Question

3 the quilt shown has a border made of right triangles and trapezoids.
a. graph one triangle with vertices at d(-4, 0), e(0, 0), and f(0, -8). to locate another triangle, follow these steps:

• reflect △def across the y-axis to form △def.
• translate △def 8 units up to form △def.

is △def the same size and shape as △def? show your work.
grid image
solution
b. check your answer to problem 3a. show your work.

Explanation:

Part a

Step1: Reflect over y - axis

To reflect a point \((x,y)\) across the \(y\) - axis, the transformation rule is \((x,y)\to(-x,y)\).
For point \(D(-4,0)\), after reflection across the \(y\) - axis, \(D'=(4,0)\).
For point \(E(0,0)\), after reflection across the \(y\) - axis, \(E'=(0,0)\) (since \(x = 0\)).
For point \(F(0,-8)\), after reflection across the \(y\) - axis, \(F'=(0,-8)\) (since \(x = 0\)).
So, \(\triangle D'E'F'\) has vertices \(D'(4,0)\), \(E'(0,0)\), \(F'(0,-8)\).

Step2: Translate 8 units up

To translate a point \((x,y)\) 8 units up, the transformation rule is \((x,y)\to(x,y + 8)\).
For point \(D'(4,0)\), after translation 8 units up, \(D''=(4,0 + 8)=(4,8)\).
For point \(E'(0,0)\), after translation 8 units up, \(E''=(0,0 + 8)=(0,8)\).
For point \(F'(0,-8)\), after translation 8 units up, \(F''=(0,-8 + 8)=(0,0)\).

Step3: Compare side lengths

• For \(\triangle DEF\):

The length of \(DE\): Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), for \(D(-4,0)\) and \(E(0,0)\), \(DE=\sqrt{(0 + 4)^2+(0 - 0)^2}=\sqrt{16}=4\).
The length of \(EF\): For \(E(0,0)\) and \(F(0,-8)\), \(EF=\sqrt{(0 - 0)^2+(-8 - 0)^2}=\sqrt{64}=8\).
The length of \(DF\): For \(D(-4,0)\) and \(F(0,-8)\), \(DF=\sqrt{(0 + 4)^2+(-8 - 0)^2}=\sqrt{16 + 64}=\sqrt{80}=4\sqrt{5}\).

• For \(\triangle D''E''F''\):

The length of \(D''E''\): For \(D''(4,8)\) and \(E''(0,8)\), \(D''E''=\sqrt{(0 - 4)^2+(8 - 8)^2}=\sqrt{16}=4\).
The length of \(E''F''\): For \(E''(0,8)\) and \(F''(0,0)\), \(E''F''=\sqrt{(0 - 0)^2+(0 - 8)^2}=\sqrt{64}=8\).
The length of \(D''F''\): For \(D''(4,8)\) and \(F''(0,0)\), \(D''F''=\sqrt{(0 - 4)^2+(0 - 8)^2}=\sqrt{16 + 64}=\sqrt{80}=4\sqrt{5}\).

Since the corresponding side lengths of \(\triangle DEF\) and \(\triangle D''E''F''\) are equal, by the SSS (Side - Side - Side) congruence criterion, the two triangles are congruent. Congruent triangles have the same size and shape.

Step1: Recall rigid transformations

Reflection and translation are both rigid transformations. A rigid transformation is a transformation that preserves the distance between points (and thus preserves the size and shape of a figure).

Step2: Analyze the transformations on \(\triangle DEF\)

• Reflection across the \(y\) - axis: This transformation maps \(\triangle DEF\) to \(\triangle D'E'F'\). For any two points \(A\) and \(B\) in \(\triangle DEF\), the distance between their images \(A'\) and \(B'\) after reflection across the \(y\) - axis is equal to the distance between \(A\) and \(B\) (because the distance formula \(\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) is invariant under the reflection \((x,y)\to(-x,y)\) since \((-x_2)-(-x_1)=-(x_2 - x_1)\) and squaring eliminates the negative sign).
• Translation 8 units up: This transformation maps \(\triangle D'E'F'\) to \(\triangle D''E''F''\). For any two points \(A'\) and \(B'\) in \(\triangle D'E'F'\), the distance between their images \(A''\) and \(B''\) after translation \((x,y)\to(x,y + 8)\) is equal to the distance between \(A'\) and \(B'\) (because \((x_2-(y_2 + 8))-(x_1-(y_1 + 8))=(x_2 - y_2)-(x_1 - y_1)\) and the \(y\) - components cancel out in the distance formula).

Since both reflection and translation are rigid transformations, the composition of these two transformations (first reflect, then[SSE Completed, Client Connection Error][SSE Completed, Client Connection Error][LLM SSE On Failure]

Answer:

Yes, \(\triangle DEF\) is the same size and shape as \(\triangle D''E''F''\) because the reflection and translation transformations are rigid motions (they preserve side lengths and angle measures), and we have shown that the corresponding side lengths of the two triangles are equal ( \(DE = D''E''=4\), \(EF = E''F'' = 8\), \(DF=D''F'' = 4\sqrt{5}\) ), so the triangles are congruent.

Part b