Question

Determine whether ΔA B C ≅ ΔX Y Z . Explain. A(3,-1), B(3,7), C(7,7), X(-7,0), Y(-7,4), Z(1,4)

Answer 1

By SSS (Side, side, side) congruency;

⇒ ΔA B C ≅ ΔX Y Z .

Given that;

The vertices of ΔA B C are;

A (3, -1) , B (3, 7) and C (7, 7)

The vertices of  ΔX Y Z  are;

X (-7, 0), Y (-7, 4), Z (1, 4)

We have to prove that ΔA B C ≅ ΔX Y Z .

What is congruency in triangle?

If all three corresponding sides are equal and all the three corresponding angles are equal in measure, then Two triangles are said to be congruent.

Now,

The vertices of ΔA B C are;

A (3, -1) , B (3, 7) and C (7, 7)

The vertices of  ΔX Y Z  are;

X (-7, 0), Y (-7, 4), and Z (1, 4)

To prove ΔA B C ≅ ΔX Y Z we use distance formula and prove all sides are equal.

In ΔA B C;

Distance between A and B;

$D_{AB} = \sqrt{(3-3)^2 + (7 -(-1))^2} = \sqrt{64} = 8$

Distance between B and C;

$D_{BC} = \sqrt{(7-3)^2 + (7 -7)^2} = \sqrt{16} = 4$

Distance between C and A;

$D_{CA} = \sqrt{(7-3)^2 + (7 -(-1))^2} = \sqrt{80}$

In ΔX Y Z;

Distance between X and Y;

$D_{XY} = \sqrt{(-7-(-7))^2 + (4 -0)^2} = \sqrt{16} = 4$

Distance between Y and Z;

$D_{YZ} = \sqrt{(-7-1)^2 + (4 -4)^2} = \sqrt{64}= 8$

Distance between Z and X;

$D_{ZX} = \sqrt{(1-(-7))^2 + (4 -0)^2} = \sqrt{80}$

Clearly, Distance between sides of  ΔA B C and  ΔX Y Z are;

$D_{AB} = D_{YZ} \\\\D_{BC} = D_{XY}\\\\D_{CA} = D_{XZ}$

Hence, By SSS congruency;

⇒ ΔA B C ≅ ΔX Y Z .

Learn more about the congruency in triangle visit:

brainly.com/question/1675117

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