Question

Q1.. 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 π /3.

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Answer 1

Answer:

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

Hence its base is

x

2

−y

2

by applying Pythagoras theorem.

Hence its area =

2

1

×base×height

Area=

2

1

×

x

2

−y

2

×y

But it is given x+y=p (say)

Substituting this in the area, we get

Area =

2

1

×

(p−y)

2

−y

2

×y

=

2

1

y

p

2

+y

2

−2py−y

2

=

2

1

y

p

2

−2py

Squaring on both the sides, we get

(Area)

2

=

4

1

y

2

(p

2

−2py)

i.e., A=

4

1

y

2

(p

2

−2py)

=

4

1

p

2

y

2

−

2

1

py

3

For maximum or minimum area,

dA

dy

=0

Here the area of the triangle is maximum when x=

3

2p

and y=

3

p

cosθ=

x

y

=

2.

3

p

3

p

Therefore, cosθ=

2

1

⇒θ=

3

π

or 60

o

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

o

.