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# 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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