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Dimas [21]
3 years ago
6

Square ABCD and square EFGH share a common center on a coordinate plane. is parallel to diagonal . Across how many lines of refl

ection can the combined figure reflect onto itself?.

Mathematics
2 answers:
Oduvanchick [21]3 years ago
7 0
Square ABCD and square EFGH will reflect onto themselves across 8 lines of reflection
Ludmilka [50]3 years ago
6 0

Answer:

<h2>4 lines of reflection.</h2>

Step-by-step explanation:

The complete problem states:

<em>Square ABCD and square EFGH share a common center on a coordinate plane. EH is parallel to diagonal AC. Across how many lines of reflection can the combined figure reflect onto itself?</em>

<em />

Basically, all diagonals work as lines of reflection of the figure, because we have squares which are complete symmetrical, and their diagonals always divide them in halves, creating a reflection.

So, the first two lines of reflection are the diagonals of the big square AC and BD. Due to square EFGH share a common center and EH is parallel to diagonal AC, bigger diagonals divide its side equally, creating reflections.

Also, based on the parallelism given, the diagonals of the small square will divide in half the sides of the big square, if we prolong them. So, there we have two more lines that create reflections.

In sum, there are 4 lines of the figure combined that reflect onto itself.

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Let's use Gaussian elimination. Consider the augmented matrix,

\left[\begin{array}{ccc|ccc}1 & -1 & -1 & 1 & 0 & 0\\-1 & 2 & 3 & 0 & 1 & 0\\1 & 1 & 4 & 0 & 0 & 1\end{array}\right]

• Add row 1 to row 2, and add -1 (row 1) to row 3:

\left[\begin{array}{ccc|ccc}1 & -1 & -1 & 1 & 0 & 0\\0 & 1 & 2 & 1 & 1 & 0\\0 & 2 & 5 & -1 & 0 & 1\end{array}\right]

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\left[\begin{array}{ccc|ccc}1 & -1 & -1 & 1 & 0 & 0\\0 & 1 & 2 & 1 & 1 & 0\\0 & 0 & 1 & -3 & -2 & 1\end{array}\right]

• Add -2 (row 3) to row 2:

\left[\begin{array}{ccc|ccc}1 & -1 & -1 & 1 & 0 & 0\\0 & 1 & 0 & 7 & 5 & -2\\0 & 0 & 1 & -3 & -2 & 1\end{array}\right]

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\left[\begin{array}{ccc|ccc}1 & 0 & 0 & 5 & 3 & -1\\0 & 1 & 0 & 7 & 5 & -2\\0 & 0 & 1 & -3 & -2 & 1\end{array}\right]

So the inverse is

\begin{bmatrix}1&-1&-1\\-1&2&3\\1&1&4\end{bmatrix}^{-1} = \boxed{\begin{bmatrix}5&3&-1\\7&5&-2\\-3&-2&1\end{bmatrix}}

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