Question

Given: ΔABC ≅ ΔEDF

Answer 1

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 $3.2\ units$

Answer 2

I think the figure is that of a right triangle.

Its short leg = 2 - 1 = 1 unit     This is the value of y.
Its long leg = 3 - 0 = 3 units    This is the value of x.

Use the Pythagorean theorem to solve for the hypotenuse.

a² + b² = c²
1² + 3² = c²
1 + 9 = c²
10 = c²
√10 = √c²
3.16 = c

The length of the hypotenuse is C.) 3.2   rounded to the nearest tenth.