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

OR

If the sum of lengths of hypotenuse and a side of a right-angled triangle is given, show that area of triangle is maximum, when the angle between them is

To prove: the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is 6√3 r

Let ABC is an isosceles triangle with AB = AC = x and BC = y

and a circle with center O and radius r is inscribed in triangle ABC

Since, O is incenter of the triangle. It divides the medians into 2:1

⇒ AO = 2r and OF = r

Using Pythagoras theorem in ∆ ABF:

(AF)² + (BF)² = (AB)²

Using Pythagoras theorem in ∆ AEO:

(AE)² + (OE)² = (AO)²

BF = BE and CF = CD (Tangents from same external points are equal)

Now, AE + EB = x

Perimeter of the triangle

= 2x + y

This shows that the least perimeter of an isosceles triangle in

which a circle of radius r can be inscribed is 6√3 r

Hence Proved

OR

Given: sum of lengths of hypotenuse and a side of a right-angled triangle is given

To prove: area of triangle is maximum, when the angle between them is .

Let ABC is a triangle with base BC = x and hypotenuse AB = y, θ be the angle between hypotenuse and base, A be the area of triangle

Sum of hypotenuse and one side i.e. base is given say k

⇒ x + y = k

⇒ y = k – x

Using Pythagoras theorem,

(AB)² = (BC)² + (AC)²

⇒ y² = x² + (AC)²

Differentiating both sides with respect to x:

Again, differentiating with respect to x:

For maximum or minimum, first derivative = 0:

Since second derivative is negative, A is maximum when

y = k – x

θ is angle between AB and BC

Hence Proved