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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
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]
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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