SOLUTION: Square ABCD has area of 320 square units. Points K, L, M, N are midpoints of the sides of ABCD, P is the point of intersection of KC and BN , and Q is the point of intersection of

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Question 1067985: Square ABCD has area of 320 square units. Points K, L, M, N are midpoints of the sides of ABCD, P is the point of intersection of KC and BN , and Q is the point of intersection of AM and BN . Find PQ.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
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Set up a coordinate system with B at the origin (0,0) and s is the length of the side of the square.
So then,
N is located at (s%2F2,s)
K is located at (0,s%2F2)
C is located at (s,0)
M is located at (s,s%2F2)
Find the line BN,
m=%28s-0%29%2F%28s%2F2-0%29=2
y%5BBN%5D-0=2%28x-0%29
y%5BBN%5D=2x
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Find the line KC,
m=%28s%2F2-0%29%2F%280-s%29=-1%2F2
y%5BKC%5D-s%2F2=-%281%2F2%29%28x-0%29
y%5BKC%5D=-%281%2F2%29x%2Bs%2F2
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Find the line AM,
m=-1%2F2
y%5BAM%5D=-%281%2F2%29x%2Bs
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P is the intersection of BN and KC,
2x=-%281%2F2%29x%2Bs%2F2
%285%2F2%29x=s%2F2
x%5BP%5D=s%2F5
and
y%5BP%5D=%282%2F5%29s
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Q is the intersection of BN and AM,
2x=-%281%2F2%29x%2Bs
%285%2F2%29x=s
x%5BQ%5D=%282%2F5%29s
and
y%5BQ%5D=%284%2F5%29s
So then the distance PQ is,
PQ%5E2=%28s%2F5-%282%2F5%29s%29%5E2%2B%28%282%2F5%29s-%284%2F5%29s%29%5E2
PQ%5E2=%28-%281%2F5%29s%29%5E2%2B%28-%282%2F5%29s%29%5E2
PQ%5E2=s%5E2%2F25%2B%284%2F25%29s%5E2
PQ%5E2=%285%2F25%29s%5E2
PQ=%28sqrt%285%29%2F5%29s
In this case,
s%5E2=320
s=8%2Asqrt%285%29
So,
PQ=%28sqrt%285%29%2F5%298%2Asqrt%285%29
PQ=8