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Question

The sum of lengths of the hypotenuse and a side of a right angled triangle is given. Show that the area of the triangle is maximum when the angle between them is 60.

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Solution

Let the hypotenuse of the right angled triangle be x and the height be y
Hence its base=x2y2 by pythagoras theorem
Hence its area=12×base×height
=12×x2y2×y
But it is given that x+y=p(say)
Substituting this in the area we get
Area=12×(py)2y2×y
=y2×p2+y22pyy2
=y2×p22py
Squaring on both the sides we get
A2=y24(p22py) where A is the area of the given triangle.
A2=y2p24py32
For maximum or minimum area dAdy=0
Hence A2=y2p24py32
Differentiating both sides w.r.t y we get
2AdAdy=2yp24p23y2
2yp24p23y2=0
p2y3py2=0
py(p3y)=0
y=0,y=p3
When y=p3x+y=p
x=py=pp3=2p3
cosθ=yx=p32p3=12
θ=π3=60
Hence the area is maximum if the angle between the hypotenuse and the side is 60

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