The area of a right angle triangle is half of product of legs of triangle, such that:

`A = (c_1*c_2)/2`

Since the problem provides that the right triangle is isosceles, hence, the legs have equal lengths, such that:

`A = (x*x)/2`

The problem provides the information that the area of triangle is of `24 cm^2` , such that:

`24 = x^2/2 => x^2 = 2*24 => x = sqrt(24*2) => x = sqrt(2^4*3)`

`x = 4sqrt3`

Since the problem provides the information that the hypotenuse of triangle is of `12 cm` , you need to test if the lengths of the legs verify the Pythagorean theorem, such that:

`12^2 = (4sqrt3)^2 + (4sqrt3)^2`

`144 = 2*16*3 => 144!=96`

**Hence, there exists no such a right isosceles triangle whose hypotenuse is of `12 cm` and area of `24 cm^2` .**

**Further Reading**

The area of an isosceles right triangle is 24 with hypotenuse 12. Let the length of the equal sides be L, `2*L^2 = 24`

=> L = `sqrt 12`

The area of the triangle is `(1/2)*L^2 = 6`

**It is not possible for an isosceles right triangle to have area 24 and hypotenuse 12**

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