Use the Pythagorean theorem:
[tex]c^2=a^2+b^2[/tex]
We have:
[tex]a=1;\ c=\sqrt3[/tex]
Substitute:
[tex]1^2+b^2=(\sqrt3)^2\\\\1+b^2=3\ \ \ |-1\\\\b^2=2\to b=\sqrt2[/tex]
We calculate the area of the triangle by two methods:
[tex]A_\Delta=\dfrac{ab}{2}\ and\ A_\Delta=\dfrac{ch}{2}[/tex]
[tex]\dfrac{h\sqrt3}{2}=\dfrac{1\sqrt2}{2}\ \ \ \ |\cdot2\\\\h\sqrt3=\sqrt2\ \ \ |\cdot\sqrt3\\\\3h=\sqrt6\ \ \ |:3\\\\h=\dfrac{\sqrt6}{3}[/tex]
