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]
