r)


PER = 6r


Thus proved


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