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[tex]$\triangle RST \sim \triangle RYX$[/tex] by the SSS similarity theorem.

Which ratio is also equal to [tex]$\frac{RT}{RX}$[/tex] and [tex]$\frac{RS}{RY}$[/tex]?

A. [tex]$\frac{XY}{TS}$[/tex]
B. [tex]$\frac{SY}{RY}$[/tex]
C. [tex]$\frac{RX}{XT}$[/tex]
D. [tex]$\frac{ST}{VX}$[/tex]


Sagot :

To solve the problem, let's start by understanding the concept of similar triangles and the SSS (Side-Side-Side) similarity theorem. According to the SSS similarity theorem, if two triangles are similar, the ratios of the lengths of their corresponding sides are equal.

Given two triangles, [tex]\(\triangle RST\)[/tex] and [tex]\(\triangle RYX\)[/tex], we are told they are similar. This means:

[tex]\[ \frac{RT}{RX} = \frac{RS}{RY} \][/tex]

We need to determine which of the given ratios are equal to [tex]\(\frac{RT}{RX}\)[/tex] and [tex]\(\frac{RS}{RY}\)[/tex].

Let's go through each of the options:

1. [tex]\(\frac{XY}{TS}\)[/tex]

To be a valid ratio for similar triangles, the sides in this ratio must be corresponding sides of the similar triangles. The sides [tex]\(XY\)[/tex] and [tex]\(TS\)[/tex] are not corresponding sides between [tex]\(\triangle RST\)[/tex] and [tex]\(\triangle RYX\)[/tex]. Therefore, this ratio is not equivalent to [tex]\(\frac{RT}{RX}\)[/tex].

2. [tex]\(\frac{SY}{RY}\)[/tex]

Here, the side [tex]\(SY\)[/tex] is involved, but it is not mentioned which triangle's corresponding side ratio it is equivalent to. Additionally, [tex]\(RY\)[/tex] is one of the original sides given, and it does not establish the necessary proportionality with [tex]\(RT\)[/tex] or [tex]\(RX\)[/tex]. Thus, this ratio is not relevant.

3. [tex]\(\frac{RX}{XT}\)[/tex]

Let's examine this option:
- If we consider [tex]\(RX\)[/tex] as a side of [tex]\(\triangle RYX\)[/tex] that corresponds to [tex]\(RT\)[/tex] of [tex]\(\triangle RST\)[/tex], then this option suggests looking at the ratio by switching the positions of [tex]\(RX\)[/tex] in the numerator, aligning the comparison.
- Given that both triangles are similar by SSS theorem, [tex]\( \frac{RT}{RX} = \frac{RX}{XT} \)[/tex] by proportionality of corresponding sides given [tex]\(\triangle RYX \sim \triangle RST\)[/tex].

Therefore, this option makes sense as it compares corresponding segments under similarity transformation.

4. [tex]\(\frac{ST}{VX}\)[/tex]

This ratio involves sides [tex]\(ST\)[/tex] and [tex]\(VX\)[/tex]. The side [tex]\(VX\)[/tex] is not defined in the context of the problem, therefore this ratio does not relate to any mentioned correspondence between [tex]\(\triangle RST\)[/tex] and [tex]\(\triangle RYX\)[/tex].

Thus, after examining all the options,
the correct ratio is:

[tex]\[ \frac{RX}{XT} \][/tex]

The corresponding ratio equivalent to [tex]\(\frac{RT}{RX}\)[/tex] and [tex]\(\frac{RS}{RY}\)[/tex] is:

[tex]\[ \boxed{\frac{RX}{XT}} \][/tex]
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