ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 m, prove that area of 
is one-sixteenth of the area of 
.
AP=1 cm, PB=3 cm,AQ=1.5cm,and QC=4.5 m
In 
APQ and 
ABC
∠A=∠A [Common]
AP/AB=AQ/AC [Each equal to 1/4]
APQ
ABC [By SAS]
By area of similar triangle theorem
Area (
) =12
Area (
) 42
Area (
) =1
Area (
) 16 x ar(
)
19