If $(x+y)^{2}-(x-y)^{2}=20$, then $x y$ is equal to:
(a) 0 ;
(b) 1 ;
(c) 2 ;
(d) 5 ;
(e) 10 .
The correct option is (d).

Since $(x+y)^{2}=x^{2}+2 x y+y^{2}$ and $(x-y)^{2}=x^{2}-2 x y+y^{2}$, we have

$$
20=(x+y)^{2}-(x-y)^{2}=x^{2}+2 x y+y^{2}-x^{2}+2 x y-y^{2}=4 x y
$$

therefore $x y=5$.