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


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


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