Answer:
Triangle KLM is not similar to triangle KBC.
Step-by-step explanation:
In similar triangles, corresponding sides are always in the same ratio. Therefore, if ΔKLM ~ ΔKBC then:
[tex]\dfrac{KL}{KB}=\dfrac{LM}{BC}=\dfrac{KM}{KC}[/tex]
From inspection of the given diagram:
- KL = 88
- KM = 25 + 107 = 132
- KB = 16
- KC = 25
Substitute these values into the proportion:
[tex]\dfrac{KL}{KB}=\dfrac{KM}{KC}\\\\\\\\\dfrac{88}{16}\overset{?}=\dfrac{132}{25}\\\\\\\\\dfrac{88}{16}\overset{?}=\dfrac{132}{25}\\\\\\\\\dfrac{11}{2}\overset{?}=\dfrac{132}{25}[/tex]
Rewrite the ratios with the common denominator of 50:
[tex]\dfrac{11\times 25}{2\times 25}\overset{?}=\dfrac{132\times 2}{25\times 2}\\\\\\\\\dfrac{275}{50}\neq\dfrac{264}{50}[/tex]
Since the two ratios are not equal, the triangles are not similar based on the given side lengths.