Let ABC be a right angle triangle at B:

Then, AC is the hypotenuse and AB and BC are the legs.

Given that AC = 9

and one of the sides (BC) = 3

We need to find the area.

We know that the area of the triangle is given by:

A = (1/2)* base * height

= (1/2)* BC * AB

We need to calculate the length of AB

We know that:

AC^2 = AB^2 + BC^2

==> AB^2 = AC^2 - BC^2

= 9^2 - 3^2 = 81 -9 = 72

==> AB = sqrt72 = 6sqrt2

Now we will substitute into the area.

==> A = (1/2)*3 * 6sqrt2

= 9sqrt2 = 12.73 square units.

**Then, the area of the triangle is 9sqrt2 = 12.73 square units.**

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