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S In the diagram, JKLM ~ EFGH. Find the values of x.y, and z.

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S In The Diagram JKLM EFGH Find The Values Of Xy And Z class=

Sagot :

Answer:

[tex]x = \boxed{\:27.5},\:\: y = \boxed{\:12},\: \:z = \boxed{\:65}[/tex]

Step-by-step explanation:

The information given states that JKLM ~ EFGH

This means that the two figures are similar to each other

What are similar figures?

Two geometric figures are similar if they have the same shape but not necessarily the same size and have the following properties1234:

  • Corresponding angles are congruent.
  • Corresponding sides are proportional.
  • The ratios of the lengths of their corresponding sides are equal.
  • The common ratio is called the scale factor.

Let's determine which sides of EFGH correspond to which sides of JKLM

I will use the symbol ⇔ for correspondence

The shortest side of EFGH is GH and so it corresponds to LM

  • GH ⇔ LM

Other correspondence is as follows

  • EH ⇔ JM
  • EF ⇔ JK
  • FG ⇔ LK

and

  • ∠HEF ⇔ ∠KJM

Using proportionality rule for two similar figures ,

[tex]\bullet \quad\dfrac{GH}{LM} = \dfrac{EH}{JM} = \dfrac{EF}{JK} = \dfrac{FG}{LK}\\\\[/tex]

Plugging in the given values for these sides, some in numbers and some as variables we get

[tex]\bullet \quad\dfrac{3}{LM} = \dfrac{y}{30} = \dfrac{8}{20} = \dfrac{11}{x}\\\\[/tex]

Ignoring the first ratio  since we are not asked to find LM, and taking each pair of the other three ratios we can determine the missing values

[tex]\bullet \quad \dfrac{y}{30} = \dfrac{8}{20}\\\\\longrightarrow y = 30 \cdot \dfrac{8}{20}\\\\y = 12[/tex]

[tex]\bullet \quad \dfrac{8}{20} = \dfrac{11}{x}\\\\\longrightarrow x = \dfrac{20 \times 11}{8}\\\\x = 27.5\\[/tex]

The angles having measure of z° in EFGH and 65° in JKLM correspond to each other and therefore they must be equal

Hence [tex]z = 65[/tex]

Answer
[tex]x = \boxed{\:27.5}\\\\y = \boxed{\:12}\\\\z = \boxed{\:65}[/tex]

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