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Question

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 solve for the length of \( LQ \), we can use the property of similar triangles since \( \triangle JKL \sim \triangle PKQ \).

From the problem, we know the following:

• The lengths \( JL = 15 \) inches.
• The lengths \( KQ = 17 \) inches.
• The lengths \( PK = 15 \) inches.

Since the triangles are similar, the ratios of their corresponding sides are equal. This gives us the following proportion based on the sides:

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

Substituting the known values into the proportion, we have:

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

This simplifies to:

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

Now, solving for \( LQ \):

\[ LQ = 17 \cdot 1 = 17 \text{ inches} \]

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

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