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In PQR,PQ=13,QR=14 and PR=15. A altitude is drawn from P and a parallel line is drawn as LM which divides the triangle into two equal areas, then the length of PL is 

  1. 5
  2. 5(36)
  3. 5(37)
  4. 5(38)


Solution

The correct option is C 5(37)
Using heron's formula to find the area of the triangle,
s=13+14+152=21Area=21×(2113)×(2114)×(2115)              =21×8×7×6=84
Now the area of the triangle can be written as,
12×PT×14=84PT=12
QT=132122=5
Assuming RM=y and LM=x


Area of LMR,
12×xy=42xy=84
Area of PQT12×5×12=30
Area of trapezium PTML12×(12+x)×(9y)=423010812y+9xxy=12×2108412y+9x=249x=12y9×84y=12yy=37x=8437=47 
So LR=x2+y2=57PL=15LR=1557PL=5(37)
 

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