Given,

• The triangle is right angled triangle.

• Hypotenuse is 5cm.

Let us consider,

• The base of the triangle is ‘a’.

• The adjacent side is ‘b’.

Now AC² = AB² + BC²

As AC = 5, AB = b and BC = a

25 = a² + b²

b² = 25 – a² ---- (1)

Now, the area of the triangle is

Squaring on both sides

Substituting (1) in the area formula

----- (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with a and then equating it to zero. This is because if the function Z (x) has a maximum/minimum at a point c then Z’(c) = 0.

Differentiating the equation (2) with respect to a:

[Since ]

----- (3)

To find the critical point, we need to equate equation (3) to zero.

a=0 (or)

[as a cannot be zero]

Now to check if this critical point will determine the maximum area, we need to check with second differential which needs to be negative.

Consider differentiating the equation (3) with a:

----- (4)

[Since ]

Now let us find the value of

As , so the function A is maximum at

Substituting value of A in (1)

Now the maximum area is