Draw the right triangle, with the short leg vertical, the long leg horizontal, and the hypotenuse slanty diagonal.
The short leg then is the height, and the long leg is the base.
The area "A" for a triangle with base "b" and height "h" is given by:
. . . . .A = (1/2)bh
You are given a value for A. Plug this in, and solve for one of the remaining variables.
For any right triangle, you have:
. . . . .a<sup>2</sup> + b<sup>2</sup> = c<sup>2</sup>
...or, in this case:
. . . . .h<sup>2</sup> + b<sup>2</sup> = 41<sup>2</sup>
You solved for one of the two variables in terms of the other. Substitute that into the above equation, and solve.
If you get stuck, please reply showing how far you have gotten. Thank you.
Eliz.