Question

If the sum of length of hypotenuse and a side of right angled triangle is given then show that the area of triangle is maximum when the angle between them is pie/3

Answer 1

If the sum of the lengths of the hypotenuse and a side of a right triangle is given, show that the area of the triangle is maximum, when the angle between them is 60∘60∘.


1 Answer

Let the hypotenuse of the right triangle be x, and the height be y



Hence its base is x2−y2−−−−−−√x2−y2 by applying phythagorous theorem.

Hence its are = 12×base×height12×base×height

Area = 1212×x2−y2−−−−−−√×y×x2−y2×y

But it is given x+y=p(say)x+y=p(say)

Substituting this in the area we get

Area = 1212×(p−y)2−−−−−−−√−y2×y×(p−y)2−y2×y

1212yp2+y2−2py−y2−−−−−−−−−−−−−−−√yp2+y2−2py−y2

=12=12yp2−2py−−−−−−−√yp2−2py

Squaring on both the sides we get

(Area)2=14(Area)2=14y2(p2−2py)y2(p2−2py)

i.e., A=14A=14y2(p2−2py)y2(p2−2py)

=14=14p2y2=12py3p2y2=12py3

For maximum or miniumu area

dydAdydA=0=0

Here the area of the triangle is maximum when x=2p3andy=p3x=2p3andy=p3

cosθ=yxcos⁡θ=yx

=p32p3=p32p3

∴cosθ=12∴cos⁡θ=12

⇒θ=π3⇒θ=π3 or 60∘60∘

Hence the area is maximum if the angle between the hypotenuse and the side is 60∘


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