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Sagot :

I can only see the similar triangle problem, with sides 6cm, 9cm, and 15cm. Since the triangles are similar, the ratio for side lengths is the same. 6:9 simplifies to 2:3. If we try to solve ?:15 we get 10 as 10:15 is equivalent to 2:3, so the length is 10 cm
TSO
JKL is similar to PQR

so they want PR 

JKL is similar to PQR

So PR is similar to JL

And they also give us the sides KL and QR

JKL is similar to PQR

Both of those sides are similar, so we can form a proportion.

[tex]\sf { \frac{PR}{JL} = \frac{QR}{KL} }[/tex]

Plug in the numbers:

[tex]\sf { \frac{PR}{6} = \frac{15}{9} }[/tex]

Cross multiply:

[tex]\sf{ PR \times 9 = 6 \times 15}[/tex]

Isolate PR

[tex]\sf{ PR = \frac{6 \times 15}{9}}[/tex]

Simplify it

[tex]\sf{ PR = 10}[/tex]

So your final answer is

[tex]\boxed{\bf{10 centimeters}}[/tex]
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