In a rectangle ABCD, AB = 9 cm and BC = 6 cm. P and Q are two points on BC such that the areas of the figures ABP, APQ, and AQCD are in geometric progression. If the area of the figure AQCD is four times the area of triangle ABP, then BP : PQ : QC is
Started 1 month ago by Shashank in
Explanatory Answer
Since Ar(ABP),Ar(APQ)&Ar(AQCD) are in GP and Ar(AQCD)=4×Ar(ABP)
Ar(ABP):Ar(APQ):Ar(AQCD)=1:2:4
9x/2:9y/2:27+9z/2=1:2:4
x:y:6+z=1:2:4
y=2x
6+z=4x
x + y + z = 6
x + 2x + 4x – 6 = 6
x = 127127
z = 4x – 6 = 48−427=6748−427=67
∴ x = 2z
Therefore, x:y:z=2:4:1
BP : PQ : QC = 2 : 4 : 1
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