Answer :

**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}

⇒ y^{2} = x^{2} + (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**

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