Question

Which statement correctly names the congruent triangles and justifies the reason for congruence? a coordinate plane with triangles ABC and FED at A 0 comma 0, B 4 comma 0, C 1 comma 3, D 5 comma 2, E 8 comma 5, and F 8 comma 1, segment AC measures 3 and 2 tenths units, angle A measures 72 degrees, segment DF measures 3 and 2 tenths units, and angle F measures 72 degrees Î"ABC ≅ Î"FDE by HL Î"ABC ≅ Î"FED by SAS Î"ABC ≅ Î"DEF by HL Î"ABC ≅ Î"DEF by SAS.

Answer 1

Use the distance formula to justify the congruence. The correct option is B. $\Delta ABC\cong \Delta FED$.

Given,

the coordinates of $\Delta ABC$ and $\Delta FED$ are $A(0,0)\ B(4,0)\ C(1,3)\ D(5,2)\ E(8,5)\ F(8,1).$

The length of segment AC is $\frac{3}{2}$ and the length of segment DF is also $\frac{3}{2}$ .

$\angle A = 72^{\circ}$ and $\angle F =72^{\circ}$.

We have to find the correct statement regarding the triangles to be congruent.

Distance formula

We know that, The distance formula to calculate the distance between two points $( x_{1} , y_{1})$ , and $( x_{2}, y_{2})$ is given as,

$D=\sqrt{(x_{2}-x_{1} ) ^{2}+(y_{2} -y_{1} )^2 }$

So, The distance between $F(8,1)$ and $E(8,5)$ Will be,

$FE=\sqrt{(8-8)^2+(5-1)^2} \\FE=\sqrt{4^2} \\FE=4$

Similarly, the distance between $A(0,0)$ and $B(4,0)$ will be,

$AB=\sqrt{(4-0)^2+(0-0)^2} \\AB=\sqrt{4^2} \\AB=4$

So, $AB=FE$.......(1)

Now, in $\Delta ABC$ and $\Delta FED$,

$AB=FE$ ( from equation 1 ).

$\angle A=\angle F$ ( Given )

$AC=DF$ ( Given )

So, By Side angle Side congruence rule,

$\Delta ABC\cong \Delta FED$.

Hence the correct option is B. $\Delta ABC\cong \Delta FED$.

For more details on SAS congruence rule follow the link:

brainly.com/question/11804042