Question

Plot point F so that △ABC ≅ △FGH. Identify a sequence of rigid motions that maps △ABC onto △FGH and use a theorem to complete the explanation of why the triangles are congruent.

Answer 1

Explanation:

Point F is plotted at (5, 4)

Translate ΔABC so that point B maps to point G.

B(1, -1) → G(0, 3)

Rule for the translation,

(x, y) → [x - 1, y + 4]

By this rule,

Point C will map the point,

C(-1, -2) → C'[(-1 - 1), (-2 + 4)]

→ C'(-2, 2)

A(-4, 0) → A'[(-4 - 1), (0 + 4)]

→ A'(-5, 4)

Now reflect these points across y-axis.

Rule for the reflection across y-axis,

(x, y) → (-x, y)

Points after reflection will be,

A'(-5, 4) → F(5, 4)

B'(0, 3) → G(0, 3)

C'(-2, 2) → H(2, 2)

Distance formula,

d =
`√((y_2-y_1)^2+(x_2-x_1)^2)`

AB =
`√((1+4)^2+(-1-0)^2)` =
`√(26)`

BC =
`√((-2+1)^2+(-1-1)^2)` =
`√(5)`

CA =
`√((-4+1)^2+(-2-0)^2)` =
`√(13)`

FG =
`√((5-0)^2+(4-3)^2) =√(26)`

GH =
`√((2-0)^2+(2-3)^2)=√(5)`

HF =
`√((4-2)^2+(5-2)^2)=√(13)`

AB ≅ FG, BC ≅ GH, CA ≅ HF

By the SSS property Triangle congruence theorem, ΔABC ≅ ΔFGH