Proving Similar Triangles Practice

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Question
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Triangle upper J upper K upper L is divided into triangle upper P upper K upper Q and quadrilateral upper J upper P upper Q upper L by vertical line upper P upper Q. Point upper P is on line upper J upper K. Point upper Q is on diagonal upper L upper K.

The two triangles are right triangles. If △JKL∼△PKQ
, and JL¯¯¯¯¯¯¯=15
inches, KQ¯¯¯¯¯¯¯¯=17
inches, and PK¯¯¯¯¯¯¯¯=15
inches, then what is the length of LQ¯¯¯¯¯¯¯
?

(1 point)
LQ¯¯¯¯¯¯¯=
inches
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Answer 1

To find the length of \( LQ \) in the triangles \( \triangle JKL \) and \( \triangle PKQ \), we can use the property of similar triangles. Since \( \triangle JKL \sim \triangle PKQ \), the ratios of their corresponding sides will be equal.

Given:

• \( JL = 15 \) inches
• \( KQ = 17 \) inches
• \( PK = 15 \) inches

Let's denote \( LQ \) as \( x \).

Using the similarity of the triangles, we can set up the following proportion based on the corresponding sides:

\[ \frac{JL}{PK} = \frac{LQ}{KQ} \]

Substituting in the values we know:

\[ \frac{15}{15} = \frac{x}{17} \]

This simplifies to:

\[ 1 = \frac{x}{17} \]

Now we can solve for \( x \):

\[ x = 17 \]

Thus, the length of \( LQ \) is:

\[ \boxed{17} \text{ inches} \]