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