we know that
the triangle ABC and triangle EDF are congruent triangles-----> given problem
therefore
[tex]AC=EF\\AB=ED\\BC=DF[/tex]
we know that
the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
Step [tex]1[/tex]
Find the distance AB
[tex]A(0,1)\\B(0,2)[/tex]
substitute in the formula of the distance
[tex]d=\sqrt{(2-1)^{2}+(0-0)^{2}}[/tex]
[tex]d=\sqrt{(1)^{2}+(0)^{2}}[/tex]
[tex]dAB=1\ unit[/tex]
Step [tex]2[/tex]
Find the distance BC
[tex]B(0,2)\\C(3,2)[/tex]
substitute in the formula of the distance
[tex]d=\sqrt{(2-2)^{2}+(3-0)^{2}}[/tex]
[tex]d=\sqrt{(0)^{2}+(3)^{2}}[/tex]
[tex]dBC=3\ units[/tex]
Step [tex]3[/tex]
Find the distance AC
we know that
the triangle ABC is a right triangle
so
Applying the Pythagorean Theorem
[tex]AC^{2}=AB^{2} +BC^{2}[/tex]
substitute the values in the formula
[tex]AC^{2}=1^{2} +3^{2}[/tex]
[tex]AC=\sqrt{10}\ units=3.16\ units[/tex]
round to the nearest tenth
[tex]AC=3.2\ units[/tex]
therefore
[tex]EF=3.2\ units[/tex]
[tex]ED=1\ unit[/tex]
[tex]DF=3\ units[/tex]
the answer is
The length of the hypotenuse is equal to [tex]3.2\ units[/tex]