Sent a picture of the solution to the problem (s).
Help Please Fast, simple
2 answers:
Answer:
D
Step-by-step explanation:
We can't state that because we don't know if ∠NPK is 90º
Triangle KPN is similar to triangle KQN
It is the same case as the alternative a, we can't say the triangles are similar because we don't the angles for KPN. We know that LN is perpendicular to NP, but we don't if NP is perpendicular to PK, ∠NPK might be 89,9º, that is almost 90º.
KP = 0.5 LN
Once the four sides of the square are equal, LQ = QN. But from the picture we can only state that KP length is approximately the length of NQ (or 0.5 LN)
PN || KM
That's the correct answer! Once ∠QNP is 90º, PN is perpendicular to KM.
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